Class 10

Haryana State Board HBSE 10th Class Science Important Questions Chapter 12 Electricity Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 12 Electricity

Question 1.
What do you mean by electric charge?
Answer:
Electric charge:

• ‘Charge’ or ‘electric charge’ is the fundamental quantity of electricity.
• Electric charge or simply charge is of two types, (1) Positive (+) charge and (2) negative (-) charge.
• If electric charge flows through a conductor such as metal wire, we say that there is electric current in the conductor.
• Equal amount of positive (+ve) charge + Equal amount of negative (-ve) charge = zero (O)
• The SI. unit of electric charge is columb ‘C’.
• An electron possesses a ‘negative charge of 1.6 x 10^-19 C’ whereas a proton possesses a ‘positive charge of 1.6 x 10^-19 C’.

Question 2.
State the properties of electric charge.
Answer:
(A) Like charges repel each other i.e. when a proton-proton or electron-electron interact, they repel.
(B) Unlike charges attract each other ie. when a proton and an electron interact, they attract each other.

Question 3.
What is a free electron?
Answer:
1. Metals have a property to lose electrons and form positive ions.

2. The outer most electrons move randomly around the nucleus.

3. These electrons work as glue and bind the protons at their place and thus make them immovable.

4. Under normal circumstances, in metallic materials the force of attraction between these outermost electrons (or say valence electrons) and the nucleus (which is positively charged) is quite less.

5. These electrons separate from their parent atom and move randomly.

6. These electrons are known as free (or conducting) electrons and are responsible for conducting electricity or say conduction.

7. Electric current can flow very easily through material which contain a large number of free electrons. Hence such a material is called a conductor. For example, metals such as copper, silver and aluminium are conductors.

Question 4.
Resolve the mystery behind the conventional current and actual flow of electric current. OR ‘In a circuit diagram, we put arrows on connecting wires pointing from the positive terminal of the cell to the negative terminal of the cell’. Explain. OR Explain conventional current and its flow.
Answer:
1. In the earlier times, when scientists did not have knowledge about electrons, it was thought that electric current was a flow of positive charge. Hence, the direction of flow of positive charges was considered as the direction of electric current. This flow came to be known as ‘conventional current’.

2. Later, negatively charged electrons were discovered and it was fund that electrons were responsible for the flow of electric current.

3. Thus, it was proved that the actual flow of electric charge was not from ‘positive to negative’ but from ‘negative to positive’.

4. However, in practice we still take direction of electric current as ‘positive to negative’ i.e. the path of conventional current. We do this by putting the arrows from positive terminal of batteries to negative terminal.

5. Thus, flow of conventional current (positive to negative) is opposite to the actual direction of flow of electrons (negative to positive).

Question 5.
Define electric current and state and explain its unit.
Answer:
Electric current:
1. The rate of flow of electric charge is known as electric current. In other words, the net quantity of electric charge that flows through any cross-section of a conductor is known as the electric current.
Thus? electric current \(=\frac{\text { Quantity of electric charge }}{\text { Time }}\)

2. Thus, if Q is the amount of electric charge passing through any cross-section of a conductor in time t then, electric current (I) = \(\frac{Q}{t}\)

3. If a quantity of one Coulomb (1Q) electric charge passes through the conductor in 1 second, then we can say that an electric current of one ampere (1A) is flowing through the conductor.

4. SI unit of electric current is Coulomb/second (C/s).

5. Electric current is also measured in Ampere (A), after the French scientist Andre-Marie Ampere.

6. Milliampere (mA) and microampere (.iA) are smaller units to measure electric current.

7. 1 mA = 10^-3 A and 1 μA = 10^-6 A.

8. An instrument called Ammeter is used to measure the electric current.

Question 6.
What is ampere? Explain.
Answer:
1. Ampere (A) is the main unit in which electric current is measured. The unit is Ampere (A) after the French scientist Andre Ampere.
2. Milliampere (mA) and microampere (pA) are smaller units to measure electric current.
3. 1 mA =10^-3 A and 1 μA = 10^-6 A.
4. An instrument called Ammeter is used to measure the electric current.
5. Ammeter is ‘connected in series’ in the electric circuit.

Question 7.
Draw a schematic diagram of an electric circuit and explain it.
Answer:
1. The figure shown here is a basic electric circuit. It comprises of a cell, an electric bulb, an ammeter (connected in series) and a key or say switch to control the circuit.

2. When key (switch) is turned on, the electric current starts flowing from positive terminal of the cell to the negative terminal, through the bulb and the ammeter.

3. The bulb will glow and ammeter will show a reading which ascertains flow of electric current in the circuit.

Question 8.
Define electric circuit and switch.
(a) Electric circuit:
A continuous and closed path along which electric current flows is called an electric circuit.

(b) Switch:
A switch is a component that provides a connecting link between the cell i.e. battery and other electrical components of electric circuit.

Question 9.
State the relationship between 1 ampere and 1 coulomb.
Answer:
If 1 coulomb charge passes through the conductor in 1 second, then we say that electric current of 1 ampere is flowing through the conductor.
1 ampere = \(\frac{1 \text { Coulomb }}{1 \text { Second }}\)
∴ \(1 A=\frac{1 C}{1 s}=1 \mathrm{Cs}^{-1}\)

Question 10.
State and explain the relation between quantity of charge and number of electrons flowing through a conductor.
Answer:
1. If the number of electrons passing through the cross-section of conductor in time ‘t’ equals to ‘n’, then the quantity of charge passing through the cross-section will be Q = ne.

2. In this sense, equation \(I=\frac{Q}{t}\) can be represented as \(I=\frac{n e}{t}\),
where e = 1.6 x 10^-9 C charge of electron.

Question 11.
How many electrons will be required to produce 1 Ampere electric current in a conductor?
Answer:
1. If 1 coulomb electric charge i.e. Q passes through a conductor in 1 second, then 1 Ampere electric current will flow through the conductor.

2. Now, current I = 1A, charge Q = 1C and time t = 1s

3. We know that Q = ne (where e = 1.6 x 10^-19 C)
Also we know that I\(=\frac{Q}{t}=\frac{n e}{t}\)
∴ No.of electrons \(\mathrm{n}=\frac{\mathrm{l} . t}{\mathrm{e}}=\frac{1 \times 1}{1.6 \times 10^{-19}}\)
= 6.25 x 10¹⁸

4. Thus, 6.25 x 10¹⁸ electrons will be required to produce 1 Ampere electric current in a conductor.

Question 12.
A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
Answer:
We are given, I = 0.5 A, t = 10 min = 600 s.
Now, Q = It = 0.5 A x 600 s= 300 C

Question 13.
While connecting a torch with battery, an electric current of 64 mA flows through the bulb. If this torch glows for 10 mm, how many electrons will pass through the bulb? (charge of electron =e = 1.6 x 10^-19 C)
Answer:
I = 64 mA = 64 x 10^-3A,t = 10min = 10 x 60
=600s, e=1.6 x 10^-19 C

= 24000 x 10¹⁶
n = 24 x 10¹⁹ electrons

Question 14.
1800 C electric charge is passing through an electric bulb in one hour. How much current will pass through the electric bulb?
Answer:
Q = 1800 C, time t = 1 hr = 3600s
Electric current \(I=\frac{Q}{t}=\frac{1800}{3600}=0.5\) current will pass through the electric bulb.

Question 15.
Give a general idea of electric potential and electrical potential difference.
Answer:
1. As shown in the figure, pour larger quantity of water in vessel A and lesser quantity of water in vessel B. Connect these two vessels with a tube.

2. Vessel A contains more water than vessel B and so water in vessel A will have higher pressure.

3. Hence, water will flow from vessel A to vessel B. In other words, water flows through the tube due to ‘water pressure difference’ or say ‘water potential difference’.

4. Similarly, if an ‘electric potential difference’ is created between two points, then the electric current can also be made to flow from one point to another.

5. The difference in electric potential that exists between the two points is known as electric potential difference.

Electric potential:

• When you bring some electric charge near another charge, the former charge may experience a force of attraction or repulsion. So, work has to be done to move it against the force of repulsion or towards the force of attraction.
• In other words, in order to maintain equilibrium between these two charges, some work has to be done.
• This work is stored in the form of potential energy.
• The work done on the charge is called electric potential.

Question 16.
What is electric potential difference? Explain the unit used for measuring potential difference.
Answer:
(a) Electric potential difference:
The amount of work done to take a unit positive charge from a given point say A to point B in a circuit carrying some current is known as the electrical potential difference between the two points.
→ Thus, electric potential difference \((V)=\frac{\text { Work done }(W)}{\text { Electric charge }(Q)}\)
\( \quad V=\frac{W}{Q}\)

(b) Voltage:
(1) In practice, electric potential difference is known as voltage. Its SI unit is joule/coulomb or volt (V).
(2) If the work done to bring 1-coulomb electric charge from one point to another is 1 joule, then the potential difference between these two points is called 1 volt. (V)
Thus, 1 volt \(=\frac{1 \text { joule }}{1 \text { Coulomb }}\) \(1 \mathrm{~V}=1 \mathrm{JC}^{-1}\)
(3) Electric potential difference (V) is measured with a device called voltmeter.
(4) The voltmeter is connected in parallel across the two points of which the potential difference is measured.

Question 17.
Define electric potential and state its unit.
Answer:
Electric potential difference:
The amount of work done to take a unit positive charge from a given point say A to point B in a circuit carrying some current is known as the electrical potential difference between the two points.
→ Thus, electric potential difference \((V)=\frac{\text { Work done }(W)}{\text { Electric charge }(Q)} \)
\(\quad V=\frac{W}{Q}\)

Question 18.
Define 1 volt and state its unit. OR what is voltmeter? How is It connected in an electric circuit?
Answer:
Voltage:
(1) In practice, electric potential difference is known as voltage. Its SI unit is joule/coulomb or volt (V).

(2) If the work done to bring 1-coulomb electric charge from one point to another is 1 joule, then the potential difference between these two points is called 1 volt. (V)
Thus, 1 volt \(=\frac{1 \text { joule }}{1 \text { Coulomb }}\) \(1 \mathrm{~V}=1 \mathrm{JC}^{-1}\)

(3) Electric potential difference (V) is measured with a device called voltmeter.

(4) The voltmeter is connected in parallel across the two points of which the potential difference is measured.

Question 19.
Name the device used to measure electric potential difference and state how it is connected in an electric circuit.
Answer:
Voltmeter:
1. The device used to measure potential difference between any two points of a circuit is called voltmeter (Volt + meter).
2. Voltmeter is connected in a parallel connection.

Question 20.
List two differences between voltmeter and ammeter.
Answer:

Question 21.
How much work Is done in moving a charge of 2 C across two points having a potential difference 12 V?
Answer:
The amount of charge Q, that flows between two points at potential difference V (= 12 V) is 2 C.
Thus, the amount of work W, done in moving the charge is W = VQ = 12 V x 2C = 24J.

Question 22.
How much work is to be done to take 2 C electric charge from the potential of 6 V to the potential of 12 V?
Q=2C
Electric potential difference V = 12V – 6V = 6V
Now \(V=\frac{W}{Q}\)
∴ Work W = VQ = 6 x 2 = 12J

Symbols of some commonly used components in circuit diagrams

Question 23.
Perform an activity to verify Ohm’s law
Answer:
Aim: Verification of Ohm’s law
Apparatus and material : 0.5 meter long nichrome wire (or an electric bulb), batteries of 1.5 V, voltmeter, ammeter and key.

Procedure:

1. Prepare an electric circuit by connecting XV nichrome wire (or an electric bulb) as resistance first with one 1.5 V battery, ammeter and a key.
2. Then connect voltmeter between two ends of wire as shown.
3. When you switch on the key, the electric current will flow through the circuit.
4. Measure the current (I) in ammeter and potential difference (V) in the voltmeter.
5. Similarly, connect two, three and finally four batteries in the circuit and each time measure the values of current I and voltage V.
6. Record all the values of I and V in the table.

Observation table:

Plot values of V and I on the graph.

Conclusion:

1. On plotting these values we observe that the graph is a straight line passing through the origin.
2. As you increase the number of batteries, the electric current in the conductor increases in a constant proportion with the increase in voltage V.
3. The ratio of V and I i.e. VII always remains constant.

Question 24.
State Ohm’s law and explain its unit.
Answer:
Ohm’s law:
1. In a definite physical situation, the electric current (I) flowing through the conductor is directly proportional to the potential difference (V) applied across it, provided its temperature and other physical conditions remain same.

2. Thus, I α V or say V α I.
∴ V = I R (where, R is the proportionality by constant and it represents resistance R of the circuit)
V/I = Constant Also, Resistance \((\mathrm{R})=\frac{\text { Voltage }(\mathrm{V})}{\text { Electric current }(\mathrm{l})}\)

3. The SI unit of resistance is volt/ampere which is called ohm and it is denoted by Ω (Omega).
∴ 1 ohm \((\Omega)=\frac{1 \text { volt }(V)}{1 \text { ampere }(i)}\)

4. When 1 volt potential difference (V) is applied across the conductor and 1 ampere (I) current flows through it, then resistance of the conductor is said to be 12.

5. The symbol represents resistance R in the circuit.

Question 25.
Discuss the factors on which strength of electric current ¡n a given conductor depends. OR How can you say that the current flowing in a circuit is strong or weak? Discuss.
Answer:
Ohms law says, current \((\mathrm{I})=\frac{\mathrm{V} \text { (Potential difference) }}{\mathrm{R} \text { (Resistance) }}\)

From this relation we draw two conclusions. They are –
(i) I α V (if R = constant) (ii) I α\(\frac{1}{R}\)(if V = constant)

From 1st:
1. Since, current (I) is directly proportional to potential difference V, if value of V is doubled, the current (I) flowing through it will also get doubled.
2. If V is halved, the current will also get halved,

From 2nd:
1. Current (I) is inversely proportional to resistance (R). So, if the value of resistance is doubled, the current (I) flowing will get halved.
2. If resistance is halved, the current (I) will get doubled.

Conclusion:
The strength of electric current in a given conductor depends on —
(i) Potential difference (V) across the ends of the conductor (ii) Resistance (R) of the conductor

Question 26.
Define resistance and variable resistance and bring out the difference between resistance and resistor
Answer:
(a) Resistance:
1. Resistance refers to the property of the conductor by virtue of which it opposes the flow of current through it.

2. Resistance is equal to the ratio of potential difference applied across the conductor to the current flowing through it.
Thus, Resistance \((\mathrm{R})=\frac{\text { Potentialdifference }(\mathrm{V})}{\text { Current }(\mathrm{l})} \quad\)
∴ \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{l}}\)

(b) Vanable resistance:
1. A component used for regulating (i.e. increasing or decreasing) the current without changing the voltage source is called a variable resistor and the concept is called variable resistance.

2. In other words, variable resistance allows us to adjust the resistance between two points in a circuit. Rheostat is one such device that helps to change resistance in the circuit. Thus, resistor creates resistance.

Question 27.
Explain the cause of resistance in conductors and insulators.
Answer:
1. Flow of free electrons In a fixed direction results in flow of electric current in a conductor.
2. These freely moving electrons collide with atoms or molecules of conductor. This retards their motion.
3. Motion of electrons means that the electric current gets opposed. This is called resistance (R) of the conductor.
4. Metals such as aluminum, steel, copper, etc. have a large number of free electrons and so they offer low resistance to the flow of electric current. Metals are good conductors.
5. Insulators such as wood, plastic, rubber, etc. do not have free electrons. Hence, no current flows through them. So, their resistance is very high. Insulators are poor conductors.

Question 28.
Define resistance (resistivity) of a substance and explain its unit.
Answer:
1. The resistance (R) of a conductor is directly proportional to the length of the conductor and inversely proportional to the area of its cross-section.

2.  Let the length of the conductor be I and its cross-sectional area be A, then,
Resistance R α length l arid also R α \(\frac{1}{A}\)
∴ \(R \alpha \frac{I}{A}=\rho \frac{I}{A}\) = (where ρ (rho) is constant and called the resistivity of conducting material.)

3. We can also say, ρ = \(R \frac{A}{l}\)

4. The S.l unit of resistivity \(\rho=\frac{\text { Unit of resistance } x \text { Unit of area }}{\text { Unit of length }}=\frac{\Omega \mathrm{xm}^2}{\mathrm{~m}}=\Omega \mathrm{m} \text { (Ohm meter) }\)

Question 29.
List the factors on which the resistance of a conductor depends. Write the formula showing relation of resistance with these factors.
Answer:
At constant temperature, the resistance of a conductor depends on the following factors:
(i) Length: Resistance R of a conductor is directly proportional to its length L i.e. R α L
(ii) Area of cross-section: Resistance R of a conductor is inversely proportional to the area of cross-section A i.e, \(R \propto \frac{1}{A}\)
(iii) Nature of material: Resistance depends on the nature of material. For example, resistance of copper wire is quite less as compared to nichrome wire.

Combining all these factors we get —
R α \(\frac{L}{A}\)  or say R = ρ\(\frac{L}{A}\) (Where p is the proportionality constant and is called ‘resistivity’ or ‘specific resistance’ which depends on the nature of the material.)

Question 30.
Classify solids on the basis of their resistivity values.
Answer:
On the basis of resistivity values, we can classify solids as:
(i) Conductors (metals and alloys):

• Metals and their alloys have very low resistivity in the range of 10^-8 to 10^-6 Ω m.
• These are known as good conductors of electricity.
• Copper, aluminium, iron, etc. are metals. Nichrome, constantan, manganin, etc. are metallic alloys.

(ii) Insulators:

• Insulators are substances that have very large resistivity. They have resistivity of more than 1 m. Insulators such as glass, hard rubber, etc. have very high resistivity range of 10¹² to 10¹⁷ Ωm.

Electrical resistivity of some substances at 200 C. (For information only)

You need not memorize these values. You can use these values for solving numerical problems.

Question 31.
Differentiate between Conductors and Insulators.
Answer:

Question 32.
(a) How much current will an electric bulb draw from a 220 V source, If the resistance of the bulb filament is 1200 Ω? (b) How much current will an electric heater coil draw from a 220 V source, If the resistance of the heater coil is 100 Ω? (Text book example 12.3)
Answer:
(a) We are given V = 220 V; R = 1200Ω We have current I = 220 V/1200 Ω = 0.18 A.
(b) We are given, V = 220 V, R = 100Ω. We have current I = 220 V/100 Ω = 2.2 A.

Question 33.
The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw If the potential difference is increased to 120 V? (Text book example 12.4)
Answer:
We are given, potential difference V = 60 V, current I = 4 A.
According to Ohm’s law, \(R=\frac{V}{l}=\frac{60 \mathrm{~V}}{4 \mathrm{~A}}=15 \Omega\)
When the potential difference is increased to 120 V the current is given by —
current = \(\frac{V}{R}=\frac{120 V}{15 \Omega}=8 A\)
The current through the heater becomes 8 A.

Question 34.
Resistance of a metal wire of length 1 m is 26Ω at 20°C. lithe diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 12.2, predict the material of the wire. (Text book example 12.5)
Answer:
We are given the resistance R of the wire = 26Ω, diameter = 0.3 mm = 3 x 10 m, and the length I of the wire = 1m.
Therefore, the resistivity of the given metallic wire is ρ= (RA/I) = (Rπd²/4l)
Substitution of values in this gives ρ = 1.84 x 10^-6 Ωm
The resistivity of the metal at 20°C is 1.84 x 10^-6 Ωm.

We are given, potential difference V = 60 V, current I = 4 A.
According to Ohm’s law, \(R=\frac{V}{l}=\frac{60 \mathrm{~V}}{4 \mathrm{~A}}=15 \Omega\)
When the potential difference is increased to 120 V the current is given by —
current = \(\frac{V}{R}=\frac{120 V}{15 \Omega}=8 A\)
The current through the heater becomes 8 A.

Question 35.
A wire of given material having length I and area of cross-section A has a resistance of 4 ? What would be the resistance of another wire of the same material having length 1/2 and area of cross-section 2A? (Text book example 12.6)
Answer:
For first wire

The resistance of the new wire is 1Ω.

Question 36.
If an electric bulb connected to 220 V line draws an electric current of 0.5 A, then what will be the resistance of the filament of the bulb?
Answer:
I = 0.5A,V=220V
According to Ohm’s law, \(R=\frac{V}{l}=\frac{220}{0.5}=440 \Omega\)

Question 37.
The resistance of a resistive wire having length l and area of cross-section A is 4Ω. If the length of same type of wire is 1/2 and the area of cross-section 2A, what will be the resistance of wire?
Answer:
For the first wire \(R=\rho \frac{l}{A}\)-, for the second wire,\(\mathrm{R}^{\prime}=\rho^{\prime} \frac{I^{\prime}}{\mathrm{A}^{\prime}}\)

Question 38.
There are more than one ways of combining resistors in a circuit. Explain them very briefly.
OR Why does the need of combining resistors in more than one way arises in a circuit?
Answer:
1. Apart from potential difference, current in circuit also depends on resistance R.

2. All electrical circuits do not require the same amount of resistance.

3. In order to obtain the desired resistance, resistors need to be connected in two different ways.
They are: Connecting resistors in (A) series and in (B) parallel.

4. If we want to increase total resistance, then we need to connect the resistors in series and if we want to decrease the resistance, we need to connect individual resistors in parallel.

Question 39.
With the help of suitable diagrams state the two methods of connecting resistors in a circuit. There are two methods of joining the resistors. They are:

(1) Joining resistors In a series connection:

• When two or more resistors are connected end-to-end consecutively, such a connection is called a series connection.
• As shown in the figure, three resistors R₁, R₂ and R₃ are connected end-to-end.

(2) Joining resistors in a parallel connection:

• When two or more resistors are connected between the same two points of a circuit, they are said to be connected in a parallel connection (because they are connected in parallel and not end-to-end).
• The diagram here shows resistors R₁, R₂ and R₃ connected in parallel between the two points A and B.

Question 40.
Explain the series connection of resistors and derive the formula of equivalent resistance.
Answer:
Series connection of resistors:
1. When two or more resistors are connected end-to-end consecutively such a connection is called a series A connection.

2. As shown in figure, consider three resistances R₁, R₂ and R₃ connected in senes. Suppose a current I flows through the circuit when a cell of voltage V is connected across the combination.

By Ohm’s law, the potential differences across the three resistances will be,
V₁ = IR₁, V₂ = IR₂, V₃ = IR₃,

If R be the equivalent resistance of the senes combination, then on applying a potential difference V across it, the same current I must flow through it. Therefore,
V = IR
But V=V₁ + V₂ +V₂
∴ IR_S = IR₁ + IR₂ + IR₂ OR R_s = R₁ + R₂ + R₃

Question 41.
State the characteristics of series connection of resistors.
Answer:
1. In series connection of resistors, the current (I) flowing through each resistance is same.
2. The total voltage drop across all the resistances connected in series equals to the sum of voltage drop across each resistance Le. V = V₁ + V₂ + …………………………. + V_n.
3. The equivalent resistance of resistors connected in series is equal to the algebraic sum of all resistors.
4. The magnitude of equivalent resistance i.e. R is always larger than the largest resistance of the circuit. In other words, equivalent resistance R is larger than any other resistance of the circuit.

Question 42.
An electric lamp, whose resistance is 20Ω and a conductor of 4Ω resistance are connected to a 6 V battery (as shown In figure). Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.
Answer:

The resistance of electric lamp, R₁ = 20Ω
The resistance of the conductor connected in series, R₂ = 4Ω
Then the total resistance in the circuit
R =R₁ + R₂
R_s =20Ω + 4Ω=24Ω
The total potential difference across the two terminals of the battery
V = 6V.
Now by Ohm’s law, the current through the circuit is given by
I =V/Rs=6V/24.Ω = 0.25A.
Applying Ohm’s law to the electric lamp and conductor separately, we get potential difference across the electric lamp,
V₁ =20Ω 2 x 0.25A = 5V;
Potential difference across the conductor, V₂ = 4Ω x 0.25 A = 1 V.
Suppose we replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance must be such that a potential difference of 6 V across the battery terminals will cause a current of 0.25 A in the circuit. The resistance R of this equivalent resistor would be
R =V/l = 6V/0.25A = 24Ω
This is the total resistance of the series circuit. It is equal to the sum of the two resistances.

Question 43.
In order to get the current 0.5 A in the circuit by connecting a bulb of resistance 20Ω with 12 V battery, how much should be the value of resistance connected in series? What will be the voltage drop across the bulb?
Answer:
If the resistance connected in series with bulb is R₁ and resistance of bulb is R₂ then the circuit will be as shown in the figure here.
V=12 V, l=0.5A, R₂=20Ω, R₁=?

Using Ohm’s law,
Here, bulb and R1 are in series,\(R=\frac{V}{J}=\frac{12}{0.5}=24 \Omega\)
therefore equivalent resistance
R=R₁+R₂ ……………………… (1)
Substituting the value of R and R₂ in equation (1)
24=20+R₂
∴ R₂ = 24 – 20 = 4Ω
The voltage drop across the bulb, V₂ = IR₂ = (0.5) (4) = 2 V

Question 44.
When an electric heater is given a voltage 120 V, an electric current of 2 A passes through It. If the heater is given a voltage of 240 V, how much electric current will flow through it? What will be the resistance of the coil of the heater?
V₁ = 120V, I₁ = 2A, V₂ = 240V, I₁ =?
Answer:
Resistance of heater coil \(R=\frac{V_1}{L_1}=\frac{120}{2}=60 \Omega\)
Now, according to Ohms’ law,

Question 45.
Define parallel connection of resistances and derive the resistance equation for equivalent resistance.
Answer:
1. When two or more resistances are connected between the same two points of a circuit, the resistances are said to be connected in parallel.

2. As shown in figure, consider three resistances R₁, R₂ and R₃ connected in parallel. Suppose a current I flows through the circuit when a cell of voltage V is connected across the combination.

3. The current I at point A is divided into three parts l₁, l₂ and l₃ through the resistances R₁, R₂ and R₃ respectively.
These three parts recombine at point B to give the same current I. Total current flowing in the circuit,
l=l₁+ l₂ + l₃

As all the three resistances have been connected between the same two points A and B, so voltage V across each of them is same. By Ohm’s law,
\(I 1=\frac{V}{R_1}, \quad 12=\frac{V}{R_2}, \quad I 3=\frac{V}{R_3}\)

If Rp be the equivalent resistance of the parallel combination, then,

Question 46.
State the characteristics of parallel connection of resistors.
Answer:
1. The sum of the current flowing through each resistor equals to total current flowing ¡n the circuit
i.e. l=l₁+ l₂ + …………………. +l_n
2. The voltage drop across each resistor remains the same.
3. The magnitude of equivalent resistance i.e. R is always smaller than the smallest resistance.
4. The reciprocal of equivalent resistance R is equal to the sum of reciprocal of individual resistors
i.e \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\ldots \ldots \ldots \ldots+\frac{1}{R_n}\)

Question 47.
In the circuit diagram given in figure  suppose the resistors R₁, R₂ and R₃ have the values 5Ω, 10Ω, 30Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current In the circuit, and (c) the total circuit resistance. (Textbook Example – 12.8)
Answer:
R₁=5Ω, R₂=10Ω and R₃=30Ω
Potential difference across the battery, V = 12 V.
This is also the potential difference across each of the individual resistor; therefore, to calculate
the current in the resistors, we use Ohm’s law.
The current l., through R₁ = V/R₁
l₁ =12 V/5Ω = 2.4 A.
The current 12, through R₁ = V/R₁
I₂ =12 V/10Ω = 1.2 A.
The current I₃, through R₃ = V/R₃
I₃ =12 V/30Ω = 0.4A.
The total current in the circuit,
I =I₁ +I₂+I₃=(2.4+1.2+0.4)A = 4A
The total resistance is given by
\(\frac{1}{R_p}=\frac{1}{5}+\frac{1}{10}+\frac{1}{30}=\frac{1}{3}\)
Thus, R_P = 3Ω.

Question 48.
lf in given figure R₁=10Ω, R₂=40Ω, R₃=30Ω, R₄=20Ω, R₅=60Ω, and a 12V battery is connected to the arrangement. Calculate (a) the total resistance In the circuit, and (b) the total current flowing in the circuit. (Textbook Example 12.9)
Answer:
Suppose we replace the parallel resistors R₁ and R₂ by an equivalent resistor of resistance, R’. Similarly we replace the
parallel resistors R₃, R₄ and R₅ by an equivalent single resistor of resistance R₁. Then, we have
1/R’ = 1/10 + 1/40 = 5/40; that is R’ = 8Ω.
Similarly,
1/ R” = 1/30 + 1/20 + 1/60 = 6/60; that is, R” = 10 Ω.
Thus, the total resistance,
R= R’+ R”= 18Ω
To calculate the current, we use Ohm’s law, and get
I =V/R= 12V/18Ω = 0.67A.

Question 49.
Three resistors are connected In parallel with the 30 V battery. The electric current of 7.5 A flows through circuit from battery. If the values of two resistors are 10 Ω and 12 Ω, determine the value of the third resistor.
Answer:
V = 30 V, 1= 7.5 A, R₁= 10Ω, R₂ = 12Ω, R₃ = ?
The equivalent resistance of circuit R =\(\frac{\mathrm{V}}{\mathrm{l}}=\frac{30}{7.5}!\) = 4Ω
Now, R₁, R₂ and R₃ are connected in parallel.
∴ R₃ = 15Ω

Question 50.
Find the electric current ¡n the following circuit:
Answer:
(A) Here, resistance R₁ is connected in parallel where as R₂ and R₃ are connected in series
Equivalent resistance of R₂ and R₃ is, R’ = R₂ + R₃ = 30 + 30 = 60Ω
Now R₁ and R’ are connected in parallel and so their equivalent resistance is, \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R^{\prime}}\)
∴ \(R=\frac{R_1 \times R^{\prime}}{R_1+R^{\prime}}=\frac{30 \times 60}{30+60}=\frac{1800}{90}=20 \Omega\)

(B) Total current flowing in the circuit \(I=\frac{V}{R}=\frac{2}{20}=0.1 \mathrm{~A}\)

Question 51.
State advantages and disadvantages of a series connection.
Answer:
Advantages of series connection:

• Series connection of resistors helps to control the current in the circuit.
• The fuse is connected in series with AC mains as well as with the electrical appliance. Thus, when a short circuit occurs in any electrical appliance, the fuse wire melts and stops electric current. Thus, senes connection prevents damage to electrical appliances.

Disadvantages of series connection:

• In series connection, the voltage gets divided. Due to this, the appliances give less output.
• If fault occurs in one appliance or at one part of the circuit, the current stops flowing in the entire circuit and other appliances also stop working.

Question 52.
State advantages and disadvantages of a parallel connection.
Answer:
Advantages of parallel connection:

• Voltage drop does not occur in parallel connection. As a result, appliances operate with full efficiency.
• If more than one appliances are connected in parallel and if an appliance stops working, even then the circuit will not break, and the other appliances will keep on working efficiently.
• The equivalent resistance in the parallel connection of resistor decreases, hence more current can be obtained.

Disadvantages of a parallel connection:

• In parallel connection, total resistance R of the circuit decreases, so total current increases. Hence, the current flow cannot be controlled.
• Bulbs having different voltages lets say 220 V, and 240 V, when connect in parallel and given voltage lets say 220 V, then only that bulb which has voltage rating of 220 V will light up with full efficiency.

Question 53.
Differentiate between series connection of resistors and parallel connection of resistors.
Answer:

Question 54.
Differentiate between resistance and resistivity.
Answer:

Question 55.
What is heating effect of electric current?
Answer:
1. When electric current is passed through a high resistance wire, the resistance wire becomes hot and produces heat.
2. Here, electrical energy ¡s converted into heat energy which is known as the heating effect of electric current.
3. The heat produced in an electrical appliance depends on electric current and resistance.

Question 56.
Explain electrical energy and derive its formula. OR Obtain the equation of heat energy for current flowing through a resistor for a specific time period.
Answer:
1. Suppose electric current is flowing through some resistor (R).
2. This results in continuous production of electric charges (Q).
3. The work done (W) by the battery of voit (V) for keeping the electric charges (Q) in continuous flow is,
W = V x Q
∴ W = V x I x t ( ∵ Q = I x t as per the definition of electric current)
Also, as per Ohm’s law, V = I. R
∴ W = l.R x lt = I²Rt
4. Thus, current flowing through a resistor R for time t is I.
5. W is the electrical energy consumed when current I flows through a resistor R in the circuit for time t.
6. This electrical energy gets converted into heat energy.
Thus, heat energy (H) = I² Rt
7. This equation is also called Joule’s law of heating.
8. The SI unit of electrical energy (W) or say heat energy (H) is joule (J).
9. The other units are Watt second (Ws) as well as Kilowatt – hour (kWh) or simply unit.

Question 57.
100 J of heat is produced each second in a 4Ω resistance. Find the potential difference across the resistor.
Answer:
H=100J, R=4Ω, t=1 s, V=?
We have the current through the resistor as

Thus the potential difference across the resistor, V is
V =IR
=5A x 4Ω
= 20V.

Question 58.
5A current flows through an electric Iron. If the resistance of electric iron is 44Ω then how much energy will be consumed in 5 minutes?
Answer:
l = 5A; R = 44Ω,t = 5min =5 x 60 = 300s.
Electricalenergy w = l²Rt
= (5)²(44) (300) = 330000 = x 10⁵ J

Question 59.
Discuss some practical applications of heating effect of electric current.
Answer:
Practical (daily life) applications of heating effect of electric current:
(i) Household heating appliances: Electric iron, toaster, sandwich maker, room heater, electric kettle, etc. ail such appliances make use of heating effect of electric current.

(ii) Electric bulbs: When electricity s passed through the filament of the electric bulbs, the bulbs light up. Here, heating effect is used for emitting light.

(iii) Electric fuse: Electric fuse is a safety device which works on the heating effect of electric current.

Question 60.
How is the heating effect of electric current used to light an electric bulb? Explain.
Answer:
1. The incandescent lamp or electric bulb is a glass which is filled with inactive nitrogen and argon gas.
2. A strong metal such as tungsten which has a very high melting point of 33800 C is used for making the filaments of these bulbs.
3. On passing electric current the filament becomes hot and emits light.

Question 61.
What is a fuse? How does the heating effect of electric current works on fuse? OR How does the fuse work?
Answer:
Fuse:

• A fuse is a safety device which works on the heating effect of electric current. It is connected in series with the device.
• It consists of a piece of wire made of metal or alloy having appropriate melting point.
• If the current larger than the specified amount flows through the circuit, the fuse wire becomes hot. This melts the fuse wire and the circuit gets broken. The connected device does not get electric supply and so it gets safeguarded against over-supply of electricity.
• Fuse for domestic purposes have rating of lA, 2A, 3A, 5A, 10A, etc.

Question 62.
Define electric power and state and explain its unit.
Answer:
Electric power:
1. The electrical energy consumed (or heat energy generated) in unit time ¡s called electric power.
2. In other words, electric power is the rate of electric energy.
3. It is denoted by R

4. The SI unit of power is joule/second or watt (W).

1 Unit power:

• If 1A current flows through the circuit through a battery of 1V, the power consumed will be 1 W.
• We have, P = IV
  ∴ 1 watt = 1 voIt x 1 ampere = 1 VA

Question 63.
Explain the practical unit of electrical energy. What do you mean by 1 unit power?
Answer:
1. \(\text { Power } P=\frac{\text { electric energy }(\mathrm{W})}{\text { time.(t) }}\)
∴ Electrical energy (W) = Power (P) x time (t) watt x second = joule
2. Thus, the unit of electrical energy is watt. second.
3. The larger unit of electrical energy is kWh.
1 kWh = 1000 watt x 3600 seconds
= 3.6 x 10⁶ watt.seconds
= 3.6 x 10⁶ joules (J)
4. kWh is a practical unit of energy which is used in every day life.
5. It is also called a ‘unit’.
6. Thus, 1 unit 1kWh = 3.6 x 10⁶ joules.
7. When a 1000 W bulb is on for 1 hour, the energy consumed is equal to 1 unit.

Question 64.
An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?
Answer:
We know that the power input is P = Vl
Thus, the current I = P/v
(a) When heating is at the maximum rate,
I = 840 W/220 V = 3.82 A; and the resistance of the electric iron is R = V/I = 220 V/3.82 A = 57.60 .

(b) When heating is at the minimum rate,
1= 360 W/220 V= 164 A; and the resistance of the electric iron is
R= V/l= 220 V/1 .64 A= 134.15Ω

Question 65.
An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
Answer:
P =Vl=220V x 0.50A=110 J/S=110 W.

Question 66.
An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kWh?
Answer:
The total energy consumed by the refrigerator in 30 days would be
400 W x 8.0 hour/day x 30 days = 96000 Wh = 96 kWh
Thus the cost of energy to operate the refrigerator for 30 days is
96 kWh x ₹ 3.00 per kWh = ₹ 288.00

Question 67.
In a house, if three bulbs of 100 W, 60 W and 30 W are used 2 hours per day, how many units of electrical energy will be consumed in 30 days?
Answer:
Energy consumed per sec P = 100 W + 60 W + 40 W= \(\frac{100}{200}\) W = kWh
P = 0.2 kWh
∴ Energy consumed per day = P x t = 0.2 kWh x 2 = 0.4 kWh
∴ Energy consumed in 30 days = 0.4 x 30 = 12 kWh
Now 1 unit = 1 kWh
∴ Energy consumed in 30 days = 12 units

Question 68.
For the circuit shown in figure below, determine the equivalent resistance between A and B. Also find the current flowing from the battery.

