Class 10

Haryana State Board HBSE 10th Class Maths Notes Chapter 15 Probability Notes.

Haryana Board 10th Class Maths Notes Chapter 15 Probability

Introduction
In class IX, we have learnt about experimental (or empirical) probabilities of events which were bases on the results of actual experiments.
Suppose we toss a coin 1000 times and get head say, 430 times and tail 570 times. Then we would say that in a single throw of a coin the probability of getting a head is \(\frac{430}{1000}\) i.e, 0.430 and getting a tail is \(\frac{570}{1000}\) i.e, 0.570. These probabilities are based on the results of an actual experiment of tossing a coin 1000 times. For this reason, these are called experimental or empirical probabilities.

Let us recall some of the basic terms and results that we have studied in the class IX along with a few new ones which are normally used in study of probability.
1. Experiment: Any process that yields a result or an observation is called experiment.
2. Trial: Performing an experiment once is called a trial.
3. Outcome: A particular result of an experiment is called an outcome.
4. Sample space: The set of all possible outcomes of an experiment is called sample space. The individual outcomes in a sample space are called sample points.
e.g. one toss of a coin results in the outcomes (H. T). If the coin in fair, then each out come is equally likely. Two tosses of a coin results in the outcomes (H H, H T, T H, T T).

5. Event: Any subset of the sample space is called an event. If A is an event, then n (A) is the number of same points that belongs to A. e.g. consider the experiment of tossing a die here S = (1, 2, 3, 4, 5, 6)
If A is the event that an odd number occurs, then A = {1, 3, 5}
If E is the even number greater than 4 occurs then E = {6}
6. Equally likely events: If one event cannot be expected in preference to other event then they are said to be equally likely.
For example when we throw a die once, each of the numbers (1, 2, 3, 4, 5, 6) has the of showing up. So, 1, 2, 3, 4, 5, 6 are outcomes of throwing a die.
7. Impossible Event: An event which cannot occur is called an impossible event. eg. throwing a die, 7 will never comes up. So, getting 7 is an impossible event. The probability of an impossibe event is zero.
8. Sure Event: An event which is certain to occur is called a sure event eg. in a single throwing of die, the event to get a number less than 7 is a sure event. The probability of a sure event is 1.
9. Complimentary event: If E denotes the happening of an event and ‘not E’ its not happening the event, then E and “not E” are the complimentary events.
∴ P(E) + P(not E) = 1 P(E) = 1 – P (not E)

10. Probability: The probability of an event A, denoted by P(A), is a measure of the possibility of the event occuring as the result of an experiment.
11. Empirical probability: The probability that a fair die will show a four when thrown is \(\frac{1}{6}\), using an argument based on equally likely outcomes.
12. Theoretical probability: It is P(E) of an event is the fraction of times we expect E to occur.
13. Elementary Event: An event having only one outcome is called an elementary event.
14. Random Experiments: The experiments which have not fixed results are called random experiments.
15. Favourable outcome: The possible outcomes for a given event are called favourable outcomes.
16. Die (Dice): A small cube with its faces numbered from 1 to 6. When the die is thrown, the probability that any particular number from 1 to 6 is obtained on the face landing upper most is \(\frac{1}{6}\).
17. At least: As much as.
18. At most: Not more than.

Probability-A Theoretical Approach
In mathematics probability is the numerical value assigned to the likelihood that a particular event will take place. For instance, if we throw an unbiased die, we have equal chances of scoring any of the numbers 1, 2, 3, 4, 5, and 6. Since there is one chance in six of throwing a 3, the probability of the event occuring is said to be \(\frac{1}{6}\). Similarly when tossing a coin the probability that it lands head is consider to be \(\frac{1}{2}\).

Theoretical probability (also called classical probability) of an event E, written as P(E), is defined as
P(E) \(=\frac{\text { Number of outcomes favourable to } \mathrm{E}}{\text { Number of all possible outcomes of the experiment }}\)

Remark: The value of probability of an event cannot be negative or greater than 1.
Some information related to the playing cards

• A deck of playing cards has in all 52 cards.
• 52 cards divided into 4 suits (spades, clubs, hearts and diamonds). Each suit has 13 cards.
• Cards of heart and diamond are red cards.
• Cards of spades and clubs are black cards.
• King, Queen and Jack are called face cards. Thus, there are in all 12 face cards.
• The total number of non face card is 52 – 12 = 40.

Haryana State Board HBSE 10th Class Maths Notes Chapter 14 Statistics Notes.

Haryana Board 10th Class Maths Notes Chapter 14 Statistics

Introduction
In the earlier classes, we have already learnt about the representation of given data into ungrouped as well as grouped frequency distributions and its representation through various graphs such as bar graphs, histograms, frequency polygons etc and you have also studied about measures of central tendency such as mean, median and mode of an ungrouped data. In this chapter, we shall learn how to calculate mean, median and mode for the grouped data. We shall also discuss the concept of cumulative frequency distribution and learn how to draw cumulative frequency curves, called Ogives.

1. Statistics: The branch of mathematics in which we study to extract meaningful information from the collected data. It is the area of study dealing with the presentation, analysis and interpretation of the data. It seems to have been derived from the Latin word ‘status’ or German word ‘statistik’ or Italian word ‘statista’.
2. Data: The facts or figures, which are numerical or otherwise, collected with a definite purpose are called data. Data is the plural form of the Latin word datum.
3. Observation: Every factor figure of the data is called an observation.
4. Frequency: The number of times a particular observation occurs is called the frequency of the observation.
5. Grouped frequency distribution: If the data are very large and the range is large, we put the data in groups of suitable size and mention the frequency of each group. Such a distribution is called grouped frequency distribution.

6. An inclusive frequency distribution: The upper limit of one class does not coincide with the lower limit of the next class. Such as, 1 – 10, 11 – 20, ……… is known as an inclusive frequency distribution.
7. An exclusive frequency distribution: The upper limit of one class coincides with the lower limit of the next class such as 1 – 10, 10 – 20, ……… is known as an exclusive frequency distribution.
8. Measures of Central Tendency: The numerical expressions which represent the characteristic of a group are called Measures of Central Tendency or Average. Mean, Median and Mode are three measures of central tendency (averages).
9. Class interval: Each group into which the raw data is condensed is called a class interval. Each class is bounded by two figures, which are called the class limits. The figures on the left side of the classes are called lower limits while figures on the right are known as upper limits.
10. Class size: The difference between the true upper limit and true lower limit of a class is called its class size.
11. Class Mark: The class mark of the class interval is the value midway between its true lower limit and true upper limit.
Class mark of a class = \(\frac{\text { True upper limit + True lower limit }}{2}\)
12. Cumulative Frequency: The cumulative frequency of a class interval is the sum of frequencies of all classes up to that class (including the frequency of that particular class).
13. Mean of grouped data: We know that if x₁, x₂, x₃, ………., x_n be n observations with respective frequencies f₁, f₂, f₃, …….., f_n, their mean is given by
\(\bar{x}=\frac{f_1 x_1+f_2 x_2+f_3 x_3+\ldots \ldots+f_n x_n}{f_1+f_2+f_3+\ldots \ldots+f_n}\)
We can write this in short form
\(\bar{x}=\frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n f_i}\)
It is more briefly written as \(\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}\), It is understood that varies from 1 to n. The Greek letter ‘Σ’ (capital sigma) is particularly used for writing summations.
With this assumption we can have the following three methods to calculate the mean of grouped data.