Answer:
As shown in the figure (part-a), R₂ and R₃ are connected in series. Their equivalent resistance
R’ =R₂+R₃=5+5=10Ω
Now R’ is parallel with R₄ (Figure Part b) equivalent resistance,

Now R” and R₁ are connected in series (Figure Part c). equivalent resistor across A and B,

∴ R =R”+R₁
= 5+5
= 10Ω
The current flowing from the battery,
\(I=\frac{V}{R}=\frac{10 V}{10 \Omega}=1 \mathrm{~A}\)

Question 69.
As shown in the figure the resistance are connected with a 12 V battery. Determine (a) Equivalent circuit resistance (b) Current flowing through the circuit.
(A) To determine equivalent resistance of the circuit, we fill first obtain :
(i) Equivalent resistance of R₁ and R₂ which are connected in parallel and
(ii) Equivalent resistance of R₃, R₄ and R₅
Answer:

(ii) Equivalent resistance of R₃, R₄ and R₅
\(\frac{1}{R^{\prime \prime}}=\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}=\frac{1}{30}+\frac{1}{20}+\frac{1}{60}=\frac{1}{10}\)
∴ R=\(\frac{10}{1}\) = 10Ω
Now, R’ and R” are connected in series and so equivalent resistance
R =R’+R”=8+10=18Ω
Thus, the equivalent resistance of the entire circuit is 18Ω

(C) Current flowing through the circuit
\(I=\frac{V}{R}=\frac{12}{18}\) = 0.66 A
Thus, 0.66 A current flows in the circuit.

Question 70.
A child has drawn the electric circuit to study Ohm’s law as shown in the figure here. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.
Answer:
The circuit diagram given in the question has three errors. They are:
(i) Ammeter is connected in parallel with R and voltmeter connected in series. In reality, ammeter should be connected in series and voltmeter in parallel.
(ii) The current is drawn from negative terminal and enters the battery through positive which
(iii) Cells are not connected in series combination in the battery of the circuit.
The correct diagram is figure 12.31.

Question 71.
Three 2Ω resistors’, A, B and C, are connected as shown in the figure. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors.
Answer:
Resistance, R = 2Ω
Maximum power, P_max = 18W
Maximum current, I_max =?
We know that P = I²R
\(I=\sqrt{\frac{P}{R}}=\sqrt{\frac{18}{2}}=3 A=I_{\max }\)

Maximum current that can flow through 2Ω resistor is 3A. This current divides along B and C because in parallel combination, voltage across B and C remain same and so,\(I \propto \frac{1}{R}\)
Since, B and C have same resistance i.e. 2Ω each, same current i.e., \(\frac{3}{2}\) = 1.5 A flows through B as well as C.

Question 72.
What is electrical resistivity? in a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?
Answer:
Resistivity: The property of a material to resist the flow of electric current is called its resistivity. It is denoted as ρ. The SI unity of resistivity is Ωm
The resistance of a uniform conductor is:
\(R=\rho \frac{1}{A}\)
Where,
I = the length of the conductor
ρ = the electrical resistivity of the material
A = the cross-sectional area
Now, if I is doubled, then R also gets doubled.

When, R is doubled, I becomes \(\frac{1}{2}\)

Question 73.
B₁, B₂ and B₃ are three identical bulbs connected as shown in Figure. When all the. three bulbs glow, a current of 3A is recorded by the ammeter A.
Answer:
(i) What happens to the glow of the other two bulbs when the bulb B₁ gets fused?
(ii) What happens to the reading of A₁, A₂, A₃ and A when the bulb B₂ gets fused?
(iii) How much power is dissipated in the circuit when all the three bulbs glow together?
Resistance of combination of three bulbs in parallel

(i) When bulb B₁ gets fused, the currents in B₂ and B₃ remain same. This means, I₂ = I₃ = 1A. The reason for this is that the voltage across the B₂ and B₃ bulb remains same. As a result, they keep illuminating with same intensity as before.

(ii) When bulb B₂ gets fused, the current in B₂ becomes zero and current in B₁ and B₃ remains 1 A. The reason for this is that the voltage across blub B₁ and B₃ remain same.
Total current I = 1 + 12 + 13 = 1 + 0 + 1 = 2A
Current in ammeter A₂ = 0 Current in ammeter A₁ = 1A
Current in ammeter A₃ = 1A Current in ammeter A = 2A

(iii) When all the three bulbs are connected.
Power dissipated, \(P=\frac{V^2}{R_{e q}}=\frac{(4.5)^2}{1.5}=13.5 \mathrm{~W}\)

Question 74.
A given length of a wire is doubled on itself and this process is repeated once again. By what factor does the resistance of the wire change?
Answer:

Question 75.
A wire of length I and area of cross-section A was drawn Into a wire of double its length by melting It. If Its original resistivity and resistance were p and R respectively, what will be its new resistivity and resistance?
Answer:
The volume of the wire remains same in both cases.
Volume = Area of cross-section x Length

Thus, the resistivity ρ remains the same.

Question 76.
A household uses the following electric appliances:
(i) Refrigerator of rating 400 W for two hours each day.
(ii) Two electrIc fans of rating 80 W each for twelve hours each day.
(iii) Six electric tubes of rating 18 W each for 6 hours each day.
Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is 3.00.
Answer:
1. Energy consumed by refrigerator of rating 400 W for ten hours each day
= P x t = 400W x 2h =\(\frac{400}{1000}\) kW x 2h = 0.8kWh

2. Energy consumed by two electric fans of rating 80 W each for twelve hours each day
= 2 x P x t = 2×80 W x 12 h = \(\frac{160}{1000}\) kW x 12 h = 1.92 kWh

3. Energy consumed by six electric tubes of rating 18 W each for six hours each day
=6 x P x t = 6 x 18 W x 6h = \(\frac{108}{1000}\) W x 6h = 0.648 kWh

4. Total energy consumed in the month of June (30 days)
= (0.8 + 1.92 + 0.648) x 30 kWh
= 6.568 x 30 = 197.04 kWh

5. Electricity bill for the month of June
= ₹ 197.04 x 3 = ₹ 591.12 = ₹ 591 (approx)

Very Short Answer Type Question :

Question 1.
What is electric charge? State Its types.
Answer:
Electric charge is the basic property of a matter carried by some elementary particles. The charge can be of two types namely positive charge and negative charge.

Question 2.
State the basic difference between static electricity and current electricity.
Answer:
Static electricity refers to the physical effect caused by the stationery charges whereas current electricity refers to physical effect caused by the moving charges.

Question 3.
What is conventional current? State the direction of flow of conventional current.
Answer:
The existence of electric current due to the flow of positive charges is known as conventional current. The direction of flow of conventional current is from the positive terminal to the negative terminal of a battery.

Question 4.
What is an open circuit?
Answer:
An electric circuit through which no current flows is called an open circuit.

Question 5.
What is a closed circuit?
Answer:
An electric circuit through which current flows continuously is called a closed circuit.

Question 6.
What is the function of an ammeter? How should it be connected in a circuit?
Answer:
Ammeter is used to measure the electric current in a circuit. It should be connected n series with the circuit.

Question 7.
Should the resistance of an ammeter be low or high? Give reason.
Answer:
Ammeter is used to measure the current and is connected in series with the circuit. If, ammeter has a high resistance, it will reduce the current flowing through the circuit and give an inaccurate result of measure of electric current. Hence, ammeter should have as low resistance as possible.

Question 8.
A charge of 900 C flows through a cross section of a conductor for 5 minutes. Calculate the electric current drawn by the conductor.
Answer:
Q = 900 t = 5 (mEnutes) x 60 (second) = 300 s
We know that \(I=\frac{Q}{t}\) ∴ \(I=\frac{900}{300}=3 A\)
Thus, electric current drawn by this circuit is 3 A.

Question 9.
Provided here is an electric circuit consisting of a bulb, battery and a key. I want to measure the potential difference potential difference between the two terminals of the bulb. Redraw the circuit by adding required device which can measure the potential difference.

Answer:

Question 10.
How much amount of charge can be moved from point to B If the work of 12 joule is done with the potential difference of 4 volt between these two points?
Answer:
W = 12J V = 4V
We know that V Volt \(=\frac{\mathrm{W}(\text { Joule })}{\mathrm{Q}(\text { Coulomb })}\)
∴ \(4=\frac{12}{Q}\)
∴ Q = 3C
Thus, 3C of charge can be moved from point A to B if work of 12 joule is done with potential difference of 4 volt between these points.

Question 11.
Heena Madam asked Rohit to draw a circuit containing Resister ‘R’ connected with a battery along with an ammeter to measure the current flowing through R. Rohit has drawn the circuit provided here and it has some errors in it. Rectify and redraw the circuit.

Answer:

Question 12.
Match the following components with their correct symbols.

Answer:
(1-b), (2-c)

Question 13.
Point out the errors in the circuit drawn below which measures the volt across resistor R.

Error 1 : Voltmeter is connected in series. It should be connected in parallel in order to measure the voltage.
Error 2 : The key is open. Hence the electric current cannot flow. It should be closed.

Question 14.
Two students perform experiments on two given resistors R₁ and R₂ and plot the V-l graphs as shown in diagram 1 and 2 respectively. If R₁> R₂, which of two diagrams correctly represent the situation on the plotted curves? Justify your answer.

Resistance is the slope of graph V-l. The graph with higher slope indicates that the resistance for that particular graph is higher. We can see from diagram 1 that R₁> R₂ hence diagram 1 correctly represents the situation.

Question 15.
What is the relationship between the potential coefficient and electric current flowing through a conductor in a circuit Who kind out this relation?
Answer:
The potential difference across a conductor is directly proportional to the electric current flowing through it in a circuit. This relation was found out by German physicist George Simon Ohm.

Question 16.
Calculate the potential difference of the battery in circuit provided here.
Answer:
l = 2.5mA = 2.5 x 10^-3A, R = 20 KΩ
= 20 x 10³ Ω
We know that V = IR
∴ V = (2.5 x 10^-3A) x (20 x 10³Ω) = 50V
Thus, the battery connected in this circuit has a voltage rating of 50 V.

Question 17.
What is the nature of a graph plotted with electric current at X-axis and potential difference at Y-axis?
Answer:
The graph plotted with electric current at X-axis and potential difference at Y-axis is a straight line passing through origin.

Question 18.
In an electric circuit, what change in electric current can we observe if we double the resistance keeping potential difference constant?
Answer:
1. Suppose, initially current and resistance of the circuit is I₁ α R₂ respectively.
2. After the resistance is doubled, current and resistance are I₂ and R₂ (R₂ = 2R₁)

Hence, if we double the resistance, current gets halved.

Question 19.
What is the relation between the resistance of a conductor and electric current flowing through it?
Answer:
In an electric circuit, the current is inversely proportional to the resistance of the conductor.

Question 20.
Classify below provided materials as (1) Good conductor, (2) Resistor, (3) Poor conductor and (4) Insulator.
Nichrome, Drinking water, Tungsten, Diamond.
Answer:
Good conductor: Tungsten, Resistor: Nichrome,
Poor conductor: Drinking water, Insulator: Diamond

Question 21.
On which factors does the resistance of a conductor depend?
Answer:
Resistance of a conductor depends on:

• Cross-sectional area,
• Length,
• Nature of material.

Question 22.
What is the relation of resistance with length and cross-sectional area of a conductor?
Answer:
Resistance is directly proportional to the length and inversely proportional to the area of cross-section of a conductor.

Question 23.
Choose the combination with highest resistance.
(a) A wire of length 2m and cross sectional area 2m²
(b) A wire of length 4m and cross sectional area of 2m²
(c) A wire of length 2m and cross sectional area 4m²
(d) A wire of length 4m and cross sectional area of 4m²
Answer:
The combination with a wire of length 4m and cross sectional area of 2m² will provide maximum resistance.

Question 24.
A wire of resistivity ρ is stretched to three times its length. What will be Its new resistivity?
Answer:
The resistivity of a material depends on the nature of the material of the wire and not its dimensions. Hence, the resistivity p will remain unchanged even if it is stretched three times.

Question 25.
Based on the resistivity of three distinct materials provided here, classify the materials Into conductors, alloys and insulators:
(a) 100 x 10^-6
(b) 1.84 x 10^-6
(c) 1.36 x 10^-10
Answer:
(a)-Alloy, (b)-Conductor, (c)-lnsulator

Question 26.
What happens to the resistance of a conductor. when its area of cross-section ¡s increased?
Answer:
\(R \propto \frac{1}{A}\) So, as the area of cross-section is increases, the resistance decreases.

Question 27.
The following table gives the resistivity of three samples:

Which of these is suitable for making heating elements of electrical appliances and why?
Answer:
Sample C has highest resistivity. Hence, it is most suitable for making heating elements of electrical appliances.

Question 28.
Haresh has purchased a coil of nichrome for his school project. He wants to cut out a segment from this coil such that resistance of the segment is 2 ohm. If the cross-sectional area of this coil is 3.14 x 10 m² and the resistivity of nichrome Is 100 x 10 ohm m, what should be the length of segment that Haresh needs to cut from coil?
Answer:
R=2Ω
A =3.14 x 10^-6 m²
ρ =100 x 10^-6  Ωm
I  = ?

Question 29.
Two wires of equal length, one of copper and the other of nichrome have the same resistance. Which wire will be thicker?
Answer:
Nichrome wire is thicker than copper wire.

Question 30.
Arrange the below-mentioned metals In order such that the metal with lowest conductivity comes first and the one with highest conductivity comes last.
(a) Copper
(b) Mercury
(c) Silver
(d) Iron
Answer:
(c), (a), (d), (b)

Question 31.
Which are the methods of joining resistors In a circuit?
Answer:
There are two combination of joining resistors in a circuit. They are
(1) Series connection and
(2) Parallel connection

Question 32.
Name the physical quantities which remain same in all the resistors when (i) they are connected In series, (ii) they are connected In parallel.
Answer:
(i) When the resistors are connected in series, the electric current is same for all the resistors.
(ii) When the resistors are connected in parallel, the voltage across each of the resistors is same.

Question 33.
Calculate the equivalent resistance in the combination of resistance shown here.

Answer:
Here, R₁ and R₂ are connected in parallel connection and R₃ is in series.
∴ equivalent resistance \(R=\frac{R_1 R_2}{R_1+R_2}+R_3\)
∴ R = \(\frac{5 \times 5}{5+5}\)+2.5=2.5+2.5 R =5Ω

Question 34.
Calculate the voltage of battery connected in circuit provided here.

Answer:
We know that V = IR
∴ V = 2A x (6+2+3)Ω
V = 22V

Question 35.
Draw a schematic diagram of an electric circuit comprising of 3 cells, a parallel combination of two bulbs and plug key in the ON mode connected with an ammeter and a voltmeter.
Answer:

Question 36.
Two resistors are connected in a parallel combination such that the equivalent resistance s 1 ohm. If one of the resistors has a value of 2 ohm, calculate the value for the other resistor.
Answer:
We know that for parallel combination,
\(R=\frac{R_1 R_2}{R_1+R_2}\)
(But equivalent resistance = 1Ω and R₁ = 2Ω)
∴ \(1=\frac{2 R_2}{2+R_2}\)
∴ 2+R₂=2R₂
∴ R₂ = 2Ω

Question 37.
Out of the two wires X and Y shown below, which one has greater resistance? Justify.

Answer:
The resistance is directly proportional to the length of the wire. Now we can see that wire Y is longer than X and hence it will offer higher resistance.

Question 38.
What would be the equivalent resistance of a parallel combination of resistors made by cutting a wire of resistor R in three equal parts. What will be the effect on resistivity of each of the segments?
Answer:
Resistance of wire (with resistance R) when cut in 3 equal parts would have resistance R/3 for each segment.
Let’s say resistances are r₁, r₂ and r₃
∴ equivalent resistance = \(\frac{r_1 r_2 r_3}{r_1+r_2+r_3}\)
But, r₁ = r₂ = r₃ = R/3
∴ equivalent resistance \(=\frac{\frac{\mathrm{R}^3}{27}}{\mathrm{R}}\)
∴ equivalent resistance \(=\frac{\mathrm{R}^2}{27}\)
Resistivity of the segments will remain the same as it was in the wire before cutting it because resistivity does not depend on the length of material.

Question 39.
Why do we use copper and aluminium wires for transmission of electric current?
Answer:
Because both copper and aluminium have low resistivity and hence high conductivity

Question 40.
Which one is having lesser resistance? A 220 V, 60 W bulb or a 220 V, 40 W bulb?
Answer:
1.  We know that P = V²/R
∴ R=V²/P
2. Now, Voltage applied is same for both the bulbs, hence relative value of resistances of both the bulbs are dependent on the value of powers.
Now, R α 1/P
Therefore, more the power less the resistance.
Hence, the bulb with 60W rating will have lesser resistance than the bulb with 40W rating.

Question 41.
Give one reason which justifies that parallel circuits are preferred in domestic wiring.
Answer:
In parallel circuit, if one device gets damaged and stops working, there is no effect in the functioning of other connected devices. Hence, parallel circuits are preferred in domestic wiring.

Question 42.
Write relation between heat energy produced In a conductor when a potential difference V is applied across its terminals and a current I flows through in time t. What is this relation known as?
Answer:
Heat energy can be expressed as H = Vit. This is also known as Joule’s law of heating

Question 43.
State the relationship between 1 Volt and 1 Joule.
Answer:
1 Volt \(=\frac{\text { 1Joule }}{1 \text { Coulomb }} \text { or } 1 \mathrm{~V}=\frac{1 \mathrm{~J}}{1 \mathrm{C}}\)

Question 44.
As per the Joule’s law of heating, on which parameters does the heat produced In a resistor depend?
Answer:
The heat produced in a resistor is directly proportional to (a) square of current, (b) resistance and (c) time for which the current flows through resistor.

Question 45.
State the difference between the wire used In the element of an electric heater and in a fuse wire.
Answer:
Wire used in electric heater should have a high resistivity and high melting point allowing the wire to produce heat without melting down where as a fuse wire should have a low resistance and low melting point so that it can protect the appliance connected with it due to fluctuating current.

Question 46.
List two characteristics of the material used in electric heating devices.
Answer:
Material used in heater wire should have

• High Resistivity and
• High melting point.

Question 47.
An electric iron draws a current of 3.5 A when connected with a voltage source of 220V. Calculate the energy consumed by iron if it is kept on for 2 hour.
Answer:
I = 3.5A
V = 220V
t = 2hr
energy consumed E =?
We know that E = Vit
=220 x 3.5 x 2
=1540 Wh
∴ E= 1.54 kWh

Question 48.
Why do we use tungsten for making bulb filaments?
Answer:
The filament in bulb should have the capacity to retain extremely hot temperature so that it can emit light. As tungsten has a very high melting point (3380°C). It can retain heat without melting down and so it is used for making bulb filaments.

Question 49.
How does use of a fuse wire protect electrical appliance?
Answer:
Fuse wire has a very low melting point. A fuse is connected in series with the appliance. Hence if more than an acceptable amount of current flows in the circuit, the fuse wire melts down due to heating effect and breaks the circuit. This helps
in protecting the appliance against the damage caused by fluctuating current.

Question 50.
Melting points of three metals are provided here: (a) 3380 °C, (b) 1085 °C, (c) 232°C. Which of these metal can be used to produce a fuse wire, a bulb and an electric transmission line.
Answer:
Metal with melting point 3380°C, should be used for producing bulb, metal with melting point 1085 °C should be used for producing electrictransmission line and the metal with melting point 232 °C should be used for producing fuse wire.

Question 51.
400 J of heat is produced in 4s in a 4Ω resistor. Find potential difference across the resistor.
Answer:
H = 400J, t = 4s and R = 4Ω.
We know that,
H=I²Rt
∴ I²=H/Rt
∴ I =5A
We Also know that,
H = V/t
∴ V=H/It
∴ V = 20V

Question 52.
How much electric energy can be consumed by a 150W toaster in lo minutes?
Answer:
P = 150 W, t = 10 minutes = 10 x 60 seconds
= 600 seconds
We know that electric energy E = Pt
∴ E = 150 x 600 = 90,000 J
∴ E =90 x 10³ J

Question 53.
Can a fuse wire of capacity 3A protect a circuit consuming 1 kW electric power when connected with a voltage source of 220V?
Answer:
The electric circuit that has a 1 kW power at 220 V will draw electric current I = P/V
∴ \(I=\frac{1000}{220}\) = 4.54 A
Hence, the fuse wire of 3A current rating cannot protect this circuit as it will require a fuse wire of More than, 4.54 A rating.

Question 54.
What is the commercial unit of electrical energy? Represent it in terms of joules. Commercial unit of energy is kW h. It can be represented in Joules as follows:
Answer:
1 Kilowatt hour = 1000 watt hour
We know that 1 watt = 1 joule / seconds
∴ 1 kWh = \(\frac{1000 \text { joule }}{\text { second }}\) x 3600 (second) second
∴ 1 kWh = 3.6 x 10⁶ joules

Question 55.
An electric device operates at 24 V and has a resistance of 8Ω. Calculate the power consumed by the device and current flowing through it.
Answer:
Voltage = 24V and R = 8Ω
We know that I = V/R
∴ I = 3A
And, P=Vl
∴ P = 24 x 3
∴ P = 72W

Question 56.
An electric current of 5.0 A flows through a 12Ω resistor. What is the rate at which heat energy is produced in the resistor?
Answer:
l = 5A and R=12Ω
Rate of consumption of energy is the power and
we know that,
P = I²R
∴ P=(5)² x 12
∴ P = 300W

Fill in the blanks:

1. …………………. is expressed by the amount of charge flowing through a particular area in unit time.
Answer: Electric current

2. Coulomb charge constitute electrons.
Answer:  6 x 10¹⁸

3. In ………………… combination of resistance, the magnitude of equivalent resistance is greater than the largest resistance.
Answer: Series

4. The work done on the electric charge to maintain its equilibrium with the other charge is called ………………
Answer: Electric potential

5. A/An ………………. (instrument) measures electric current in a circuit whereas a/an (instrument) is used to measure the potential difference.
Answer: ammeter, voltmeter

6. Chemical reaction within a cell is responsible for generating ……………………. across its terminals.
Answer:  potential difference

7. A device called ………………… is used to change electric current flowing in a circuit by keeping potential difference constant.
Answer: rheostat

8. Joule/Coulomb is the SI unit for ………………………
Answer:  potential difference

9. Voltmeter is connected in …………………….. and ammeter is connected in ………………..
Answer: Parallel; Series

10. The graph of potential difference v/s electric current is a and it validates Law.
Answer: straight line, Ohm’s

11. The unit Volt/Ampere can also be represented as and is symbolized as ……………………..
Answer: Ohm

12. If we increase the cross sectional area of a conductor, the amount of current flowing through it ………. whereas the resistance …………………….. .
Answer: increases, decreases

13. The resistance of any conduction substance is …………….. proportional to length of the substance.
Answer: Directly

14. The resistance of a conductor is proportional to area of cross-section of the substance.
Answer: Inversely

15. ………….. is widely used as a filament in electric bulbs.
Answer: Tungsten

16. In a combination of resistance, all the resistors are simultaneously connected together between two points in a circuit, whereas in a combination, resistors are connected end to end consecutively between two points in a circuit.
Answer: parallel, series

17. In an electric circuit, a part of the source energy is consumed in the form of useful work whereas a fraction of this energy gets expended in form of…………………
Answer:  heat

18. Heat produced in a resistor is directly proportional to ………………….., and ……………………..  of electric current flowing through the circuit and also time during which the current flows through circuit.
Answer: resistance, square

19. Devices such as electric iron and electric heater works on the principle of …………… Law.
Answer:  Joule’s

20. A porcelain cartridge encases a ……………… which protects the appliances from fluctuating currents.
Answer: fuse wire

21. 1 kWh =………………….
Answer:  3.6 x 10⁶ joule

22. If you run ……………… watt bulb for 1 hour, the energy consumed will be 1 unit.
Answer: 10³

23. A fuse wire should have relatively …………………… (higher/lower) melting point than the conductor it is connected with.
Answer: lower

24. ………………………………. can be termed as the rate of consumption of energy.
Answer: Power

25. Electric energy consumed is the product of power and …………….
Answer:  time

True OR False:

1. The direction of electric current flowing through a conductor is from the negative terminal to positive terminal of a battery which is opposite to the direction of conventional current. — True
2. The electric current in a circuit gets doubled as we double the length of a conductor through which the current flows. — False
3. 6 x 10¹² electrons flowing through a conductor in one second constitutes one microampere of electric current. — True
4. In a series combination of resistors, equal amount of electric current flows through each resistor whereas, in a parallel circuit, the current gets divided proportionately among each resistors. — True
5. A voltmeter should be connected in a series combination and an ammeter should be connected in a parallel combination with the conductor through which we want to calculate current and voltage. — False
6. The V-l graph provided here is a straight line with constant resistance and it validates the Ohm’s law — False
7. Two resistors of same material with different length and same cross-sectional area will have different resistivity, — False
8. Resistance as well as resistivity of a material are dependent on the temperature of that material. — True
9. Material A has a resistivity of 5.2 x 10^-8 and material B has a resistivity of 49 x 10^-8. In this case, A can be chosen over B to produce electric transmission lines. — True
10. In a series combination of a resistors 1 equal amount of potential difference is observed across each resistor whereas in a parallel combination, potential difference gets divided based on the value each resistor. — False
11. Equivalent resistance in a parallel combination is higher than the highest individual resistance of the resistor connected in circuit. — False
12. Electric bulbs are filled with non-reactive noble gas to prolong the life of filament. — True
13. A large amount of power consumed by an electric bulb is used in emitting light and the remaining amount gets dissipated in the form of heat. — False
14. A fuse wire of 4A current rating is incapable of protecting an appliance with a rating of 5A. — True
15. Fuse wires are always connected in parallel to the circuit because they have very low resistance hence any high fluctuating amount of current would pass through the fuse without harming the appliance connected with it. — False
16. Rate of doing work is the product of voltage applied in the circuit and the current flowing through it. — True
17. 1 kW h is the energy consumed when one Watt of power is utilized for one hour.– False

Match the Following:
Question 1.

Answer:
(1-c), (2-d), (3-b), (4-a)

Question 2.

Answer: (1-c), (2-a), (3-e), (4-d)

Question 3.

Answer:
(1-c),(2-e),(3-d),(4-b)

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 1 वास्तविक संख्याएँ Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 1 वास्तविक संख्याएँ

परीक्षोपयोगी अन्य महत्त्वपूर्ण प्रश्न

प्रश्न 1.
240 और 6552 का HCF यूक्लिड विभाजन एल्गोरिथ्म का प्रयोग करके ज्ञात कीजिए ।
हल :

यूक्लिड विभाजन एल्गोरिथ्म से-
6552 = 240 × 27 + 72
240 = 72 × 3 + 24
72 = 24 × 3 + 0
क्योंकि यहाँ पर शेषफल शून्य तथा भाजक 24 है।
इसलिए 240 और 6552 का HCF = 24

प्रश्न 2.
दर्शाइए कि कोई भी धनात्मक पूर्णांक 3q या 3q + 1 या 3q + 2 के रूप का होता है, जहाँ q कोई पूर्णांक है।
हल :
माना a एक धनात्मक पूर्णांक है तथा b = 3 हो तो यूक्लिड विभाजन एल्गोरिथ्म से-
a = 3q + r
क्योंकि 0 ≤ r < 3 है, इसलिए संभावित शेषफल 0, 1 या 2 हो सकते हैं ।
अर्थात् a संख्याओं 3q, 3q + 1 या 3q + 2 के रूप का हो सकता है जहाँ q कोई पूर्णांक है ।

प्रश्न 3.
निम्नलिखित संख्याओं को \(\frac {p}{q}\) के रूप में व्यक्त कीजिए-
(i) 0.375
(ii) 0.104
(iii) 0.15
(iv) 15.75
हल :

प्रश्न 4.
एक मिठाई विक्रेता के पास 420 काजू की बर्फियाँ और 130 बादाम की बर्फियाँ हैं । वह इनकी ऐसी ढेरियाँ बनाना चाहती है कि प्रत्येक ढेरी में बर्फियों की संख्या समान रहे तथा ये ढेरियाँ बर्फी की परात में न्यूनतम स्थान घेरें । इस काम के लिए प्रत्येक ढेरी में कितनी बर्फियाँ रखी जा सकती हैं?
हल :
प्रश्नानुसार, यहाँ पर हमें 420 और 130 का HCF यूक्लिड एल्गोरिथ्म द्वारा ज्ञात करना है-

420 = 130 × 3 + 30
130 = 30 × 4 + 10
30 = 10 × 3 +0
क्योंकि यहाँ पर शेषफल शून्य तथा भाजक 10 है ।
इसलिए 420 और 130 का HCF = 10
अतः प्रत्येक प्रकार की बर्फी के लिए मिठाई विक्रेता 10-10 की ढेरी बना सकता है

प्रश्न 5.
8232 को अभाज्य गुणनखंडों के गुणनफल के रूप में व्यक्त कीजिए ।
हल :

8232 = 2 × 2 × 2 × 3 × 7 × 7 × 7
= 2³ × 3 × 7³

प्रश्न 6.
सिद्ध कीजिए कि \(\sqrt{2}\) एक अपरिमेय संख्या है।
हल :
माना \(\sqrt{2}\) एक परिमेय संख्या है जो कि दिए गए के विपरीत है।
अब \(\sqrt{2}\) = \(\frac {a}{b}\) जहाँ a और b सह-अभाज्य पूर्णांक हैं तथा b ≠ 0 है।
⇒ \(\sqrt{2}\)b = a
दोनों ओर का वर्ग करने पर, 2b² = a² ………..(i)
इससे पता चलता है कि a², 2 से विभाज्य है, इसलिए a भी 2 से विभाज्य होगी। ………..(ii)
⇒ a = 2m जहाँ m एक पूर्णांक है ।
a का मान समीकरण (i) में रखने पर,
2b² = (2m)²
⇒ 2b² = 4m²
या b² = 2m²
इसका अर्थ यह है कि 2 से b² विभाजित हो जाता है इसलिए b भी 2 से विभाजित होगा । ………….(iii)
इसी प्रकार समीकरण (ii) व (iii) से a और b में कम-से-कम एक उभयनिष्ठ गुणनखंड 2 है। इससे हमारी कल्पना गलत होती है कि a और b सह-अभाज्य हैं । अतः सिद्ध है कि \(\sqrt{2}\) एक परिमेय संख्या नहीं है।
अतः \(\sqrt{2}\) एक अपरिमेय संख्या है।

प्रश्न 7.
संख्या 40, 36 और 126 का अभाज्य गुणनखंडन विधि द्वारा HCF और LCM ज्ञात कीजिए ।
हल :

40 = 2 × 2 × 2 × 5 = 2³ × 5
36 = 2 × 2 × 3 × 3 = 2² × 3²
126 = 2 × 3 × 3 × 7 = 2 × 3² × 7
∴ HCF(40, 36, 126) = अभाज्य गुणनखंडों की उभयनिष्ठ सबसे छोटी घातों का गुणनफल = 2 उत्तर
LCM(40, 36, 126) = अभाज्य गुणनखंडों की सबसे बड़ी घातों का गुणनफल = 2³ × 3² × 5 × 7
= 8 × 9 × 5 × 7 = 2520

प्रश्न 8.
छः अंकों की सबसे बड़ी संख्या ज्ञात कीजिए जो 24, 15 तथा 36 से पूर्णतया विभाज्य हो ।
हल :
यहाँ पर हम पहले 24, 15 व 36 का LCM ज्ञात करेंगे।

24 = 2 × 2 × 2 × 3 = 2³ × 3
15 = 3 × 5
36 = 2 × 2 × 3 × 3 = 2² × 3²
LCM(24, 15, 36) = 2³ × 3² × 5 = 8 × 9 × 5 = 360
हम जानते हैं कि छः अंकों की बड़ी से बड़ी संख्या = 999999
अतः अभीष्ट संख्या = 999999 – 279 = 999720

प्रश्न 9.
यदि दो संख्याओं का LCM व HCF क्रमशः 180 व 6 हो तथा इनमें से एक संख्या 30 हो तो दूसरी संख्या ज्ञात कीजिए ।
हल :
यहाँ पर,
LCM = 180
HCF = 6
पहली संख्या = 30
∴ दूसरी संख्या = LCM × HCF / पहली संख्या
= \(\frac{180 \times 6}{30}\)
= 6 × 6 = 36

प्रश्न 10.
वह बड़े से बड़े धनात्मक पूर्णांक ज्ञात कीजिए जो 398, 436 और 542 को विभाजित करने पर क्रमशः 7, 11 व 15 शेषफल छोड़े?
हल :
यहाँ पर

अब हम 391, 425 व 527 का HCF ज्ञात करेंगे-

391 = 17 × 23
425 = 5 × 5 × 17 = 5² × 17
527 = 17 × 31
HCF (391, 425, 527) = 17
अतः वांछित पूर्णांक = 17

प्रश्न 11.
सिद्ध कीजिए कि निम्नलिखित संख्याएँ अपरिमेय है-
(i) 3 – 2\(\sqrt{5}\)
(ii) 2\(\sqrt{3}\)
हल :
(i) माना 3 – 2\(\sqrt{5}\) एक परिमेय संख्या है जो कि दिए गए के विपरीत है ।
अब 3 – 2\(\sqrt{5}\) = \(\frac {a}{b}\) जहाँ a और b सह-अभाज्य पूर्णांक हैं तथा b ≠ 0 है।
⇒ 2\(\sqrt{5}\) = 3 – \(\frac {a}{b}\)
या 2\(\sqrt{5}\) = \(\frac{3 b-a}{b}\)
या \(\sqrt{5}\) = \(\frac{3 b-a}{2 b}\)
क्योंकि a और b पूर्णांक हैं जिस कारण \(\frac{3 b-a}{2b}\) एक परिमेय संख्या होगी ।
इसलिए \(\sqrt{5}\) एक परिमेय संख्या होगी जो कि असत्य है क्योंकि \(\sqrt{5}\) एक अपरिमेय संख्या है।
अतः हमारी कल्पना गलत है। इससे सिद्ध होता है कि 3 – 2\(\sqrt{5}\) एक अपरिमेय संख्या है।

(ii) यदि संभव हो तो माना 2\(\sqrt{3}\) एक परिमेय संख्या है।
तो 2\(\sqrt{3}\) = \(\frac {a}{b}\) जहाँ a और b सह-अभाज्य पूर्णांक हैं तथा b ≠ 0 है।
⇒ \(\sqrt{3}\) = \(\frac {a}{2b}\)
क्योंकि a और b पूर्णांक हैं जिस कारण \(\frac {a}{2b}\) एक परिमेय संख्या होगी ।
⇒ \(\sqrt{3}\) एक परिमेय संख्या होगी जो कि असत्य है क्योंकि \(\sqrt{3}\) एक अपरिमेय संख्या है।
अतः हमारी कल्पना गलत है। इससे सिद्ध होता है कि 2\(\sqrt{3}\) एक अपरिमेय संख्या है।

प्रश्न 12.
सिद्ध कीजिए कि \(\sqrt{3}\) एक अपरिमेय संख्या है।
हल :
माना \(\sqrt{3}\) एक परिमेय संख्या है जो कि दिए गए के विपरीत है ।
अब \(\sqrt{3}\) = \(\frac {a}{b}\) जहाँ a और b सह-अभाज्य पूर्णांक हैं तथा b ≠ 0 है ।
⇒ \(\sqrt{3}\)b = a
दोनों ओर का वर्ग करने पर,
3b² = a² ………..(i)
इससे पता चलता है कि a², 3 से विभाज्य है, इसलिए a भी 3 से विभाज्य होगी। …………..(ii)
⇒ a = 3m जहाँ m एक पूर्णांक है।
a का मान समीकरण (i) में रखने पर,
3b² = (3m)²
⇒ 3b² = 9m²
या b² = 3m²
इसका अर्थ यह है कि 3 से b² विभाजित हो जाता है इसलिए b भी 3 से विभाजित होगा । …………..(iii)
इसी प्रकार समीकरण (ii) व (iii) से a और b में कम-से-कम एक उभयनिष्ठ गुणनखंड 3 है। इससे हमारी कल्पना गलत होती है कि a और b सह-अभाज्य हैं । जिस कारण \(\sqrt{3}\) एक परिमेय संख्या नहीं है ।
अतः \(\sqrt{3}\) एक अपरिमेय संख्या है।

प्रश्न 13.
सिद्ध कीजिए कि 3\(\sqrt{2}\) एक अपरिमेय संख्या है।
हल :
यदि सम्भव हो तो माना 3\(\sqrt{2}\) एक परिमेय संख्या है।
तो 3\(\sqrt{2}\) = \(\frac {a}{b}\) जहाँ a और b सह – अभाज्य पूर्णांक हैं तथा b ≠ 0 है।
⇒ \(\sqrt{2}\) = \(\frac {a}{3b}\)
क्योंकि a और b पूर्णांक हैं जिस कारण \(\frac {a}{3b}\) एक परिमेय संख्या होगी ।
⇒ \(\sqrt{2}\) एक परिमेय संख्या होगी जोकि असत्य है क्योंकि \(\sqrt{2}\) एक अपरिमेय संख्या है।
अतः हमारी कल्पना गलत है। इससे सिद्ध होता है कि 3\(\sqrt{2}\) एक अपरिमेय संख्या है।

प्रश्न 14.
\(\frac {7}{160}\) का दशमलव प्रसार लिखिए ।
हल :

प्रश्न 15.
परिमेय संख्या 34.12345 के हर के अभाज्य गुणनखंडों के विषय में आप क्या कह सकते हैं?
हल :
क्योंकि 34.12345 एक सांत दशमलव संख्या है। इसलिए इस हर के अभाज्य गुणनखंडन 2ⁿ × 5^m के रूप के होंगे, जहाँ n और m पूर्णांक होंगे।

बहुविकल्पीय – प्रश्न :

प्रश्न 1.
निम्नलिखित में से कौन-सी संख्या अपरिमेय नहीं है?
(A) \(\sqrt{2}\)
(B) \(\sqrt{3}\)
(C) \(\sqrt{4}\)
(D) \(\sqrt{5}\)
हल :
(C) \(\sqrt{4}\)

प्रश्न 2.
प्रत्येक भाज्य संख्या को एक अद्वितीय रूप से …………….. संख्याओं के गुणनफल के रूप में व्यक्त किया जा सकता है ।
(A) अभाज्य
(B) सम
(C) विषम
(D) भाज्य
हल :
(A) अभाज्य

प्रश्न 3.
यूक्लिड विभाजन प्रमेयिका a = bq + का प्रयोग 17 व 6 के लिए करने पर का मान होगा-
(A) 6
(B) 2
(C) 17
(D) 5
हल :
(D) 5

प्रश्न 4.
यूक्लिड विभाजन प्रमेयिका a = bq + r का प्रयोग 20 व 4 के लिए करने पर का मान होगा-
(A) 5
(B) शून्य
(C) 4
(D) 20
हल :
(B) शून्य

प्रश्न 5.
12 व 5 का HCF क्या होगा ?
(A) 5
(B) 12
(C) 1
(D) 60
हल :
(C) 1

प्रश्न 6.
किन्हीं दो अभाज्य संख्याओं का HCF क्या होगा ?
(A) दोनों संख्याओं का गुणनफल
(B) बड़ी संख्या
(C) छोटी संख्या
(D) एक
हल :
(D) एक

प्रश्न 7.
20 व 5 का HCF क्या होगा ?
(A) 1
(B) 5
(C) 20
(D) 4
हल :
(B) 5

प्रश्न 8.
10 व 3 को यूक्लिड विभाजन प्रमेयिका के रूप में लिखा जा सकता है-
(A) 10 = 3 × 3 – 1
(B) 10 = 3 × 3 + 1
(C) 10 = 3 × 1 + 7
(D) 3 = 10 × 3 + 1
हल :
(B) 10 = 3 × 3 + 1

प्रश्न 9.
4 व 19 का HCF क्या होगा ?
(A) 4
(B) 19
(C) 1
(D) 76
हल :
(C) 1

प्रश्न 10.
510 व 92 का HCF क्या होगा ?
(A) 2
(B) 81
(C) 27
(D) 243
हल :
(A) 2

प्रश्न 11.
दो धनात्मक पूर्णांकों के LCM और HCF में सम्बन्ध होगा-
(A) HCF > LCM
(B) HCF = LCM
(C) LCM > HCF
(D) इनमें से कोई नहीं
हल :
(B) HCF = LCM (यदि धनात्मक पूर्णांक समान हों) तथा (C) LCM > HCF (यदि धनात्मक पूर्णांक भिन्न हों)

प्रश्न 12.
प्रत्येक धनात्मक विषम पूर्णांक को किस रूप में लिखा जा सकता है, जबकि p कोई पूर्णांक हो ?
(A) 2p+ 1
(B) 2p
(C) 2p + 2
(D) 4p
हल :
(A) 2p + 1

प्रश्न 13.
निम्नलिखित में से किस युग्म का HCF, 4 होगा ?
(A) 23, 4
(B) 28, 4
(C) 31, 4
(D) 10, 4
हल :
(B) 28, 4

प्रश्न 14.
निम्नलिखित में से किस युग्म का HCF, 4 नहीं होगा ?
(A) 12, 4
(B) 20, 4
(C) 22, 4
(D) 28, 4
हल :
(C) 22, 4

प्रश्न 15.
निम्नलिखित में से अपरिमेय (irrational) संख्या छाँटिए-
(A) \(\sqrt{36}\)
(B) \(\sqrt{121}\)
(C) \(\sqrt{9}\)
(D) \(\sqrt{8}\)
हल :
(D) \(\sqrt{8}\)

प्रश्न 16.
यदि 306 और 657 का HCF 9 हो, तो उसका LCM क्या होगा ?
(A) 2482
(B) 22338
(C) 2754
(D) 5913
हल :
(B) 22338

प्रश्न 17.
26 व 91 का HCF क्या होगा ?
(A) 26
(B) 867
(C) 13
(D) एक
हल :
(C) 13

प्रश्न 18.
1232 व 32 का HCF क्या होगा ?
(A) 16
(B) 616
(C) 8
(D) 2464
हल :
(A) 16

प्रश्न 19.
336 व 54 का HCF क्या होगा ?
(A) 9
(B) 3
(C) 6
(D) 18
हल :
(C) 6

प्रश्न 20.
यदि q कोई पूर्णांक हो तो एक धनात्मक विषम पूर्णांक को लिखा जा सकता है-
(A) 4q
(B) 4q + 1
(C) 4q + 2
(D) 4q + 4
हल :
(B) 4q + 1

प्रश्न 21.
यदि q कोई पूर्णांक हो तो एक धनात्मक विषम पूर्णांक को किस रूप में नहीं लिखा जा सकता?
(A) 6q +1
(B) 6q + 2
(C) 6q + 3
(D) 6q + 5
हल :
(B) 6q + 2

प्रश्न 22.
एक मिठाई विक्रेता के पास 420 काजू की बर्फियाँ और 130 बादाम की बर्फियाँ हैं। वह इनकी ऐसी ढेरियाँ बनाना चाहता है कि प्रत्येक ढेरी में बर्फियों की संख्या समान रहे तथा ये ढेरियाँ बर्फी की परात में न्यूनतम स्थान घेरें । इस काम के लिए, प्रत्येक ढेरी में कितनी बर्फियाँ रखी जा सकती हैं?
(A) 30
(B) 130
(C) 20
(D) 10
हल :
(D) 10

प्रश्न 23.
किसी परेड में 616 सदस्यों वाली एक सेना की टुकड़ी को 32 सदस्यों वाले एक आर्मी बैंड के पीछे मार्च करना है। दोनों समूहों को समान संख्या वाले स्तंभों में मार्च करना है। उन स्तंभों की अधिकतम संख्या क्या होगी जिसमें वे मार्च कर सकते हैं?
(A) 8
(B) 32
(C) 4
(D) 2
हल :
(A) 8

प्रश्न 24.
प्रत्येक भाज्य संख्या को अभाज्य संख्याओं के एक गुणनफल के रूप में व्यक्त (गुणनखंडित) किया जा सकता है तथा यह गुणनखंडन अद्वितीय होता है, इस पर कोई ध्यान दिए बिना कि अभाज्य गुणनखंड किस क्रम में आ रहे हैं। इसे कहा जाता है-
(A) यूक्लिड विभाजन प्रमेय
(B) अंकगणित की आधारभूत प्रमेय
(C) बीजगणित की आधारभूत प्रमेय
(D) रेखागणित की आधारभूत प्रमेय
हल :
(B) अंकगणित की आधारभूत प्रमेय

प्रश्न 25.
सबसे छोटी अभाज्य संख्या है-
(A) 2
(B) 3
(C) 1
(D) 5
हल :
(A) 2

प्रश्न 26.
सबसे छोटी विषम अभाज्य संख्या है-
(A) 1
(B) 2
(C) 3
(D) 5
हल :
(C) 3

प्रश्न 27.
30 के अभाज्य गुणनखंड होंगे-
(A) 6 × 5
(B) 1 × 30
(C) 2 × 15
(D) 2 × 3 × 5
हल :
(D) 2 × 3 × 5

प्रश्न 28.
दो पूर्णांकों 336 तथा 54 का LCM होगा-
(A) 12
(B) 3024
(C) 6
(D) इनमें से कोई नहीं
हल :
(B) 3024

प्रश्न 29.
72 के अभाज्य गुणनखंड होंगे-
(A) 2³ × 3³
(B) 2³ × 3³
(C) 2³ × 3³
(D) 2³ × 3²
हल :
(D) 2³ × 3²

प्रश्न 30.
यदि 96 और 404 का HCF 4 हो तो उनका LCM होगा-
(A) 38784
(B) 9696
(C) 2274
(D) इनमें से कोई नहीं
हल :
(B) 9696

प्रश्न 31.
140 के अभाज्य गुणनखंड होंगे-
(A) 2² × 5² × 7²
(B) 2² × 5² × 7
(C) 2² × 5 × 7
(D) 2 × 5 × 7
हल :
(C) 2² × 5 × 7

प्रश्न 32.
156 के अभाज्य गुणनखंड होंगे-
(A) 2² × 3 × 13
(B) 2² × 3² × 13
(C) 2² × 3² × 13²
(D) 2 × 3² × 13
हल :
(A) 2² × 3 × 13

प्रश्न 33.
12 और 20 के HCF और LCM में सम्बन्ध होगा-
(A) HCF > LCM
(B) HCF < LCM
(C) HCF = LCM
(D) इनमें से कोई नहीं
हल :
(B) HCF < LCM

प्रश्न 34.
6, 72 व 120 का LCM होगा-
(A) 240
(B) 360
(C) 120
(D) 152
हल :
(B) 360

प्रश्न 35.
12, 15 व 21 का LCM होगा-
(A) 420
(B) 315
(C) 410
(D) इनमें से कोई नहीं
हल :
(A) 420

प्रश्न 36.
किन्हीं तीन विभिन्न अभाज्य संख्याओं का HCF होगा-
(A) तीनों संख्याओं का गुणनफल
(B) सबसे छोटी संख्या
(C) सबसे बड़ी संख्या
(D) एक
हल :
(D) एक

प्रश्न 37.
किन्हीं तीन विभिन्न अभाज्य संख्याओं का LCM होगा-
(A) सबसे छोटी संख्या
(B) सबसे बड़ी संख्या
(C) तीनों संख्याओं का गुणनफल
(D) एक
हल :
(C) तीनों संख्याओं का गुणनफल .

प्रश्न 38.
17, 23 व 29 का LCM होगा-
(A) 1140
(B) 1139
(C) 1239
(D) इनमें से कोई नहीं
हल :
(D) इनमें से कोई नहीं

प्रश्न 39.
510 व 92 का LCM होगा-
(A) 23460
(B) 3460
(C) 2346
(D) 1800
हल :
(A) 23460

प्रश्न 40.
पूर्णांक 8, 9 और 25 का LCM है-
(A) 1500
(B) 1700
(C) 1800
(D) इनमें से कोई नहीं
हल :
(C) 1800

प्रश्न 41.
यदि 36 व x का HCF व LCM क्रमशः 6 व 180 हो तो x का मान होगा-
(A) 180
(B) 36
(C) 30
(D) 6
हल :
(C) 30

प्रश्न 42.
यदि दो संख्याओं का गुणनफल 240 तथा उनका HCF 4 हो तो LCM होगा-
(A) 12
(B) 20
(C) 240
(D) 60
हल :
(D) 60

प्रश्न 43.
यदि दो संख्याओं का HCF व LCM क्रमशः 24 व 360 हो तथा इनमें से एक संख्या 72 हो तो दूसरी संख्या होगी-
(A) 120
(B) 60
(C) 30
(D) 15
हल :
(A) 120

प्रश्न 44.
किसी खेल के मैदान के चारों ओर एक वृत्ताकार पथ है। इस मैदान का एक चक्कर लगाने में सोनिया को 18 मिनट लगते हैं, जबकि इसी मैदान का एक चक्कर लगाने में रवि को 12 मिनट लगते हैं। मान लीजिए वे दोनों एक ही स्थान और एक ही समय पर चलना प्रारंभ करके एक ही दिशा में चलते हैं। कितने समय बाद वे पुनः प्रारंभिक स्थान पर मिलेंगे?
(A) 6 मिनट
(B) 12 मिनट
(C) 18 मिनट
(D) 36 मिनट
हल :
(D) 36 मिनट

प्रश्न 45.
किसी स्कूल की नौवीं कक्षा के सैक्शन-ए में 36 विद्यार्थी तथा सैक्शन -बी में 32 विद्यार्थी हैं । इस स्कूल की लाइब्रेरी में कम से कम कितनी किताबें होनी चाहिएँ ताकि सैक्शन-ए या सैक्शन-बी के विद्यार्थियों को बराबर-बराबर दी जा सके ?
(A) 144
(B) 288
(C) 32
(D) 36
हल :
(B) 288

प्रश्न 46.
\(\sqrt{5}\) हैं :
(A) सम
(B) विषम
(C) परिमेय
(D) अपरिमेय
हल :
(D) अपरिमेय

प्रश्न 47.
………………… और अपरिमेय संख्याएँ मिलकर वास्तविक संख्याएँ बनाती हैं?
(A) परिमेय
(B) सम
(C) विषम
(D) अभाज्य
हल :
(A) परिमेय

प्रश्न 48.
\(\sqrt{3}\) है :
(A) परिमेय
(B) अपरिमेय
(C) मिश्रित संख्या
(D) इनमें से कोई नहीं
हल :
(B) अपरिमेय

प्रश्न 49.
निम्नलिखित में से कौन-सी संख्या अपरिमेय नहीं है?
(A) \(\sqrt{2}\)
(B) \(\sqrt{3}\)
(C) \(\sqrt{13}\)
(D) \(\sqrt{16}\)
हल :
(D) \(\sqrt{16}\)

प्रश्न 50.
\(\sqrt{2}\) है-
(A) अपरिमेय
(B) परिमेय
(C) मिश्रित संख्या
(D) इनमें से कोई नहीं
हल :
(A) अपरिमेय

प्रश्न 51.
यदि \(\frac {p}{q}\) एक ऐसी परिमेय संख्या हो, जिसमें q = 2ⁿ5^m जहाँ n और m ऋणेतर पूर्णांक है तो इसका दशमलव प्रसार होगा-
(A) असांत आवर्ती
(B) असांत अनावर्ती
(C) सांत
(D) असांत
हल :
(C) सांत

प्रश्न 52.
निम्नलिखित में से किस संख्या का दशमलव प्रसार असांत आवर्ती होगा ?
(A) \(\frac{3}{2^3}\)
(B) \(\frac{13}{5^3}\)
(C) \(\frac{23}{2^3 5^2}\)
(D) \(\frac{129}{2^2 5^7 7^5}\)
हल :
(D) \(\frac{129}{2^2 5^7 7^5}\)

प्रश्न 53.
निम्नलिखित में से किस संख्या का दशमलव प्रसार सांत होगा-
(A) \(\frac{29}{7^3}\)
(B) \(\frac{2}{5^1}\)
(C) \(\frac{11}{3^1 5^1 7^1}\)
(D) \(\frac{13}{2^3 7^3}\)
हल :
(B) \(\frac{2}{5^1}\)

प्रश्न 54.
निम्नलिखित में से किस संख्या का दशमलव प्रसार असांत आवर्ती होगा ?
(A) \(\frac {3}{8}\)
(B) \(\frac {13}{125}\)
(C) \(\frac {17}{9}\)
(D) \(\frac {7}{80}\)
हल :
(C) \(\frac {17}{9}\)

प्रश्न 55.
\(\frac {3}{8}\) का दशमलव प्रसार होगा-
(A) 0.375
(B) 0.0375
(C) 3.75
(D) 0.00375
हल :
(A) 0.375

प्रश्न 56.
निम्नलिखित में से अपरिमेय (irrational) संख्या छाँटिए-
(A) \(\sqrt{9}\)
(B) \(\sqrt{14}\)
(C) \(\sqrt{25}\)
(D) \(\sqrt{36}\)
हल :
(B) \(\sqrt{14}\)

प्रश्न 57.
यदि 124 और 148 का HCF 4 हो, तो उसका LCM क्या होगा ?
(A) 1147
(B) 18352
(C) 4588
(D) इनमें से कोई नहीं
हल :
(C) 4588

प्रश्न 58.
\(\frac {17}{8}\) का दशमलव प्रसार होगा-
(A) 0.2125
(B) 0.02125
(C) 2.125
(D) 21.25
हल :
(C) 2.125

प्रश्न 59.
निम्नलिखित में से अपरिमेय (irrational) संख्या छाँटिए-
(A) \(\sqrt{8}\)
(B) \(\sqrt{16}\)
(C) \(\sqrt{49}\)
(D) \(\sqrt{81}\)
हल :
(A) \(\sqrt{8}\)

प्रश्न 60.
यदि 135 और 225 का HCF 45 हो, तो उसका LCM क्या होगा ?
(A) 405
(B) 1125
(C) 675
(D) इनमें से कोई नहीं
हल :
(C) 675

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 15 Probability Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 15 Probability

Short Answer Type Questions

Question 1.
If the probability of winning a game is 0.07, what is probability of lossing it?
Solution :
P (winning a game) = 0.07 (given)
P (winning a game) + P (lossing a game) = 1
0.07 + P (lossing a game) = 1
P (lossing a game) = 1 – 0.07
= 0.93

Question 2.
If probability of ‘not E’ = 0.95, then find P (E).
Solution :
Probability of “not E’ = 0.95 (given)
P (E) + P (not E) = 1
P (E) + 0.95 = 1
P (E) = 1 – 0.95
P (E) = 0.05

Question 3.
A die is thrown once. Find the probability of getting ‘at most’?
Solution :
A die is thrown once, the possible numbers are: 1, 2, 3, 4, 5, 6
The numbers ‘at most’ are = 1, 2
Let E be event of getting at most 2
Number of outcomes to favourable event E = 2
P (E) = \(\frac{2}{6}=\frac{1}{3}\)

Question 4.
A dice is thrown once, find the probability of getting
(i) composite number
(ii) a prime numbers.
Solution :
A dice is thrown once, the possible numbers are: 1, 2, 3, 4, 5, 6
The numbers of all possible outcomes = 6
(i) The composite numbers are 4, 6
Let E₁ be event of a getting of composite number
Number of outcomes to favourable event
E₁ = 2
P (E₁) = \(\frac{2}{6}=\frac{1}{3}\)

(ii) The prime number are 2, 3, 5
Let E₂ be event of a getting of a prime number
Number of outcomes to favourable to event
E₂= 3
P (E₂) = \(\frac{3}{6}=\frac{1}{2}\)

Question 5.
A child has a die whose six faces show the letters as shown below:

The die is thrown once. What is the probability of getting
(i) A?
(ii) C?
Solution :
In throwing the die any one of the six faces may come upward
The number of all possible outcomes = 6
(i) Since, there are two faces with letter A
Let E₁ be event of getting faces with letter A
Number of outcomes to favourable to event E₁ = 2

(ii) The prime numbers from 1 to 20 are 2, 3, 12, 15, 18 i.e. 6 numbers
Let E₃ be event of drawn card a divisible by 3
Number of outcomes to favourable to event
P(E₃) = \(\frac {3}{10}\)

Question 7.
If a number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that x² ≤ 4?
Solution :
A number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3
x² = 9, 4, 1, 0, 1, 4, 9
The number all possible outcomes = 7
The numbers x² ≤ 4 are 4, 1, 0, 1, 4, 9 i.e. 5 numbers
Let E be event of chosen at random x² ≤ 4
Number of outcomes to favourable to event E = 5
∴ P (E) = \(\frac {5}{7}\)

Question 8.
A pair of dice is thrown once. What is probability of getting a doublet?
Solution :
A pair of dice is thrown once. The number of all possible outcomes = 6 × 6 = 36

Some number on both dice are:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) i.e. 6 faces are doublet.
Let E be event of getting the same number on both dice.
Number of outcomes to favourable to event E = 6
∴ P (E) = \(\frac{6}{36}=\frac{1}{6}\)

Question 9.
A piggy bank contains hundred coins of Rs. 1, twenty coins of Rs. 2, fifteen coins of Rs. 5 and ten coins of Rs. 10. If it is equally likely that one coin will fall, when the bank is turned upside down,
(i) will be a Rs. 2 coin,
(ii) will not be Rs. 5 coin.
Solution :
Total number of coins = 100 + 20 + 15 = 145
The number of all possible outcomes 145
(i) Number of Rs 2 coins = 20
Number of favourble outcomes = 20
∴ P (Rs. 2 coins) = \(\frac{20}{145}=\frac{4}{29}\)

(ii) Number of coins other than Rs. 5 coins = 100 + 20 + 10 = 130
Number of favourable outcomes = 130
∴ P (will not be a Rs 5 coins) = \(\frac{130}{145}=\frac{26}{29}\)

Question 10.
A die is thrown twice. Find the probability that (i) 5 will come up at least once. (ii) 5 will not come up either time.
Solution :
A die is thrown twice. The number of all possible outcomes 36

(i) The number 5 will come up at least once are (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) i.e. 11 faces
Let E be the event of getting 5 will come up at least once
∴ Number of outcomes to favourable to event E = 11
∴ P (E) = \(\frac {11}{36}\)

(ii) P (5 will not come up either time)
= 1 – P(E)
= 1 – \(\frac {11}{36}\)
= \(\frac {25}{36}\)

Question 11.
The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is. The prolability of selecting a black marble at random from the same jar is. If the jar contains 11 green marbles, find the total number of marbles in the jar.
Solution :
Let the number of marbles of blue and black be x and y respectively.
Total number of marbles = x + y + 11
Number of all possible outcomes = x + y +11
According to question
P (black marbles) = \(\frac {1}{4}\) (given)

Putting the value of y in the equ. (2), we get
x = \(\frac{5+11}{4}=\frac{16}{4}\) = 4
Hence, total number of marbles in the jar
= 4 + 5 + 11
= 20

Question 12.
A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is :
(i) a card of space or an ace
(ii) a black king
(iii) neither a jack nor a king
(iv) either a king or a queen
Solution :
Total playing cards = 52
Number of all possible outcomes = 52
(i) Number of cards of spade or an ace = 13 + 3 = 16
Number of favourable outcomes = 16
∴ P (spade or an ace) = \(\frac {16}{52}\) = \(\frac {4}{13}\)

(ii) Number of black king cards = 2
Number of favourable outcomes = 2
∴ P (a black king) = \(\frac {2}{52}\) = \(\frac {1}{26}\)

(iii) Number of cards of Jack or King = 4 + 4 = 8
Number of cards other than Jack and King = 52 – 8 = 44
Number of favourable outcomes = 44
∴ P (neither jack or king) = \(\frac {44}{52}\) = \(\frac {11}{13}\)

(iv) Number of cards of either a king or a queen = 4 + 4 = 8
Number of cards other than Jack amd King = 8
Number of favourable outcomes = 8
∴ P (either a king or a queen) = \(\frac {8}{52}\) = \(\frac {2}{13}\).

Question 13.
The king, queen and jack of clubs are removed. The remaining cards are mixed together and then a card is drawn at random from it. Find the probability of getting (i) a face card (ii) a card of heart (iii) a card of clubs (iv) a queen of diamond.
Solution :
Since, king, queen and jack of clubs are removed from a deck of 52 cards.
∴ Number of all possible outcomes
= 52 – 3 = 49
(i) Number of face cards = 12 – 3 = 9
Number of favourable outcomes = 9
∴ P (a face card) = \(\frac {9}{49}\)

(ii) Number of cards of heart in deck = 13
Number of favourable outcomes = 13
∴ P (a card of heart) = \(\frac {13}{49}\)

(iii) Number of cards of clubs 13 – 3 = 10
Number of favourable outcomes = 10
∴ P (a card of clubs) = \(\frac {10}{49}\)

(iv) There is only one queen of diamond number of favourable outcomes = 1
∴ P (a queen of diamond) = \(\frac {1}{49}\)

Question 14.
All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and them a card is drawn at random from them. Find the probability that the drawn card is (i) a red card (ii) a face card and (iii) a card of clubs.
Solution :
Since, all red face cards are removed from a pack of playing cards.
Remaining cards = 52 – 6 = 46
Number of all possible outcomes = 46
(i) Remaining red cards = 26 – 6 = 20
Number of favourable outcomes = 20
P (a red card) = \(\frac {20}{46}\) = \(\frac {10}{23}\)

(ii) Number of face cards = 6
Number of favourable outcomes = 6
P (a face card) = \(\frac {6}{46}\) = \(\frac {3}{23}\)

(iii) Number of cards of clubs = 13
Number of favourable outcomes = 13
P (a card of club) = \(\frac {13}{46}\)

Fill in the Blanks

Question 1.
Experiments which have not fixed results experiment are called ______ experiment.
Solution :
Random

Question 2.
An event having only one _______ called an elementary event.
Solution :
outcome

Question 3.
The possible outcomes for a given event are called _______ outcomes.
Solution :
favourable

Question 4.
King, Queen and Jack are called ______ cards.
Solution :
face

Question 5.
The total number of non-face cards are ______ .
Solution :
40

Question 6.
Cards of spaced and clubs are ______ cards.
Solution :
black

Question 7.
Value of probability of an event cannot be _______ or greater than 1.
Solution :
negative.

Multiple Choice Questions

Choose the correct answer each of the following :

Question 1.
Cards bearing numbers 2, 3, 4, ……. 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is :
(a) \(\frac {1}{2}\)
(b) \(\frac {2}{5}\)
(c) \(\frac {3}{10}\)
(d) \(\frac {5}{9}\)
Solution :
(a) \(\frac {1}{2}\)

Numbers are 2, 3, 4, 5, 6, 7, 8, 9, 10, 1
Prime numbers are, 2, 3, 5, 7, 11 = 5
p(E) = \(\frac{5}{10}=\frac{1}{2}\)
So, correct choice is (a)

Question 2.
If an event cannot occur, then its probability is :
(a) 1
(b) \(\frac {3}{4}\)
(c) \(\frac {1}{2}\)
(d) 0
Solution :
(d) 0

The event which can not occur is said to be impossible event and probability of impossible event is zero.
So correct option (d) is correct.

Question 3.
An event is very unlikely to happen. Its probability is closest to:
(a) 0.0001
(b) 0.001
(c) 0.01
(d) 0·1
Solution :
(a) 0.0001

In an event is vary unlikely to happen, its probability should be quite small.
Hence the option (a) is correct.

Question 4.
Which of the following cannot be probability of an event?
(a) 5%
(b) 0.9
(c) 1·1
(d) 0.1
Solution :
(c) 1·1

The probability lies between 0 and 1 so correct choice is (c).

Question 5.
The probability of certain event is :
(a) 1
(b) 0
(c) \(\frac {1}{2}\)
(d) None of these
Solution :
(a) 1

Question 6.
The probability of an impossible event is :
(a) \(\frac {1}{3}\)
(b) 0
(c) 1
(d) None of these
Solution :
(b) 0

Question 7.
The chance that a non leap year contains 53 sundays is:
(a) \(\frac {2}{7}\)
(b) \(\frac {1}{7}\)
(c) \(\frac {3}{7}\)
(d) \(\frac {1}{365}\)
Solution :
(b) \(\frac {1}{7}\)

Question 8.
The probability that in a family of 3 children there will be at least one boy is:
(a) \(\frac {1}{8}\)
(b) \(\frac {6}{8}\)
(c) \(\frac {4}{8}\)
(d) \(\frac {7}{8}\)
Solution :
(d) \(\frac {7}{8}\)

Haryana State Board HBSE 10th Class Science Important Questions Chapter 11 Human Eye and Colourful World Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 11 Human Eye and Colourful World

Question 1.
Explain the functions of the main parts of a human eye along with a diagram.
Answer:
The human eye is one of the most valuable and sensitive sense organs. It enables us to see the wonderful world and its colours.
Structure and working of human eyes:
1. The human eye is the best natural optical instrument whose construction and working can be compared with a camera.

Working of eyes:

1. The light rays coming from the object enter the eye through cornea. The cornea forms a transparent bulge in front of the eye-balls. The eye-ball is almost spherical shaped and has a diameter of about 2.3 cm.

2. Just behind the cornea, there lies a dark muscular diaphragm called ins which controls the amount of light that enters the eye.

3. There is a hole in the iris which is the aperture of the eye i.e. the pupil of eye.

• The size of this aperture or pupil is controlled by iris.

4. After passing through the pupil, the light rays are incident on the eye-lens.
The eye lens is a convex lens made of transparent jelly like material.

5. Ciliary muscles hold the eye lens.

• Ciliary muscles change the thickness of the eye lens while focusing. In other words, the focal length of eye lens can be changed by changing its shape with the help of ciliary muscles. This helps in proper viewing of the objects.

6. The screen on which the image is formed in the eye is called retina.

• The retina is a delicate membrane having a large number of light sensitive cells.
• When light rays falls on retina, its light sensitive cells generate electrical signals.

7. The retina sends these signals to the brain through optic nerve.

• The brain interprets the image of the object. The eye lens forms an inverted real image of the objects on the retina.
• This interpretation allows us to see the objects.

Question 2.
Explain accommodation power of an eye.
Answer:
Accommodation power of an eye:
1. The ability of the eye lens to adjust its focal length as per requirement so that objects can be seen clearly is called accommodation power of an eye.
2. With the help of ciliary muscles, an eye can focus the images of the distant objects as well as the near by objects on its retina by changing the focal length of its lens.
3. When the eyes are looking at a distant object, the ciliary muscles are relaxed, the lens is thin and focal length is more. This enables eye to see the distant object clearly.
4. Similarly when the ciliary muscles contract, the curvature of the eye lens increase and focal length decreases. This enables to see nearby objects clearly.

Question 3.
Explain Near point of an eye and far point of an eye.
Answer:
(i) Near point of an eye :

• The minimum distance at which the objects can be seen clearly without contracting the eye lens
  i. e. without any strain is called the least distance of the distinct vision or near point of an eye.
• For a young adult having normal vision, the near point of the eye is 25 cm.

(ii) Far point of an eye :

• The farthest distance up to which the eye can see objects clearly is called far point of an eye.
• The farthest point of a person with normal vision lies at an infinite distance. This is so because the maximum distance an individual can see cannot be measured.
• Thus, a person with normal vision can see objects clearly from 25 cm to infinite distance.

Question 4.
How do ciliary muscles help to see nearby and far objects?
Answer:
1. The lens of our eyes is made up of a fibrous jelly-like material. Because the lens is made with such flexible material its shape or say curvature can be changed while seeing the object. The ciliary muscles have the ability to change the curvature and hence the focal length of the eyes.

2. When the ciliary muscles are in relaxed position, the lens remains thin. So, the focal length of the lens increases and we can see far-off objects clearly.

3. When we look at objects closer to us (for example, while reading), the ciliary muscles contract. This increases the curvature of the eye lens and the lens become thicker. The focal length decreases and so we are able to see nearby objects clearly.

Question 5.
What is cataract?
Answer:
1. Sometimes, in old aged people, the crystalline lens of the eyes becomes milky and cloudy.
2. As a result, the vision becomes hazy or even opaque due to the formation of a membrane over the lens. This condition is called cataract.
3. Cataract causes partial or complete loss of vision. The vision can be restored through cataract surgery-

Question 6.
How does defect of vision occur? Name the different types of defects of vision.
Answer:
1. The light rays coming from an object passes through the eye lens and forms an image on the retina of the eye.
2. To see objects clearly, image should be formed exactly on retina.
3. For this, the ciliary muscles in the eye change the thickness of the eye lens which results in the change of focal length of the eye.
4. When thickness of this lens does not change with respect to the object distance, the object cannot be seen properly. This is known as defect of vision.
5. The vision becomes blurred due to such refractive defects.

Types of defect of vision:

• Near sightedness or Myopia
• Far sightedness or Hypermetropia
• Presbyopia

Question 7.
What is the cause of short (near) sightedness in eye ? How can short sightedness defect be removed ? OR Write a short note on near sightedness or myopia.
Answer:
1. To see distant objects, the eye lens should become thin.
2. When the lens is unable to do so, the light rays converge more than they should. So, the image gets formed before the retina rather than on it. Hence, distant objects cannot be seen clearly.
3. This defect is known as near sightedness or myopia. Myopia arises due to

• Excessive curvature of the lens or
• Elongation of the eye ball.

4. To correct this defect, concave lens of appropriate focal length is used.

Question 8.
Write a short note on far sightedness (hypermetropia).
Answer:
1. If eye lens does not become thick as per the requirement, then the rays coming from nearby objects gets less converged and hence are focused behind the retina.
2. Due to this, the image is formed behind the retina and not exactly on the retina. Hence, nearby objects cannot be seen clearly.
3. The near point for person suffering from such defects is more than the normal near point of 25 cm. So, he has to keep a reading material at a distance much beyond 25 cm for reading comfortably.
4. This type of defect is known as far sightedness or hypermetropia.
5. This defect occurs due to less convergence of the light rays.
6. To correct this defect, convex lens of appropriate focal length is used.

Question 9.
Write a short note on aging-eye (presbyopia).
Answer:
Presbyopia:
1. Presbyopia which literally means aging eye’, is an age related eye condition which makes it difficult to see objects that are too close.
2. As a person grows older, the power of accommodation of an eye usually decreases.
3. The near point of aged people recedes and they find It difficult to see nearby objects clearly without spectacles. Such a defect is called presbyopia.
4. This defect arises due to weakened ciliary muscles and loss of elasticity of eye lens.

5. At times, people also find it difficult to see even distant objects without spectacles i.e. they suffer from both myopia as well as hypermetropia. Vision of such people can be cured using spectacles of bifocal lens.
6. The upper part of bifocal lens is made up of concave lens and its lower part is made up of convex lens.

Question 10.
Enlist the defects of vision in human eyes and their remedies.
Answer:

Question 11.
What is a prism? Draw its diagram.
Answer:
Prism:
1. A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a certain angle.
2. As shown in the figure, the prism has two triangular bases and three rectangular lateral surfaces (or faces). These surfaces are inclined to each other.
3. The angle between the two lateral faces is called the angle of prism.

Question 12.
Draw a ray diagram to show the path of a light ray that enters the glass prism obliquely. Label on it the angle of incidence and angle of deviation.
Answer:
1. The figure here shows the principle section ABC of a glass prism.
2. Ray PQ is an incident ray on face AB.
3. On entering the denser medium (grass) from rarer medium (air), the ray PQ bends towards the normal along the path QR. This means ray PQ got refracted.

4. Ray QR again undergoes refraction when it hits face AC. The ray now travels from denser medium to rarer medium (air). Hence, it bends away from the normal. It emerges out as ray RS.
5. Angle ‘D’ between the incident ray PO and the emergent ray is called the angle of deviation i.e. the angle at which the incident ray deviates.
6. Angle ‘e’ made by the emergent ray with the normal to the retracting face ‘AC’ is called the angle of emergence. An important result is – i + e = A + D i.e. Angle of incidence + Angle of emergence = Angle of prism + Angle of deviation

Question 13.
What is dispersion of light? Which are the colours of the spectrum obtained from the dispersion through a glass prism?
Answer:
Dispersion of light:
1. Splitting of white light into its seven constituent colours on passing through a transparent medium like a glass prism is called dispersion of light.
2. When white light is incident on a prism, the prism decomposes the white light and hence a band of seven colours is obtained on the screen.
3. The band of these colours is known as spectrum.
On the screen, we get a band of seven colours in the following order from bottom to top:
Violet, Indigo, Blue, Green, Yellow, Orange and Red (VIBGYOR)

Question 14.
What is the reason behind formation of spectrum and refraction of light into seven colours?
Answer:
1. All the constituent colours of a white light have saine velocity in vacuum. However, when white light passes through a transparent medium like glass, water, glycerin, etc., velocity of the constituent colour changes and due to this, deviation of the different constituent colours occurs at different angles.
2. Each colour has a unique refractive index. Hence, the seven colours of the spectrum get refracted in different proportions.

Example:

• The velocity of violet light is least and so it bends the most.
• The velocity of red light is highest and so it bends the least.
•  Hence, in a spectrum of prism, red colour is at the top while violet colour is at the bottom of the spectrum.

Question 15.
Explain the dispersion of white light by a glass prism using necessary figure. OR Explain
Answer:
1. Take two identical prisms P₁ and P₂ having same angles.
2. Arrange these prisms as shown in the figure.
3. Incident a beam of white light on prism P₁.
4. The light coming out of the prism P₁ will consist of seven colours.

5. Arrange prism P₂ if such a way that the band of seven colours of prism P₁ falls properly on P₂ and finally gives white beam of light on a screen.
6. From this experiment, Newton established that a white light is composed of seven constituent colours.
7. This way we can understand the structure of white light through the phenomenon of dispersion of light.

Question 16.
Explain formation of a rainbow with a neat diagram.
Answer:
Rainbow:
1. A rainbow is a natural spectrum visible in the sky after rain shower.
2. Rainbow is formed when the water droplets present in the atmosphere disperse the sunlight tailing on diem.
3. Note that a rainbow is always formed in the direction opposite to that of the sun.