(a) Direct Method
1. For each class, find the class mark x_i, as
\(x_i=\frac{\text { lower limit }+\text { upper limit }}{2}\)
2. Find the product of each x_i with the corresponding fi, and find the algebraic sum of these products, i.e. Σf_ix_i.
3. Find the sum of all the frequencies i.e., Σf_i
4. Calculate the value of \(\bar{x}\), using the formula.
\(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)

(b) Assumed Mean Method (Shortcut Method)
In case the values of the variable are very large in magnitude ie., the values of xi are very large in magnitude then computation of the mean \(\bar{x}\) becomes rather tedious and lengthy. To make calculation easier we use assumed mean method to find \(\bar{x}\).

Here, we choose an arbitrary constant a, also called assumed mean and subtract it from each of the value x_i. The reduced value d_i = x_i – a is called the deviation of x from a.

While using this method, we go through the following steps:
1. For each class interval find class mark x_i, as x_i = \(\frac{1}{2}\)(lower limit + upper limit)
2. Assume a suitable value of x_i in the middle of x_i‘ s as the assumed mean.
3. Find out the deviations of the mid value of each from the assumed mean (d_i = x_i – a).
4. Calculate the product of deviation (d_i) with corresponding frequency (f_i) for each class.
5. Find the algebraic sum of these products ie., Σf_id_i.
6. Find the sum of all the frequencies ie., Σf_i
7. Calculate the value of \(\bar{x}\), using the formula
\(\bar{x}\) = \(a+\frac{\sum f_i d_i}{\sum f_i}\)

(c) Step Deviation Method
The shortcut method discussed is further simplified or calculations are reduced to a great extent by adopting step deviation method. Scaling down the deviation (from the assumed mean) by a step (further dividing by a common factor), will reduce the calculation to a minimum.

Here we choose an arbitrary constant a (also called assumed mean) and subtract it from each of the value x_i. The reduced value (x_i – a) is called the deviation of x_i from ‘a’. These deviations are then divided by constant h, where h is the suitable divisor of all the d_i‘s.

In this method; we go through the following steps:
1. For each class interval, calculate the class mark x_i by using the formula,
x_i = \(\frac{1}{2}\)(lower limit + upper limit)
2. Choose the assumed mean ‘a’ in the middle of x_i.
3. Calculate the values of d_i (d_i = x_i – a)
4. Calculate the values of u_i {u_i = \(\frac{x_i-a}{h}\), where h is the class width}
5. Find the product of each u_i with the corresponding f_i.
6. Find the algebraic sum of these products i.e. Σf_iu_i
7. Find the sum of all the frequencies i.e., Σf_i.
8. Calculate the mean \(\bar{x}\) by using formula.
\(\bar{x}\) = \(a+\left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) h\)

Mode of Grouped Data
Recall from class IX that the mode of statistical data is the value among the observations which occurs most frequently. In other words, mode of a statistical data is the value of the observations which has maximum frequency.

In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode of grouped data is a value inside the modal class, and is given by the formula.
Mode = \(l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h\)
Where,
l = Lower limit of the modal class
f₁ = Frequency of the modal class
f₀ = Frequency of the class preceding the modal class
f₂ = Frequency of the class succeeding the modal class
h = Size of the class interval (assuming all class sizes to be equal)

Remark: In some cases, It is possible that more than one value may have the same maximum frequency. In such a case the data is said to be multimodal. Though grouped data can also be multimodal, we shall restrict ourselves to unimodal data only i.e. the data having a single mode.

Median of Grouped Data
Median of a distribution is the value of the middle-most observation which divides it exactly in two equal parts when the data are arranged in ascending (or descending order).

(a) Median of an ungrouped data:
Arrange the data in ascending or descending order. Let the total number of observations be n
(i) If n is odd, the median is the value of the \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation.
(ii) If n is even, the median is mean of the \(\left(\frac{n}{2}\right)^{\text {th }}\) and \(\left(\frac{n}{2}+1\right)^{\text {th }}\) observations.

(b) Median of discrete series:
Arrange the terms in ascending or descending order. Then prepare a cumulative frequency distribution table. Let the total frequency be n
(i) If n is odd, then median = size of the \(\left(\frac{n+1}{2}\right)^{\text {th }}\) term.
(ii) If n is even then
median = \(\frac{1}{2}\)[size of the \(\left(\frac{n}{2}\right)^{\text {th }}\) term + size of the \(\left(\frac{n}{2}+1\right)^{\text {th }}\) term]

(c) Cumulative frequency distribution:
The frequencies are expressed as cumulative total against the class intervals in cumulative frequency distribution. It is of two types:
For example
(i) Less than type :

(ii) More than type :

(d) Median of grouped or continuous frequency distribution:
In order to calculate the median of the grouped data or continuous frequency distribution, we go through the following ahead steps:
(1) Prepare a cumulative frequency distribution and obtain n = Σf_i
(2) Find \(\frac{n}{2}\)
(3) Locate the class whose cumulative frequency is greater than (and nearest to) \(\frac{n}{2}\). This class is the median class.
(4) Calculate the median using the formula given by:
Median = \(l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h\)
Where l = lower limit of median class
n = number of observations
c_f = Cumulative frequency of class preceding the median class
f = frequency of median class
h = class size (assuming class size to be equal)

(e) Empirical relation between mean, median and mode.
3 Median = Mode + 2 Mean

Graphical Representation of Cumulative Frequency Distribution
In class IX, we have learnt about representing statistical data by using bar graphs, histograms and frequency polygons. In this section, we shall learn about to represent cumulative frequency distribution through cumulative frequency curves or ogives.

As we already know that cumulative frequency distribution are of two types, namely, less than type and more than type, accordingly there are two types of cumulatives frequency curves (or ogives).
(a) Less than ogive:
To draw a less than ogive, we go through the following steps:

• Prepare a less than cumulative frequency distribution from the given ordinary frequency distribution.
• Mark the upper class limits along the x-axis choosing a suitable scale.
• Mark the cumulative frequencies along the y-axis choosing a suitable scale.
• On joining these points successively by a free hand smooth curve, we get a cumulative frequency curve or an ogive (of less than type).

(b) More than ogive:
To draw a more than ogive, we go through the following steps:

• Prepare a more than frequency distribution from the given ordinary frequency distribution.
• Mark the lower class limits along the x-axis choosing a suitable scale.
• Mark the cumulative frequencies along the y-axis choosing a suitable scale.
• On joining these points successively by free hand smooth curve, we get a cumulative frequency curve or an ogive of more than type.

Remark 1: Draw any one of the two types of ogives for the given distribution. Take a point P(0, \(\frac{n}{2}\)) on the y-axis and draw PM || x-axis cutting the above curve at a point M. Again draw MN perpendicular to x-axis, cutting the x-axis at point N. Then median = x co-ordinate of point N.

Remark 2: Draw both types of ogives i.e., less than type and more than type for the given distribution on the same graph paper. Mark A as the point of intersection of these two ogives. Draw AP perpendicular to x-axis, cutting x-axis at P. Then median = x co-ordinate of point P.

Remark 3: For drawing ogive, it should be ensured that the class intervals are continuous.

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 11 Constructions Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 11 Constructions

Short/Long Answer Type Questions

Question 1.
Construct a triangle with sides 4 cm, 5 cm and 7 cm and then another triangle whose sides are \(\frac {3}{4}\) of the corresponding sides of the first triangle.
Solution :
Steps of Construction :
1. Draw a line segment BC = 5 cm.
2. With B as centre and radius 4 cm draw and arc.
3. With C as centre and radius 7 cm draw another are to intersect the previous are at A.
4. Join AB and AC to get ΔABC
5. Draw any ray BX making an acute angle with BC on the opposite to the vertex A.