Formation:

1. When sunlight falls on the atmospheric water drops, they first disperse the incident light and then reflect it internally (not necessarily total internal reflection).
2. Finally the light gets refracted again while it is coming out of rain drops.
3. We see different colours in a rainbow because light enters into our eye through dispersion and internal reflection.
4. In a rainbow, water droplets act as small prisms.
5. The colour at the bottom of the rainbow is violet while the top most colour is red. Such a rainbow is called a primary rainbow.

Question 17.
Draw the structure of rainbow formation.
Answer:
Rainbow:
1. A rainbow is a natural spectrum visible in the sky after rain shower.
2. Rainbow is formed when the water droplets present in the atmosphere disperse the sunlight tailing on diem.
3. Note that a rainbow is always formed in the direction opposite to that of the sun.

Formation:

1. When sunlight falls on the atmospheric water drops, they first disperse the incident light and then reflect it internally (not necessarily total internal reflection).
2. Finally the light gets refracted again while it is coming out of rain drops.
3. We see different colours in a rainbow because light enters into our eye through dispersion and internal reflection.
4. In a rainbow, water droplets act as small prisms.
5. The colour at the bottom of the rainbow is violet while the top most colour is red. Such a rainbow is called a primary rainbow.

Question 18.
What is atmospheric retraction? Which phenomena results from it?
Atmospheric refraction :
1. Atmosphere consists of layers of different densities spread across it. This means the density of atmosphere is not same everywhere.
2. The layer at lower altitude from the earth has more density than that at the higher altitude.
3. Due to this difference, the refractive index of atmosphere continuously decreases as one moves from lower to higher altitudes.
4. Moreover, the physical conditions of the refracting medium i.e. the atmospheric air keeps on changing. As a result, we feel that the position of an object in the atmosphere is changing.
5. This phenomenon is called atmospheric refraction or the refraction of light by earth’s atmosphere.
6. Phenomena such as twinkling of stars, early sunrise and delayed sunset occur due to this effect.

Question 19.
Write a note on twinkling of stars. OR Stars seem higher than they actually are. Explain. OR Explain twinkling of stars in detail.
Answer:
1. Density of atmosphere is not uniform everywhere.
2. There are different layers of atmosphere with different refractive indices.
3. Atmospheric layer at lower altitude is colder and denser compared to layer at higher altitude.
4. When the light travels from star at rarer medium towards earth at denser medium, in bends towards normal.
5. Thus, due to refraction towards normal,the position of star appears higher from its actual position.

Reason for twinkling of stars:
1. The physical condition of refracting medium .e. the earth’s atmosphere is not stationary. Hence, the position of star seems to be changing continuously.
2. This results in the continuous change in the path of the rays coming from the stars and continuous change in the intensity of light. This leads to twinkling of stars.
3. Light from a star is refracted or say bent as it leaves the space and enters the earth’s atmosphere.
4. Air at higher altitude is rare whereas near the earth’s surface it is dense.
5. As a result, when light from the star comes down, the dense air bends the light more.
6. Due to this, refraction of star’s light, the star appears to be at a higher position than it actually is.

Question 20.
Why do we experience early sunrise and delayed sunset when actually it is not so? OR Our day is longer by four minutes. Give reason.
Answer:
1. Actual sunrise means actual appearance of the sun at the horizon.
2. When the sun is slightly below the horizon,the sun rays pass from less dense air to more dense air in the atmosphere and get refracted downwards.
3. Due to this atmospheric refraction, the sun appears to be raised above the horizon when actually it is still slightly below the horizon. Thus, we can see sunrise two minutes before it actually comes to horizon.

4. In the same way, when sun is setting i.e. when it moves below the horizon, it is seen to us for two minutes.
5. In total, 2 minutes early sunrise and 2 minutes delayed sunset make our day longer by four minutes.
6. Thus, we experience early sunrise and delayed sunset when actually it is not so.

Question 21.
What is scattering of light? On what factors does it depend?
Answer:
Scattering of light:

• The deflection of light by minute particles and molecules in all the directions is known as scattering of light.
• The colour of scattered light depends upon the size of scattering particles.
• For example, minute particles scatter light of small wavelength such as blue colour.
• Whereas bigger particles scatter light of larger wavelength such as red colour.
• If the size of scattering particles is much bigger, the scattered light appears white.

Question 22.
Describe Tyndall effect.
Answer:
Tyndall effect:
1. The earth’s atmosphere is a heterogeneous mixture of smoke particles, tiny water droplets and air particles.
2. When light falls on such colloidal particles, a path of light beam becomes visible.
3. This phenomenon is known as Tynciall effect.
4. The light rays reach us after getting deflected in all directions from these particles.
5. Commercially, Tyndall effect helps in determining the density of aerosol and other colloidal particles that are emitted.

Examples:

• When a fine beam of sun light enters a room filled with smoke through a small hole, a path of light beam can be seen due to Tyndall effect.
• When sunlight enters a canopy of dense forests, Tyndall effect can be seen due to scattering of light through tiny water droplets of the mist.
• Sometimes smoke emitted by the combustion of engine oil appears blue in colour.

Question 23.
Why does clean sky appear blue in colour?
Answer:
1. The sunlight is made up of seven colours.
2. When sunlight passes through the atmosphere, most of the longer wavelength lights such as red, orange, yellow, etc. present in it do not get scattered much by the molecules of the air and other fine particles and hence pass through straight.
3. The shorter wavelength blue light is however scattered all around the sky by air molecules in the atmosphere.
4. The wavelength of red light is about 1.8 times more than that of blue colour.
5. Thus, when sunlight passes through the atmosphere the tine particles in the air scatter blue colour more strongly than red.
6. As a result, the sky appears blue.

Question 24.
Why does sun appear reddish at sunrise and sunset?
Answer:
1. White light coming from the sun has to travel a large distance in the atmosphere before reaching to the observer.
2. During sunrise or sunset, most of the blue colour present in sunlight has been scattered out and it is away from our sight.
3. As a result, only red light remains present in the beam of sunlight and so only red colour reaches our eye.

Question 25.
Why do some people use spectacles with bifocal lenses?
Answer:
1. Some people suffer from near-sightedness (myopia) as well far-sightedness (hypermetropia).
2. Myopia can be cured with concave lens whereas hypermetropia with convex lens. Spectacles with bifocal lens have two lenses namely concave and convex to overcome this problem.
3. Hence, people suffering from both myopia as well as hypermetropia wear bifocal lens.

Question 26.
In the figure, a narrow beam of white light is shown to pass through a triangular glass prism. After passing through the prism it produces a spectrum XV on a screen.
(a) State the colour seen at X and Y.
(b) Why do different colours of white light bend through different angles with respect to the incident beam of light?
Answer:

(a) ‘X’ represents violet colour and ‘Y’ represents red.
(b) The refractive index of glass is different for different colours.
Hence, the different colours present in the white light bend at different angles at the prism.

Question 27.
A beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7 as shown in the diagram.
A student makes the following statements about the spectrum observed on the screen:
Answer:

(a) The colours at positions marked 3 and 5 are similar to the colour of the sky and the core of a hard boiled egg respectively. Is the statement made by the student correct or incorrect? Justify.
(b) Which two positions correspond closely to the colour of the colour of —
(i) A solution of potassium permanganate? (ii) ‘Danger’ or stop signal lights?
Answer:
The colours of the spectrum observed on the screen are in the order:

(a) The statement made by the student is incorrect. The colour at position 3 is yellow while the sky is blue. The colour at position 5 is blue while the core of the boiled egg is yellow.
(b) (i) Solution of potassium permanganate is violet coloured. Position 7 corresponds to this colour.
(ii) Danger or stop signal lights are of red colour. Position 1 corresponds to this colour.

Question 28.
Stars and planets both belong to space, even then stars twinkle but planets do not. Give reason.
Answer:
1. The stars are very far from us and so they may be considered as the point sources of light.
2. Compared to stars, planets are much nearer to the earth and so planets appear quite big.
3. Hence, planets cannot be considered a single point source of light but collection of a very large number of point sources of light.
4. Thus, on the whole, the brightness of the planet always remains the same and it does not appear to twinkle. But the atmospheric refraction affects stars more since they are point sources of light and so stars twinkle.

Question 29.
Why are danger signal lights red in colour? OR Signal lights used to indicate danger are red in colour. Give reason.
Answer:
1. Wavelength of red colour is quite large. It does, not allow light to scatter much.
2. Thus when red coloured light is used, it gets scattered least in fog or smoke and so it can be seen from a long distance.
3. Hence, red coloured light is used to indicate danger.

Question 30.
There was a beautiful village in a Himalayan valley. When trains passed from the village, the whistle and the sound of train, mixed with the sound of waterfall, seemed to be very pleasant to everyone. Children loved this and so they used to play near the railway track. On one foggy day, a group of children found that a fishplate was missing from the track. This worried all the villagers. One of the children Raghav placed his ear on the railway track and tried to know if a train was coming. He asked his friends to inform the nearest railway station and he himself put-off his red shirt and started running towards the train, waving his red shirt. The engine driver sensed some danger and stopped the train. A major accident was averted.

Questions:

1. Name the two physical phenomena of science used by Raghav.
2. Why did Raghav use his red shirt instead of any other coloured shirt or cloth?
3. What moral values do you learn from Raghav?

Answers:

1. (a) Sound travels through a medium. (b) Scattering of light
2. The red light is least scattered by fog or smoke. Hence, it can be seen from a large distance.
3. (a) Proper knowledge of science and its right application.
   (b) Concern for others and awareness about things.

Question 31.
Four friends went bicycling in a park. It was monsoon and the weather was extremely pleasant. They played several games. Suddenly, Shweta observed seven colours in the sky. She said to others, “Look! What a beautiful rainbow!” Shivani, one of the four friends, asked her what a rainbow was. Shweta explained about it and gave other details to all. Everyone got delighted and felt thankful.

Questions:

1. What would have been the position of the sun when Shweta was facing the rainbow?
2. Which device can be used to obtain such a phenomenon at home?
3. What moral value do you learn from Shweta?

Answers:

1. The sun was behind Shweta.
2. A small prism can be used to form a rainbow at home.
3. Shweta delivered the values of sharing knowledge and helping others.

Question 32.
Bhavan studies in class 5. He sits in 10th bench. He is extremely fond of junk food and mostly carries food items such as pasta, maggi, etc. in his lunch. Off late his teacher finds that Bhavin faces difficulty In reading the blackboard text. The teacher advises Bhavin’s mother that he should only bring healthy lunch with loads of green vegetables and fruits.

Questions:

1. Name the eye defect Bhavin is suffering from.
2. What are two possible deformities related to his eye defect?
3. What values do you learn from this?

Answers:

1. Bhavin is suffering from myopia (near-sightedness)
2. Increased thinness of lens due to excessive curvature of the eye lens and eyeball defect due to elongation of the eyeball.
3. The teacher displayed the values of concern and awareness.

Very Short Answer Type Question :

Question 1.
State the structure of cornea.
Answer:
Cornea has a transparent bulged structure. It is located at the front of the eye ball.

Question 2.
Give an Idea about structure of eye ball.
Answer:
Eye ball is nearly a spherical shaped structure having a diameter of about 2.3 cm.

Question 3.
What is the role of eye lens?
Answer:
The eye lens does the task of adjusting focal length required to focus objects at different distances on retina.

Question 4.
What is Iris?
Answer:
Iris is a dark muscular structure lying behind the cornea. It controls the size of the pupil.

Question 5.
How are functions of cornea and pupil related?
Answer:
It is cornea that allows the light to enter but it is pupil that regulates and controls the amount of light that should enter the eye.

Question 6.
What kind of image is created on retina?
Answer:
Real and inverted.

Question 7.
What do the light-sensitive cells located on retina do?
Answer:
When light falls on light-sensitive cells, the cells get activated. They then generate electric signals and send them to the brain.

Question 8.
What role does the brain play In vision?
Answer:
The brain interprets the electric signals sent by the retina. It then processes them so that we can see the objects.

Question 9.
What is power of accommodation?
Answer:
The ability of the eye lens to adjust its focal length as per the requirement so that objects can be seen clearly is called accommodation power of the eye.

Question 10.
What is the farthest distance upto which the eye can see objects clearly called?
Answer:
Far point of an eye

Question 11.
State the distance of near point and far point of the eye.
Answer:
Near point: Minimum distance 25 cm, Far point: Infinite

Question 12.
How many types of common defects of vision arise in eye? Name them.
Answer:
Myopia, Hypermetropia, Presbyopia

Question 13.
What is cataract?
Answer:
Sometimes, the crystalline lens of people of old age becomes milky and dowdy which results in partial or complete loss of vision. This is called cataract.

Question 14.
What causes refractive defect of vision?
Answer:
When the thickness of the eye lens cannot be adjusted to clearly focus the object, the object cannot be seen properly. This is the cause of refractive defect of vision.

Question 15.
State the causes of hypermetropia.
Answer
(i) The focal length of the eye lens is too long.
(ii) The eye ball has become too small.

Question 16.
What is presbyopla? What causes it?
Answer:
Presbyopia is an age related eye condition which makes it difficult to see the objects that are too close; Weakened ciliary muscles and loss of elasticity of eye lens.

Question 17.
How can one get rid of presbyopia?
Answer:
By wearing bi-focal lens i.e. a lens consisting of both concave and convex lens.

Question 18.
How does the focal length of the eye lens change when we shift looking from a distant object to a nearby object?
Answer:
The focal length of the eye lens decreases.

Question 19.
Why does it take some time to see the objects In a dim room when we enter the room from bright sunlight outside?
Answer:
The light.sensitive cells of retina are less sensitive in dim light. Hence, it takes time to get the cells activated and see the object.

Question 20.
What is angle of prism?
Answer:
The angle between two rectangular lateral faces of the prism is called the angle of prism.

Question 21.
What is dispersion of light?
Answer:
The splitting of white light into its various components or say 7 colours is called dispersion of light.

Question 22.
What made Newton believe that the sunlight is made up of seven colours?
Answer:
Newton allowed sunlight to pass through one prism. He then focused the spectrum on an inverted prism. On this he got the sunlight again. This made him believe that sunlight is made of seven colours.

Question 23.
What determines the colour of an object?
Answer:
The colour of light which Is reflected from the object

Question 24.
Why does green leaf looks green?
Answer:
(A) It absorbs all colours except green

Question 25.
Answer the following questions:
(i) An object reflects all the colours of incident light. What should be the colour of that object? An object absorbs all the colours of incident light. What should be the colour of that object?
Answer:
(i) White; (ii) Black

Question 26.
Why danger signals and signs are of red colour?
Answer:
Red colour has the largest wavelength and it is least scattered by fog and dust. Hence, its visibility is the highest. As a result

Question 27.
What Is a rainbow? What causes It?
Answer:
A rainbow is a natural spectrum appearing in the sky after the raw. Dispersion of sunlight by tiny water droplets.

Question 28.
State two examples where in you can see the rainbow even on a sunny day i.e. other than during monsoon.
Answer:
We can see a rainbow through a waterfall and through a fountain when the sun is behind us.

Question 29.
What is internal reflection?
Answer:
When a ray of light enters a medium from another (example: from air to water droplet) and gets reflected within that second medium before moving out of that medium then such a reflection is called internal reflection.

Question 30.
What is atmospheric retraction?
Answer:
Atmospheric retraction is the deviation of light or other electromagnetic waves from a straight line as it passes through the atmosphere due to the variation in air density.

Question 31.
Why is the density of earth’s atmosphere non-uniform everywhere?
Answer:
Because it consists of layers of different densities

Question 32.
Define Tyndall effect.
Answer:
The scattering of light in the nature due to small particles present in the atmosphere is called Tyndall effect.

Question 33.
Why can we see sun about two minutes before the actual sunrise?
Answer:
Because of the refraction of the light by atmosphere

Question 34.
Why does the sky appear dark to passengers flying at very high altitudes?
Answer:
Scattering of light is not possible at such heights. Hence, ………………….

Fill in the Blanks:

1. The light rays coming from the object first enter the eye through ……………
Answer:
Cornea

2. Retina is a …………. type of screen.
Answer:
Light-sensitive

3. An aperture of an eye behind the cornea at the center is known as ………………
Answer:
Pupil

4. Through the optic nerves, the electrical signals are sent to the …………..
Answer:
Brain

5. The focal length of tens changes with the change in the ……….. of an eye lens.
Answer:
Curvature

6. The ability of an eye to adjust focal length of the eye lens as per requirement is known as of ……………. an eye.
Answer:
Accommodation power

7. Least distance of distinct vision is also called ……………
Answer:
Near point of an eye

8. Myopia is also known as ……………….
Answer:
Near sightedness

9. If lens remain thin and does not become thick as per requirement, rays coming from nearby object are less refracted and are focused behind…
Answer:
The retina

10. The angle of incidence at which the angle of refraction is 900 is called …………
Answer:
Critical angle

11. Splitting of white light into seven colours is known as …………….
Answer:
Dispersion

12. In ………….. medium, violet coloured light is deviated maximum.
Answer:
Transparent

13. Recombining the seven colours of spectrum gives …………. colour.
Answer:
White

14. The deflection of light by minute particles and molecules in all the directions is known as …………….
Answer:
Scattering of light

15. If the size of scattering particle is much bigger the scattering light appears ………….. coloured.
Answer:
White

16. When Newton further tried to split the colours of the spectrum of white light by using another prism
Answer:
He could not get any more colours.

17. Any light tat gives a spectrum similar to that of sunlight is often referred as ………….
Answer:
White light

18. In total refractions and reflections take place in the water droplet for the formation of a rainbow.
Answer:
Three

19. In Tyndall effect, the light rays reach us after deflection of light in all the direction from …………….
Answer:
Colloidal particles

20. The time difference between the actual sunset and apparent sunset is about …………….
Answer:
2 minutes

True Or False

1. A muscular diaphragm behind the cornea is known as pupil. — False
2. The ciliary muscles helps in changing the thickness of the eye lens. — True
3. If eye lens do not become thin as per requirement but remain thick only, then rays coming from distant object after being refracted by lens can be focused before the retina. — True
4. In bifocal lens, the lower part of a small circular section is made up of convex lens. — True
5. Contact lens is widely used to remove the defect of eye. — True
6. Cataract causes only partial loss of vision. — False
7. Myopia can be cured with convex lens. — False
8. A water droplet can act like a small prism. — True
9. Atmospheric layer at lower altitude is colder and denser as compared to layer at higher altitude. — True
10. When the sun is below the horizon, sunhiit reaches to our eyes after being reflected in the atmosphere. — False

Match the Following 

Answer: 1-q, 2-p, 3-r

Answer: 1-d, 2-c, 3-a, 4-b

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 14 Statistics Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 14 Statistics

Short/Long Answer Type Questions

Question 1.
Per day expenses of 25 families of the frequency distribution of a Dhani of a village is given as follows:

Find the mean expense of families by direct method.
Solution :
We prepare the cumulative frequency table as given :

∴ mean = \(=\frac{\Sigma f_i x_i}{\Sigma f_i}\)
= \(\frac {1240}{25}\)
= ₹ 49.6
Hence, mean expense of families = ₹ 49.6

Question 2.
The following distribution shows the daily pocket allowance of children of a locality.

Find the mean daily pocket allowance by using appropriate method.
Solution :

∴ mean = \(=\frac{\Sigma f_i x_i}{\Sigma f_i}\)
= \(\frac {955}{25}\) = ₹ 38.2
Hence, mean daily pocket allowance = ₹ 38.2

Question 3.
The marks obtained by 110 students in an examination are given below :

Find the mean marks of students.
Solution :

Hence, mean marks of students = 44.82

Question 4.
The following data gives the information on the observed life times (in hours) of 200 electrical components.

Determine the modal lifetimes of the components.
Solution :
The class interval 80 – 100 has maximum frequency. So, it is the modal class.
∴ l = 80, f₁ = 65, f₀ = 38, f₂ = 24, h = 20

Hence, modal lifetimes of components = 87.94 hours.

Question 5.
Find the mode of the following distribution.

Solution:
The class interval 35 – 40 has maximum frequency. So, it is the modal class.
∴ l = 35, f₁ = 50, f₀= 34, f₂ = 42, h = 5

Question 6.
Find the mode of the following frequency distribution.

Solution :
The class interval 30 – 40 has maximum frequency. So, it is the modal class.
∴ l = 30, f₁ = 16, f₀ = 10, f₂ = 12, h = 10

Question 7.
If the median of the following frequency distribution is 32.5. Find the values of f₁ and f₂.

Solution :
Let us prepare the cumulative frequency distribution table as given below.

Here, n = 40
⇒ f_i + f₂ + 31 = 40
⇒ f_i + f₂ = 9
Median is 32.5, which lies in the class interval 30 – 40, So the median class is 30 – 40.
l = 30, f = 12, cf = f₁ + 14, h = 10

Putting the value of f₁ in equation (1), we get
3 + f₂ = 9
f₂ = 6
Hence, f₁ = 3, f₂ = 6

Question 8.
On the annual day of school agewise participation of students is given in the following distribution table.

Find the median of students and get the median graphically.
Solution :
We convert the cumulative frequency table into depicting class intervals with their respective frequencies as given below :

\(\frac {n}{2}\) = \(\frac {76}{2}\) = 38
But 38 comes under the cumulative frequency 42 and the class interval against cumulative frequency 42 is 12 – 14. So, it is the median class.
Here, l = 12, f = 20, cf = 22, h = 2

We plots the points (6, 2), (8, 6), (10, 12), (12, 22), (14, 22), (16, 67), (18, 76) on the graph paper joining these points with a free hand to get less than curve as shown in a graph. We have n = 38.

Now, locate the point on the ogive where, ordinate is 38. The n coordinate corresponding to this ordinate is 13.6. Therefore required median on the graph is 13.6.

Question 9.
The following table gives production on yield per hectare of wheat of 100 farms of a village.

Change the distribution of a more than type distribution and draw its ogive.
Solution :
We prepare the cumulative frequency table by more than type method as given :

\(\frac {n}{2}\) = \(\frac {100}{2}\) = 50
We plots the points (40, 100), (45, 96), (50, 90), (55, 74), (60, 54) and (65, 24) on the graph paper. Joining these points with a free hand to obtain more than type ogive curve as shown in a graph.

Question 10.
By changing the following frequency distribution to less than type distribution, draw idts ogive.

Solution :
We prepare the cumulative frequency table by less than type method as given :

We plots the points (15, 6), (30, 14), (45, 24), (60, 30) and (75, 34) on the graph paper. Joining these points with a free hand to obtain less than type ogive curve as shown in graph.

Question 11.
Given below is a frequency distribution table showing daily income of 100 workers of a factory.

Convert this table to a cumulative frequency distribution table of more than type.
Solution :
We prepare the cumulative frequency table by more than type method as given :

We plots the points (200, 100), (300, 88), (400, 70), (500, 35) and (600, 15) on the graph paper. Joining these points with a free hand to obtain more than type ogive curve as shown in a graph.

Fill in the Blanks

Question 1.
The data having more than one mode is called _______ data.
Solution :
multimodal

Question 2.
The cumulative frequency table is useful in determining the _______ .
Solution :
median

Question 3.
Each group into which the raw data is condensed is called a _______ .
Solution :
class interval

Question 4.
The ______ between the true upper limit and true lower point of a class is called its class size.
Solution :
difference

Question 5.
_______ of observation is the sum of the values of all the observations divided by the total number of observations.
Solution :
mean

Question 6.
The _______ frequency of a class is the frequency obtained by adding the frequency of all classes preceding the given class.
Solution :
cumulative

Question 7.
The data having only one mode is called _____ data.
Solution :
Unimodal

Multiple Choice Questions

Choose the correct answer each of the following:

Question 1.
If x_is are the midpoints of the class intervals of grouped data, f_is are the corresponding frequencies and \(\bar{x}\) is the mean, then Σ(f_ix_i – \(\bar{x}\)) is equal :
(a) 0
(b) – 1
(c) 1
(d) 2
Solution :
(a) 0

We know that the mean of the data is given by
\(\bar{x}\) = \(\frac{Σfixi}{x}\)
where n = Σf_i
∴ Σ(f_ix_i – \(\bar{x}\)) = Σf_ix_i – Σ\(\bar{x}\)
= n\(\bar{x}\) – n\(\bar{x}\)
= 0 [∵ x\(\bar{i}\) = n\(\bar{x}\)]
Hence, correct choice is (a).

Question 2.
If the mode of a data is 18 and mean is 24, then median is
(a) 18
(b) 24
(c) 22
(d) 21
Solution :
(c) 22

mode = 18, mean = 24,
∵ 3 median = mode + 2 mean
= 18 + 2 (24)
= 18 + 48 = 66
∴ median = \(\frac {66}{3}\) = 22
Hence, correct choice is (c).

Question 3.
Mode of data 2, 3, 5, 2, 3, 6, 5, 2, 2, 5, 7, 4, 4, is:
(a) 5
(b) 3
(c) 2
(d) 4
Solution :
(c) 2

Here frequency of 2 is 4 so mode = 2.
Hence, correct choice is (c).

Question 4.
For the following distribution :

The modal class is:
(a) 10-20
(b) 20-30
(c) 30-40
(d) 50-60
Solution :
(c) 30-40

Here the heighest frequency is 30, which likes in the interval 30 – 40.
Hence correct choice is (c).

Question 5.
In an arranged discrete series in which total number of observations n is even, median is:
(a) (\(\frac {n}{2}\))^th terms
(b) (\(\frac {n}{2}\) + 1)^th terms
(c) The mean of (\(\frac {n}{2}\))^th term and (\(\frac {n}{2}\) + 1)^th term
(d) none of these
Solution :
(c) The mean of (\(\frac {n}{2}\))^th term and (\(\frac {n}{2}\) + 1)^th term

When n is even then median
The mean of (\(\frac {n}{2}\))^th term and (\(\frac {n}{2}\) + 1)^th term
So correct choice is (c).

Question 6.
While computing mean of grouped data, we assume that the frequencies are :
(a) evenly distributed over all the classes
(b) centred at the class marks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes
Solution :
(b) centred at the class marks of the classes

In computing the mean of grouped data, the frequencies are centered at the class marks of the classes.
Hence, the option (b) is correct.

Question 7.
The cumulative frequency table is useful in determining the:
(a) mean
(b) median
(c) mode
(d) all of these
Solution :
(b) median

Question 8.
For the following distribution :

The sum of lower limits of the median class and modal class is:
(a) 15
(b) 25
(c) 30
(d) 35
Solution :
(b) 25

Now \(\frac {N}{2}\) = \(\frac {66}{2}\) = 33
Which lies in the interval 10 – 15.
Therefore lower limit of the median class is 10.
The highest frequency is 20, which lies in the interval 15 – 20, therefore lower limit of modal class is 15. Hence required sum is 10 + 15 = 25.
Hence correct choice is (b).

Haryana State Board HBSE 10th Class Science Important Questions Chapter 10 Light Reflection and Refraction Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 10 Light Reflection and Refraction

Question 1.
What is light? How does it enables us to see things?
Answer:
1. Light consists of electro-magnetic waves. These waves produce sensation in our eyes.
2. When light falls on an object let us say a book, or a table, etc, it gets reflected. The reflected light rays reach our eyes and enable us to see the object.

Question 2.
Discuss the nature of light. OR State the basic properties of light.
Answer:
1. Light waves do not require a material medium (such as solid, liquid or gas) for their propagation and hence are called non-mechanical waves.
2. Light waves travel at a speed of 3 x 10⁸ m/s in vacuum.
3. The speed of light waves depend upon the nature of medium through which they pass.

4. In practical terms, a light wave can travel from one point to another. Such a propagation of light waves in one straightline and from one point to another is known as a ‘ray’. Bundle of such rays is known as ‘beam of light’.
5. When light is incident on a surface separating two’ medias such as air and water, then some part of incident light is reflected, some part is transmitted and some part of it is absorbed by that surface.
6. Light which is incident on a completely polished shining plane surface gets reflected whereas light incident on a transparent medium (such as glass) gets refracted and passes through that medium.
7. Due to these properties of light i.e. reflection and refraction, the light rays can be focused by mirrors and lenses respectively.
8. Moreover, due to reflection and refraction of light, real or virtual images are formed.

Question 3.
What is reflection of light? State its types.
Answer:
Reflection :

• The phenomenon of sending back the light rays which fall on the surface of an object when light is incident on it, is called reflection of light.
• For example, when light falls on a plane mirror, the mirror sends it back i.e. reflects it.

Types of reflection:
(A) Regular reflection and
(B) Irregular reflection

Question 4.
Explain the types of reflection.
Answer:
(A) Regular reflection :
When a parallel beam of light falls on a smooth or shining surface, then the reflected beam is also parallel and directed back in a fixed direction. This type of reflection is known as regular reflection. For example, reflection by a mirror.

(B) Irregular (diffused) reflection :
When a parallel beam of light falls on a rough surface i.e. irregular surface, the reflected light is not parallel but spreads over a wide area. This type of reflection is known as irregular reflection. For example, light falling on chair, book, etc. gets reflected and helps us to see these objects.

Question 5.
Define incident angle and reflection angle. Also state the laws of reflection.
Answer:
Angle of incidence (θ_i):
The angle that the incident ray makes with normal drawn at the point of incidence is known as incident angle or angle of incidence. It is denoted by θ_i (θ = theta).

Angle of reflection (θ_r):

The angle that the reflected ray makes with normal drawn at the point of reflection is known as reflection angle or angle of reflection, It is denoted by θ_r

Laws of reflection:
1. The angle of incidence (θi) = the angle of reflection (θr).

2. (a) The incident ray,
(b) the normal to the mirror at the point of incidence and
(c) the reflected ray – all lie in the same plane.

(Note: The laws of reflection are applicable equally to plane as well as spherical mirrors. Moreover they are also applicable to regular as well as irregular surfaces.)

Question 6.
What is an image? Explain the types of images.
Answer:
Image: When a number of rays emerging from a point of an object, after reflection or refraction, ‘meet’ or ‘appear to meet’ at another point, then the point of meeting is called the image of the first point.

Types:
(1) Real image and
(2) Virtual image

(1) Real image: When the rays of light after reflection or refraction ‘actually meet’ at a point, the image formed is called a real image.

• Real image can be obtained on a screen.
• Real image is always inverted. For e.g., image formed on cinema screens.

(2) Virtual image: When the rays of light after reflection or refraction ‘do not actually meet’ but ‘appear to meet’ at a point, the image formed is called a virtual image.

• Virtual image cannot be obtained on a screen.
• Virtual image is always erect. For e.g., our image formed when we stand in front of a mirror is a virtual image.

Question 7.
What is a mirror? State Its types.
Answer:
A polished shining surface which reflects almost all the light incident on it is called a mirror.
Types of mirror:
(1) Plane mirror and
(2) Spherical mirror. Further, spherical mirrors are of two types.

They are:

• Concave mirror and
• Convex mirror.

Question 8.
Discuss the different types of mirrors.
Answer:
(A) Plane mirror:
1. A plane mirror is a thin, flat and smooth sheet of glass having a shining coat of silver metal on one side.
2. For example, mirror used in vehicles, in dressing tables, etc.
3. When a parallel beam of light rays falls on a plane mirror, it is reflected as a parallel beam.
4. Thus, plane mirror changes only the direction of the incident light rays. It does not converge or diverge them.
(Note: To bring the parallel light rays closer is called converging them where as to spread out is called diverging.)

(B) Spherical mirror:

• Unlike plane mirrors, spherical (curved) mirrors converge or diverge the parallel light rays incident on them.
• A spherical mirror is a mirror whose reflecting surface is in fact a part of a hollow sphere of a glass.
•  When light rays fall on spherical mirrors, they converge or diverge i.e. the incident parallel rays come closer or spread-out unlike plane mirrors where they are just reflected as parallel beams.

Spherical mirrors are of two types :
(i) Concave mirrors and
(ii) Convex mirrors.

Question 9.
Give a brief idea about how a spherical mirror is formed.
Answer:
1. A spherical mirror is a curved mirror (i.e. not a plane mirror).
(Note: Since the spherical mirror is curved, we consider the mirror as cut-out from a sphere. Doing so helps us to understand various terms related to spherical mirrors and image formation through these mirrors.)
2. The simplest spherical mirror is a spoon that we have in our kitchen. The front and back of the spoon behaves like two different types of spherical mirrors giving upward and inverted images.

Types: Spherical mirrors are of two types.

(a) Concave mirror: The spherical mirror in which light rays converge to form an image is called a concave mirror (or converging mirror).

(b) Convex mirror: A spherical mirror in which light rays diverge to form an image is called a convex mirror (or diverging mirror).

Question 10.
With the help of a suitable diagram define the following terms for concave and convex mirrors (Spherical mirrors).
Answer:
(1) Centre of curvature (C)
(2) Radius of curvature (R)
(3) Pole (P)
(4) Principal axis
(5) Aperture (MN)
(6) Principle focus (F)
(7) Focal length (f)

1. Centre of curvature (C): The centre of spherical sphere from which a mirror is made is known as centre of curvature (C). (Note: Centre of curvature of a concave mirror lies in front where as in convex mirror it lies behind.) Point C is also called 2F since it lies at double the distance of PR

2. Radius of curvature (R): The radius of spherical sphere i.e. from centre of curvature C to pole P i.e. CP from which a mirror is made is known as radius of curvature (R). Thus CP = R. (Note: CP = 2 times PF and so is also written as 2F.)

3. Pole (P): The centre of reflecting surface i.e. MN is known as pole of mirror (P).

4. Principal axis: The imaginary line passing through the pole (P) and centre of curvature (C) is known as principal axis. A

5. Aperture (MN): The portion of the reflecting surface (mirror) from which the reflection of light actually takes place is known as its aperture MN.

6. PrincIple focus (F): The point (F) on the principle axis where the beam of light parallel to the principle axis either actually converges to or diverges from, after reflection from a mirror is called the principle focus (F).

7. Focal length (f): The distance between the pole (P) and principle focus (F) is called focal length (f). Thus, f = PF.

Question 11.
How can you study image formed by a mirror (or a lens)?
Answer:
1. The image of an object formed by a spherical mirror (or a (ens) can be obtained and studied by constructing a ray diagram.
2. Two rays are sufficient to draw a ray diagram because by intersecting two reflected rays we can obtain the position of the image formed.

Question 12.
State the various types of rays that hit a spherical (concave) mirror and direction of their reflecton. Draw ray-diagram to explain how the image will form through these rays.
Answer:
Types of Rays and their Reflection in a Concave Mirror

Question 13.
How do parallel rays converge or diverge in spherical mirrors? Explain with the help of diagrams. OR With necessary diagrams, explain the formation of principal focus and focal length of concave and convex mirrors.
(A) Concave mirror:

• When parallel rays are incident on concave mirror, on reflection, they converge i.e. come closer to each other and meet at a point called principal focus (F). Hence, concave mirror is also called a converging mirror.
• The distance between the pole (P) of the concave mirror and focus (F) is called its focal length (1).
• Note that since all the reflected rays actually pass from the focus (F), a concave mirror has a real focus.
• The focus (F) of concave mirror is in the front of the mirror.

(B) Convex mirror:

• When parallel rays are incident on convex mirror, on reflection, they diverge i.e. spread out from each other. Hence, convex mirror is also called a diverging mirror.
• If you look at the mirror from left side, the reflected (divergent) rays appear to be coming from a point (F) behind the convex mirror.
• The distance between the pole (P) and principal focus (F) is called the focal length (f) of the convex mirror.
• This point F is called the principal focus of the convex mirror.
• Note that since the reflected rays do not actually pass through the focus (F) of the convex mirror, it is said that convex mirror has a virtual (non-real) focus.
• Also note that this virtual focus lies behind the convex mirror.

Question 14.
State and explain the relation between radius of curvature (R) and focal length (f) of a spherical mirror
Answer:
1. For a spherical mirror (both concave and convex), the principal focus (F) lies exactly at the mid-way between pole (P) and center of curvature (C).
2. We know that distance between pole (P) and center of curvature (C) is called the radius of curvature and is denoted by R.
∴ Focal length \((f)=\frac{1}{2} R=\frac{R}{2}\)
Also, R = 2f i.e. radius is twice the focal length (f).

Question 15.
Radius of curvature of a convex minor is 15 cm. Find out its focal length.
Answer:
R = 15cm
F= ?
We know that
Focal length f = \(\frac{\mathrm{R}}{2}\)
∴ F = \(\frac{15}{2}\) = 7.5cm
Thus, focal length of the convex mirror is 7.5 cm.

Question 16.
Focal length of a concave mirror used in a laboratory is 12 cm. What will be the radius of its curvature?
Answer:
Answer:
F = 12cm
R = ?
We know that, Focal length f =\(\frac{\mathrm{R}}{2}\)
∴ R = 2f = 2 x 12 = 24cm
Thus, the radius of curvature of the mirror is 24 cm.

Question 17.
What do you mean by focal length of a spherical mirror is 20 cm?
Answer:
1. Focal length (f) is the point at which all the parallel rays meet after getting reflected from the spherical mirror.
2. If the focal length of a spherical mirror is 20 cm it means all the parallel rays that hit the spherical mirror will meet at 20 cm away from the spherical mirror.

Question 18.
At how many positions can we place an object to obtain Images from a concave mirror?
Answer:

As shown in figure, we can place the objects at six positions. They are:
1. Much beyond C i.e. at infinity
2. Beyond C
3. On center of curvature (C)
4. Anywhere between focus (F) and center of curvature (C)
5. On focus (F)
6. Anywhere between the pole (P) and focus (F)

Question 19.
Draw the ray diagram for a concave mirror when an object s placed at following points. Also give the position, nature and size of image.
Answer:
Object is placed at the following positions:
(1) At infinity
(2) Beyond C
(3) At C
(4) Between C and F
(5) At F
(6) Between P and F
(Note: Each diagram along with its detail carry 2 marks and hence the questions will be asked accordingly. However, here we have given detail of all the positions.)

Images formed by a Concave mirror

Summary of images formed in a concave mirror

Question 20.
State the uses of concave mirrors.
Answer:
Uses of concave mirrors:

• Concave mirrors are used in torches, search-lights and vehicle head-lights to obtain powerful parallel beam of lights.
• They are also used in shaving mirrors to see a larger image of the face.
• Dentists use concave mirrors to see enlarged view of teeth an patients.
• To concentrate sunlight for producing heat in solar furnace.

Question 21.
Why are focal length and centre of curvature negative for convex mirrors?
Answer:
1. In a convex mirror, the rays parallel to principle axis, instead of converging at a point in the front of the mirror, appear to diverge from a point behind the mirror. This point is called the focal point or focus F of the mirror. Similarly, the centre of curvature also lies behind the mirror.

2. As per the rule, positions in the front of any mirror will have positive sign and those behind the mirror will have negative signs. Since, F and C both lie behind the convex mirror, their signs are negative.
(Note: (1) The focus of convex mirror is also called virtual focus. (2) f = R/2 holds true for convex mirrors also.)

Question 22.
At how many positions can we place an object to obtain images from a convex mirror? Also explain why we place the object at fewer positions in convex mirror than concave mirror.
Answer:
To obtain an image on a convex mirror we can place our object only at two positions. They are:
(1) At infinity
(2) Between pole P and infinity

Reason:
1. In concave mirrors, the focus F and center of curvature C lie before the mirror. The object is also placed before the mirror and so it can be placed at six positions in concave mirror.

2. In a convex mirror, the focus F and center of curvature C lie behind the mirror. Now, the image is formed only when an object is placed before the mirror and not behind it. Hence, in convex mirror, there can be only two positions for placing the object before the mirror.

Question 23.
Draw and explain the type of Images formed by convex mirror.
Answer:
Images formed by a convex mirror

Summary of images formed in a convex mirror

Question 24.
List out the properties of the image formed by a convex mirror.
Answer:
Properties of image formed by a convex mirror:
(1) The image formed is always virtual and erect.
(2) The image is highly diminished i.e. point sized.
(3) Image is always formed between F and R
(4) As you move the object closer to the pole, the image also moves closer to the pole. It grows in size and at one point becomes equal to the object size.

Question 25.
Explain New Cartesian Sign Convention for reflection by spherical mirror.
Answer:
1. For studying the reflection from spherical mirrors, the method or sign convention called the New Cartesian Sign Convention is used.
2. Here, pole P of the mirror is taken as origin and The principal axis XX’ is taken along the X-axis of the New Cartesian co-ordinate system. The axis drawn perpendicular to principal axis at pole P is considered as Y-axis. According to the New Cartesian Sign

Convention:
(i) The object AB always lies on the left side of mirror and hence light is incident from left side of the mirror.
(ii) All the distances parallel to the principal axis are measured from the pole of the mirror P.
(iii) Distances measured in the same direction as that of incident light (left to right) along the +ve x-axis are taken as positive. x-axis are taken as positive.
→ Distances measured against the direction of incident light (right to left) along the -ve x-axis are taken as negative.
(iv) Distances measured perpendicular to and above the principal axis (along +ve y-axis) are taken as positive.
→ Distances measured perpendicular to and below the principal axis (along -ve y-axis) are taken as positive.

Summary of signs of various distances for concave and convex mirrors.

Note: Refer the New Cartesian Sign Convention while reading the table.

Question 26.
State the mirror formula for concave and convex mirrors.
Answer:
The relationship between the object distance (u), image distance (v) and focal length (f) of a mirror is called the mirror formula.
As per mirror formula:\(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
In words: \(\frac{1}{\text { Object distance }}+\frac{1}{\text { Imagedis tance }}=\frac{1}{\text { Focal length }}\)
Object distance image distance Focal length
Mirror formula is same for concave as well as convex mirror.

Question 27.
What do you mean by magnification? State Its definition and give to formula.
Answer:
Magnification:
Whenever we observe the image of an object after reflection by a spherical mirror, there can be a change in the size of the object. This means that the image formed can be larger than the object size, smaller or even of the same size as the object. This phenomenon of change in the image size is known as magnification (m).

Definition:
1. The ratio of image height (h’) to object height (h) is called the magnification (m).
2. Thus, for spherical mirrors, magnification (m) \((m)=\frac{\text { Height of image }\left(h^{\prime}\right)}{\text { Height of object }(h)} \text { i.e. } m=\frac{h^{\prime}}{h}\)
3. In terms of object distance (u) and image distance (y), magnification (m) can be expressed as,
\(m=\frac{h^{\prime}}{h}=-\frac{v}{u}\)
4. If value of m = positive, image is virtual.
5. If value of m = negative, image is real.

The two important aspects of magnification that determines the sign and value of m.
We know magnification\((m)=\frac{\text { Image height }\left(h^{\prime}\right)}{\text { Object height }(h)}\)

There are two important aspects of magnification. They are:
(a) Sign of m, (b) Value of m

(a) Sign of m:

• The sign of m i.e. magnification can be either positive or negtive.
• The sign of m depends on the signs of h’ and h.

(i) Sign of h:  h represents the height of the object. Since the object is always placed above the principal axis, value of h will always be positive.

(ii) Sign of h’:

• represents the height of image.
• If the image is virtual, it will always be erect and hence above the principal axis. Hence, h’ will be positive.
• If the image is real, It will always be inverted and hence below the principal axis. Here, h’ will be negative.

Conclusion:

• m will be positive for virtual images (Since both h and h’ will be positive.)
• m will be negative for real images. (Since h will be positive but h’ will be negative.)
• Concave mirrors can produce both virtual and real images. So, magnification m of a concave mirror can either be positive or negative and vice-versa.
• Convex mirrors and plane mirrors can produce only virtual images. So, their magnification will always be positive and vice-versa.

(b) Value of m:

• The magnification i.e. the value of m can either be greater than 1, equal to 1 or less than 1.
  (i) m> 1
•  If value of m> 1, it means that the size of image > size of object.

Example:
h’ = 10 cm and h = 5 cm
\(m=\frac{h^{\prime}}{h}=\frac{10}{5}=2\)
∴ image size > object size.

(ii) m = 1
If value of m = 1, it means size of image = size of object.

Example:
h’ = 10 cm and h= 10 cm
\(M=\frac{h^{\prime}}{h}=\frac{10}{10}=1\)∴ image size = object size

(iii) m < 1
→  If value of m < 1, it means size of image < size of object.

Example:
h’ = 5, h = 10
\(\mathrm{m}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=\frac{5}{10}=\frac{1}{2}=0.5\)
∴ image size <object size

Conclusion:

• Image produced by a plane mirror is always same size as that of object. Thus, m = 1.
•  Image produced by a convex mirror is always diminished i.e. smaller than the object. Thus, m < 1.
• Image produced by concave mirror can be both larger or smaller than the object. Thus, m> 1 or m < 1.

Question 28.
Obtain the position, nature and size of an image formed by a plane mirror from the formula of magnification.
Answer:
1. We know that magnification m =\(\frac{\mathrm{h}^{\prime}}{\mathrm{h}}\)
2. Now in a plane mirror, image height (h’) = object height (h).
3. So we get value of m=+1.
4. The image formed in its case by the plane mirror will be virtual, erect and of same size as the object.
5. Also, magnification m = \(\frac{\mathrm{v}}{\mathrm{u}}\).
6. But, m = 1 and so 1 = \(\frac{\mathrm{v}}{\mathrm{u}}\)
∴ u = -v i.e. y = – u.
7. Since y = – u, the size of the image y will be same as that of the object u but negative sign indicates that the image will be behind the mirror.

Summary of size of image, type of image and magnification for various mirrors

Question 29.
Textbook illustration 10.1: A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the Image.
Solution:
Radius of curvature, R = + 3.00 m;
Object-distance, u = – 5.00 m;
Image-distance, v=?
Height of the image, h’ = ?
Focal length, f = R/2 = +\(\frac{3.00 \mathrm{~m}}{2}\) = +1.50m (Since the principal focus of a convex mirror is behind the mirror)

The image is virtual, erect and smaller in size by a factor of 0.23.

Question 30.
Textbook illustration 10.2: An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.
Answer:
Object-size, h= +4.0cm;
Object-distance, u = – 25.0 cm;
Focal length, f = -15.0 cm;
Image-distance, v = ?
Image-size, h’ = ?

or \(\frac{1}{v}=\frac{-5.0+3.0}{75.0}=\frac{-2.0}{75.0}=\text { or, } v=-37.5 \mathrm{~cm}\)
The screen should be placed at 37.5 cm in front of the mirror. The image is real.
Also, magnification, \(m=\frac{h^{\prime}}{h}=-\frac{v}{u}\)
\(\text { Or, } h^{\prime}=-\frac{v h}{u}=-\frac{(-37.5 \mathrm{~cm})(-4.0 \mathrm{~cm})}{(-25.0 \mathrm{~cm})}\)
Height of the image, h’ = – 6.0 cm. The image is inverted and enlarged.

Question 31.
Differentiate between regular reflection and irregular reflection.
Answer:

Question 32.
Differentiate between concave mirror and convex mirror.
Answer:

Question 33.
Differentiate between real image and virtual image.
Answer:

Question 34.
Give a brief Introduction about refraction.
Answer:
Refraction:
1. Light travels on a same path i.e. without bending if it is travelling in same transparent medium such as air or water.
2. While travelling if light comes across another transparent medium, its velocity changes and so it bends.
3. For example, if light is travelling in air (medium 1) and then it strikes water (medium 2) obliquely, it bends. This bending of light is called refraction of light.

Question 35.
Define refraction of light. State its laws.
Answer:
Refraction:
When a ray of light enters obliquely from one transparent medium to another transparent medium, its velocity changes due to which it gets deviated from its original direction at the surface separating two medias. This is called refraction.

Laws of refraction:
1. Incident ray, refracted ray and normal to the surface separating two medias at the point of incidence, all lie in the same plane.
2. The ratio of the sine of angle of incidence to the sine of angle of refraction is constant for the light of a given colour (wave length) and for the given pair of media. This law is known as Snell’s Law of Refraction.
\(\frac{\sin \theta_1}{\sin \theta_2}=\text { constant }\)
Here, θ₁ = angle of incidence and θ₂= angle of refraction.
Example:
Refraction takes place when light travels from air (medium 1) into glass (medium 2).

Question 36.
With respect to travel of light and its refraction, explain medium and comparative types of mediums. Also explain behaviour of light in these mediums.
Answer:
Medium:

• A transparent substance in which light can travel is called a medium.
• Air, water gas, glass, kerosene, etc. are examples of mecums.
• The densities of each medium are different and so they are classified as follows —

(i) Optically rarer medium:
A medium in which the speed of light is more is called optically rarer medium.

Example:
Air is optically rarer than water and glass.

(ii) Optically denser medium:
A medium in which the speed of light is less (compared to an optically rarer medium) is called an optically denser medium.

Example:
1. Glass is optically denser than air. Kerosene is optically denser than water. Refraction of light in various mediums:
2. When light travels from a rarer medium (e.g. air) to a denser medium (e.g. glass), It bends towards normal at the point of incidence.
3. When light travels from an optically denser medium (e.g. glass) to a rarer medium (e.g. air), it bends away from normal at the point of incidence.

Question 37.
What do you mean by refractive index of a medium?
Answer:
1. A ray that travels obliquely from one transparent medium into another will change its direction in the second medium. This means that the light has refracted.
2. The extent of the change in direction that occurs within the given pair of media (let us say air and glass) can be expressed in the terms of refractive index.

Question 38.
Explain relative refractive index.
Answer:
Relative Refractive Index :
1. The ratio of speed of light v₁ in medium 1 to the speed of light v₂ in medium 2 is known as the relative refractive index of medium 2 with respect to medium 1 (except the medium of vacuum).
2.  It is denoted by η₁₂ i.e. refractive index of medium 2 with respect to medium 1.
∴ Refractive index η₁₂ = \(\eta_{21}=\frac{\text { Speed of light in medium } 1}{\text { Speed of light in medium } 2}=\frac{v_1}{v_2}\)
3. Similarly, the ratio of speed of light 2 in medium 2 to the speed of light y1 in medium 1 is known as the relative refractive index of medium 1 with respect to medium 2.
∴ Refractive index \(\eta_{12}=\frac{\text { Speed of light in medium } 2}{\text { Speed of light in medium } 1}=\frac{v_2}{v_1}\)

Question 39.
Define and explain absolute refractive index.
Answer:
Absolute refractive index:
1. The ratio of speed of light in vacuum (or air) (C) to the speed of light of any other medium is known as the absolute refractive index or simply refractive index.
2. It is denoted by η_m
∴ Absolute refractive index \(\eta_m=\frac{\text { Speed of light in vacuum (or air) }}{\text { Speed of light in medium }}=\frac{\mathrm{c}}{\mathrm{v}}\)

Question 40.
Give Snells law.
Answer:
Snells law:
1. The ratio of sine of angle of incidence to the ratio of sine of angle of refraction is constant.
2. This constant is known as relative refractive index of medium 2 with respect to medium 1 and is denoted by η₂₁.
∴ \(\eta_{21}=\frac{\sin \theta_1}{\sin \theta_2}\)
3. In terms of ratio of velocity of light In two media, the refractive index of light is represented as under –
The ratio of velocity of light v₁ in medium 1 to the velocity of light v₂ in medium 2 is called relative refractive index of medium 2 with respect to medium 1.
∴ \(\eta_{21}=\frac{v_1}{v_2}\)
4. The refractive index of the transparent medium with respect to vacuum is called the absolute refractive index of a medium. It is commonly known as refractive index.

Question 41.
Why do we take absolute refractive index of air = 1?
Answer:
1. Actually the velocity of light in vacuum is slightly more than the velocity of light in air.
2. But for calculating purpose, magnitude of velocity of light in vacuum is taken as equal to the velocity of light in air =
3 x 10⁸m/s
Absolute refractive index of air = \(\frac{\text { velocity of light in vaccum }}{\text { velocity of light in air }}=\frac{3 \times 10^8}{3 \times 10^8}=1\) gives us the value of absolute refraction index of air = 1.
4. Thus, absolute refractive index of air is taken as 1.

Question 42.
Why absolute refractive index Is always greater than 1? Explain with the help of an example.
Answer:
1. Velocity of light is highest in a vacuum.
2. Hence, velocity of light in any other medium will always be lesser than the velocity of light in vacuum or air.
3. In practice, we take the velocity of light in vacuum = velocity of light in air = 3 x 10⁸ rn/s
4. Thus, absolute refractive index
\(\eta=\frac{\text { velocity of light in vacuum }}{\text { velocity of light in medium } 2 \text { i.e. air }}=\frac{3 \times 10^8}{3 \times 10^8}=1\)
5. Now, when we consider absolute refractive index of vacuum or air with any other medium, the answer will always be greater than 1.

Example:
Velocity of light in vacuum (air) = 3 x 10⁸ m/s
Velocity of light in water = 2.25 x 10⁸ m/s
∴ Absolute refractive index of water = \(\frac{\text { velocity of light in vacuum }}{\text { velocity of light in water }}=\frac{3 \times 10^8}{2.25 \times 10^8}=1.33\)
Thus, absolute refractive index will always be greater than 1.

Question 43.
(Activity 10.10) Perform an activity to demonstrate refraction of light through a glass slab.
Answer:
1. Fix a sheet of white paper on a drawing board using drawing pins.
2. Place a rectangular glass slab over the sheet In the middle.
3. Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
4. Take four identical pins.
5. Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
6. Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
7. Remove the pins and the slab.

8. Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
9. Join O and O’. Also produce EF up to P, as shown by a dotted line in figure.

Observation:
1. At O, light ray along EF enters from air into glass. It bends towards the normal NN’. This is the first refraction.
2. At O’, the light ray enters from glass into air. It bends away from the normal MM’ and travels along GH. This is second refraction.
3. Here, angle of emergence r₂ is equal to angle of incidence t₁, i.e., the emergent ray is parallel to the original direction of the incident ray. This is because there is an identical medium, air in this case, adjacent to both edges AB and CD.
4. However, the light ray is shifted sideways slightly. This is lateral displacement and is represented by O’L.

Conclusion:

• The ray of light travelling from a rarer medium (air) to a denser medium (glass) bends towards the normal.
• The ray of light travelling from a denser medium (glass) to a rarer medium (air) bends away from the normal.
• The emergent ray is parallel to the incident ray but is slightly displaced sideways.

Types: (A) Convex lens:
A lens which is thick at the center and tapered at the upper and lower ends is called a convex lens.

(B) Concave lens:
A lens which is thin in the middle but thicker at the edges is called concave lens.

Uses of lens:
In magnifying glass, cameras, repairing watches, microscope, telescope, etc.

Question 45.
What do you mean by converging and diverging lens?
Answer:
Converging lens:
1. A lens that converges (brings closer) the rays of light travelling parallel to Its principal axis to a point called focus (F) is called a converging lens.
2. Convex lens is a converging lens.

Diverging lens:
A lens that diverges (spreads out) the rays of light travelling parallel to its principal axis is called a diverging lens. If we plot these rays backwards, they appear to diverge from the focus (F) of the lens.
(Note: Lenses have two faces. So light can pass from both sides i.e. left and right.)

Question 46.
Define various terms related to lenses along with suitable diagrams.
Answer:
1. As shown in the diagram, a convex lens (or a concave lens) can be thought of as formed from two spheres.
2. Since there are two spheres and the light can pass from both the sides of a lens, there are two radii, two focuses and two centers of curvature.

1. Centres of curvature:

• The centres of respective spheres of which the surface of lens form parts are called centres of curvature.
• Lens has two centres of curvature denoted by C₁ and C₂.

2. Radii of curvature:

• The radii of spheres of which surfaces of lens form parts are called radii of curvature
• Lens has two radii of curvature denoted by R₁ and R₂.

3. Principal axis: A straight imaginary line passing through the centres of curvature C₁ and C₂ of lens is called the principal axis of lens.

4. Optical centre: The centre of lens, located on principal axis, is known as the optical centre. It is denoted by O.

5. Principal focus: When rays parallel to the principal axis of convex lens are refracted, they converge at a point on the principal axis. This point is called principal focus of the convex lens. A convex lens has two principal foci F₁ and F₂ on either side of lens.

→ The rays parallel to the principal axis of concave lens are refracted such that they appear to be diverging from a point on principal axis. Such a point is called focus of concave lens. Concave lens also has two focii F₁ and F₂ on either side of lens.

6. Focal length
→ The distance between the optical centre (O) and the principal focus (F) is called the focal length of the lens (f).
→ We also mark another point called 2F on the principal axis. This point lies at twice the distance of F.
(Note: In a lens, 2F and C may or may not be at same position.)

Question 47.
Explain with the help of ray diagram how an image can be formed with convex lens. For obtaining an image through convex lens, we consider the following rays:

Ray 1: It moves parallel to the principal axis and after refraction passes through the principal focus (F₂) on the other side of the lens.

Ray 2: It passes through the principal focus (F₁) and after refraction becomes parallel to the principal axis on the other side of the F lens. 1A

Ray 3: The ray passing through optical centre O will emerge from other side of the lens without any deviation.

Question 48.
List out the various positions at which we can place an object in front of convex lens.

Positions at which we can place an object in front of a convex lens.

(Note: 1. Normally, for both concave and convex lens, the rays are shown coming from the left side of the lens. After refraction they move towards the right side of the lens. 2. The focus F₁ lies on the left side i.e. towards surface 1 of the lens and focus F₂ on the right side i.e., towards surface 2 of the lens.

In a convex lens, an object can be placed at the following positions to obtain its image.
(1) At Infinite distance
(2) Beyond 2F₁
(3) At 2F₁
(4) Anywhere between F₁and 2F₁
(5) At principal focus (F₁)
(6) Anywhere between principal focus (F₁) and optical center (O)

Question 49.
Draw the diagrams for a convex lens when an object is placed at the following points. Also give the position, nature and size of the image.
(1) At infinity
(2) Beyond 2F₁
(3) At 2F₁
(4) Between F₁ and 2F₁
(5) At focus F₁
(6) Between focus F₁ and optical center (O).

(Please note : Each diagram along with its detail carry 2 marks and hence the questions will be asked accordingly. However, we have given detail of all the positions.)

Images formed by a convex lens

Summary of images formed by convex lens

Question 50.
Draw ray diagram for a concave lens when an object Is placed at (1) Infinity and when object is placed (2) Between infinity and optical centre (O) of the lens.

Images formed by a concave lens

Question 52.
State the New Cartesian Sign convention to be followed for refraction of light through spherical lenses.
Answer:
New Cartesian Sign convention for refraction of light through spherical lenses:
(i) The object is always placed on the left side of the lens. This means that the light is incident on the lens from left side.
(ii) All distances parallel to the principal axis are measured from the optical centre (O) of the lens.
(iii) The distance measured in the same direction as the incident light is taken positive.
(iv) The distance measured in the direction opposite to the direction of incident light is taken negative.

(v) Height measured above the principal axis of the lens and perpendicular to it i.e. in the upward direction is taken as positive.
(vi) Height measured below the principal axis of the lens and perpendicular to it in the downward direction is taken as negative.

New Cartesian Sign Convention followed for refraction of light through spherical lenses
Sign convention for a convex lens:

Object distance = – u
Object height = + h
Image distance = + v for real image
Image height = – h’ for real image
Focal length = + f
Note: If the image is virtual, then image distance
= -v and image height = + h’

Sign convention for a concave lens:

Object distance = – u
Object height = – v
Image distance = – v
Image height = + h’
Focal length = – f

Note: If the image is virtual, then image distance
= – v and image height = + h’
→ Finally, the linear magnification, m = h’/h is positive for a virtual image and negative for a real image

Question 53.
State and explain the lens formula.
Answer:
Lens formula:
(i) The lens formula is a mathematical relation between the object distance (u), image distance (v) and focal length (f) of a spherical lens.
As per lens formula,\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
In other words,\(\frac{1}{\text { Image distance }}-\frac{1}{\text { Objectdistance }}=\frac{1}{\text { Focallength }}\)
Image distance Object distance Focal length
(ii) Lens formula is applicable to both concave lens as well as convex lens.

Question 54.
State the equation for magnification produced by a lens.
Answer:
1. The magnification produced by a lens is defined as “The ratio of the height of the image formed by the lens to the height of the object”.
Thus, magnification m = Height of the image = h’
, Height of the object h \(m=\frac{\text { Height of the image }}{\text { Height of the object }}=\frac{h^{\prime}}{h}\)
Also, \(m=\frac{h^{\prime}}{h}=\frac{v}{u}\)
2. ‘m’ is positive for virtual image formed by the lens.
3. ‘m’ is negative for real image formed by the lens.

Question 55.
Textbook Illustration 10.3: A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens.
Solution:
A concave lens always forms a virtual, erect image on the same side of the object.
Image-distance v = -10 cm;
Focal length f = -15 cm;
Object-distance u = ?

\(\frac{1}{u}=\frac{-2+2}{30}=\frac{1}{-30}\)
Thus, the object – distance is 30 cm.
Magnification m = v/u
\(\mathrm{m}=\frac{-10 \mathrm{~cm}}{-30 \mathrm{~cm}}=\frac{1}{3}=+0.33\)
The positive sign shows that the image is erect and virtual. The image is one-third of the size of the object.

Question 56.
Textbook illustration 10.4: A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens Is 15 cm. Find the nature, position and size of the image. Also find its magnification.
Answer:
Height of the object h =+ 2.0 cm;
Focal length f =+ 10 cm;
object-distance u = – 15 cm;
Image-distance v = ?
Height of the image h’ = ?
The positive sign of v shows that the image is formed at a distance of 30 cm on the other side of the optical centre. The image is real and inverted.
Magnification \(m=\frac{h^{\prime}}{h}=\frac{v}{u}\)
or, h’ = h (v/u)
Height of the image, h’ = (2.0) (+30/-15)
= – 4.0 cm
Magnification m = v/u
or \(\mathrm{m}=\frac{+30 \mathrm{~cm}}{-15 \mathrm{~cm}}=-2\)
The negative signs of m and h’ show that the image is inverted and real. It is formed below the principal axis. Thus, a real, inverted image, 4 cm tall, is formed at a distance of 30 cm on the other side of the lens. The image is two times enlarged.

Question 57.
Write a short note on : Power of lens. OR Write definition of power of lens. Give Its SI unit. Name the Instrument used to measure the power of lens.
Answer:
1. The ability of a lens to converge or diverge light rays depends on its focal length.
2. The convex lens of short focal length bends the light rays at large angles by focusing them closer to the optical centre.
3. Similarly concave lens of very short focal length causes higher divergence than the one with larger focal length.
4. The efficiency with which a lens converges or diverges the light rays is expressed in terms of its power.
5. Thus, the power of lens (P) is the “Reciprocal of the focal length (I) of lens”.
Power of lens \((\mathrm{P})=\frac{1}{f}\)
6. The SI unit of power of a lens is diopter and it is denoted by D.
7. If fis expressed in meters, then \(1 D=\frac{1}{1}=1 m^{-1}.\)
8. We can measure the power of lens by using diopter meter.
9. As per Cartesian sign system, focus of convex lens lies on right side of the lens and so power of convex lens is positive, while focus of concave lens lies on left side of lens and so power of concave lens is negative.

Question 58.
Give the sign convention for powers of converging and diverging lenses.
Answer:
1. Converging lens has a positive focal length and so its power (P) is positive.
2. Diverging lens has a negative focal length and so its power (P) is negative.

Question 59.
A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine. Support your answer with reason.
Answer:
1. A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. This happens because the light reaching out from the portion of the pencil inside seems as if it is coming from a different direction as compared to the part above water due to refraction of light.

2. The pencil will look bent at different extents when materials such as kerosene or turpentine are used because their refractive indices are different. This produces deviation from incident ray by different extents.

Question 60.
The film or CCD chip must be placed where the image is focused In order to record a sharp image. Give reason.
Answer:
1. Light from very distant objects reaches the camera in the form of parallel beams.
2. This will allow the light to focus right at the focal point of the lens.
3. Anything closer will have rays that will not be parallel but diverging and so they will be focused at a distance further from the lens than the focal point of the lens.
4. Hence, the film or CCD chip must be placed where the image is focused properly in order to record a sharp image.

Question 61.
Convex mirrors are used In car headlights where as concave mirrors are used in street lights. Give reason.
Answer:
1. Concave mirrors have an ability to form a parallel and powerful beam of light focusing on a particular area.
2. On keeping the bulb of the car head light at the focal point of the reflector, a powerful beam of light is produced. This allows better focusing of the lights and hence the visibility of the driver improves.
3. Whereas, convex mirrors have the ability to diverge the rays over a large area. This brightens a larger area compared to concave mirror.
4. Hence, convex mirrors are used in car headlights where as concave mirrors are used in street lights.

Question 62.
You are given three mirrors of equal sizes — concave, convex and plane. How will you identify them without touching their surfaces?
Answer:
Take one mirror at a time and watch your face in it. First keep the face quite close and then slowly and gradually move the face away from the mirror. Do this in each mirror and decide the type of mirror based on the following observations —

• If the image formed is of same size as our face but laterally inverted i.e. left portion looks right and right looks left for all positions, then it is a plane mirror.
• If the image formed is erect and enlarged initially but gets inverted as the face is moved further, the mirror is a concave mirror.
• If the image formed is erect and smaller in size for all the position, the mirror is convex.

Question 63.
The phrase “Objects in mirror are closer than they appear” is written on the side mirrors of the vehicles. Give reason.
Answer:
1. The side mirrors of the vehicles are convex mirrors.
2. The image obtained through convex mirrors is always diminished.
3. As a result the actual object appears smaller.
4. Since smaller-appearing objects seem farther than they actually are, a driver might change a lane assuming an adjacent vehicle is at a safe distance behind, when in fact it may be quite closer.
5. Thus, the phrase Objects in mirror are closer than they appear” serves as a warning/reminder to the driver of this potential problem.

Question 64.
A diamond kept under water sparkles less as compared to when kept in air. Give reason.
Answer:
1. The sparkling of a diamond actually occurs because of repeated total internal reflection.
2. When light enters a diamond, it is difficult for it to leave because of the high index of refraction so it tends to totally internally reflect inside the diamond.
3. The multiple internal reflection of light in the diamond in different directions, allow it to sparkle.
4. Putting a diamond underwater means that the difference in speeds (or index of refraction) is much less which leads to less total internal reflection. This decreases the sparkle of the diamond.

Question 65.
State the type of mirror preferred as
(i) Rear view mirror In vehicles and
(ii) Shaving mirror. Justify your answer giving two reasons in each.
Answer:
(i) A convex mirror is preferred as a rear-view mirror because —

• Convex mirror always forms a virtual, erect and a diminished (point-like) image of an object placed anywhere in front of it. This means the driver can see the image of person even at a very distant position.
• Convex mirror has a wider field of view. So it covers a large rear distance. This helps the driver as he can see more.

(ii) A concave mirror Is preferred for shaving because —

• When such a mirror is held near the face, the person can see enlarged image of his face. This helps to have a cleaner shave.
• An erect image is formed.

Question 66.
A person in a dark room looking through a window can clearly see a person outside in the daylight, whereas the person outside cannot see the person inside. Give reason.
Answer:
1. When light falls on an object, it absorbs some light and reflects some.
2. If you are inside in the dark, a person outside in bright sunlight is sending out (reflecting) lots of light
3. Most of this light would come through the window to you, so you see them clearly.
4. Since it is so bright outside, there is also a good amount of light, which reflects back towards you.
5. In your case, you are standing in dark and so very little light falls on you. As a result, you do not reflect much light. This makes it difficult for the person standing outside to see you clearly.

Question 67.
Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp Image of the building? What Is the approximate focal length of this lens?
Answer:
Sudha should move the screen towards the lens to for obtaining a sharp Image of the building.

Reason:
1. The window pane was lying beyond focus 2F i.e. center of curvature and convex lens forms its image, on the other side between F and 2F
2. The building that Sudha sees from her window can be considered as a distant object i.e. at infinity. When Sudha tries to focus the building, the lens forms the Image of building at a distance of focal length. The approximate focal length of this lens is 15 cm.

Question 68.
Raj along with his friend Dipti visited Nagina-wadi in Kankaria, Ahmedabad which has got a variety of mirrors. Raj showed her a mirror In which Dipti’s image showed the upper half of the body very fat and lower half of the body very thin. Dipti thought there is a problem with the mirror. She got embarrassed and also angry. Raj laughed and explained not to worry as the mirror is designed that way.
(1) Can you name the two types of mirrors used?
(2) Name the mirror In which the size of Image ¡s small.
(3) What values are displayed by Raj?
Answer:
(1) Concave and convex mirror
(2) Convex mirror
(3) Compassion and empathy towards Dipti in consoling her and making her calm.

Question 69.
Under what condition in an arrangement of two plane mirrors, the incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.
Answer:
When we place two plane mirrors at right angle with each other, then the incident ray and reflected ray will always be parallel to each other, whatever the angle of incidence.

Question 70.
Four friends were on a picnic. Ranjit was driving the car. Suddenly he noticed that in his side mirror that the car which was behind their car had met with an accident. He immediately stopped his car and the friends decided to help the injured person. All of them took the Injured person to the nearest hospital. After taking first-aid from the hospital, the victim thanked them for saving his life.
(1) Name the type of mirror from which Ranjit saw the accident.
(2) Why is this type of mirror used as a side mirror in vehicles?
(3) Which values are displayed by Ranjit?
Answer:
(1) A convex mirror
(2) Convex mirrors, have wider field of view, as they are curved outwards. Therefore, a convex mirror enables the driver to view much larger area.
(3) Ranjit displayed alertness as well as duty of a citizen towards other. The friends also displayed selflessness and empathy.

Question 71.
A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top. Which combination do you think is correct? Justify your reason.
(1) Plane, convex and concave,
(2) Convex, concave and plane
(c) Concave, plane and convex,
(d) Convex, plane and concave
Answer:
1. Combination (c) Concave, plane and convex is the correct combination.
2. Concave mirrors of large focal length can be used to see a larger image of the head, the plane mirror for middle portion to see the body of the same size whereas convex mirror to see the diminished image of leg.
3. Hence, the selected combination is correct.

Question 72.
A pencil when dipped in water in a glass tumbler appears to be bent at the Interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine. Support your answer with reason.
Answer:
1. A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. This happens because the light reaching out from the portion of the pencil inside seems as if it is coming from a different direction as compared to the part above water due to refraction of light.
2. The pencil will look bent at different extents when materials such as kerosene or turpentine are used because their refractive indices are different. This produces deviation from incident ray by different extents.

Question 73.
If an object of 4 cm height Is placed at a distance of 18 cm from concave mirror having focal length 12 cm, find the position, nature and height of the image.
Answer:
Object height h = 4 cm
Object distance u = -18 cm
Focal length f = -12 cm
(Both, principal focus F and u lie to the left of concave mirror, hence negative.)
From mirror formula,

Conclusion:
1. Magnification is positive, h’ is negative which indicates image is real and inverted. Also, it is enlarged compared to object.
2. The image is formed 36 cm on the left side of pole P beyond the center of curvature C.

Question 74.
A convex mirror Is fitted on an automobile with focal length of 3 m. If a vehicle behind is at a distance of 5 m, determine the position and nature of an image.
Answer:
Object distance u = – 5 m
Focal length f = 3 m
(f is positive because focus of convex mirror lies on right side of pole P)
From mirror formula,

\(v=\frac{15}{8}\) = 1.875m
Here, y is positive and y < u
∴ m<1
Therefore, virtual, erect and diminished image behind the convex mirror is obtained at a distance 1.875 m from the mirror.

Question 75.
An object of height 5 cm is placed at a distance of 10 cm from convex mirror of focal length 15 cm. Find the position, nature and size of an image.
Answer:
Object distance u = -10 cm
Focal length f = 15 cm
Object height h = 5 cm
According to mirror formula,

Here, image distance v is positive and h’ is positive so, we get a virtual image of 3 cm height behind the mirror which is erect and diminished.

Question 76.
Refractive index of media A, B, C and D are

In which of the four media is the speed of light (i) maximum and (ii) minimum? Find the refractive index of medium C with respect to medium B.
Answer:
(i) Speed of light is maximum is medium A for which n (= 1.33) is minimum
(ii) Speed of light is minimum in medium D for which n (= 1.65) is maximum. Refractive index of medium C with respect to medium B.
\(\mathrm{n}_{\mathrm{CB}}=\frac{\mathrm{n}_{\mathrm{C}}}{\mathrm{n}_{\mathrm{B}}}=\frac{1.52}{1.44}=1.05\)

Question 77.
(a) The refractive index of Ruby is 1.71. What is meant by this statement?
(b) The refractive index of some medium are given below:
Crown glass – 1.52, Water – 1.33, Sapphire  – 1.77
In which of the medium is the speed of light:
(i) Maximum (ii) Minimum?
(c) Calculate speed of light in sapphire.
Answer:
(a) The refractive index of ruby is 1.71. It means that the ratio of the speed of light in vacuum to the speed of light in Ruby is 1.71.

(b) (i) Speed of light is maximum in water for which n (= 1.33) is minimum
(ii) Speed of light is minimum in sapphire for which n (= 1.77) is maximum

(c)

Question 78.
Light enters from air to glass having refractive index 1.5. What is the speed of light in glass? The speed of light in vacuum is
3 x 10⁸ m/s.
Answer:
Absolute refractive index of glass
n = \(\frac{\mathrm{c}}{\mathrm{v}}\) and c = 3 x 10⁸ m/s^-1
\(v=\frac{c}{n}=\frac{3 \times 10^8}{1.5} m/s^-1\) (n = 1.5) = 2 x 10⁸ m/s^-1[/latex]

Question 79.
A light ray enters from air to the water medium having the absolute refractive index 1.33. If the angle of refraction of light is 17°30’, what will be the angle of incidence at the surface separating the two media ? Take absolute refractive index of air as 1.00.
Answer:
Taking air as medium 1 and water as medium 2,
n₁ = 1, n₂ = 1.33, θ₂ = 17°30’
According to Sneil’s law,
n1 sin θ1 = n2 sin θ2
sin θ₁ = 1.33 X 17°30’
The value of sin 17°30’ is determined from the Mathematical table of natural sine.
sin 17°30’ = 0.3
∴ sin θ₁ =1.33 x 0.3 = 0.4
∴ From the table of natural sine
sin 23° 36’ = 0.4007
∴ θ₁ = 23° 36’
∴ Angle of incidence = 23° 36’

Question 80
The refractive index of light entering glass from water is 1.12. Find the absolute refractive index of water if the absolute refractive index of glass is 1.5.
Answer:
Here, we take water as medium 1 and glass as medium 2.
Absolute refractive index of glass η₂= 1.5
Absolute refractive index of glass with respect to water η₂₁ = 1.12
Therefore, absolute refractive index of water η₁ =?