6. Along BX mark 4 points B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
7. Join B₄C
8. From B₃ draw B₃C’ || B₄C meeting BC at
9. From C’ draw A’C’ || AC meeting AB at A’. Then A’BC’ is required triangle, each of whose side is \(\frac {3}{4}\) of corresponding sides of ΔABC.

Justification: Since
A’C’ || AC
Therefore, A’BC’ ~ ΔABC
\(\frac {A’B}{AB}\) = \(\frac {BC’}{BC}\) = \(\frac {A’C’}{AC}\) = \(\frac {3}{4}\)

Question 2.
Construct a triangle with sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac {3}{5}\) times of the corresponding sides of the given triangle.
Solution:
Steps of Construction:
1. Draw a line segment QR = 4 cm.

2. With Q as centre and radius 5 cm draw an
3. With R as centre and radius 6 cm draw another are to intersect the previous arc at P.
4. Join PQ and PR to get ΔPQR
5. Draw any ray QX making an acute angle with QR on the opposite to the vertex P.
6. Along QX mark 5 points Q₁, Q₂, Q₃, Q₄ and Q₅ such that QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₄Q₅
7. Join Q₅R
8. From Q₃ draw Q₃R’ || Q₅R meeting QR at R’
9. From R’ draw R’P’ || PR. Then P’ QR’ is required triangle.
Justification: Since
P’R’ || PR
Therefore, P’QR’ ~ ΔPQR
\(\frac {P’Q}{PQ}\) = \(\frac {QR’}{QR}\) = \(\frac {P’R’}{PR}\) = \(\frac {3}{5}\)

Question 3.
Construct a ΔABC with AB = 6 cm, BC = 5 cm and ∠B = 60°. Now construct another triangle whose sides are \(\frac {2}{3}\) times the corresponding sides of ΔABC.
Solution :
Steps of Construction :
1. Draw a line segment BC = 5 cm.
2. At B, draw ∠CBY = 60°
3. From B, draw and are AB = 6 cm meeting by at A

4. Join AC. Thus, ΔABC obtained
5. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
6. Along BX mark 3 points B₁, B₂, and B₃ such that BB₁ – B₁B₂ = B₂B₃
7. Join B₃C
8. From B₂ draw B₂C’ || B₃C meeting BC at C
9. From C’ draw A’C’ || AC meeting AB at A’. Then A’BC’ is required triangle, each of whose side is  \(\frac {2}{3}\) of corresponding sides of ΔABC

Justification: Since
A’C’ || AC
Therefore, A’BC’ in ΔABC
\(\frac {A’B}{AB}\) = \(\frac {BC’}{BC}\) = \(\frac {A’C’}{AC}\) = \(\frac {2}{3}\)

Question 4.
Construct a ΔABC in which CA = 6 cm, AB = 5 cm and ∠BAC = 45°. Then construct a triangle whose sides are \(\frac {3}{5}\) of the corresponding sides of ΔABC.
Solution:
Steps of Construction:
1. Draw a line segment AB = 5 cm.

2. At A, draw ∠BAY = 45°
3. From A draw an arc AC = 6 cm meeting A at C
4. Join BC. Thus, ΔABC obtained.
5. Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.
6. Along AX mark 5 points A₁, A₂, A₃, A₄ and A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅
7. Join A₅B
8. From A₃ draw A₃B’ || A₅B meeting AB at B’
9. From B’ draw B’C’ || BC meeting AC at C’.
Then AB’C’ is required triangle, each of whose side is \(\frac {3}{5}\) of corresponding sides of ΔABC.
Justification: Since
B’C’ || BC
Therefore, AB’C’ ~ ΔABC
\(\frac {AB’}{AB}\) = \(\frac {B’C’}{BC}\) = \(\frac {AC’}{AC}\) = \(\frac {3}{5}\)

Question 5.
Draw a ΔABC in which BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct another triangle whose sides are \(\frac {3}{4}\) times corresponding sides of ΔABC.
Solution:
Steps of Construction:
1. Draw a line segment BC = 7 cm.

2. In ΔABC, ∠B = 45°, 2A = 105°
∴ ∠C = 180° – (45° + 105°) = 30°
At B draw and angle ∠B = 45° and draw ∠C = 30° intersecting each other at A to get ΔABC
3. At B draw any ray BX making an acute angle with BC on the side opposite to the vertex A
4. Along BX mark 4 points B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
5. Join B₄C
6. From B₃ draw B₃C’ || B₄C meeting BC at C’
7. From C’ draw C’A’ || AC meeting AB at A
Then A’BC’ is required triangle, each of whose side is \(\frac {3}{4}\) of corresponding sides of ΔABC.

Justification: Since
A’C’ || AC
Therefore, A’BC’ ~ ΔABC
\(\frac {A’B}{AB}\) = \(\frac {BC’}{BC}\) = \(\frac {A’C’}{AC}\) = \(\frac {3}{4}\)

Question 6.
Construct a triangle ABC with side BC = 6 cm, ∠B = 45°, ∠A = 105°. Then construct another triangle whose sides are \(\frac {3}{4}\) times the corresponding sides of ΔABC.
Solution :
Steps of Construction:
1. Draw a line segment BC = 7 cm.
2. In ΔABC, ∠B = 45°, 2A = 105°
∴ ∠C = 180° – (45° + 105°) = 30°
At B draw and angle ∠B = 45° and draw ∠C = 30° intersecting each other at A to get ΔABC
3. At B draw any ray BX making an acute angle with BC on the side opposite to the vertex A
4. Along BX mark 4 points B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
5. Join B₄C
6. From B₃ draw B₃C’ || B₄C meeting BC at C’
7. From C’ draw C’A’ || AC meeting AB at A
Then A’BC’ is required triangle, each of whose side is \(\frac {3}{4}\) of corresponding sides of ΔABC.

Justification: Since
A’C’ || AC
Therefore, A’BC’ ~ ΔABC
\(\frac {A’B}{AB}\) = \(\frac {BC’}{BC}\) = \(\frac {A’C’}{AC}\) = \(\frac {3}{4}\)

Question 7.
How many tangents can be constructed to any point on the circle of radius 4 cm ?
Solution :
We know that, if a point lies on the circle, then there is only one tangent to the circle at this point. So, one tangent can be constructed to any point on the circle of radius 4 cm.

Question 8.
Draw a circle of radius 3.5 cm. From a point P, 6 cm from its centre, draw two tangents to the circle.
Solution :
Steps of Construction :
1. Draw a circle with O as the centre and radius 3.5 cm
2. Mark a point P outside the circle such that OP = 6 cm.
3. Join OP and draw prependicular bisector of PO meeting PO at M.,
4. Draw a circle with M as the centre and radius equal to PM = OM intersecting the given circle at points Q and R.

5. Join PQ and PR then PQ and PR are required tangents.

Question 9.
Draw a circle of radius 2 cm with centre O and take a point P outside the circle such that OP = 6.5 cm. From P, draw two tangents to the circle.
Solution :
Steps of Construction :
1. Draw a circle with O as the centre and radius 2.0 cm
2. Mark a point P outside the circle such that OP = 6.5 cm.
3. Join OP and draw perpendicular bisector of PO meeting PO at M

4. Draw a circle with M as the centre and radius equal to PM = OM intersecting the given circle at points Q and R.
5. Join PQ and PR then PQ and PR are required tangents.

Question 10.
Draw a pair of tangents to a circle of radius 5 cm, which are inclined to each other at an angle of 70°.
Solution :
Steps of Construction:
1. Draw a circle with centre O and radius 5 cm
2. Construct raddi OA and OB such that ∠AOB = 360° (90° + 90° + 70°) = 110°.
3. Draw AL ⊥ OA at A and BM ⊥ OB at B. They intersect at P.
4. Then PA and PB are the required tangents inclined to each other at 70°.

Justification: PA ⊥ OA and PB ⊥ OB and ∠AOB = 110°
∠APB = 360° – (90° + 90° + 110°)
∠APB = 360° – 290°
∠APB = 70°
Therefore, tangents PA and PB are inclined at 60°

Question 11.
Draw a circle of radius 5 cm. From a point 13 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Also verify the measurement by actual calculation.
Solution:
Steps of Construction:
1. Draw a circle with centre O of radius 5 cm
2. Mark a point A, 13 cm away from the centre.
3. Join AO and bisect it at M.
4. Draw a circle with M as the centre and radius equal to AM intersects the given circle at points P and Q.
5. Join AP and AQ

Then AP and AQ are the required tangents lengths of AP and AQ are 12 cm.

Justification:
Join OP
∴ ∠APO = 90°
[Angle is a semicircle]
AP ⊥ OP
Therefore, AP is a tangent to the given circle. Similarly AQ is a tangent to the given circle.
In right triangle APO, we have
AO² = PO² + AP²
13² = 5² + AP²
169 – 25 = AP²
AP = \(\sqrt{144}\)
AP = 12 cm

Fill in the Blanks

Question 1.
The part of a curve between two given points on the curve is known as……….
Solution :
Curve

Question 2.
……. is a straight line which touches the circle at one point only.
Solution :
tangent

Question 3.
There can be ……. tangents drawn to the circle from point in side it.
Solution :
no

Question 4.
The angle between radius and tangents is …… a angle.
Solution :
right

Question 5.
To draw a ……. at a given point to the circle, We simply draw radius at this point and draw a line perpendicular through this point.
Solution :
tangent.

Multiple Choice Questions

Choose the correct answer in each of the following:

Question 1.
In the given figure, if PA₁ =A₁A₂ = A₂A₃ = A₄Q and A₁B₁ || A₂B₂ || A₃B₄ || A₄B₄ || QR. Then A₂ divides the line segment PR in the ratio:

(a) 2 : 2
(b) 2 : 3
(c) 1 : 2
(d) 3 : 2
Solution :
From fig it is clear that A₂ divides the line segment PR in the ratio 2 : 3.

So correct choice is (b)

Question 2.
In the above figure, A₃ divides the line segment PR in the ratio :
(a) 3 : 5
(b) 2 : 3
(c) 3 : 2
(d) 3 : 6
Solution :
A₃ divides the line segment PR in the ratio 3 : 2
So, correct choice is (c)

Question 3.
In the above figure A, divides the line segment in the ratio :
(a) 1 : 1
(b) 1 : 2
(c) 1 : 4
(d) 1 : 5
Solution :
Here, A₁ divides the line segment in the ratio 1 : 4.
So correct choice is (c).

Question 4.
To divide a line segment AB in the ratio 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A₁, A₂, A₃, … are located at equal distances on the ray AX and the point B is joined to:
(a) A₁₂
(b) A₁₁
(c) A₁₀
(d) A₉
Solution :
4 + 7 = 11
∴ A₁₁
Hence correct choice is (b).

Question 5.
To construct a triangle similar to a given ΔABC with its sides \(\frac {3}{7}\) of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B₁, B₁, B₁, …. on BX at equal distance and next step is to join :
(a) B₁₀ to C
(b) B₃ to C
(c) B₇ to C
(d) B₄ to C
Solution :
B₇ to C
So correct choice is (c).

Question 6.
To divide a line segment AB in the ratio 3 : 4. We draw a ray ∠BAX, so that is an acute angle and then marks the points on the ray AX at equal distances such that the minimum number of these points is :
(a) 3
(b) 7
(c) 4
(d) 12
Solution :
3 + 4 = 7
So correct choice is (b).

Question 7.
To construct a triangle similar to a given ΔABC with its sides \(\frac {8}{5}\) of the corresponding sides of ΔABC draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is :
(a) 5
(b) 8
(c) 13
(d) 3
Solution :
8
so correct choice is (b).

Question 8.
To divide a line segment AB in the ratio 5 : 6, draw a ray AX such that ∠BAX is an acute angle, then draw a ray by parallel to AX and the points A₁, A₂, A₃, and B₁, B₂, B₃, … are located at equal distances on the ray AX and BY, respectively. Then the points joined are :
(a) A₅ and B₆
(b) A₆ and B₅
(c) A₄ and B₅
(d) A₅ and B₄
Solution :
So, the points joined are A₅ and B₆ [By Prop. of similar Δs]

So correct choice is (a).

Haryana State Board HBSE 10th Class Maths Notes Chapter 13 Surface Areas and Volumes Notes.

Haryana Board 10th Class Maths Notes Chapter 13 Surface Areas and Volumes

Introduction
In previous classes, we have learnt about surface areas and volumes of solids like cuboid, right circular cylinder, right circular cone and sphere. In this chapter we shall use our previous knowldege about these solids and learn about such solids which are combinations of two or more solids. For example a test tube which is combination of a cylinder and a hemisphere, an ice cream cone is a combination of a cone and a hemishpere, conical circus tent with cylindrical base is combination of a right circular cylinder and right circular cone etc. We shall find the surface area and volume of such solids.

In this chapter, we shall also discuss problems on surface area and volume of frustum of a cone for example bucket, glass tumbler, a friction clutch etc.

Some important terms are following:
1. Solids: Bodies which have three dimensions in space are called solids.
2. Volume: The amount of space occupied by a solid or bounded by a closed surface is known as the volume of solid.
3. Surface area: Surface area is the total sum of all the areas of all the shapes that cover the surface of solid.
4. Lateral surface area: Lateral surface in a solid is the sum of surface areas of all its faces excluding the bases of solid.
5. Cuboid: A cuboid is a solid bounded by six rectangular plane regions.
6. Cube: When all the edges of cuboid are equal in length, it is called a cube.
7. Right circular cylinder: If a rectangle is revolved about one of its sides, the solid thus formed is called right circular cylinder.
8. Right circular cone: If a right angled triangle is revolved about one of the sides containing a right angle, the solid then generated is called a right circular cone.
9. Sphere: The set of all points in space which are equidistant from a fixed point, is called a sphere.
10. Hemisphere: A plane through the centre of a sphere divides the sphere into equal parts, Each part is called a hemisphere.
11. Spherical shell: A spherical shell is the difference between the two solid concentric spheres.
12. Slant height: The slant height of a right circular cone is the distance measured along the lateral surfaces from any point on the circular base to top apex of the cone.
13. Frustum of a right circular cone: If a right circular cone is cut off by a plane parallel to its base. The portion of the cone between the plane and base of the cone is called a frustum of the cone. The shape of this part (Frustum of cone is like that of a bucket, funnels cans, etc.)

Surface area of a Combination of Solids Cuboid
If l, b and h denote respectively the length, breadth and height of a cuboid.

Then
(i) Volume of cuboid = l × b × h cubic units
= Area of the base × height
(ii) Total surface area of a cuboid = 2(lb + bh + hl) square units
(iii) Surface area of 4 walls of a room = 2(l + b) × h square units
(iv) Surface area of cuboid, in which top face is open = lb + 2(bh + hl) square units
(v) Diagonal of cuboid = \(\sqrt{l^2+b^2+h^2}\) units

Cube:
If all lengths of the edges of a cuboid are equal, then it is called a cube.
Let the length of each edge of cube be “a” units.