Thus, absolute refractive index of water is = 1.34

Question 81.
A swimmer lights a torch under the surface of sea water. Light from the torch is incident on water surface In such a way that incident light makes an angle 370 with water surface. Find the angle of refraction. Absolute refractive index of water is 1.33 and absolute refractive index of air is 1.0.
Answer:
Taking medium 1 as water, we get η₁ = 1.33
Taking medium 1 as water, we get η₂ = 1.00
Incident angle θ₁ = 37°
Angle of refraction θ₂ = ?
Using the formula, =

Angle of refraction θ₁ = 53° (from log table).
∴ The value of angle of reflection is equal to 53°

Question 82.
A convex lens forms a real and inverted image of an object at a distance of 40 cm from it. What will be the distance of an object if size of an image is same as the object? Determine the power of lens.
Answer:
Given:
Lens = Convex
Image = Real and inverted
Size of image h’ = object size h
Now, \(m=\frac{h^{\prime}}{h}\)
But, h’= h ∴ m=1.
The image is inverted. ∴ h’ will be negative.
∴ magnification on will be -1.
(Note: You can also think this way. For a convex lens, if the image is real and inverted, the magnification will be negative.)

Thus, object distance = – 40 cm and also image distance = 40 cm.
Substituting, values of y and u in lens formula we get,
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Question 83.
The object of 5 cm height is placed at a distance 25 cm from the centre of convex lens of focal length lo cm. Draw a ray diagram and find position, nature and the size of an image formed.
Answer:
1. Take a scale 2.5 cm = 1 unit on principal axis and draw a ray diagram. The ray diagram is shown below.
2. This shows that real, inverted and diminished image of an object is formed on other side of lens at a distance of 16.67 cm from the lens.
3. Also, h’ = – 3.33 cm

As per lens formula,

Image will be formed on the other side of the lens at a distance of 30 cm.

Question 84.
An object Is placed perpendicular to the principal axis of concave lens focal length 30 cm. Find the position of an image when the object is at a distance 20 cm from the lens.
Answer:
The focal length of lens f = -30 cm
Object distance u = -20 cm, image distance v = ?
As per the lens formula,

The image will be obtained on the same side of the object at a distance of 12 cm from lens.

Question 85.
The magnification produced by a spherical lens is +2.5. What is the
(a) Nature of image
(b) Nature of lens?
Answer:
(a) When the magnification is positive, then the image is virtual and erect. In this case, the magnification has a positive sign, so the nature of image is virtual and erect.
(b) The value of magnification given here is 2.5 (which is more than 1). So, the image is larger than the object. A virtual, erect and magnified can be formed only by a convex lens, therefore, the nature of lens is convex.

Question 86.
Power of a convex lens is +4.0 D. At what distance should the object from the lens be placed to obtain its real and inverted image of the same size on the screen?
Answer:
1. The focal length of this convex lens is + 4D. As per formula \(p=\frac{1}{f(\text { in meters })}\) we get
\(f=\frac{1}{p}=\frac{1}{4}\) = 0.25m = 25cm
2. Now, in a convex lens, when the image size is same as object size, then the object should be placed from the lens at a distance of 2f.
3. Thus, object distance = 2f = 2 x 25 = 50 cm.

Question 87.
Two thin lenses of power +3.5 D and – 2.5 D are placed in contact. Find the power and focal length of the lens combination.
Answer:
Power of combination of lenses, p = p₁ + p₂
So, p = + 3.5 + (-2.5)
p=+3.5 – 2.5
p = + 1.0 D
Thus, the power of this combination of lenses is, + 1.0 diopter.
We will now calculate the focal length of this combination of lenses. We know that:
Power, \(P=\frac{1}{f} \text { or }+1=\frac{1}{f}\)= and f= +1 m
So the focal length of this combination lens is + 1 meter.

Question 88.
You have two lenses A and B of focal lengths + 10 cm and -10 cm respectively. State the nature and power of each lens. Which of the two lenses will form a virtual and magnified Image of an object placed 8 cm from the lens? Draw a ray diagram to justify your answer.
Answer:
Here, = + 10 cm, = – 10 cm

Lens A is convex and lens B is concave. Lens A will form a virtual and magnified image when the object is placed as 8 cm from it.

Question 89.
If the absolute refractive indices of water, benzene, and sapphire are 1.33, 1.50 and 1.77 respectively, then which medium has maximum relative refractive index?
(i) Sapphire relative to water
(ii) Sapphire relative to benzene
(iii) Benzene relative to water
(iv) Water relative to benzene
Answer:
Absolute refractive index of water = 1.33,
Absolute refractive index of benzene = 1.5, and Absolute refractive index of sapphire r = 1.77,
(A) Relative refractive index of sapphire η₃ to water η₁  i.e. = η₃₁ =\(\frac{\eta_3}{\eta_1}= 1.77/1.33\) = 1.33
(B) Relative refractive index of sapphire η₃ to benzene η₂ i.e. = η₃₂= \(\frac{\eta_3}{\eta_2}\)= 1.77/1.5 = 1.18
(C) Relative refractive index of benzene η₂ to water η₁ i.e. = η₂₁ =\(\frac{\eta_2}{\eta_1}=1.5/1.33\)= 1.12
(D) Relative refractive index of water η₁ to benzene η₂ i.e. η₁₂ =\(\frac{\eta_1}{\eta_2}= 1.33/1.5\)= 0.88
Thus, sapphire to water has the highest relative refractive index.

Question 90.
It he absolute refractive indices of water and glass are 4/3 and 3/2 respectively, then what will be the ratio of velocity of light In water to that of glass?
→ 8/9
Solution:
Absolute refractive index of water = \(\eta_1=\frac{4}{3}\)
Absolute refractive index of glass = \(\eta_2=\frac{3}{2}\)
∴ Ratio of velocity of light in water to glass will be \(\eta_{21}=\frac{v_1}{v_2}\)

Very Short Answer Type Question

Question 1.
What is reflection? State its types.
Answer:
The phenomenon of sending back the light rays which fall on the surface of an object when light is incident on it, is called reflection of light.
Types:

• Regular reflection and
• Irregular (diffused) reflection.

Question 2.
What is regular reflection?
Answer:
1. When a parallel beam of light falls on a smooth or shining surface, then the reflected beam is also parallel and directed back in a fixed direction.
2. This type of reflection is known as regular reflection. E.g. reflection by a mirror.

Question 3.
What is irregular or diffused reflection?
Answer:
1. When a parallel beam of light falls on a rough surface or irregular surface, then reflected light is not parallel but spreads over a wide area.
2. This type of reflection is known as irregular reflection.

Question 4.
State the laws of reflection.
Answer:
(i) The angle of incidence (θi) is equal to the angle of reflection θr i.e. θi = θr.
(ii) The incident ray, the normal to the mirror at the point of incident and the reflected ray, all lie in the same plane.

Question 5.
Define angle of incidence.
Answer:
The angle that incident ray makes with normal drawn at the point of incidence is known as incident angle or angle of incidence. It is denoted by θ_i.

Question 6.
Define angle of reflection.
Answer:
The angle that reflected ray makes with normal drawn at the point of reflection is known as reflection angle or angle of reflection. It is denoted by θ_r.

Question 7.
What is a plane mirror?
Answer:
A plane mirror is a thin, flat and smooth sheet of glass having a shining coating of silver metal on one side.

Question 8.
What happens when a parallel beam of light is Incident on shining plane or smooth surface?
Answer:
The beam remains parallel after reflection a specific direction

Question 9.
What is a spherical mirror?
Answer:
1. A spherical mirror is a mirror whose reflecting surface is in fact a part of a hollow sphere of a glass.
2. Unlike plane mirrors, a spherical (curved) mirror converges or diverges parallel light rays incident on them.

Question 10.
Draw a labeled diagram showing various parts of a concave mirror.
Answer:

Question 11.
State the relation between radius and focal length of a spherical mirror.
Answer:
Radius of curvature (R) = 2 times focal length (f)
∴ \(f=\frac{R}{2}\)

Question 12.
Radius of a mirror is 25 cm. What will be its focal length?
Answer:
R = 2f \(f=\frac{R}{2}=\frac{25}{2}\) = 12.5 cm.

Question 13.
What should be the position of an object with respect to focus of a convex lens of focal length 20 cm, so that its real and magnified image is obtained? Draw a ray diagram to show the image formation.
Answer:
1. Given that f = 20 cm.
2. To obtain real and magnified image, the object should be placed between F₁ and 2F₂. Hence, the range will be between 20 cm to 40 cm.

Question 14.
Draw a labeled diagram showing various parts of a convex mirror.
Answer:

Question 15.
How does a ray parallel to the principle axis behaves In a concave mirror? Show with diagram.
Answer:
Ray parallel to principal axis which after reflection passes through principal focus F.

Question 16.
How does a ray that passes through the principle focus of a concave mirror behaves? Show with diagram.
Answer:
Ray passing through the principle focus will become parallel to the principle axis after reflection.

Question 17.
Draw the diagram for a ray directed towards principle focus and its reflection through a convex mirror.
Answer:

Question 18.
Look at the diagram and describe the position, nature and size of the image formed.
Answer:

Position: Beyond centre of curvature C
Nature: Real and inverted
Size: Magnified

Question 19.
For a concave mirror, draw the diagram for image formed when the object is placed beyond C.
Answer:

Question 20.
Fill the table with respect to concave mirror.
Answer:
(1) Position of image — On c, Nature — Real and inverted,
(2) Position of object — Behind the mirror, Nature — Virtual and erect

Question 21.
State one use of convex mirrors.
Answer:
It is used as rear-view mirrors in vehicles.

Question 22.
ROM wants to have an erect Image of an object using a converging mirror of focal length 40 cm.
(i) Specify the range of distance where the object can be placed in front of the mirror. Give reason for your answer.
(ii) Will the image be bigger or smaller than the object?
Answer:
(i) To have an erect image of an object, the position of object should be between pole of the concave mirror and its focus, i.e. between 0 to 40 cm.
(ii) The image formed will be magnified i.e. image will be bigger than the object.

Question 23.
Raghav drew the following diagram in which a ray of light is incident on a convex mirror. You need to redraw the diagram and complete the path of the ray after It gets reflected from the mirror.
Answer:

Question 24.
The focal length of a concave mirror is 20 cm. Where will you place the object if you want an image which is real, inverted and of same size as the object?
Answer:
40cm

Question 25.
A concave mirror produces a real image which is five times magnified as compared to the object. The object is placed at 15 cm in the front of the mirror. At what distance will the image be produced?
Answer:
75cm
(Hint: magnification m = -v/u.
Since image is real, take – m instead of m. Object lies to the left of the mirror and so u = -u
∴ – m = -v/-u)

Question 26.
What is the angle of incidence for a ray of light passing through the centre of curvature of a concave mirror?
Answer:
0°

Question 27.
State the mirror formula.
Answer:
The formula which gives relation between object distance (u), image distance (v) and focal length (f) of a spherkal mirror is known as the mirror formula.
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Question 28.
Define magnification of spherical mirror.
Answer:
The ratio of image height (h’) to the object height (h) of the spherical mirror is called the magnification of the mirror.

Question 29.
State the formula of magnification.
Answer:
Magnification \(m=\frac{\text { Image height }\left(h^{\prime}\right)}{\text { Object height }(h)}=-\frac{v}{u}\)

Question 30.
For a mirror, h’ = 10 cm and h = 5 cm. Give its magnification.
Answer:
\(M=\frac{h^{\prime}}{h}=\frac{10}{5}=2\)

Question 31.
If m < 1, state the relation between image size and object size.
Answer:
If m < 1, image size < object size.

Question 32.
What will be the magnification of an object whose height is 4 cm and which produces a real image of 6 cm height?
Answer:
-1.5cm

Question 33.
What is refraction?
Answer:
When a ray of light enters obliquely from one transparent medium to another transparent medium, its velocity changes due to which it gets deviated from its original direction at the surface separating two medias. This is called refraction.

Question 34.
Look at the diagram given here and watch the alphabets. Now write the correct sequence of (a) angle of incidence, (b) angle of emergence, (c) angle of refraction and (d) lateral displacement shown in the diagram as a, b, c and d.

Answer:
The correct sequence is b, a, d, c

Question 35.
What is relative refractive index?
Answer:
1. The ratio of speed of light v₁ in medium 1 to the speed of light v₂ in medium 2 is known as the relative refractive index of medium 2 with respect to medium 1 (except the medium of vacuum).
2. It is denoted by 21 i.e. refractive index of medium 2 with respect to medium 1.

Question 36.
State the two factors that determine lateral displacement of a ray of light passing through a rectangular glass slab.
Answer:
(i) Lateral displacement is directly proportional to the thickness of optical glass slab.
(ii) Lateral displacement is directly proportional to the angle of incidence.

Question 37.
A student traces the path of a ray of light passing through a rectangular glass slab. Which angle is correctly marked?
Answer:
→ r

Question 39.
What is absolute refractive index?
Answer:
1. The ratio of speed of light in vacuum (or air) (c) to the speed of light of any other medium is known as the absolute refractive index or simply refractive index.
2. It is denoted by ‘i’

Question 40.
State the equation for absolute refractive Index.
Answer:
∴ Absolute refractive index η
\(=\frac{\text { speed of light in vacuum }}{\text { speed of light in medium }}=\frac{\mathrm{c}}{\mathrm{v}}\)

Question 41.
What is Snell’s law? Give its equation.
Answer:
1. The ratio of sine of angle of incidence to the ratio of sine of angle of retraction is constant.
2. This constant is known as relative,e refractive index of medium 2 with respect to medium 1 and denoted’ by η₂₁
∴ \(\eta_{21}=\frac{\sin \theta_1}{\sin \theta_2}\)

Question 42.
Observe the following incomplete ray diagram, where the image AB’ is formed after refraction from a convex lens:
On the basis of the above information, fill in the blanks:
(1) The position of object AB (not shown In the figure) must be ………………….
(2) The size of the object must be Than the size of the image.
Answer:

(1) beyond 2F₁ (2) greater.

Question 43.
Akash took a candle and focused the image of its flame on a white screen with the help of a convex lens.
He recorded the following distances —
(1) Position of the candle (object) = 26 cm,
(2) Position of the convex lens = 50 cm, (3)
Position of the screen = 74 cm. Based on the given information calculate the focal length of the convex lens.
Answer:
Here, u = – (50 – 26) = – 24.0 cm
From the lens formula

∴ focal length f = 12 cm

Question 44.
What will you call a point inside a lens through which the light passes without deviation?
Answer:
Optical centre

Question 45.
What causes more bending of light rays passing through it – (1) A convex lens of long focal length or (2) A convex lens of short focal length?
Answer:
A convex lens of short focal length

Question 46.
State the equation for power of lens
Answer:
Power (P) = \(\frac{1}{f}\)

Question 47.
The magnification of an image formed by a convex lens is -1. Find the object distance in terms of its focal length.
Answer:

Question 48.
Look at the two ray diagrams given here. In which diagram will magnification be positive? Why?
Answer:
Magnification is positive in diagram (ii).

Reason: The image formed is virtual, erect and magnified. This means both the object and the image have positive heights.

Question 49.
Kruti says a thick convex lens has more power as compared to a thin convex lens. Is she correct? State your argument.
Answer:
Yes she is correct because thick convex lens has shorter focal length and hence more power.

Question 50.
What is the SI unit of power of lens?
Answer:
Dioptre
(Hint: Use magnification formula m Image height h’.
\(=\frac{\text { Image height } \mathrm{h}^{\prime}}{\text { object height } \mathrm{h}}\) Real images are negative object height h images.)

Question 51.
A lens has a focal length of -25 cm. What will be the power and nature of this lens?
Answer:
4D; concave

Fill in the Blanks:

1. We cannot see things through regular reflection because …………..
Answer:
The light is reflected back only in one direction.

2. As per the law of reflection, the incident ray, the normal and the reflected ray are …………….
Answer:
In same plane

3. Image formed by plane mirror is ………… (Answer in terms of image property).
Answer:
Virtual and erect.

4. The size of image formed by plane mirror is same as object but is ……………….
Answer:
Laterally inverted

5. Light is ……………..
Answer:
Electromagnetic radiation which produces sensation in our eyes

6. If there is no reflection, the world would appear ……………
Answer:
Dark

7. ……….. and ……….. are the different types of reflection.
Answer:
Regular; Irregular

8. If the …………. surface is made reflecting, one can get a convex mirror.
Answer:
Outer

9. The centre of spherical glass ball is called of ……………… mirror.
Answer:
Centre of curvature

10. The …………… of the reflecting surface of the mirror is known as aperture.
Answer:
Diameter

11. The distance between pole and of mirror is called
Answer:
Focal length

12. …………….. Surfaces which are light in large amount.
Answer:
Completely polished

13. …………… image will be formed actually meet at some point.
Answer:
Real

14. …………….. mirror is used for close shaving.
Answer:
Concave

15. You can see a full length image of a tall building/tree in a ………………. mirror.
Answer:
Convex

16. If you visit Agra Fort you can see the distant full length view of the Taj Mahal in …………………….. mirror fitted there.
Answer:
Convex

17. In a concave mirror, a ray which is parallel to principal axis after reflection will ……………….
Answer:
Pass through the principal focus

18. In a concave mirror, a ray which is passing through centre of curvature will …………….
Answer:
Reflect back along the same path

19. The image formed by a concave mirror is at centre of curvature, real and inverted and same size as object. The position of the objects is …………………
Answer:
At centre of curvature principal focus reflect incident if the light rays

20. For a convex mirror, if the position of the object is at infinity, then the position, nature and size of the image will be ………..
Answer:
At focus, virtual and erect, highly diminished

21. An object of size 1 cm is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. The image distance will be ………………
Answer:
30cm
(Hint: Use mirror formula and follow sign convention.)

22. As per New Cartesian Sign Convention, the object is always placed to the of the mirror.
Answer:
Leftside

23. All the distances parallel to the principle axis are measured from
Answer:
The pole of the mirror.

24. A negative sign in the value of magnification indicates that the image is …………………..
Answer:
Real

25. In mirror formula, as per the sign convention, the image is taken as ……………….
Answer:
-v

26. In mirror formula, as per the sign convention, radius of curvature is taken as ……………….
Answer:
-R

27. Mirror formula can be represented as ………………
Answer:
\(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

28. The ratio of image height to the object height is called …………….
Answer:
Magnification

29. The positive value of magnification represents ……………….. image of an object.
Answer:
Virtual

30. ……… medium retracts most of the light incident on it.
Answer:
Transparent

31. Suppose that the radius of curvature R of a spherical mirror is 30 cm. In this case, its focal length (t) will be …………..
Answer:
15cm

32. The ratio of sine of angle of incidence to the sine of angle of refraction remains constant, subjected lo certain conditions. This law is known as …….
Answer:
Snell’s law of refraction

33. A medium in which speed of light is more is called ……………
Answer:
Optically rarer medium.

34. The sideways Shift of emergent ray from the direction of incident ray is called Lateral shift
Answer:
Leternal shifts

35. According to Snell’s Law constant.
Answer:
\(\frac{\sin \theta_1}{\sin \theta_2}\)

36. When a ray of light travels from optically denser to optically rarer medium,………………
Answer:
It moves away from normal

37. The formula for absolute refractive index is ……………………
Answer:
Velocity of light in vacuum (or air)! Velocity of light in medium

38. The refractive index of water in Which light travels from air is 1.33. Thus, the refractive index of light going from water to air will be ……………….
Answer:
0.75

39. For a convex lens. if an object is positioned at infinity, then the position, nature and size of the image will be and …………….
Answer:
Opposite side at F, real and Inverted, highly diminished (point sized)

40. For a convex lens, ¡f the image is between F and 2F, real and inverted and diminished, then the object should be …………
Answer:
Beyond 2F

41. In a convex lens, If we put an object at 2F, we get its image also on other side of lens at 2F but real and inverted and
Answer:
Of same size

42. In a concave lens, If we wish to have an image between focus F and optical centre O we should place the object…………..
Answer:
Between infinity and optical centre O of the lens

43. ………………………….. the power of lens having 1m kcal length.
Answer:
1D

44. A person having myopic eyes uses a concave lens of local length 75 cm. The power of his lens will be ………………………

True or False

1. The incident ray, normal to the mirror at the point of incidence and the reflected ray, all lie in the same plane. — True
2. A shining spoon can be considered as a plane mirror. — False
3. Focal length (f) = 2R. — False
4. Minimum three rays are needed to construct a ray diagram. — False
5. SpherIcal mIrrors are the most commonly used curved mirrors. True or false. — True
6. Convex mirror is commonly used for shaving. — False
7. You can put an object at only two positions to obtain an image through a convex mirror. — True
8. Concave mirrors are commonly used as rear-view mirrors in vehicles. True or false? — False
9. Concave mirrors have wider field of view. — False
10. For W > h,m > 1 — True
11. Speed of light in a medium / Speed of light in air will give you absolute refractive Index. — True
12. Concave lens is a converging lens. — False
13. Lens formula and mirror formula differ only by a mathematical sign. — True
14. Sideways shifting of the light In a glass slab is called lateral shift. — True

Match the Following 

Question 1.

Answer: (1-b), (2-d), (3-a), (4-c)

Question 2.

Answer: (1-b), (2-d), (3-a), (4-c)

Question 3.

Answer:
(1-c), (2-b), (3-d), (4-a)

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 13 Surface Areas and Volumes Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 13 Surface Areas and Volumes

Short/Long Answer Type Questions

Question 1.
A copper rod of radius 1 cm and length 2 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.
Solution :
We have,
Radius of copper rod (r₁) = 1 cm
Length of copper rod (h₁) = 2 cm
Volume of copper rod = πr₁² × h₁
= π × 1² × 2
= 2π cm³
Let r² be radius of wire
Lenght of wire (h²) = 18 m
= 1800 cm

So, the diameter of the crossection i.e.
thickness of wire = \(\frac{2}{30}=\frac{1}{15}\)cm
= 0.67 mm (approx.)

Question 2.
A solid metallic sphere of diameter 16 cm is melted and recast into smaller solid cones, each of the radius 4 cm and height 8 cm. Find the number of cones so formed.
Solution :
We have,
Radius of sphere (R) = \(\frac {16}{2}\) = 8 cm
Volume of sphere = \(\frac {4}{3}\)πR³
= \(\frac {4}{3}\) × π × 8³
= \(\frac{4 \times 512 \pi}{3}\)cm³
radius of cone (r) = 4 cm
and height of cone (h) = 8 cm
Let the number of recast x cones be x
Volume of recast x cones
= x × \(\frac {1}{3}\)πr²h
= x × \(\frac {1}{3}\)π × 4² × 8
= \(\frac {128πx}{3}\)cm³

Since, solid sphere is melted and recast into smaller cones
∴ Volume of recast x cones = Volume of sphere
⇒ \(\frac{128 \pi x}{3}=\frac{4 \times 512 \pi}{3}\)
⇒ x = \(\frac{4 \times 512 \pi \times 3}{3 \times 128 \pi}\)
⇒ x = 16
Hence, number of recast cones = 16

Question 3.
A solid metallic cuboid of dimensions 9 m × 8m × 2 m is melted and recent into solid cubes of edges 2 m. Find the number of cubes so formed.
Solution :
We have,
Dimensions of cuboid = 9m × 8m × 2m
Volume of cuboid = 9 × 8 × 2
= 144 m³
edge of recast cube (a) = 2m
Volume of recast cube = a³
= (2)³ = 8m³
Let the number of recast cube be x
Since, cuboid is melted and recast into small x cubes
So, volume of x recasted cubes = Volume of cuboid
⇒ 8 × x = 144
⇒ x = \(\frac {144}{8}\) = 18
Hence, number of recast cubes = 18

Question 4.
A solid mettalic cylinder of radius 3.5 cm and height 14 cm is melted and recast into a number of small solid mettalic balls, each of radius \(\frac {7}{12}\)cm. Find the number of bass so formed.
Solution :
We have,
Radius of cylinder (R) = 3.5 cm
Height of cylinder (h) = 14 cm
Volume of cylinder = πr² h
= π × (3.5)² × 14 cm³
Andradius of recasted ball (r) = \(\frac {7}{12}\)cm
Volume of recasted small ball = \(\frac {4}{3}\)πr³
Let the number of recasted small balls be x
= \(\frac {4}{3}\)π(\(\frac {7}{12}\))³ cm³
since, cylinder is melted and recasted into x small balls
∴ Volume of x recasted = Volume of cylinder small balls

Hence, number of recasted smaller balls = 648

Question 5.
Find the number of coins of 1.5 cm diameter and 0.2 cm thickness to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution :
We have,
Radius of coins (r) = \(\frac {1.5}{2}\) = 0.75 cm
Thickness of coin (h) = 0.2
Volume of coin = πr²h
= π × (0.75)² × 0.2
= 1.125 π cm³
Let the number of recasted coins be x
∴ Volume of x coins = x × 1.125 πcm³
Radius of cylinder (R) = \(\frac {4.5}{2}\) = 2.25 cm
and Height of cylinder (H) = 10 cm
Volume of cylinder = πR²H
= π × (2.25)² × 10
= 50.625π cm³
Since, x coins to be melted to form a right circular cylinder
∴ Volume of x coins = Volume of cylinder
⇒ x × 1.125 π = 50.625π
⇒ x = \(\frac {50.625π}{1.125π}\)
⇒ x = 45
Hence, number of coins melted = 45

Question 6.
A sphere of diameter 12 cm is Iropped in a right circular cylindrical vessel, partly illed with water. If the sphere is completely ubmerged in water, the water level in the cylindrical vessel rises by 3\(\frac {5}{9}\) cm. Find the diameter of the cylindrical vessel.
Solution :
We have,
Diameter of sphere = 12 cm
Radius of the sphere (r) = \(\frac {12}{2}\) = 6 cm
Volume of the sphere = \(\frac {4}{3}\)πr³
= \(\frac {4}{3}\) × π × 6 × 6 × 6
= 288π

If sphere dropped into the cylinder, water rises by 3\(\frac {5}{9}\) cm
= \(\frac {32}{9}\)cm
∴ Height of the water rises in the cylinder (h) = \(\frac {32}{9}\) cm
Let radius of cylinder = R cm
Volume of the water rises that contained cylinder = πR²h
= π × R² × \(\frac {32}{9}\)
Volume of rises water of cylinder = Volume of the sphere
⇒ π × R² × \(\frac {32}{9}\) = 288π
⇒ R² = \(\frac{288 \pi \times 9}{32 \times \pi}\)
⇒ R² = 9 × 9
⇒ R = \(\sqrt{9 \times 9}\)
⇒ R = 9 cm.
∴ Diameter of the cylinder
= 2 × 9 = 18 cm.
Hence, diameter of the cylinder
= 18 cm.

Question 7.
Water is flowing at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank of dimensions 50 m × 44 m. Find the time in which the level of water in the tank will rise by 7 cm.
Solution :
We have,
Speed of water in pipe = 5 km/hr
∴ Length of water flow in pipe in 1 hours.
= 5000 m
Let time consumes in it = t hour
Radius of pipe (r) = \(\frac {14}{2}\) = 7cm = \(\frac {7}{100}\)m
Volume of water flown in t hours through pipe = πr²H
= π × (\(\frac {7}{100}\))² × 5000 × t
= π × \(\frac {49}{100}\) × 5000 × t m ³
= 24.5 π × t m³
length of rectangle tank (l) = 50 m
Breadth of rectangle tank (b) = 44 m
Level of water rise (h) = 7 cm = \(\frac {7}{100}\)m
Volume of water in tank = l × b × h
= 50 × 44 × \(\frac {7}{100}\)
= 154 m³
Since, water flown through a pipe in the rectangular tank, so
Volume of water flawn through pipe in t hours = Volume of water in the tank
⇒ 24.5 π × t = 154
⇒ \(\frac{24.5 \times 22}{7}\) × t = 154
⇒ 77 × t = 154
⇒ t = \(\frac {154}{77}\)
⇒ t = 2 hours
Hence, time taken in which the level of water in the tank will rise by 7 cm = 2 hours.

Question 8.
Two spheres of same metal weight 1 kg and 7 kg. The radius of smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.
Solution :
Radius of small sphere (r) = 3 cm
Volume of smaller sphere = \(\frac {4}{3}\)π × (3)³
= 36π cm³
Weight of smaller sphere is 1 kg and that of larger sphere is 7 kg
Since, 1 kg metal sphere occupies 36π cm³ sapse
How, weight of recasted metal sphere = 1 + 7 = 8 kg
∴ 8 Kg metal sphere occupies 36π × 8 cm³
Two sphere 1 kg and 7 kg mass melted and recast single sphere
Let the R be radius of recasted single sphere
∴ Volume of recasted single sphere = Volume of two 8 kg spheres
⇒ \(\frac {4}{3}\)πR³ = 36π × 8
⇒ R³ = \(\frac{36 \pi \times 8 \times 3}{4 \pi}\) = 216
⇒ R = 3\(\sqrt{6 \times 6 \times 6}\)
⇒ R = 6 cm
Hence, diameter of recasted new sphere = 2 × 6 = 12 cm.

Question 9.
504 cones each of diameter 3.5 em and hight 3 cm are melted and recast into a metallic sphere. Find the diameter of sphere and hence find its surface area. (use π = 22/7)
Solution :
We have,
Radius of each cone (r) = \(\frac {3.5}{2}\)
and its height (h) = 3 cm
Volume of 1 cone = \(\frac {1}{3}\)π × r²h
= \(\frac {1}{3}\)π × (\(\frac {3.5}{2}\))² × 3
= \(\frac {12.25}{4}\)π cm³
Volume of 504 cones = \(\frac{12.25 \pi \times 504}{4}\)
= 1543.5 π
Let the R be radius of recasted sphere volume of recasted sphere = \(\frac {4}{3}\)πR³
Since, 504 cones are melted and recast into metallic sphere
∴ Volume of recasted sphere = Volume of 504 cones

Hence, diameter of sphere 21 cm and its surface area = 1386 cm²

Question 10.
From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into at cylindrical wire of 1.4 cm thickness. find the length of wire. (use π = 22/7)
Solution :
We have,
Base radius of cylinder (R) = 4.2 cm
Height of cylinder (h) = 10 cm
Volume of cylinder = πR²h
= π × (4.2)² × 10
= 176.4π cm³
Volume of metal scooped out = 2 × volume of hemisphere

= 98.784π cm³
Volume of rest of cylinder
= 176.4π × 98.784π
= 77.616 π cm³
let the length of wire be l cm
∴ Volume of wire = πr²l [here H = 1 = π × (0.7)² × l]
[∴ radius of wire (r) = \(\frac {1.4}{2}\) = 0.7 cm]
= 0.49 πl
Since, rest of cylinder is melted and converted into a cylindrical wire.
∴ Volume of wire = Volume of rest of cylinder
⇒ 0.49πl = 77.616π
⇒ l = \(\frac {77.616л}{0.49π}\)
⇒ l = \(\frac {77.616}{0.49}\) = \(\frac {77.616}{49}\)
⇒ l = 158.4 cm
Hence, length of wire = 158.4 cm

Question 11.
A well of diameter 7 m is dug and earth from digging is evenly spread out to form a platform 22 m × 14 m × 2.5 m. Find the depth the well.
Solution :
We have,
Diameter of well = 7 m
∴ Its radius (r) = \(\frac {7}{2}\)m
Let depth of well be h m
Volume of soil dugout from well spread out to form a plateform.
So, volume of soil dugout from well = Volume of soil to form plateform
⇒ π × \(\frac {49}{4}\) × h = 770
⇒ h = \(\frac{770 \times 4}{49 \times \pi}\)
⇒ h = \(\frac{110 \times 4 \times 7}{7 \times 22}\)
⇒ h = 5 × 4 = 20 m
Hence, depth of well = 20 m

Question 12.
A well of diameter 4 m dug 21 m deep. The earth taken out of it has beed spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment.

Solution :
We have,
Radius of the well (r) = \(\frac {4}{2}\) = 2 m
Depth of the well (h) = 21 m
Volume of the earth dug out = πr²h
= π × 2² × 21
= 84π m³
Radius of the embankment (R)
= 2 + 3 = 5 m
Let height of embankment be h’ m.
Volume of the earth used for making embankment = volume of the earth dugout
⇒ πr²h’ – πr²h’ = 84π
⇒ πh’ (R² – r²) = 84π
⇒ h’ (5² – 2²) = \(\frac {84π}{π}\)
⇒ h’ (25 – 4) = 84
⇒ h’ × 21 = 84
⇒ h’ = \(\frac {84}{21}\) = 4m
Hence, height of embankment is 4 cm.

Question 13.
Find the curved surface area of the frustum of a cone, the diameters of whose circular ends are 20 m and 6 m and its height is 24 m.

Solution :
We have,
Radii of two circular ends are :
r₁ = \(\frac {20}{2}\) = 10m, r₂ = \(\frac {6}{2}\) = 3m
And its height (h) = 24 m
Slant height of frustum of cone (l)
= \(\sqrt{h^2+\left(r_1-r_2\right)^2}\)
= \(\sqrt{24^2+(10-3)^2}\)
= \(\sqrt{576+49}\)
= \(\sqrt{625}\)
= 25 m
Curved surface area = πl(r₁ + r₂)
Frustum of a cone = 3.14 × 25 (10 + 3)
= 78.5 × 13
= 1020.5 m²

Question 14.
A bucket is in the form of a frustum of a cone of height 30 cm with raddi of its lower and upper ends as 10 cm and 20 cm, respectively. Find the capacity of the bucket. Also find the cost of milk which can completely fill the bucket at the rate of ₹ 40 per litre. (use π = 22/7)
Solution :
We have,
Raddi of lower and upper ends of frustum of a cone are
r₂ = cm and r₁ = 20 cm
And its height (h) = 30 cm
Capacity of bucket = Volume of bucket
= \(\frac {πh}{3}\)(r₁² + r₂² + r₁r₂)
= \(\frac{22}{7} \times \frac{30}{3}\)[(20)² + (10)² + 20 × 10]
= \(\frac {22}{7}\) × 10[400 + 100 + 200]
= \(\frac {220}{7}\) × 700
= 22000 cm³
Milk in the container = \(\frac {22000}{1000}\) liters
[∴ 1 litre = 1000 cm³]
= 22 litres
cost of milk = 22 × 40 = ₹880
Hence, capacity of bucket = 22 litres and cost of milk = ₹880

Question 15.
A metal container, open from the top, is in the shape of frustum of a cone of height 21 cm with raddi of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of ₹35 per litre.
Solution :
We have,
Radii of lower and upper circular ends of a metal container open from the top are
r₂ = 8 cm, r₁ = 20 cm
and its height (h) = 21 cm
Volume of container = \(\frac {πh}{3}\)(r₁² + r₂² + r₁r₂)
= \(\frac{22}{3} \times \frac{21}{3}\)[(20)² + 8² + 20 × 8]
= 22 [400 + 64 + 160]
= 22 [624]
= 13728 cm³
= \(\frac {13728}{1000}\)
= 13.728 litres
cost of milk = 13.728 × 35
= ₹ 480.48

Question 16.
The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of volume the two parts.
Solution :
Let VAB cone and VN be its axis, and let M be mid point of VN,

VN = 10 cm, VM = MN = \(\frac {10}{2}\)
Let VM be h’ = 5 cm
Let radii of upper and lower ends of frustum be r₁ cm and r₂ cm respectively
Δ VNB ~ Δ VMD [By AA similarity]

Hence, ratio of the required volumes = 1 : 7

Question 17.
A right circular cone is cut by two planes parallel to the base trisecting the height. Compare the volumes of the three parts into which the cone is divided.
Solution :
Let the right circular cone be cut by two parallel planes DE and FG.
∴ BC || DE || FG.
Now,
Δ ARG ~ ΔAPC
⇒ \(\frac {RG}{PC}\) = \(\frac {AR}{AP}\)

Fill in the Blanks

Question 1.
A playin top (lattu) consistr of a hemisphere and a _______
Solution :
cone

Question 2.
πl (r₁ + r₂) is _______ surface area of frusum.
Solution :
curved

Question 3.
In conversion of one solid to other solid, we have constant ________
Solution :
volume

Question 4.
Two spheres have raddi in the ratio 2 : 1, then ratio of their volume is _______
Solution :
8 : 1

Question 5.
While calculating the volumes of combi- nation of solids, the volume of common parts should be _______
Solution :
excluded

Question 6.
Volume of cylinder of radius r and height h is 27 cm³. Then the volume of cone of same radius and same height is _______ cm³.
Solution :
9

Question 7.
Bodies which have three dimensions in space are called ______.
Solution :
solids

Multiple Choice Questions

Question 1.
Volume of two spheres are in the ratio 64 : 27 the ratio of their surface areas is :
(a) 3 : 4
(b) 4 : 3
(c) 9 : 16
(d) 16 : 9
Solution :
(d) 16 : 9

So correct choice is (d).

Question 2.
A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that \(\frac {1}{8}\) space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is :
(a) 142296
(b) 142396
(c) 142496
(d) 142596
Solution :
(a) 142296

Number of marbles

So correct choice is (a)

Question 3.
A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form at cone of base diameter 8 cm. The height of the cone is:
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm
Solution :
(b) 14 cm

So correct choice is (b).