(i) Volume of cube = a³ cubic units
(ii) Total surface area of cube = 6a² square units
(iii) Diagonal of cube = \(\sqrt{3}\)a units
(iv) Edge of cube = \(\sqrt[3]{\text { Volume }}\)

Right circular cylinder:
If radius of base is r and height h, then

(i) Volume of cylinder = Area of base × Height
= πr²h cubic units
(ii) Curved surface area or lateral surface area of cylinder
= Perimeter of base × Height
= 2πrh square units
(iii) Total surface area of cylinder = Curved surface area + Area of two circular ends
= 2πrh + 2πr²
= 2πr (h + r) square units

Right circular hollow cylinder:
If R and rare external and internal radii of a right circular hollow cylinder of height h. Then

(i) Area of each end = π (R² – r²)
(ii) Volume of material = Exterior volume – Interior volume
= πR²h – πr²h
= πh (R² – r²) cubic unita
(iii) Curved surface area = External surface area + Internal surface area
= 2πRh + 2πrh
= 2πh (R + r) square units
(iv) Total surface area = curved surface area + Area of base rings
= 2πRh + 2πrh + 2π (R² – r²)
= 2πh(R + r) + 2π(R + r) (R – r)
= 2π (R + r) (h + R – r) square units

Right circular cone:
Consider a cone in which Radius of base = r, Height = h, and slant height (l) = \(\sqrt{h^2+r^2}\). Then,

(i) Volume of the cone = \(\frac{1}{3}\)πr²h cubic units
= \(\frac{1}{3}\)(Area of buse) × height
(ii) Curved surface area of cone = πrl sq. units
(iii) Total surface of the cone = Curved surface + Area of base
= πrl + πr²
= πr(l + r) sq. units

Sphere:
For a sphere of radius r. we have

(i) Volume of sphere = \(\frac{4}{3}\)πr³ cubic units
(ii) Surface area of sphere = 4πr² square units

Hemisphere:
For a hemishpere of radius r.
We have

(i) Volume of hemisphere = \(\frac{2}{3}\)πr³ cubic units
(ii) Curved surface area of hemisphere = 2πr² sq. unita
(iii) Total surface Area of hemisphere = 3πr² sq. units
In our daily life, we come across different solids which are combinations of different solids (cube, right circular cylinder, right circular cone, sphere, hemisphere, etc.).

For example: An ice-cream cone consisting a cone and a hemisphere, a circus tent which is combination of right circular cylinder and a right circular cone. A capsule is the form of a cylinder with hemispherical ends etc. Following examples will illustrate the method of finding surface areas and volumes of such combinations of solids.

Volume of a combination of solids
In the previous section, we have discussed how to find the surface area of solids made up of a combination of two basic solids. In this section, we shall explain the method of finding volumes of solids which are made by the combination of basic solids. The volume of the solid formed by joining two basic solids will actually be the sum of the volumes of the constituents, as we see in examples given below.

Conversion of solid from one shape to another
At times we may convert objects from one shape to another objects with different shapes or when a liquid which originally filled one container of a particular shape is poured into another container of a different shape or size.

For examples: A metallic sphere melted and recast into smaller cones. Spherical bullets made out of the solid cube. A sphere may be melted and recast into a wire, etc. For solving such types of problems the surface area will be change only and no change of volume takes place. Let us consider some examples to clear the method of conversion of solids from one shape to another.

Frustum of a right circular cone
If a right circular cone is cut off by a plane parallel to its base, the portion of the cone between the plane and base of the cone is called a frustum the cone. The shape of this part (Frustum of cone is like that of a bucket, funnels, flower pots, cans, jugs, lampshades etc.)

We consider a cone (VAB) which cuts by a plane CD parallel to the base AB through some point (0) on its axis and the portion containing the vertex (V) is removed. The left portion ACDB is usually called frustum of the cone.
A frustum of a right circular cone has two unequal flat circular bases and a curved surface.

The line segment OP joining the centres of two bases called the height of the frustum. Each of line segment AC and BD of frustum ACDB is called its slant height.

Volume and surface area of frustum
Let r₁ and r₂ be the radii of the circular end bases (r₁ > r₂) of the frustum ABCD of the cone (VAB). Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l. The frustum of right circular cone can be viewed as the difference of the two right circular cone VAB and VCD. Let the height of cone VAB be h₁ and its slant height be l₁, i.e, VP = h₁ and VA = VB = l₁
Height of the cone (VCD) = VP – OP = h₁ – h
Right ΔVOD ~ right ΔVPB (By AA similarity)




Volume of the frustum ACDB of the cone (VAB)
= Volume of cone (VAB) – Volume of the cone (VCD)

\(\frac{1}{3}\)πh(r₁² + r₁r₂ + r₂²)
Hence, volume of the frustum of cone = \(\frac{1}{3}\)πh(r₁² + r₁r₂ + r₂²).
Again if A₁ and A₂ (A₁ > A₂) are the surface areas of the two circular bases. Then
A₁ = πr₁² and A₂ = πr²
Volume of the frustum of cone
= \(\frac{1}{3}\)πh(r₁² + r₂² + r₁r₂)
= \(\frac{h}{3}\) (πr₁² + πr₂² + \(\left.\sqrt{\pi r_1^2} \times \sqrt{\pi r_2^2}\right)\)}
Hence, volume of frustum of cone
= \(\frac{h}{3}\)(A₁ + A₂ + \(\sqrt{\mathrm{A}_1 \mathrm{~A}_2}\))
Now form right ΔDEB,
DB² = DE² + BE².
l² = h² + (r₁ – r₂)
[where BE = r₁ – r₂]
⇒ l = \(\sqrt{h^2+\left(r_1-r_2\right)^2}\)
Again ΔVOD ~ ΔVPB (By AA similarity of criterion)

⇒ \(l_1-l=\frac{l r_1-l r_1+l r_2}{r_1-r_2}\)
⇒ \(l_1-l=\frac{l r_2}{r_1-r_2}\) ……..(4)
Curved surface of the furstum of cone

Hence, curved surface of the frustum of right circular cone
= πl(r₁ + r₂)
where l = \(\sqrt{h^2+\left(r_1-r_2\right)^2}\)
Total surface area of frustum of right circular cone = curved surface area + area of base 1 + area of base 2
= πl(r₁ + r₂) + πr₁²+ πr₂²
= π[r₁² + r₂² + l (r₁ + r₁)]
Hence, total surface area of frustum of right circular cone = π[r₁² + r₂² + l(r₁ + r₂)].

Haryana State Board HBSE 10th Class Science Important Questions Chapter 6 Life Processes Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 6 Life Processes

Question 1.
Only what is visible is alive does not hold true for a life. Explain.
Answer:
1. When we see an animal moving we call it alive. Similarly when we see a green plant we call it alive.
2. Actions that we can see such as running, breathing, laughing, changing colours of leaves, etc. also make us believe that the organism is alive.
3. So, we believe some sort of movement, either growth-related or not as an evidence of being alive.
4. But, a plant which we cannot see growing and animals that breathe without showing any movement are also alive. Hence, it would be wrong to say that what is visible is only alive.

Question 2.
How one can differentiate living organisms and non-living things? What are the exceptions here?
Answer:
1. Movement is one of the basic criterions to identify life. All living organisms show movement without any external help.
2. Animals show clear visible movements.
3. Plants show invisible movements inside their body (i.e., movements of various elements and compounds). External movement of plants is quite minute and very slow and so it is difficult to notice.
4. Viruses are an exception here.

Question 3.
What do you mean by molecular movement? Explain.
Answer:
1. The movements of molecules within living organisms which we cannot see through our naked eyes but are critical for carrying life processes such as photosynthesis, respiration, digestion, excretion, etc. are called molecular movements.
2. Molecular movement includes movement of molecules such as oxygen for respiration, movement of compounds such as enzymes, nutrients, hormones, etc.

Need of molecular movement:

• Living organisms are highly complex, but very well organized structures. Going down they are made of organs, tissues, cells and cell organelles.
• Therefore it is necessary that molecules move and reach up to cell level so that functions such as growth, maintenance and survival are maintained.
• For example, oxygen inhaled move throughout the body via, the process of breathing.

Question 4.
What is the connecting link between living organisms and non-living things? Why?
Answer:
Viruses are the connecting link between living organisms and non-living things.

Reason:
1. It is said that molecular movement is necessary in living organisms. Non-living things do not show any molecular movement.
2. In a free environment, viruses do not show any molecular movement. But, when they come in the contact of the host cells they start acting as living organisms in the form of obligate parasite.
3. Thus, viruses on one side do not show molecular movement i.e. behave as non-living things, but under favourable environment they start showing molecular movement and behave like living organisms. Hence, viruses are called a connecting link between living organisms and non-living things.

Question 5.
What are life processes? State the important life processes that perform the job of maintaining these processes.
Answer:
Life processes:
1. All living organisms perform certain important functions to maintain their survival. These main functions are called life processes.
2. In other words, the processes which together perform the job of maintenance of the body are known as life processes.
3. Maintenance process requires energy. The energy is used for—

• Growth and maintenance of organisms and
• Preventing break-down.

The energy needed to the organism comes from external sources such as food, sun, atmosphere, etc.

The main life processes are:
(1) Nutrition,
(2) Respiration,
(3) Transportation,
(4) Excretion,
(5) Control and co-ordination,
(6) Movement and
(7) Reproduction.

Question 6.
Define and explain each life process very briefly.
Answer:
1. Nutrition:

• The process of transferring a source of energy which we call food, from outside the body of the organism to the inside is called nutrition.
• The food consumed is then converted into smaller and smaller units so that it can be absorbed by the body.
• Most of the food sources are carbon-based. Depending upon the complexity of these carbon sources, different organisms can then use different kind of nutritional processes.

2. Respiration:

• The process of acquiring oxygen from outside the body, and it in the process of break-down of food-sources for cellular needs, is known as respiration.
• Through respiration, organisms break down food and release energy.

3. Transportation:

• The process through which absorbed substances are transported to various parts of the body is called transportation.
• Transportation system is important to carry food and oxygen from one place to another.
• This system is also important for carrying waste products from body cells to excretory organs.

4. Excretion:
Waste materials produced in various cells of the body are removed by the process of excretion.

5. Control and co-ordination (Chapter 7):
Control and co-ordination make the living organisms adapt to the changing environment and survive there in.

6. Movement (Chapter 7):
The process of movement makes the living organism move from one place to another.

7. Reproduction (Chapter 8):
Reproduction involves multiplication of existing organisms. This enables them to maintain the existence of their species on the earth.)

Question 7.
What are nutrients? Give examples.
Answer:
1. The substances obtained by an organism from the surroundings and used them as a source of energy are called nutrients.
2. For example, carbohydrates, lipids, proteins, mineral, salts, etc.

Question 8.
Why a unicellular organism does not need specific organs for nutrition, respiration and transportation?
Answer:
In unicellular organisms, entire body surface of the organism is in the direct contact with the environment and transfer can take place through entire body. So, a unicellular organism does not need specific organs for nutrition, respiration and transportation.

Question 9.
What is nutrition? Classify the types of nutrition.
Answer:
1. Nutrition:

• The process of transferring a source of energy which we call food, from outside the body of the organism to the inside is called nutrition.
• The food consumed is then converted into smaller and smaller units so that it can be absorbed by the body.
• Most of the food sources are carbon-based. Depending upon the complexity of these carbon sources, different organisms can then use different kind of nutritional processes.

2. Respiration:

• The process of acquiring oxygen from outside the body, and it in the process of break-down of food-sources for cellular needs, is known as respiration.
• Through respiration, organisms break down food and release energy.

3. Transportation:

• The process through which absorbed substances are transported to various parts of the body is called transportation.
• Transportation system is important to carry food and oxygen from one place to another.
• This system is also important for carrying waste products from body cells to excretory organs.

4. Excretion:
Waste materials produced in various cells of the body are removed by the process of excretion.

5. Control and co-ordination (Chapter 7):
Control and co-ordination make the living organisms adapt to the changing environment and survive there in.

6. Movement (Chapter 7):
The process of movement makes the living organism move from one place to another.

7. Reproduction (Chapter 8):
Reproduction involves multiplication of existing organisms. This enables them to maintain the existence of their species on the earth.)

Types of nutrition:
(1) Autotrophic nutrition,
(2) Heterotrophic nutrition:

• Saprophytic nutrition
• Parasitic nutrition
• Holozoic nutrition

Question 10.
Define autotrophic nutrition and give examples of autotrophes.
Answer:
Autotrophic nutrition:
1. ‘Auto’ means ‘Self’ and ‘Trophe’ means ‘Nutrition’. So, the mode of ‘nutrition’ in which the organism ‘itself’ synthesizes its own food from the simple inorganic materials like carbon dioxide and water with the help of sunÍight, is called autotrophic nutrition.
2. All green plants and some bacteria show autotrophic mode of nutrition.

Question 11.
How do autotrophs fulfill their need of carbon and energy?
Answer:
1. Carbon dioxide (CO₂), (2) Water (H₂O) and (3) Nutrients are the main food materials that the autotrophic organisms need for their growth and maintenance. These materials fulfill the carbon and energy requirement of the autotrophs.
2. Autotrophs obtain these materials through the process called photosynthesis.
3. During photosynthesis plants take up carbon dioxide from the atmosphere, water from the ground and in the presence of sunlight convert these materials into carbohydrates. The carbohydrate then provides energy to the plants.
4. The excess carbohydrate remains stored as starch within the plants.
5. Autotrophs fulfill their needs of minerals such as nitrogen, phosphorus, iron and magnesium from the soil.

Question 12.
Write a note on photosynthesis. OR Define photosynthesis and enlist Its main events.
Answer:
Photosynthesis:

• The process by which the green plants make their own food by converting carbon dioxide and water into carbohydrates in the presence of sunlight and chlorophyll is called photosynthesis.
• Thus, the basic materials used in photosynthesis are carbon dioxide and water.

Chemical equation of photosynthesis:

Following events take place during photosynthesis:
(1) Absorption of light energy by chlorophyll.
(2) Conversion of light energy to chemical energy and splitting of water molecule into hydrogen and oxygen.
(3) Reduction of carbon dioxide to carbohydrate.

Question 13.
Is it necessary that all events of photosynthesis take place one after the other Immediately?
Answer:
Give an example to explain.
No, it is not necessary that all the events of photosynthesis occur one after the other.

Example:
1. The desert plants absorb CO₂ at night from the surrounding and repare the intermediate compound before the sun rises.
2. This intermediate compound is then acted upon after the chlorophyll absorbs solar radiation in the morning.

Question 14.
How do plants obtain carbon dioxide (CO₂) from the atmosphere?
Answer:
A large number of pores are present on the leaves of plants. These pores are known as stomata. These stomata are responsible for gaseous exchange.
1. During the gaseous exchange, a large amount of water is lost. To control the water loss, the stomata is surrounded by guard cells.
2. The guard cells control the opening and closing of the stomatal pore according to the need of CO₂ in the plant.
3. As the need of CO₂ arises, the guard cells absorbs water and so due to turgidity (swelling up) the pores get opened.
4. When the need gets satisfied, the guard cells lose the water. So, they tend to shrink. As a result, the pores are closed.