Question 4.
A solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm, is moulded to form a solid slphere. The radius of the splhere is:
(a) 21 cm
(b) 23 cm
(c) 25 cm
(d) 19 cm
Solution :
(a) 21 cm

volume of sphere = volume of cuboid.
\(\frac {4}{3}\)πr³ = l × b × h
\(\frac {4}{3}\) × \(\frac {22}{7}\) × r³ = 49 × 33 × 24
r³ = 7³ × 3³
r = 7 × 3 = 21 cm
so correct choice is (a).

Question 5.
A mason constructs a wall of dimensions 270 cm × 300 cm × 350 cm with the bricks each of size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that \(\frac {1}{8}\) space is covered by the moter then the number of bricks used to construct the wall is :
(a) 11100
(b) 112000
(c) 11000
(d) 11300
Solution :
(b) 112000

So correct choice in 11200 i.e., (b).

Question 6.
Twelve solid spheres of the same size are made by melting a solid metallic cyclinder of base diameter 2 em and height 16 cm. The diameter of each sphere is :
(a) 4 cm
(b) 3 cm
(c) 2 cm
(d) 6 cm
Solution :
(c) 2 cm

volume of 12 sphere volume of cylinder
12 × \(\frac {4}{3}\)πr³ = πR²Н = π × 1 × 16
⇒ r = 1 cm
∴ d = 2r= 2 cm
So correct choice (c).

Question 7.
The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is
(a) 4950 cm²
(b) 4951 cm²
(c) 5952 cm²
(d) 4953 cm²
Solution :
(a) 4950 cm²

l = 45 cm, R = 28 cm, r = 7 cm
CSA = π(R + r)
= \(\frac {1}{2}\) × 45(28 + 7)
= \(\frac{22 \times 45 \times 35}{7}\)
= 4950 cm²
So correct choice is (a).

Question 8.
A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. the capacity of the capsule is:
(a) 0.36 cm³
(b) 0.35 cm³
(c) 0.34 cm³
(d) 0.33 cm³
Solution :
(a) 0.36 cm³

r = \(\frac {0.5}{2}\)cm
capacity of capsule


Hence correct choice is (a).

Haryana State Board HBSE 10th Class Science Important Questions Chapter 9 Heredity and Evolution Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 9 Heredity and Evolution

Question 1.
‘Every new generation accumulates variation during reproduction.’ Explain the statement.
Answer:
1. Organisms have two modes of reproduction,

• Asexual reproduction and
• Sexual reproduction. Offspring are born through any of these modes.

2. During reproduction, these offspring inherit two things from their previous generation. They are –

• A common basic body design and
• Some fine (i.e. minor) variations.

3. When the offspring grow and reproduce the next generation, the new generation offspring will again inherit two things.
They are –

• Same basic body design along with differences given by their previous generations and
• Some fine new variations.

4. This way with each new generation born, the variations get accumulated.

Question 2.
Explain: Parental generation, First generation and Second generation.
Answer:
(a) Parental generation (P):
The first set of parents crossed is called parental generation. The parental generation is denoted by ‘P’.

(b) First generation (F₁):
All the offspring born from the first set of parents crossed are said to be belonging to the First generation. The First generation is denoted by F₁

(c) Second generation (F₂):
All the offspring born from the F₁ generation are said to be belonging to the Second generation. The Second generation is denoted by F₂.

Question 3.
Explain with the help of a diagram how variation (or diversity) is created over succeeding generations.
Answer:
1. As shown in the diagram, consider a single bacteria at the top reproduces on its own asexually.
2. It gives birth to two individuals. The two individuals reproduced will have

• Similar body design like their previous (parental) generation and
• Some minor variations/ differences of their own.

3. Each of these organisms in turn may give birth to two individuals in the next i.e. first generation. So, there are in total four organisms in the second generation.

4. As can be seen, each of the four individuals in the bottom row is somewhat different from each other. Each one has

• The common body design inherited from its previous generation along with
• Some unique differences of its own. So, each organism although similar, is not the exact copy and hence shows variation.

5. Since this is an example of asexual reproduction, the differences are very minor. In case of sexual reproduction, the changes would be much diversed.

Question 4.
What is the most obvious outcome of reproduction? What do the rules of heredity determine?
Answer:
1. The most obvious outcome of reproductive process is that similar designs of individuals are maintained generation after generation.
2. The rules of heredity determine the process by which traits and characteristics get transferred from one generation to another.

Question 5.
Organisms undergo variation during reproduction. Do all these variations survive?
Answer:
1. Although organisms undergo variation, all the variations do not survive.
2. Depending upon the nature of variations, different individuals would gain different kinds of advantages from the variations.
3. An individual has to withstand several environmental effects and changes. So, organisms having variations which can adapt to the environment survive, others may not.

Question 6.
What do you mean by heredity? Explain.
Answer:
Heredity:
1. Heredity means the transmission of characters from parents to offspring. For example, eggs laid by a sparrow will hatch sparrow and not any other bird. Similarly, a dog gives birth only to pups.
2. Thus, in this sense, it can be said that heredity is continuity of features from one generation to another. This is the essence of heredity.
3. Hereditary information is present in the fertilized egg i.e. zygote.
4. The zygote develops into the same organism to which it belongs. And so the new organisms produced resemble their parents.

Question 7.
Explain the following terms:
Answer:
1. Heredity: The transmission of parental traits to their offspring is called heredity. This transmission always follows certain principles or laws.
2. Genetics: The study of how inherited traits (or say heredity) are passed-on and the variations occurring therein is called genetics.
3. Trait: A trait is a particular characteristic, quality or tendency that someone or something has. For example, height, skin or eye colour, certain behaviour, etc.
4. Inherited trait: A feature or a characteristic of an organism that it has received in its genes from its parents is called an inherited trait.

Question 8.
Why there are two versions for each trait in each child?
Answer:
In a sexual reproduction, the father as well as the mother contributes equal amounts of genetic material to the child. This means that each trait that the child has can be influenced by the DNA of the father as well as mother. Thus, for each trait there will be two versions in each child.

Question 9.
What are the fundamental characteristics of inheritance of traits?
Answer:
The fundamental characteristics of inheritance of traits are as follows:
(i) A unit of inheritance is called as a gene. Each gene controls a character(s).
(ii) Genes have two alternative forms for each inherited trait.
(iii) From the two alternative traits, the trait which is expressed is known as a dominant trait whereas, the trait which is not expressed is known as a recessive trait.
(iv) From two alternative traits for an expression of a character, one trait may be dominant over the other.
(v) The dominant trait is represented by capital letter (Generally ‘T’) while the recessive trait is represented by the small letter (‘t’).
(vi) The unexpressed/recessive trait may be expressed in any future generation.
(vii) In sexual reproduction both the parents give equal amount of chromatin material/genetic material to make a zygote. So, each inherited trait is influenced by both the parents equally.

Question 10.
Explain Mendel’s contribution through his study of inheritance of characteristics. OR In Mendel’s study, the ‘T’ gene is dominant while the ‘t’ gene is recessive. Explain in detail with an example. OR Explain Gregor Johann Mendel’s experiment showing inheritance of any one character.
Answer:
1. Mendel studied garden pea plants for the expression of a character. The character under study was the height of plants
2. He observed two contrasting expressions of height — tall plants and dwarf plants/short plants.
3. For the purpose of his study, Mendel took pure tall plants (TT) and pure short plants (tt).

(i) Cross-pollination between parent (P) generation plants:
Mendel performed cross-pollination between TT — pure tall plants and tt — pure short plants. These TT and ft are known as a parent (P) generation plants.

Observation:
The plants that got produced from Parent (P) generation are known as First (F₁ generation) plants. Mendel observed that all the plants of F₁ generations were as tall as TT of P generation i.e. as tall as their parents. So, Mendel thought that only one parental trait i.e. of tallness (instead of both trait i.e. tallness and shortness) was inherited by the F₁ generation plants.

(ii) Self-pollination of (a) Tall Parental Generation plants and (b) F₁ generation plants:
In the second phase, Mendel performed self-pollination of
(a) Tall i.e. U Parental Generation plants and
(b) F₁ generation plants.

Observation:
(a) The new plants produced by the self-pollination of Tall i.e. U Parental Generation plants were all tall.
(b) The new plants produced i.e. the second (F₂) generation plants by the self-pollination of F₁ generation plants were a mix of tall and short plants. Mendel observed that in F₂ generation, around 75°/° of the plants were tall and 25% of the plants were short. This means the ratio of tall short plant in the F₂ generation was 3:1.

Thus, in the second trial, Mendel concluded that in the first trial, the F₁ generation plants had inherited both the traits i.e. the trait of tallness and dwarfism, from their parent (P) generation but, only the trait of tallness got expressed. However, when he self-pollinated the F₁ generation plants, both the traits i.e. the trait of tallness and dwarfness could be seen in the F₂ generation plants.

Question 11.
What did Mendel conclude from his experiment on pea plants?
Answer:
Mendel’s conclusion:
1. The contrasting characters (traits) i.e. tallness and shortness of the pea plant are determined by factors.
2. Two copies (i.e. in pairs) of such factors are present in sexually reproducing organism and they are known as a ‘gene’.
3. These two may be identical, or may be different, depending on the parentage.
4. When individuals having contrasting features are crossed, the feature which gets expressed is known as a dominant feature (trait) and the one that does not get expressed is known as a recessive feature (trait).
5. Tall plants are represented by ‘TT’ as well as ‘Tt’, but short plants are represented by ‘tt’. This means that presence of even single copy of‘T’ is enough to make a plant tall, but for a plant to be short, both the copies have to be ‘t’. Hence, T is called a dominant trait whereas ‘t’ is called a recessive trait.

Question 12.
Explain Mendel’s experiment on inheritance of two characters simultaneously. OR Explain Mendel’s experiment with different seeds. OR Explain the inheritance of seed colour and shape of the seed of pea plant Pisum Sativum.
Answer:
1. Mendel performed an experiment to study garden pea plants having two different characters. The two characters that the pea plant had were –

• Colour of the seeds and
• Shape of the seeds

2. Mendel wanted to study how the traits express themselves when the plants contain two characters as mentioned above rather than one.
At first, Mendel selected the following two pure breeding pea plants:
(a) Plants with yellow colour seed and round shape seeds.
(b) Plants with green colour seed and wrinkled shape seeds.

• He considered these plants as parent plants.
• Mendel performed cross-pollination between both these parent plants and obtained the F₁ generation.

1. Observation of F₁ generation plants:

• All the plants of F₁ generation had yellow coloured seeds having round shape.
• This means that ‘yellow colour’ and ‘round seeds’ were dominating characteristics (traits) in the F₁ generation.
• In the second trial, he performed self-pollination between the F₁ plants to obtain F₂ generation.

2. Observation of F₂ generation plants:
Mendel obtained four different types of F₂ generation plants as mentioned below:
(a) Yellow and round seed plants
(b) Yellow and wrinkled seed plants
(c) Green and round seed plants
(d) Green and wrinkled seed plants

This indicates that characteristics (traits) of colour yellow and green and characteristics (traits) of shape round and wrinkled are inherited independently from each other.

The ratio in which Mendel obtained the plants was —

• (Yellow round) : (Green round) : (Yellow wrinkled) : (Green wrinkled) = 9 : 3 : 3 : 1
• The results obtained in F₂ generation clearly indicate that the inheritance of two different characters is independent to each other.

Question 13.
How does the mechanism of heredity work in terms of genes control?
Answer
1. The DNA found in the cells is the fundamental source of genetic information for making specific proteins in the cell.
2. In general, a segment of DNA which provides information to synthesize protein is known as a gene.
3. In plants as well as organisms, the characters (traits) get expressed due to specific reactions. These reactions take place with the help of enzymes.
4. The enzymes as well as several hormones are made up of proteins.
5. So, when a specific gene undertakes synthesis of a specific protein it results in expression of a specific character.

Example:
1. Let us consider height (tallness) as a characteristic of a plant.
2. Plants have hormones that trigger growth. This means height of plants depend upon the amount of a particular harmone.
3. A particular type of enzyme works as a catalyst for this hormone. And the synthesis of this enzyme is regulated by gene located on the DNA.
4. Thus, if the enzyme is synthesized in required amount and if it works efficiently, a lot of quality hormone can be produced which will then boost the height of the plant.
5. However, it the gene has altered due to some reason, the enzyme will become less efficient.
6. This will result in less amount of hormone in the plant and hence it will be of dwarf size.
7. From this example, it can be clearly said that genes control the characteristics or traits.

Question 14.
Describe sex determination in human beings in detail.
Answer:
Sex determination:
1. In sexual reproduction, the mechanism to determine the sex of an organism is known as sex determination.
2. In human beings, the sex is determined by genes located on the chromosomes.
3. These genes are inherited by the offspring from their parents.
4. In humans, each cell contains 23 pairs of chromosomes.
5. Out of these, 22 pairs are autosomes and one pair is of sex chromosomes.
6. These 22 pairs of autosomes are same in males as well as females. Only the 23 pair differs.
7. In females, the 23 pair contains two similar ‘X’ sex chromosomes i.e. the 23 pair is ‘XX’.
8. In males, the 23rd pair contains one ‘X’ sex chromosome and one ‘Y’ sex chromosome i.e. the 23’’ pair is ‘XV’. The ‘Y’ sex chromosome is smaller than the ‘X’ sex chromosome.
9. Out of all the sperms, 50% sperms contain ‘X’ chromosome while the other 50% sperms contain ‘Y’ chromosome.

Sex of the foetus:
1. If a sperm carrying ‘X’ chromosome fuses with the ‘X’ chromosome of female egg i.e. if ‘XX’ combination occurs, female will be produced.
2. However, if a sperm carrying ‘Y’ chromosome fuses with the ‘X’ chromosome of female egg, i.e. if ‘XY’ combination occurs, male will be produced.
3. Since male sperm carries both ‘X’ and ‘Y’ chromosomes and female carries only ‘X’ and ‘X’ chromosomes, males are responsible for the sex of the child born.
4. When zygote is formed and the embryo starts developing, the gonads (organs that produce sex cells) are undifferentiated.
5. Later, the gonads develop either into testes or ovaries.
6. If the zygote contains the Y sex chromosomes, the gonad differentiates into testes.
7. These testes then produce male sex hormones and stimulate development of a male individual.
8. Similar process takes place if the zygote contains only ‘X’ chromosome.

Question 15.
What are sex chromosomes? Which sex chromosomes are found in male and female human beings? State the chromosome responsible for the development of male child in human beings.
Answer:
1.The chromosomes that determine the sex of the new born baby are called sex chromosomes. X chromosome and Y chromosome together form the pair of sex chromosomes in male human beings. Thus, the male contains XV sex chromosomes.
2. In female, both the sex chromosomes are X. So, the female sex chromosomes are called XX.
3. A male child can take birth only when the zygote receives Y chromosome from the male human being.

Question 16.
Every organism resembles its parents. Give reason.
Answer:
1. When male and female sex cells fertilize, zygote is formed.
2. Zygote develops into an organism.
3. Zygote contains heredity information from the parents.
4. This information is passed on to the next generations.
5. Hence, every organism will resemble its parents.

Question 17.
What is natural selection? Explain.
Answer:
1. Natural selection is a central concept of evolution. In general terms it is also called ‘survival of the fittest’.
2. Natural selection can be considered as the one ‘selected by nature’.
3. When organisms develop certain favourable traits i.e. the traits that are more suited for nature to select, the organisms survive and hence reproduce. These organisms then pass their traits to the next generation.
4. Over time this process allows organisms to adapt to their environment. The organisms better adapted to their environment tend to survive and produce more offspring.

Question 18.
Give an example of beetles to understand the concept of natural selection.
Answer:
1. Consider a group of 12 red beetles. They live in bushes mat have green leaves.
2. The population of beetles will grow by sexual reproduction and so every generation produced will show variations.
3. Let us consider that crows feed on these beetles. The more beetles the crows eat, the fewer beetles will remain for reproduction.
4. Now, suppose, a colour variation arises during reproduction. Under this, one of the beetle reproduced attains ‘green colour’ instead of red.

5. When this green coloured beetle undergoes reproduction, it passes this green colour to its progeny. So, all the new beetles reproduced from this green coloured beetle are green.
6. Because green-coloured beetles get merged with green leaves, now the crows cannot see green coloured beetles on the green leaves and therefore cannot eat them. As a result, the green beetles do not get eaten, while the red beetles continue to be eaten.
7. In conclusion, the population of green beetles increase whereas that of red beetles decline drastically.
8. Here, the beetles that adapted the green colour became more suited to nature and so got ‘selected by nature (i.e. natural selection)’ for remaining alive.

Question 19.
Give an example of beetles to understand the concept of genetic drift.
Answer:
1. Consider a group of 12 red beetles. They live in bushes that have green leaves.

2. The population of beetles will grow by sexual reproduction and so every generation produced will show variations.

3. A colour variation arises during reproduction and few of the new beetles born are of blue colour
instead of red. The blue beetle can pass the colour on to its progeny, so that all its progeny beetles are blue.

4. Crows living in the bush feed on the beetles. They can see both red as well as blue beetles and thereiore can eat both the varieties.

5. Initially, in the population, most of the beetles are of red coloured and only few are blue. At this point, a sudden and an unexpected change takes place. An elephant comes and stamps on the bushes where the beetles live.

6. By chance, the stamping of elephant kills mostly red beetles. The few beetles that have survived are mostly blue.

7. Now, the beetle population again slowly expands, but the beetles in the population are mostly blue.

8. Here, the change in colour as a variation seen in the beetles did not increase their chance of survival, the way it happened when the beetles adopted a green colour. The blue coloured beetle survived simply by chance because the elephant foot got stamped on the red beetles and not the blue.

9. From this we conclude that accidents in small populations can change the frequency of some genes in a population, even if they give no survival advantage This concept is called genetic drift. Genetic drift resulted in variation but did not give any advantage of adaptation to the beetles.

Question 20.
What is genetic drift? Explain.
Answer:
Genetic drift:
The sudden and random change in the gene frequency that occurs by chance in a small population is known as genetic drift.’

Explanation:
1. When a large group of population migrates or dies due to natural calamity, certain traits get eliminated from the population.
2. As a result, in the remaining population the gene frequency is altered. So, in the next progenies some different kind of variations arise.
3. At times due to genetic drift, the population may get free of some unfavourable character and at times, it is also possible that some important character is lost.

Importance:
1. Genetic drift influences evolution.
2. It is possible that the population becomes different because there is a probability that each population may fix a different genotype.

Question 21.
What is evolution? Explain.
Answer:
Evolution:
1. Evolution is a type of gradual formation of new plants or organisms from the pre-existing primitive plants or animals respectively by Constant and relatively long time changes.
2. The word ‘evolution’ has been derived from Latin word ‘evolvere’ which means to untold’ or ‘unroll’.
3. Evolution is the sequence of gradual changes, which have taken place over millions of years in primitive plants and animals from which new species are formed.
4. Evolution is a constant process taking place in the organisms since life originated. In fact, all the varieties of organisms, which we see around us. have evolved from some ancestors that lived on this earth long time ago.

Question 22.
Explain acquired and Inherited traits (characteristics).
Answer:
(A) Acquired traits:
1. Acquired trait means a trait or characteristic of an organism that it has not inherited but has developed in response to the environment.
2. For example, if an organism starves for sometime and reduces its weight, then it ¡s called acquired trait.
3. The reduced weight due to starvation will not change DNA of the cells.
4. As a result, reduced weight is not a trait, which can be. inherited by future generations of organisms that will starve.
5. Other examples are cut tail of a mouse, a man who knows how to swim, or speak German or roller skate, or may have a scar on the face due to accident. None of these traits will be passed on to the future generations because these traits have not altered the genes of the organisms.
6. Thus, the changes in the non-reproductive body cells of an organism cannot be transmitted to its progeny.
7. Only those traits will be inherited to their progeny where in change has occurred in the genes in gametes of organisms during the process of reproduction.

(B) Inherited trait :
1. A trait of an organism, which is caused by a change in its DNA, is known as inherited trait.
2. For example, suppose a population of red beetles is living in bushes with green leaves.
3. Suppose a colour variation arises during reproduction in the gene of reproductive cells and one green colour beetle arises instead of red.
4. Here, the colour of this beetle is an inherited trait, which can be transmitted to the next generations.

Question 23.
Explain the terms: (a) Variations (b) Acquired variations.
(a) Variations:

• Any difference that occurs between cells, individual organisms, or groups of organisms of any species either by genetic differences or by the effect of environmental factors on the expression of the genetic potential is called variation.
• Organisms may show variation in their physical appearance, metabolism, rate of fertility, mode of reproduction, behaviour, etc.

(b) Acquired variations :

• An acquired characteristic (variation) is a change which is acquired not due to heredity but due to response to the environmental factors.
• Acquired variation may occur in the function or structure of an organism caused after birth due to reasons such as disease, Injury, accident, repeated use or misuse, or other environmental factors.
• Acquired variations are limited to the organism that acquires the variation. Such variations than vanish and with the death of that organism and hence do not get inherited.

Question 24.
Differentiate between acquired and inherited traits.
Answer:

Question 25.
What is speciation? Explain.
Answer:
Speciation: The process of evolution of new species that occurs when members of similar population no longer interbreed to produce fertile off spring is known as speciation.

Speciation due to geographical isolation:
1. Suppose due to some reason the population of a given species gets divided into two groups.
2. These groups then get separated from each other geographically by certain barriers like rivers, seas, mountain ranges, or even natural calamities.
3. This means these two groups got geographically isolated and they will not be able to interbreed and hence exchange genes.
4. Over a period, both groups might adapt to its environment through natural selection and develop their gene pool.
5. Eventually, the gene pools of both populations might become so different that the new species of these organisms would evolve in different populations.
6. Thus, due to process of random change in gene frequency (gene drift), after thousands of years individuals become so different that they cannot reproduce with each other and so new species are formed.
7. Thus, geographical isolation leads to reproductive isolation due to which there is no gene flow between two separated groups of population.
8. Thus, we may get completely new species of population.

Other reasons of speciation:
Other reasons can be changes in DNA, change in number of chromosomes, gene cells of two isolated groups of populations, which cannot fuse with each other, etc.

Question 26.
How are organisms classified? OR What do you mean by characteristics as the base of classification? Explain.
Answer:
1. When we classify organisms by grouping them in certain way, it helps us to study them properly.
2. One of the basic methods of classifying the organisms is on the basis of similarities they possess.
3. For classifying we will have to decide which characteristics show ‘fundamental differences’ or say important differences and which show ‘basic differences’ or say less important differences.

Characteristics as a base of classification:

1. A characteristic means a particular type of appearance (form) or behaviour (function).
2. We humans have four limbs (two hands and two legs) and it is our appearance characteristic.
3. Plants perform photosynthesis is a behaviour (function) characteristic.
4. Some basic characteristics are shared by most organisms. For example, cell is the basic unit of life in all organisms. However, the characteristics of the next level of classification would be shared by most, but not all organisms.
5. At cell level, the classification could be whether the cell has a nucleus or not.
For example, cells – of bacteria do not have nucleus, while the cells of most other organisms do.
6. Further, organisms that have nucleus in their cells can be classified as uni-cellular or multi-cellular.
7. Among multi-cellular organisms, classification can be done as whether they undergo photosynthesis or not.
8. The multi-cellular organisms that do not undergo photosynthesis can be further classified as,whether the skeleton is inside the body or around the body.
9. By classifying in this manner we can develop a hierarchy of organisms.

Question 27.
Giving an example of humans to explain what increases the probability of two species having a common ancestor.
Answer:
The more characteristics two species will have in common, the more closely they will be related. And the more closely they are related, the more are the chances of them being having a recent common ancestor.
Example:
A brother and a sister are closely related. They have common ancestors in the first generation before them, namely, their parents. Now, the girl and her first cousin are also related, but less than the girl and her brother. The reason for this is that although cousins have common ancestors i.e. their grandparents, in the second generation but the parents of these two cousins are different. So, the cousins are less related.

Question 28.
It is believed that non-living material must have given rise to life. How can one say this?
Answer:
1. If we keep on forming small groups of species with recent common ancestors, then eventually we can reach at a stage of forming super-groups of these groups with more distant common ancestors, and so on.
2. Theoretically, if we keep going backwards like this we may reach to a notion that originally there were a single species at the very beginning of evolutionary time.
3. If that is the case, then at some point in the history of the earth, non-living material must have given rise to life.

Question 29.
How did life originate on earth?
Answer:
It is believed that life must have developed from simple inorganic molecules which were present on earth soon after it was formed.
2. It was also speculated that the conditions on earth at that time could have given rise to more complex organic molecules that were necessary for life.
3. It is believed that the first primitive organisms would have arisen from further chemical synthesis.

Question 30.
What do you mean by evidences of evolution? In what way homologous organs and analogous organs give evidence for evolution?
Answer:
Evidence for evolution: Certain significant sources which provide proofs for evolution are called evidences for evolution.

The main ones are:
(1) Evidences through homologous organs,
(2) Evidence through analogous organs and
(3) Evidence through fossils

(1) Evidences through homologous organs:

• Those organs that have same internal structure but different functions are called homologous organs.
• For example, the basic design of internal structure of bones of forelimbs of a frog, lizard, bird, bat and man is same, even though these organs perform different functions.
• This indicates that all these forelimbs have evolved from a common ancestral animal, which had a same basic internal structure.

(2) Analogous organs:

• Those organs, which have different designs but similar appearance and carry out similar functions are called analogous organs.
• For example, wings of insects and birds have different structures but perform similar functions.
• Thus, the presence of analogous organs in different animals provide evidence that they are not evolved from common ancestors, still they perform similar functions to survive prevailing environment.

Question 31.
Differentiate between homologous organs and analogous organs.
Answer:
1. When we classify organisms by grouping them in certain way, it helps us to study them properly.
2. One of the basic methods of classifying the organisms is on the basis of similarities they possess.
3. For classifying we will have to decide which characteristics show fundamental differences’ or say important differences and which show basic differences’ or say less important differences.

Characteristics as a base of classification:
1. A characteristic means a particular type of appearance (form) or behaviour (function).
2. We humans have four limbs (two hands and two legs) and it Is our appearance characteristic.
3. Plants perform photosynthesis is a behaviour (function) characteristic.
4. Some basic characteristics are shared by most organisms. For example, cell is the basic unit of life in all organisms. However, the characteristics of the next level of classification would be shared by most, but not all organisms.
5. At cell level, the classification could be whether the cell has a nucleus or not. For example, cells
of bacteria do not have nucleus, while the cells of most other organisms do.
6. Further, organisms that have nucleus in their cells can be classified as uni-cellular or multi-cellular.
7. Among multi-cellular organisms, classification can be done as whether they undergo photosynthesis or not.
8. The multi-cellular organisms that do not undergo photosynthesis can be further classified as, whether the skeleton Is Inside the body or around the body.
9. By classifying in this manner we can develop a hierarchy of organisms.

Question 27.
Giving an example of humans to explain what increases the probability of two species having a common ancestor.
Answer:
The more characteristics two species will have in common, the more closely they will be related. And the more closely they are related, the more are the chances of them being having a recent common ancestor.
Example:
1. A brother and a sister are closely related. They have common ancestors in the first generation before them, namely, their parents.
2. Now, the girl and her first cousin are also related, but less than the girl and her brother. The reason for this is that although cousins have common ancestors i.e. their grandparents, in the second generation but the parents of these two cousins are different. So, the cousins are less related.

Question 28.
It is believed that non-living material must have given rise to life. How can one say this?
Answer:
1. It we keep on forming small groups of species with recent common ancestors, then eventually we can reach at a stage of forming super-groups of these groups with more distant common ancestors, and so on.
2. Theoretically, if we keep going backwards like this we may reach to a notion that originally there were a single species at the very beginning of evolutionary time.
3. It that is the case, then at some point in the history of the earth, non-living material must have given rise to life.

Question 29.
How did life originate on earth?
Answer:
1. It is believed that life must have developed from simple inorganic molecules which were present on earth soon after it was formed.
2. It was also speculated that the conditions on earth at that time could have given rise to more
complex organic molecules that were necessary for life.
3. It is believed that the first primitive organisms would have arisen from further chemical synthesis.

Question 30.
What do you mean by evidences of evolution? in what way homologous organs and analogous organs give evidence for evolution?
Answer:
Evidence for evolution: Certain significant sources which provide proofs for evolution are called evidences for evolution.
The main ones are:
(1) Evidences through homologous organs
(2) Evidence through analogous organs and
(3) Evidence through fossils

(1) Evidence.s through homologous organs:

• Those organs that have saine internal structure but iterent functions are called homologous organs.
• For example, the basic design of internal structure of bones of forel,mbs of a frog, lizard, bird, bat and man is same. even though these organs perform different functions.
• This Nidicates that all these forelimbs have evolved from a common ancestral animal, which had a same basic internal structure.

(2) Analogous organs:

• Those organs, which have different designs but similar appearance and carry out similar functions are called analogous organs
• For example, wings of insects and birds have different structures but perform similar functions
• Thus, the presence of analogous organs in different animals provide evidence that they are not evolved from common ancestors still they perform similar functions to survive prevailing environment.

Question 31.
Differentiate between homologous organs and analogous organs.
Answer:

Question 32.
What are fossils? How do they provide evidence for evolution? OR What are fossils?
Answer:
1. The remains of dead organisms buried under the earth for millions of years are known as fossils.
2. Fossils are impressions of dead plants or animals that lived in the past.
3. When plants or animals die, the micro-organisms get decomposed in the presence of moisture and oxygen.
4. However, sometimes due to environmental conditions, their bodies do not decompose completely.
5. Such body parts of the plants and animals become fossil and can be available on digging the earth.
6. Fossils can be in the form of imprints, burrow of a worm, or even an actual bone.

Example:

• If the dead leaf gets caught in the mud, leaf will not decompose completely.
• The mud around the leaf will set around it as a mould which will then slowly become hard to form a rock and retain the impression of the leaf. Thus fossil of a leaf is formed.
• By studying these fossils, scientists learn how organisms evolved over time.

Question 33.
Describe any three methods of tracing evolutionary relationships among organisms.
Answer:
Evidence for evolution: Certain significant sources which provide proofs for evolution are called evidences for evolution.
The main ones are:
(1) Evidences through homologous organs
(2) Evidence through analogous organs and
(3) Evidence through fossils

(1) Evidences through homologous organs:

• Those organs that have saine internal structure but different functions are called homologous organs.
• For example, the basic design of internal structure of bones of forel,mbs of a frog, lizard, bird, bat and man is same. even though these organs perform different functions.
• This Nidicates that all these forelimbs have evolved from a common ancestral animal, which had a same basic internal structure.

(2) Analogous organs:

• Those organs, which have different designs but similar appearance and carry out similar functions are called analogous organs
• For example, wings of insects and birds have different structures but perform similar functions
• Thus, the presence of analogous organs in different animals provide evidence that they are not evolved from common ancestors still they perform similar functions to survive prevailing environment.

1. The remains of dead organisms buried under the earth for millions of years are known as fossils.
2. Fossils are impressions of dead plants or animals that lived in the past.
3. When plants or animals die, the micro-organisms get decomposed in the presence of moisture and oxygen.
4. However, sometimes due to environmental conditions, their bodies do not decompose completely.
5. Such body parts of the plants and animals become fossil and can be available on digging the earth.
6. Fossils can be in the form of imprints, burrow of a worm, or even an actual bone.

Example:

• If the dead leaf gets caught in the mud, leaf will not decompose completely.
• The mud around the leaf will set around it as a mould which will then slowly become hard to form a rock and retain the impression of the leaf. Thus fossil of a leaf is formed.
• By studying these fossils, scientists learn how organisms evolved over time.

Question 34.
Mention three important features of fossils which help in the study of evolution.
Answer:
(i) Fossils represent how the ancient species were preserved.
(ii) Fossils help in establishing evolutionary relationship between organisms and their ancestors.
(iii) Fossils help in knowing the time period in which organisms lived.

Question 35.
How can you say that eyes evolved in stages?
Answer:
1. The eye is an important organ for animals.
2. It is a complicated organ which cannot be generated by a single DNA change.
3. The eyes of animals have been created in stages after many generations.
4. First of all, eye was formed in planaria (flat-worm).
5. The eyes of planaria are very simple and are just like ‘eyes-spots’ which detect light.
6. These simple eyes provide a survival advantage to planaria.
7. Thus eye seems to be a very popular adaptation. Gradually, it becomes a complex organ.
8. Most of the animals like insects, octopus, invertebrates and vertebrates have eyes.
9. The structure of eye in above organisms is different which suggests the evolution of eye and is an example of evolution taken place stages.

Question 36.
How can you say that birds are very close to reptiles? OR Give reason. Birds are closely related to reptiles.
Answer:
1. In some dinosaurs, feathers could not be used for flying but provided insulation in cold weather. But, later they might have become useful for flight. Birds however later adapted to flights.
2. Thus, presence of feathers in the birds tells us that birds are quite closely related to reptiles, since dinosaurs which had feathers were reptiles.

Question 37.
Explain how wild cabbage has evolved over the years. OR Explain evolution by stages of analogous organs with the help of artificial section of evolution method.
Answer:
1. As per the definition of analogous organs, the organs that have different basic design but similar appearance and carry out similar functions are called analogous organs.
2. From this concept, it is said that very dissimilar looking structures evolved from common ancestral body design. But all these guesses are about history, which is very old.
3. A present day example to prove this concept can be of wild cabbage.
4. If we study cabbage, we can learn that cabbage provides evidence that a completely dissimilar ‘ looking plant can evolve from it by the process of evolution.
5. Before more than 2000 years, the farmers cultivated wild cabbage as food plant.
6. With the passage of time, farmers gave various modifications to it.
7. Some farmers obtained broccoli (a kind of vegetable) while others obtained sterile flowers and developed cauliflower.
8. Some other farmers selected swollen parts of wild cabbage and developed another variety known as kohlrabi.
9. Some farmers have developed slightly large leaves of wild cabbage and their leafy vegetable is called kale.
10. Thus, now wild cabbage is the ancestor of cabbage, broccoli, cauliflower, kohlrabi and kale varieties, which are obtained by farmers by performing artificial selection. Moreover, all these varieties look different from their ancestor i.e. wild cabbage.

Question 38.
It is incorrect to consider evolution as progress. Explain.
Answer:
1. Actually, no real ‘progress’ has taken place in the idea of evolution.
2. Evolution is simply the generation of diversity and the shaping of the diversity by environmental selection.
3. The only progressive trend that is seen in evolution is that with time more and more complex body designs have emerged. This does not mean that the older designs were inefficient. Many older and simpler designs still survive.
4. One of the simplest life forms bacteria still inhabit the most inhospitable habitats like hot springs, deep-sea thermal vents and the ice in Antarctica.
5. Hence, it is incorrect to consider evolution as progress.

Question 39.
How can you say that humans have not evolved from different races and that they all belonged to only one region of the world?
Answer:
1. Across the earth, there is a great diversity in humans.
2. For quite a long time, man thought about different races of human.
3. Initially, human race was identified on the basis of skin colour and the humans were known as yellow, black, white or brown.
4. However, in the recent years, it has been proved that all the human beings have evolved from a single species called the Homo sapiens.
5. By research, it has been established that we all have emerged from Africa.
6. Our genetic foot prints can be traced back to Africa.
7. A couple of hundred thousand years ago, some of our ancestors left Africa while others stayed back.
8. Those who left Africa slowly spread over the planet from Africa to West Asia to Central Asia, Eurasia, South Asia and East Asia.
9. Humans also migrated to islandsof Indonesia and Philippines, Australia and reached America.
10. They went with groups, sometimes separating from each other and mixing with each other, even moving in and out of Africa.
11. Thus, humans have all evolved from Homo sapiens and initially belonged to Africa.

Question 40.
Differentiate between dominant trait and recessive trait.
Answer:

Question 41.
Give the pair of contrasting traits of the following characters in pea plant and mention which is dominant and which recessive.
(i) Yellow seed
(ii) Round seed
Answer:

Question 42.
Why did Mendel choose pea plant for his experiments? OR Mendel selected pea plant (Pisum sativum) for his experiments to study inheritance. Give reason.
Answer:
1. Pea plants are very small and so it is quite easy to manage them. Also, they grow easily.
2. They produce a large number of progeny and so the characters they show can be studied more accurately on a larger scale.
3. Pea plants have both male and female reproductive organs. As a result, they can either self-pollinate themselves or cross-pollinate with another plant. This helps in understanding the traits properly.
4. Pea plants have sharply defined and contrasting traits such as height (tall, dwarf), seed colour (yellow, green), etc.
5. They are annual plants. So, every year, one can obtain a new generation plant.
6. These characteristics of the pea plant make it quite favourable for performing research. Hence, Mendel choose this plant.

Question 43.
Males are responsible for the sex of the child produced. Give reason.
Answer:
1. Sex of the child is determined by the sex chromosomes that its zygote contains.
2. If the zygote contains ‘Y’ sex chromosome over and above ‘X’ chromosome, it will turn into male and if both sex chromosomes are ‘X’, the child born will be a female.
3. A female sex cell contains ‘XX’ sex chromosome whereas a male contains ‘XY’ chromosome.
4. Thus, a female can always donate ‘X’ chromosome but it is the male that can provide ‘Y’ chromosome to female ‘X’ chromosomes and produce a male.
5. However, if the male donates ‘X’ chromosome, a female will be born.
6. Thus, males and not females are responsible for the sex of the child produced.