Question 15.
Draw a labeled diagram of cross-section of a leaf.
Answer:

Question 16.
How do plants obtain water (H₂O)?
Answer:
1. Roots of plants have specialized system to absorb underground or land water.
2. The water which is absorbed by roots reaches to the organs of the plants through xylem tissue.
3. In addition to this, the entire surface of plants is also able to absorb water Generally, plants absorb water through a process called osmosis.

Question 17.
State important functions performed by roots of terrestrial plants.
Answer:
1. In terrestnal plants, the root plays a crucial role of absorbing water and salts dissolved in the land water.
2. The root system also absorbs salts of nitrogen, phosphorus iron, magnesium, etc. from the soil.

Question 18.
What is the importance of nitrogen in plants? In which form do plants absorb nitrogen?
Answer
1. In plants, nitrogen is important for protein synthesis as well as for the production of other important substances.
2. Generally, plants absorb nitrogen in form of nitrates [N0O₃] or nitrite [NO₂] compounds.
3. Sometimes, the plants also take up atmospheric nitrogen prepared by symbiotic bacteria, in the form of organic compounds.

Question 19.
How is chlorophyll important for photosynthesis?
Answer:
1. Photosynthesis occurs in the chloroplast of a plant cell.
2. The chloroplast contains a pigment called chlorophyll to trap light energy.
3. The light energy of the sun is absorbed by the chlorophyll in the form of photons.

Question 20.
Define heterotrophic nutrition, and explain it.
Answer:
1. The term, ‘Hetero’ means ‘others’ and ‘Trophe’ means ‘nutrition’.
2. The mode of nutrition in which the living organisms cannot synthesis their own food material from the simple inorganic material they have and hence have to depend on other organisms for food is known as heterotrophic nutrition.

Question 21.
Describe the types of heterotrophic organisms on the basis of the type of food they consume.

Answer:
(a) Herbivores: These organisms feed on plants.
– Generally, the food source for organisms is stationary. For example, grass and plants. Herbivores such as cow and goat have to go to these sources to obtain nutrition.

(b) Carnivores: These organisms feed on other animals.

• The food source is mobile. For example, a lion has to hunt deer which is a moving source of nutrition.

Example: Frog, tiger, lizard, etc.

(c) Omnivores: These organisms feed on plants as well as animals.
Example: human beings, dog, ant, etc.

(d) Scavengers: These organisms feed on dead materials.
Example: Vulture

Question 22.
Describe the type of heterotrophic organisms on the basis of the mode of feeding. OR Discuss the strategies used to take up the nutrition.

(a) Saprophytic nutrition:

• Some organisms break down the complex food material outside their bodies and then absorb it. Such a mode of nutrition is called saprophytic nutrition.
• Saprophytes absorb soluble organic nutrients from the dead parts of the animals, plants, dung, etc.
  Example: Bacteria, fungi, yeast, mushrooms.

(b) Holozoic nutrition:
Some organisms undertake either a part or whole of animals or plants and then break down such sources inside their bodies. Such nutrition is known as holozoic nutrition. Example: Human beings, dog.

(c) Parasitic nutrition:
The organisms that live inside or outside of other organisms and obtain nutrition from them are known as parasites. The mode of their nutrition is called parasitic nutrition.

Example: Several bacteria, lice, tape warm, ascaris, cuscuta (plant).
Parasitic organisms derive nutrition from plants or animals without killing them.

Question 23.
Describe the process of nutrition in amoeba with the help of a figure.
Answer:
Amoeba is a unicellular organism. Its mode of nutrition is holozoic.
The various processes involved in nutrition are as follows :

(a) Ingestion: When amoeba comes in contact with food particle, it ingests the food by forming temporary finger-like extensions of the cell surface called pseudopodia.

• The food is then encaptured into a bag called food vacuole.

(b) Digestion: The complex food particles are broken down into simpler forms and digested in food vacuole with the help of digestive enzymes.
(c) Absorption: The digested food is absorbed directly from food vacuole into cytoplasm by the process of diffusion.
(d) Egestion: The undigested food is moved to the surface of the cell and thrown out.

Question 24.
Differentiate between autotrophic and heterotrophic nutrition.
Answer:

Question 25.
Describe the process of nutrition In paramoecium.
Answer:
1. Paramoecium is a unicellular organism whose cell has a definite shape. There is a specific spot in its cell. The entire body is covered with cilia which helps in bring the food to this specific spot.
2. The ingested (= entered) food material is then digested by paramoecium.

Question 26.
Differentiate between saprophytic and parasitic nutrition.
Answer:

Question 27.
Define digestion. What are the main functions of the human digestive system?
Answer:
In animals, the process of breaking-down the large and complex food material into small and simple absorbable molecules is known as digestion. The main functions of human digestive system are:

• Ingestion: To take in food
• Digestion: Converting food into small, simple and absorbable molecules
• Absorption: To absorb the digested food
• Egestion: To remove the undigested food from the body

Question 28.
Draw a labeled diagram of human digestive system and list out the main parts associated with the process of digestion.
Answer:
Human digestive system:
1. The function of digestive system is to break down larger molecules of food into smaller forms so that they can be absorbed easily into the body.
2. The human digestive system consists of alimentary canal and its associated glands.

The main organs of the human digestive system are:
Mouth, oesophagus, stomach, small intestine and large intestine.

The main glands associated with digestion are:
Salivary gland, liver and pancreas.

Question 29.
Explain the process of digestion in humans.
Answer:
Process of digestion In humans:
(1) Buccal cavity

(a) Mouth:

• Food is ingested through mouth.
• Humans put food in the mouth through their hands and the digestion starts simultaneously.
• The mouth (buccal) cavity contains teeth, salivary glands and tongue.

(b) Teeth:

• Teeth cut the food into small pieces and chew and grind it.

(c) Salivary gland:

• The salivary gland secretes saliva ¡n our mouth. It is a watery liquid and so it wets (lubricates) the food in mouth. It is easy to swallow wet food.
• The salivary gland also secretes an enzyme called amylase. Amylase breaks the complex molecule called starch present in the food into sugar.

(d) Tongue:

• The tongue does the job of properly mixing the food with saliva.
• The food remains for a very short time in the mouth and hence only a part of food gets digested here.

(2) Oesophagus:

• The partly digested food comes down to oesophagus or say food-pipe.
• The main function of the oesophagus is to push the food down to stomach.

(3) Stomach:

• Stomach is a large organ which expands when the food enters it. Moreover, the stomach releases certain juices that help in digestion.
• The muscular walls of the stomach churn the food thoroughly with these digestive juices.
• During this process, the food is broken down into smaller pieces and is converted into a semi-solid paste.
• The wall of stomach contains three tubular glands which secrete gastric juice.

The gastric juice contains:
(A) Dilute hydrochloric acid, (B) An enzyme called pepsin and (C) Mucus

(a) Dilute hydrochloric acid:

• Since the stomach releases dilute hydrochloric acid, the digestive juices are acidic in nature.
• The presence of acid enables the enzyme pepsin to digest protein present in the food.
• Therefore, the function of hydrochloric acid is to create an acidic medium in the stomach.
• It also kills the bacteria that enter the stomach through food.

(b) Pepsin:
Pepsin is an enzyme that helps in digesting protein and converting food into smaller molecules.