Question 44.
Changes in the non-reproductive body cells of an organism cannot be transmitted to its progeny. Give reason.
Answer:
1. Only those traits will be inherited to their progeny wherein change has occurred in the genes in gametes of organisms during the process of reproduction.
2. For example, if an organism starves for some time and reduces its weight, then It is called acquired trait.
3. The reduced weight due to starvation will not change DNA of the cells.
4. As a result, reduced weight is not a trait, which can be inherited by future generations of organisms that will starve.
5. Thus, the changes in the non-reproductive body cells of an organism cannot be transmitted to its progeny

Question 45.
Does genetic combination of mothers play a significant role in determining the sex of a new born?
Answer:
1. The mothers have a pair of XX chromosomes. This means that all children whether boy or girl can inherit X-chromosome only from their mother.
2. Only males have XV chromosome. One of these two will be passed on to the offspring. So, whether the sex of the new born will be male or female can be decided only through the male chromosome. Hence, genetic combination of mothers does not play a significant role in determining sex of the new born.

Question 46.
Why do all the gametes formed in human females have only X-chromosome?
Answer:
The genetic making of the females is such that their sex chromosomes have only X chromosomes. This means both the chromosomes in the female germ cells have X chromosome. So, naturally, the gametes formed in human females have only X-chromosomes.

Question 47.
A woman has only daughters. Analyze the situation genetically and provide a suitable explanation.
Answer:
1. Man contains both X and Y chromosomes. This means males have XV chromosome. So, it is mans sex cells that determine the sex of the child born.

2. Women contain only XX chromosomes so she can contribute X chromosome only. A woman has only daughters. This indicates that in every fusion, the sperm of the male carrying X-chromosome got fertilized with the X chromosome of the female ovum. So, if a boy were to be born, the Y chromosome from male should have travelled and fertilized with X chromosome of the female.

Question 48.
Why reduction in weight of beetle bug cannot be considered as an inherited trait?
Answer:
1. The weight of the beetle bug got reduced because of starvation. It did not change the DNA of the germ (sex) cells. Change in non-reproductive tissues cannot be passed on to the DNA of the germ cells.

2. Since the DNA is not changed, the change i.e. low weight cannot be inherited by the progeny of a starving beetle.

3. Hence, even if some generations of beetles are low in weight because of starvation, it cannot be considered as an example of evolution because the change is not inherited over generations. When such scarcity of food gets over and the beetle get sufficient food the new beetles will be healthy.

Question 49.
‘Experience of an Individual during its lifetime cannot be passed on to its progeny and cannot direct evolution’ What do you mean by this statement? Explain giving example.
Answer:
Change in non-reproductive tissues cannot be passed on to the DNA of the germ cells. Hence the experiences of an individual during its lifetime can neither be passed on to its progeny nor can such experiences result in evolution.

Example:
1. If we breed a group of mice, all their progeny will have tails.
2. Suppose we remove the tails of each generation of these mice by surgery then it does not mean that the progeny is tailless.
3. Artificially removing the tail does not bring any change in the genes of the germ cells of the mice.
4. So, the cut-tall that a generation of mice experienced cannot become a trait to be passed on to the next generation.

Question 50.
Work out which trait would be considered dominant and which one recessive in the given figure.
Answer:
Stage 1 (F₁ generation):
When a cross occurs between a coloured flower plant and a white flower plant, the F₁ generation plant produced has all the coloured flowers. This means ‘the coloured flowered trait’ is dominant over ‘white flowered trait’.

Stage 2 (F₂ generation):
Although, the white flowered trait did not get expressed in F₁ generation. it did get expressed in F₂ generation. This means the ‘white flowered trait’ is the recessive trait.

Question 51.
Sakhaaram, a poor farmer has 2 daughters. He feels that he should have a son who can help him in his farming work. So when his wife again gets pregnant, he tells her that this time she should not give birth to a girl again. If she does so, he will abandon her. So, she insists her to go for sonography and remove foetus it is found to be of a female.
Now, answer the following questions:

Questions:
1. Who is the responsible for the sex of the children? — Sakhaaram or his wife?
2. ‘Owing to the poor condition of Sakhaaram, removing the female foetus is a good solution.’ State your comments.
3. Which sex chromosome of Sakhaaram did not do its job of giving birth to a male child in previous cases?
4. The village sarpanch advised Sakhaaram to accept the born child irrespective of the sex as it is God’s decision and also because that is the only legal way. Which values has Sarpanch shown?
Answers:
1. Sex of the child gets determined by the chromosome of the male parent. Hence, Sakhaaram is responsible for the sex of the children.
2. Removing female foetus is a crime. Sakhaaram can be jailed for doing so. Moreover, socially, such an act disturbs the sex-ratio.
3. Y-chromosome.
4. The Sarpanch has shown the values of morality, empathy towards Sakhaaram and care and concern towards his wife. Moreover, the Sarpanch also displayed the values of awareness towards the laws of our country.

Question 52.
Given below is the experiment carried out by Mendel to study inheritance of two traits in garden pea:
(a) What do the letters A, B and C represent?
(b) State the objective for which Mendel performed this experiment.
Answer:
(a) A = gamete of round green (RY) plant
B = gamete of wrinkled (ry)
C = F₁ generation (all round yellow)

(b) Mendel’s objective was to show independent inheritance of traits or to prove law of independent assortment.

Question 53.
The legendary Milkha Singh, also known as ‘The Flying Sikh’ is the source of inspiration for millions of indians who aspire to become a great runner. However, Milkha’s son Jeev is a golfer and not a runner.
Now answer the following questions:

Questions:
1. Didn’t Jeev inherit the running skill from Milkha Singh? Why or why not?
2. Which type of traits is running?
Answers:
1. Skills such as running, swimming, acting, dancing, etc. cannot be inherited because such skills are acquired skills. Milkha Singh acquired it by extreme hard work and dedication. His son acquired the skills of golfing.
2. Running is an acquired skill. It can be acquired by putting dedicated and planned effort in the field of one’s choice.

Question 54.
Explain whether the traits like eye colour or height are genically Inherited or not. Do power to lift weight and reading French also belong to same category? Justify your answer.
Answer:
(a) Colour of eye and height are genetically inherited.
Reason:
Traits such as colour of eye and height cannot be changed or acquired during a life because they are controlled by the genes which an individual inherits from the parents.

(b) Power to lift weight and reading French can be acquired during life time.
Reason:
1. These traits are achieved by an individual by experience and practice. Hence, these are acquired characters.
2. These acquired characters do not change DNA of germ cells and so they cannot be inherited or passed to the next generation.

Question 55.
Which of the following traits cannot be passed on the progeny? Justify your answer.
(i) Rudimentary eyes of planaria.
(ii) Absence of tail in a mouse (after surgical removal)
(iii) Low weight of a starving beetle.
Answer:
(i) Trait number
(ii) Absence of tail in a mouse (after surgical removal) and
(iii) Low weight of a starving beetle cannot be passed on the progeny.

Reason:
These traits are present in non-reproductive tissues. So, they do not change the DNA of the germ cells.

Very Short Answer Type Questions 

Question 1.
Define heredity.
Answer:
The process of transmission of characters from parents to the offspring is known as heredity.

Question 2.
Define genetics
Answer:
The branch of science which deals with heredity and variation is known as genetics

Question 3.
In a sexual reproduction, offspring is similar to the characteristics of parents however it is not identical to the parents. Justify this statement.
Answer:
Inheritance from previous generation includes carrying forward the basic body structure along with minor variations. Hence It is rightly said that though offspring is similar it is not identical to the parent.

Question 4.
Why would you say that asexual reproduction is not suitable for introducing variation in species?
Answer:
Asexual reproduction involves a single parent. The offspring produced has same set of chromosomes which exhibit little or no variation. Hence asexual reproduction is not suitable for introducing variation in species.

Question 5.
How can we say that variation In species Increases the probability of survival?
Answer:
Variation improves the adaptability of the species according to the climatic conditions. This increases the chances of survival.

Question 6.
State the types of earlobes that can be seen in humans.
Answer:
We can see two types of earlobes in humans. They are —
(i) Attached earlobes Le, earlobes which are attached to the side of the head.
(ii) Free earlobes i.e. earlobes which seem hanging and hence unattached to the side of the head.

Question 7.
Point out a variation that has been observed in human beings.
Answer:
Free or attached earlobes is a variation that can be observed in human beings.

Question 8.
What is meant by a pure plant (or pure line plant)?
Answer:
A breed or strain of animals or plants that maintains a high degree of consistency in certain characters as a result of inbreeding for generations is called a pure breed of plant or animal. Such a plant of animal maintains homozygous for all genes.

Question 9.
Who is considered as a ‘father of genetics’?
Answer:
Gregor Johann Mendel is considered as a ‘father of Genetics’.

Question 10.
What is the contribution of Gregor Johann Mendel?
Answer:
Gregor Johann Mendel established Mendelian Inheritance Laws. He made major contributions in genetics.

Question 11.
Which term did Mendel use to describe genes?
Answer:
Mendel used the word ‘factor’ to describe genes.

Question 12.
Name the plant on which Mendel performed his experiments.
Answer:
Mendel performed experiments on Pisum Sativum (Garden pea)

Question 13.
What do you mean by dominant and recessive traits?
Answer:
While inheriting the traits from parent, the allele of a gene that masks the other variants is called as the dominant trait whereas the one that gets masked is called as recessive trait.

Question 14.
What are F₁ and F₂ generations?
Answer:
F₁ is the first generation reproduced from two distinct parental type of same species. F₂ is the second generation reproduced from breeding of F₁ generation.

Question 15.
What did Mendel observe in F₁ progeny when he crossbred a tall and a short plant?
Answer:
Mendel observed that there were no medium height plants. All the plants in F₁ progeny were tall in height.

Question 16.
What was the second observation that Mendel make from F₂ progeny of garden peas plant?
Answer:
Mendel found that 75% of F₂ generation plant produced from F₁ progeny were tall whereas 25% of them were short heighted.

Question 17.
If TT and Tt trait combination produces tall plants, which of the traits is a dominant trait?
Answer:
In the above stated combination of traits, T is the dominant trait for tallness.

Question 18.
How did Mendel prove that the traits are independently Inherited by the progeny?
Answer:
When Mendel cross pollinated one plant with yellow and round seeds with another plant having green and wrinkled seeds, he observed that the F₂ progeny generated four different types of plants with the different combinations of the above mentioned traits. This proves that the traits are independently inherited by progeny.

Question 19.
What is the phenotype ratio of Mendel’s experiment done to study inheritance of a single character?
Answer:
3:1

Question 21.
What is the genotype ratio of Mendel’s experiment done to study InherItance of a single character?
Answer:
1:2:1

Question 22.
What is the phenotype ratio of Mendel’s experiment to study inheritance of two characters?
Answer:
9:3:3:1

Question 23.
State the correct sequence for the expression of tall trait in pea plant.
(A) A lot of hormone will be made
(B) A gene provides information for protein
(C) A hormone triggers the growth in height
(D) An enzyme works efficiently
Answer:
(B) – (D) – (A) – (C)

Question 24.
Write expanded form of DNA.
Answer:
Deoxyribonucleic acid

Question 25.
What is a gene?
Answer:
Gene is considered to be the basic unit of heredity. A section of DNA in a cell which provides information of a protein is called as gene for that protein.

Question 26.
How is the sex of an offspring determined in human beings?
Answer:
The genes inherited from the parents determines the sex of the offspring in human beings. If an X chromosome is inherited from father, the sex will be female whereas if Y chromosome is inherited from father, the sex will be male.

Question 27.
Give an example of an animal in which sex is not genetically determined.
Answer:
In snails, sex is not determined genetically.

Question 28.
Would a new born child with XX pair of chromosome be a boy or a girl?
Answer:
The new born child with XX pair of chromosome would be a girl.

Question 29.
What is the difference between the sex chromosome of a male and a female?
Answer:
Women have a perfect pair of sex chromosomes and both are called X whereas men have a mismatched pair in which one is X with a normal length and the other is Y which is of shorter length.

Question 30.
Suppose, a gene of brown eyes is dominant to a gene of blue eyes. What would be the eye colour of a child born If he inherits gene of blue eyes from mother and a gene of brown eyes from father.
Answer:
The child have brown coloured eyes.

Question 31.
It a woman with brown hair gene b marries a man with black hair gene B, what would be the hair colour of offspring born with the below mentioned combination?
(i) bb, (ii) Bb, (iii) BB
Answer:
Offspring born with contrition bb would have brown hair whereas the offspring born with the combinations Bb and BB would have black hair

Question 32.
How would you differentiate an acquired trait from an inherited trait?
Answer:
The trait of an organism that is passed on from the DNA of a parent to an offspring is an inherited flit whereas the trait that an organism develops as a response to the environment and through me experience is an acquired trait.

Question 33.
Do young once of mice whose tails have been cut are born without tails? Justify your name.
Answer:
The young ones are born with tails because cutting the tail of parent mice does not after the genes of mice

Question 34.
What is evolution?
Answer:
The change introduced in the offspring from the parents through variation in species is known as evolution.

Question 33.
Name the scientist who first came up with the theory of evolution.
Answer:
Changes Darwin first introduced the theory of evolution.

Question 38.
Which are the two basic factors responsible for evolution?
Answer:
Heredity and deflation are the two basic factors responsible for evolution.

Question 37.
What are the two population of one species who cannot reproduce called as?
Answer:
Such poputation are called as independent population.

Question 38.
What is natural selection?
Answer:
The process whereby organisms get batter adapted to their environment tend to survive and produce more oftsprng is called …….. .

Question 39.
State any three reason which cause speciation.
Answer:
Geographical isolation, changes in DNA. emergence of a variation n population. etc are few of the newly reasons which can cause speciation.

Question 40.
How can geographical isotation impact specistion in species which reproduce:
(i) asexualy,
(ii) sexually?
Answer:
In case of asexual reproduction geographical isolation may not result in speciation because the offspring are reproduced from already existing parent throught salt fertilization. it has high chance of speciation in case of sexual reproduction as geoephically isolated species may rarely get involved in cross fertilization.

Question 41.
What do you mean by genetic drift?
Answer:
The sudden and random changes in the gene frequency occurring in a small population by chance is known as a genetic drift.

Question 42.
I am able to inculcate diversity in organisms without any adaptations that too simply by chancing the frequency of certain genes in a population. identify me.
Answer:
Genetic drift

Question 43.
On what basis do we classify organisms? Give an example.
Answer:
We can classify organisms based on characteristics i.e. the appearance or behavior. The ability to do photosynthesis is a characteristic which helps us classify these organisms in plant kingdom.

Question 44.
What are the homologous characteristics of organisms? Give an example.
Answer:
The organs having similar origin and similar basic structure but may perform different functions are homologous organs. For example, frogs. lizards, birds and humans. All of these have limbs with similar structures but they are used to perform different functions.

Question 45.
Describe analogous organs with an example.
Answer:
Analogous organs are the organs which have similar functions even though they might have different origin. For example, the wings of a bat and a butterfly perform the functions of flying but they do not have common ancestry.

Question 46.
What are fossils?
Answer:
When the organisms die, their body gets decomposed and then gets absorbed in nature. But at times, the body or few parts of body remains without getting decomposed completely. Such preserved traces of organisms are called fossils.

Question 47.
Which evolutionary explanation is true for the feathers?
(1) Birds are very closely related to reptiles
(2) Birds and bat are very closely related to each other.
Answer:
Statement I

Question 48.
What Is the ‘relative way’ to find the age of a fossil?
Answer:
Relative way compares the depth of two fossils at which they are found in earth’s surface. The fossil that is found near to surface of earth is more recent than fossil that is discovered at deeper layer.

Question 49.
Which fossil dId we discover in Narmada Valley recently?
Answer:
A skull of a dinosaur was discovered in Narmada valley a few years ago.

Question 50.
How does the dating method work for calculating the age of a fossil?
Answer:
In dating method, the different isotopes (Carbon-14) of the same element found in fossil are compared to the current available element for calculating the age of a fossil.

Question 51.
What is the possible explanation of evolution of complex organs like eyes?
Answer:
It can be explained that the evolution of such complex organs took place at regular stages. For example, it may have started with a rudimentary eye which initially just detected light.

Question 52.
Give an example of an animal having rudimentary eye.
Answer:
Planaria is a type of flat-worm having rudimentary eye.

Question 53.
Mention one characteristIc which lndicates close relation between birds and reptiles
Answer;
Existence of feathers on the body of few dinosaurs (who were reptiles) and also on the body of birds indicates a close relationship between birds and reptiles.

Question 54.
Give an example of evolution through artificial selection. OR List any six vegetables which have been evolved from wild cabbage due to artificial selection.
Answer:
Humans, through artificial selection have cultivated wild cabbage by focusing on various characteristics. This resulted in evolution to multiple types of vegetables such as cauliflower, broccoli, kohlrabi, kale, cabbage and red cabbage.

Question 55.
How can variation in population help us in understanding evolution?
Answer:
We know that minor changes in DNA during reproduction is the basic event In evolution. If we compare DNA of different species over the period of time this can help us in understanding evolution of species.

Question 56.
Give an example which proves that older body designs and characteristics are not in efficient.
Answer:
Bacteria, which are the simplest and oldest body designs are successfully surviving in conditions such as hot springs, deep-sea thermal vents and ice in Antarctica. This proves that older body designs and characteristics are not inefficient.

Question 57.
Where were the earliest traces of human species found? What is the name of that species?
Answer:
The earliest traces of human beings known as Homo sapiens have been found in Africa.

Fill in the Blanks:

1. During reproduction, an offspring inherits two things from its previous generation, common basic body design and …………..
Answer: some fine (minor) variations

2. Variations in the new generation occur due to ……………
Answer: DNA copying

3. Selection of variants by environmental factors forms the basis for …………………
Answer: evolutionary processes

4. The rules of determine the process by which traits and charactenstics are reliably inherited.
Answer: Heredity

5. If the traits get expressed in the first generation, they are known as …………..
Answer: Dominant traits

6. The DNA segment which is responsible for the synthesis of a specific type of protein is called ……………..
Answer: Gene

7. The lowest part of the ear pinna is known as an …………..
Answer: Ear-lobe

8. …………… can change their sex on their own.
Answer: Snails

9. It is the chromosome of ……………….. that determines the sex of the child born.
Answer: male sex cell

10. In layman language, the phrase related to natural selection is ……………..
Answer: survival of the fittest

11. A trait developed in response to the environment is called ………………..
Answer: acquired trait

12. Acquired traits are ……. in nature.
Answer: temporary

13. Genetic drift leads to accumulation of …………………
Answer: different changes (variation)

14. The of eyes provide survival advantage to them……………..
Answer: planaria

15. Thus, presence of feathers in the birds tells us that birds are quite closely related to…………
Answer: reptiles

16. …………. used to be the most common way of identifying human races.
Answer: Skin colour

True OR False

1. Organisms have two modes of reproduction. — True
2. Suppose a bactena is able to survive better in a heat wave. This means that quite likely the next generation will be able to survive in a heat-wave — True
3. Organisms undergo variation and most variations survive resulting in high rate of survival of those organisms. — False
4. If a trait A exists in 8% of a population of an asexually reproducing species and a trait B exists in 42% of the same population, it means population with trait A must have arisen earlier and is more stable to variation. — False
5. Practically, the father contributes more amount of genetic material to the child as compared to the mother. — False
6. In Mendel’s experiment, in the first place, no ‘medium-height’ plants were produced. This means the plants gained trait only from one of the parent and not both and so his first experiment failed. — False
7. In Mendel’s experiments F₁ progeny was obtained by cross pollination. — True
8. In Mendel’s experiments F₂ progeny was obtained by cross pollination. — False
9. In Mendel’s experiments, characters of both the parents were observed in F₂ generation. — True
10. The enzyme synthesis in the cell is controlled by genes. — True
11. In general, the pea plant shows cross-pollination. — True
12. In some reptiles, change in temperature of the fertilized egg determines the sex of the newborn. — True
13. The size of both the chromosome in women is same, while it is unequal in men. — True
14. A child who inherits a Y chromosome from her father will be a girl, and one who inherits an X chromosome from him will be a boy. — False
15. If a person is on dieting and loses weight, he is said to have acquired a trait. — True
16. An inherited trait alters the DNA. — True
17. If a person from Ahmedabad who loves living in mountains migrates to Himalayas and lives there for 20 years, his child when born will develop similar love for the mountains. — False
18. Our forelimbs are similar to lizards. — True
19. It is believed that both, human beings and chimpanzees have a common ancestor. — True
20. Practically speaking, there is no real ‘progress’ in the idea of evolution. — True

Match the Following:

Question 1.

Answer:
(1- q), (2 – r), (3 – s), (4 – p)

Question 2.

Answer:
(1 – s), (2 – r), (3 – q), (4 – p)

Question 3.

Answer:
(1 – q), (2 – s), (3 – r), (4 – p)

Haryana State Board HBSE 10th Class Science Important Questions Chapter 8 How do Organisms Reproduce Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 8 How do Organisms Reproduce?

Question 1.
Explain reproduction. Why is it vital? OR Reproduction is not necessary for maintaining the life of an Individual then why do organisms reproduce?
Answer:
Reproduction:
1. The production of new organisms from the existing organisms of same species is known as reproduction.
2. Every organism remains alive for a limited time. In other words, one that takes birth has to die at some point of time.
3. Considering this fact, if an organism does not reproduce, it is quite likely that one day its species will become extinct. Not only this, even if one species vanishes, it disturbs the entire environmental cycle.
4. Hence, even though, the organism does not need to undergo reproduction for maintaining his own life, it reproduces so that it may continue the existence of its species. And this is the reason why reproduction is vital for the survival of the species.

Question 2.
Organisms of the same species look similar. What is the role of DNA copying in this? OR Discuss the meaning and need of DNA copying.
Answer:
1. All organisms of a particular species look similar. For example, dogs look like dogs and a human looks like another. This happens because the designs of their bodies are similar.
2. Naturally, if the body designs are to be similar, the blueprints for these designs should be similar. Thus, similar blue prints lead to formation of similar bodies.

Producing copies of DNA:

1. The first and foremost task of the reproduction process is to make ‘copies of blueprints of body design’. This is done by creating copies of DNk
2. The nucleus of cells contains chromosomes and chromosomes contain genetic material called DNA. DNA contains information about protein synthesis.
3. If the information is changed, there will be a change in protein synthesis, which will eventually change the body design. In such case, one organism may look quite different than the other and we would not be able to classify an organism into a specific species. Therefore, the most basic work in the process of reproduction is to create a DNA copy. Cells undergo certain chemical reactions to build the copies of their DNA.

Question 3.
Why only DNA copying is not sufficient for creating similar organisms? What else should be done? OR ‘Although DNA copying is the basic activity for creating similar organisms, It does not ensure survival’. Explain.
Answer:
1. A cell that undergoes reproduction creates two copies of DNA. Then, both these copies need to be separated from each other.
2. Production of a new cell takes place through the process called ‘cell division’.
3. It is not possible to keep one copy of DNA inside the cell and simply push out the second copy. The second copy of DNA would not survive this way.
4. In order to make the DNA survive, a complete ‘cell apparatus’ or simply a complete cell will have to be produced. The new cell will safely contain and preserve the second copy of the DNA.

Question 4.
How does DNA copying leads to variation In organisms? Is this variation harmful?
Answer:
1. During cell division, a cell divides to give rise to two cells. Although, the two cells will be similar but will not be exactly identical. It is quite obvious that the process of copying the DNA will have some variations. Hence, the generated DNA copies will be similar, but may not be identical to the original.

2. At times, some variations in DNA copying are so drastic that the new DNA copy cannot work with the cellular apparatus it inherits. Such a new born cell will eventually die.

3. In conclusion, the cells that survive after the cell division are (1) similar to each other as well as (2) different i.e. show variation. This inbuilt tendency for variation during reproduction is the basis for evolution which is useful for the organism to adapt to ever changing environment.

Question 5.
Is a reproduced cell identical? What happens If this ¡s not the case?
Answer:
1. Whether the new cell formed will be identical or not depends on how accurately the sequence of information has been copied.
2. Copy error is a rare event, still at times it may take place.
3. If the newly produced cell is not identical, there may occur changes in its characteristics.
4. If these changes are not compatible with the existing cellular system, the cell may die.

Question 6.
What is meant by creation of additional cellular apparatus?
Answer:
1. When a cell undergoes division, it must make copy of the cell DNA. However, simply making a copy of DNA is not sufficient during cell division, since the DNA molecule itself cannot survive and carry out all the functions of a cell.
2. The DNA will require all the other parts of the cell i.e. the cellular organelles such as nucleus. mitochondria, Golgi body, etc. so that the DNA can survive in the newly formed whole cell. This would be a fully functional cell and hence called a cellular apparatus or simply a cell.
3. Creating an addItional cellular apparatus means to produce a copy of all the parts of the cell (other than DNA).

Question 7.
What is a clone? Why do offspring formed by asexual reproduction exhibit remarkable similarity?
Answer:
1. A clone is an identical genetic copy of either a piece of DNA, or a cell or a whole organism. Many organisms produce clones through asexual reproduction.
2. A clone genetically identical to the parent i.e. it possesses the exact copy of DNA of its parent and hence it shows remarkable similarity to the parent as well as to one another.

Question 8.
What is niche? Explain.
Answer:
Niche:
1. A niche is the role and position a species plays in the environment, how the species meets its needs for food and shelter, how it survives and how it reproduces.
2. Different species of organisms have different niches.
3. For example, the way a lion gathers Its food or shelter or reproduces are quite different from that of a crocodile i.e. the niches of two organisms vary.

Important aspects with respect to niches are —

• Variation is seen among niches of various species.
• The niches can change because the causes for their changes are not under the control of the organisms. Changes could be due to change in the temperature of the earth, which can go up or down, variation in water level, meteor hits, etc.
• Minor variations in the body designs with respect to a changing niche increases the chances of survival of the organism.

Question 9.
Give an example where in variation in species becomes a boon for surviving in the changed niche.
Answer:
1. Suppose a population of reproducing organism was suited to a particular niche.
2. A drastic change occurs in the niche, which can wipe out the entire population because the population cannot withstand such a drastic change.
3. Assume a population of bacteria living in temperate waters. If the temperature of water increases to global warming, most of these bacteria would die. But if in some individuals of this population, some variations exist which can allow them to with stand such changes. then these individuals will survive and the entire population will not wipe off.
4. Thus, we can say variation in the species can save the existence of that species in this world. Thus, variation is useful for the survival of species.

Question 10.
What is asexual reproduction? State its types.
Answer:
Asexual reproduction :
The method of producing a new plant (or animal) without the act of fertilization of gametes is called asexual reproduction.

Types:

• Fission
• Fragmentation
• Regeneration
• Budding
• Vegetative propagation and
• Spore formation.

Question 11.
Explain the fission method of asexual reproduction.
Answer:
Fission :
1. Fission is the simplest method of asexual reproduction in unicellular organisms such as bacteria and protozoa.
2. In fission process, a unicellular organism divides to form two or more new organisms.

Types : (A) Binary fission and (B) Multiple fission

(A) Binary fission:

• Many bacteria and protozoa simply split into two equal halves during cell division.
• In binary fission, the nucleus lengthens and then divides into two parts.
• After that the cytoplasm divides into two parts. This results in two daughter cells.

(1) Binary fission in amoeba:
Amoeba has no definite shape. Hence, binary fission can take place from any place (or say plane) of the amoeba cell resulting into two daughter cells.

(2) Binary fission in leishmania:

• Some unicellular organisms like leishmania shows somewhat more organized binary fission of their bodies.
• Leishmania has a whip-like i.e. longitudinal structure at one end of the cell. So in leishmania, binary fission takes place in a definite orientation (plane) unlike amoeba.
• Leishmania causes an infectious disease called kala-azar or say black fever.

(B) Multiple fission:
1. In multiple fission, the parent organism divides to form many new organisms at the same time.
2. Malarial parasite, plasmodium, divides into many daughter cells simultaneously by multiple fission.

Question 12.
How is binary fission in amoeba different than that in lelshmania?
Answer:
1. In amoeba, binary fission can happen at any place or say plane of the cell.
2. Leishmania has a whip-like i.e. longitudinal structure at one end of the cell. So, in leishmania, binary fission takes place in a definite orientation (plane).

Question 13.
Differentiate between binary fission and multiple fission.
Answer:

Question 14.
What is fragmentation?
Answer:
1. The process wherein the body of a multi-cellular organism breaks (fragments) into two or more pieces and on maturing each piece grows to form a complete new organism is known as fragmentation.

2. Simple asexual reproduction method can work even in multi-cellular organisms provided the organisms have relatively simple body organization. For example spirogyra, a simple multi-cellular organism reproduces through fragmentation.

Question 15.
Why fragmentation reproduction method works well only for few multi-cellular organisms? Why there is a need of more complex way of reproduction for most of the multi-cellular organisms?
Answer:
1. Although multi-cellular organisms like spirogyra reproduces through fragmentation, such multi-cellular organisms have simple body organization.
2. When the body organization is complex, the multi-cellular organisms cannot simply divide and reproduce. They need to reproduce through a more complex way of reproduction.

Reason:

• Many multi-cellular organisms are not simply a random collection of cells. Such organisms have specialized cells. The cells are organized as tissues, and tissues are organized into organs. The organs are placed at definite positions in the body.
• In such a carefully organized situation, cell-by-cell division is impractical.
• Also, in multi-cellular organisms, different cell types perform different specialized functions. So, reproduction in such organisms is also the function of a specific type of cells. Hence, multi-cellular organisms need to reproduce through a more complex way of reproduction.

Question 16.
Write a note on regeneration.
Answer:
1. In some plants and animals, if some parts of their bodies get cut, they have the ability to regenerate and form a complete new organism. This method of reproduction is called regeneration.

2. If such individuals somehow get cut or broken up into many pieces, many of these pieces grow into separate individuals.
3. For example, simple animals like hydra and planaria can be cut into any number of pieces and each piece would grow into a complete organism.
4. Regeneration is carried out by specialized cells. These cells perform repeated cell division and form large numbers of cells or say cell-mass. From this mass of cells, different cells undergo changes to become various cell types and tissues.
5. These changes take place in an organized sequence which is referred as ‘development’.
6. Note that regeneration is not the same as reproduction because getting cut to reproduce is not the way to reproduce for most of the organisms.

Question 17.
Name the process by which planaria grows. Explain the basic features of this process.
Answer:
Planaria grows through regeneration.

Basic features:

• Regeneration is carried out by specialized cells. These cells perform repeated cell division and form large numbers of cells or say cell-mass.
• From this mass of cells, different cells undergo changes to become various cell types and tissues.

Question 18.
What is budding? Explain.
Answer:
1. In budding, a small part of the body of the parent organism grows out as a ‘bud’. It then detaches from parent and becomes a new organism. For example, hydra and planaria reproduce through budding.

2. During budding, hydra makes use of regenerative cells for reproduction.
3. A bud develops as an outgrowth due to repeated cell division at one specific site. These buds develop into tiny individuals and when they mature fully, they get detached from the parent body and become new independent individuals.

Question 19.
Differentiate between fragmentation and budding.
Answer:

Question 20.
Explain vegetative propagation. OR What is vegetative propagation? Draw a labeled diagram to show vegetative propagation.
Answer:
1. Vegetative propagation is an asexual method of reproduction which occurs only in plants.
2. In vegetative propagation, new plants are reproduced from the plant parts such as roots, stem and leaves of old plants, without taking help of any reproductive organs.

3. There are many plants in which parts such as root, stem and leaves develop into new plants under appropriate conditions.
4. Similarly, buds produced in the notches along the leaf margin of bryophyllum fall on the soil and develop into new plants is also an example of vegetative propagation.
5. This property of vegetative propagation is used in methods such as layering or grafting for growing plants such as sugarcane, roses and grapes.

Question 21.
State the advantages of vegetative propagation.
Answer:
Advantages of vegetative propagation:
1. Plants raised by vegetative propagation can bear flowers and gifts earlier than those produced from seeds.
2. Vegetative propagation is also helpful in growing plants such as banana, orange, rose and jasmine that have lost the capacity to produce seeds.
3. The plants produced are genetically similar to the parent plant and so have similar characteristics as the parent plant.

Question 22.
Name the plant that reproduces through leaves. List two advantages of this method of reproduction.
Answer:
Bryophyllum is a plant that reproduces through leaves.
Advantages:

• Many buds are produced in the notches along the margin of a single leaf. Each leaf bud of bryophyllum can produce a plant.
• One single leaf reproduces a large number of young plants. This helps in survival of the species.

Question 23.
What is a spore? Explain asexual reproduction through spore formation. OR Explain asexual reproduction in rhizopus.
Answer:
1. A spore is a microscopic reproductive unit of plants which is covered by a protective coat.
2. When the coat of spore bursts, the spores spread into air. These airborne spores settle on food and under favourable condition, germinate to produce new plants.

Example:
1. A bread mould such as rhizopus is a simple multi-cellular organism (fungi). Numerous spores are produced in rhizopus within sacs called sporangia (singular: sporangium).
2. The sporangia are knob (or blob) like tiny structures present at the top of the thread like structures called hyphae. The sporangia contain cells or spores.

3. When these sporangia bursts, the spores get scattered by rain, wind or insects and under suitable conditions develop into a new rhizopus plant, when they come in contact with a moist substance such as bread.

Question 24.
What is sexual reproduction? Explain very briefly how it takes place.
Answer:
Sexual reproduction :
1. The mode of reproduction that takes place with the help of both male sex cells and female sex cells is known as sexual reproduction.
2. The sex cells involved in sexual reproduction are called gametes.
3. During sexual reproduction, a male gamete unites with a female gamete to produce a ‘zygote’.
4. As time passes, the zygote develops into a new organism.

Question 25.
Discuss the important aspects of variation occurring during DNA copying.
Answer:
1. Creation of two new cells from one cell involves copying of the DNA’ as well as copying the ‘cellular apparatus’.

2. Although, the two cells will be similar but will not be exactly identical. It is quite obvious that copying of DNA will not be 100% accurate and there will be some variations in the copied version. These variations are known as copying errors. These errors or variations lead to variations in population of organisms.

3. The negative effect of variation is that at times, the variation developed in new cells lead to the death of the cells and even organisms. The positive side of variation is that the species develop resistance to adapt to various environments. This enables the population to survive when a change in the environment occurs.

4. Every individual organism cannot be protected by variations, but variations developed in a population due to DNA copying are useful for ensuring the survival of the species. Thus, ideally, variations in the reproduction are helpful for the organisms.

5. It is a known fact that DNA-copying mechanisms are not absolutely accurate. However, these mechanisms are so accurate that they maintain the speed of variation in such a manner that the DNA copies formed adapt to the cellular apparatus and hence do not die.

Question 26.
How is the process of making variants speeded up by sexual mode of reproduction?
Answer:
1. A single cell divides to form two cells and it involves copying of two things,

• DNA and
• Cellular apparatus.

2. Copying of DNA leads to variation. Moreover, the newly formed DNA copy already contains variations accumulated from previous generations.
3. Owing to these two sources of variations, the two different individuals in a population would have quite different patterns of accumulated variations.
4. When variations from two or more individuals combine, it leads to creation of new combinations of variants. Moreover, each combination would be novel, since each would have come from two different individuals.
5. The sexual mode of reproduction incorporates such a process of combining DNA from two different individuals during reproduction.

Question 27.
How in sexually reproducing organisms, the number of chromosomes and the DNA content maintained at a constant level in each generation?
Answer:
1. In humans, both male and female germ-cells have 46 chromosomes each.
2. In sexually reproducing organisms, each new generation is the combination of the DNA copies from its two parents. So, logically, the new generation organism should have double the chromosome (i.e. 46 + 46 = 92) in its body, but it does not work this way.
3. The germ-cells of sexually reproducing organisms undergo meiosis. Meiosis reduces the number of chromosome to half.
4. Under meiosis, when two haploid germ cells from two individuals combine during sexual reproduction they form diploid zygote which makes the number of chromosome and the DNA content same as its previous generation i.e. 46.

Question 28.
How are the germ cells different in higher or say complex organisms?
Answer:
As the body design becomes more complex i.e. in higher organisms, the germ cells become specialized and show some differences. They are listed below.

Question 29.
Draw the diagram of a flower and explain its reproductive parts.
Answer:
1. Flowering plants belong to the group of angiosperms. The reproductive parts of angiosperms are located in the flower. So, we can say that flower is the reproductive organ of the plant. The four main reproductive parts of a flower are:

• Stamen,
• Pistil,
• Petal and
• Sepal

2. Stamen and pistil contain germ-cells (gametes).
(i) Unisexual flower: If the flower contains only one part out of stamen or pistil then such a flower is called unisexual. For example, papaya and watermelon.

(ii) Bisexual flower: If the flower contains both stamen as well as pistil the flower is called bisexual. For example. hibiscus and mustard.

(iii) Male reproductive part: Stamen is the male reproductive part. It produces yellowish coloured pollen grains. Stamen is made up of two parts namely (a) anther and (b) filament.

(iv) Female reproductive part: Pistil (or carpel) is the female reproductive part. It is present at the centre of a flower. It is made of three parts namely, (a) ovary, (b) style and (c) stigma. The swollen bottom part is the ovary, middle elongated part is the style and the terminal part which may be sticky is the stigma. The ovary contains ovules and each ovule contains an egg cell.