(c) Mucus:
The mucus prevents the damage that the hydrocholic acki may cause to the inner lining of the stomach.

(4) Small Intestine:

• The partly digested food then moves from stomach to small intestine with the help of sphincter muscles. These muscles release the food in the small intestine in small parts.
• The small intestine is the longest part of the alimentary canal and hence an extensive coiled structure. Owing to this structure it can fit in a very compact space.
• The small intestine is the main site for complete digestion of carbohydrates, proteins and fats. It receives secretions from liver and pancreas for this purpose (whose functions are discussed below).

(a) Function of bile juice:

• Liver secretes bile juice.
• Bile is alkaline in nature and so it converts the acidic food coming from stomach into alkaline form so that enzymes of pancreas can act on it.
• In addition, the bile salts break the fats present in the food into small globules (droplets), which makes it easy for the enzymes to act and digest them.

(b) Function of pancreatic juice:
The pancreas secrete pancreatic juice, which contains enzymes such as amylase for digesting carbohydrates, trypsin for digesting protein and lipase for digesting fats.

(c) Function of intestinal juice:

• The glands of the wall of the small intestine secrete intestinal juice.
• The intestinal juice contains various enzymes, which complete the digestion of carbohydrates into glucose, proteins into amino acids and fats into fatty acids and glycerol.

Question 30.
Explain the process of absorption of food in human body.
Answer:
1. Absorption of food takes place in the small intestine after the food has been digested completely.
2. The inner wall of small intestine contains millions of small, finger-like projections called villi.
3. The presence of villi gives the inner walls of small intestine a very large surface area. This helps in the rapid absorption of digested food.
4. The villi possess numerous blood-vessels which take the absorbed food to each and every cell of the body.
5. The food absorbed in the cells is then used for obtaining energy, building new tissues and repairing old tissues.

Question 31.
Explain the process of assimilation of food and its egestion from the human body
Answer:
Assimilation :

• Blood receives food which is absorbed from the walls of small Intestine.
• Blood then carnes this food to all the parts of the body where it is assimilated by the cells.
• This assimilated food is used by all the cells for obtaining energy, growth arid repair of the body tissues.

Egestion:

• The unabsorbed food is sent to the large intestine.
• In the large intestine, most of the water of the undigested food is absorbed again by villi. This makes the undigested food almost solid.
• This solid form called faeces or stool is removed from the body through the anus.
• Sphincter muscles present in the anus regulate the removal of this waste material from the body.

Question 32.
What is the function of saliva?
Answer:
Salivary gland:
1. The salivary gland secretes saliva in our mouth. It is a watery liquid and so it wets (lubricates) the food in mouth. It is easy to swallow wet food.
2. The salivary gland also secretes an enzyme called amylase. Amylase breaks the complex molecule called starch present in the food into sugar.

Question 33.
Which gastric juices are released in our stomach? State their roles. OR Why does the stomach release hydrochloric acid?
Answer:
Stomach:
1. Stomach is a large organ which expands when the food enters it. Moreover, the stomach releases certain juices that help in digestion.
2. The muscular walls of the stomach churn the food thoroughly with these digestive juices.
3. During this process, the food is broken down into smaller pieces and is converted into a semi-solid paste.
4. The wall of stomach contains three tubular glands which secrete gastric juice.

Question 34.
Discuss the functions of Intestinal and pancreatic juices.
Answer:
Small intestine:
1. The partly digested food then moves from stomach to small intestine with the help of sphincter muscles. These muscles release the food in the small intestine in small parts.
2. The small intestine is the longest part of the alimentary canal and hence an extensive coiled structure. Owing to this structure it can fit in a very compact space.
3. The small intestine is the main site for complete digestion of carbohydrates, proteins and fats. It receives secretions from liver and pancreas for this purpose (whose functions are discussed below).

(a) Function of bile juice:

• Liver secretes bile juice.
• Bile is alkaline in nature and so it converts the acidic food coming from stomach into alkaline form so that enzymes of pancreas can act on it.
• In addition, the bile salts break the fats present in the food into small globules (droplets), which makes it easy for the enzymes to act and digest them.

(b) Function of pancreatic juice:
The pancreas secrete pancreatic juice, which contains enzymes such as amylase for digesting carbohydrates, trypsin for digesting protein and lipase for digesting fats.

(c) Function of intestinal juice:

• The glands of the wall of the small intestine secrete intestinal juice.
• The intestinal juice contains various enzymes, which complete the digestion of carbohydrates into glucose, proteins into amino acids and fats into fatty acids and glycerol.

Question 35.
What are villi (Singular — vlllus, Plural — vllli)?
Answer:
The Finger like projections or structures present on the inner surface of the small intestine are known as villi.
Function: Villi increases the surface area of intestine for better absorption of digested food.

Question 36.
How does food move In the alimentary canal?
Answer:
1. The food moves in the alimentary canal by a process called ‘peristalsis’.
2. The circular and longitudinal muscles of the alimentary canal undergo involuntary contraction and relaxation. This results in the wavy movement which is known as ‘peristalsis’.
3. Due to peristalsis, the food is pushed ahead and the passage behind the food gradually closes. This also prevents the backward flow of food.

Question 37.
Differentiate between gastric juice and bile.
Answer:

Question 38.
Explain how different organisms adopt different ways to release energy obtained from food material. OR Explain break-down of glucose by various pathways.
Answer:
1. The food material taken in during the process of nutrition is used in cells to provide energy for various life processes. The process of converting food into energy varies among organisms.

2. In any case, the first step is to break-down glucose which is a 6-carbon molecule, into two 3-carbon  molecule called pyruvate. This process takes place in the cytoplasm.

Further conversion of pyruvate:
(1) In absence of oxygen (anaerobic respiration):

• Conversion of pyruvate in absence of oxygen is called anaerobic respiration.
• In the absence of oxygen, the conversion of pyruvate may take place in two ways. They are —

(a) The pyruvate may be converted into ethanol and carbon dioxide.
This process takes place in yeast during fermentation.

(b) When our muscles lack oxygen, the pyruvate gets converted into lactic acid which is also a 3-carbon molecule.
When we over-stress our muscles they build up lactic acid which then results in cramps.

(2) In presence of oxygen (aerobic respiration):

• The conversion of pyruvate in the presence of oxygen is called aerobic respiration.
• In aerobic respiration, pyruvate breaks down in the mitochondria. This process breaks up the three-carbon pyruvate molecule to give three molecules of carbon dioxide. Water is also formed in this process.
• Aerobic process releases much greater energy as compared to aerobic process.
• The energy released during cellular respiration is immediately used to synthesize a molecule called ATP which is used to fuel all other activities in the cell.

Question 39.
What is respiration? What are its types?
Answer:
Respiration:
The process of releasing energy from food substances such as glucose, amino acids and fats under the control of enzymes so as to carry Out various life processes by an organism is known as respiration. There are two types of respiration:

• Aerobic respiration
• Anaerobic respiration

Question 40.
Explain anaerobic respiration.
Answer:
Anaerobic respiration:

• The respiration which takes place in the absence of oxygen is known as anaerobic respiration.
• It is seen in microorganisms like bacteria, yeast, fungi, endoparasites and muscle cells.

(a) In yeast (plant):

During anaerobic respiration in yeast (plants), carbon dioxide and ethanol are formed as end products.

(b) In animal medium (In muscle cells):

Question 41.
Explain aerobic respiration.
Answer:
Aerobic respiration :

• The respiration which takes place in the presence of oxygen is known as aerobic respiration.
• Most of the organisms show aerobic respiration.
•  It takes place in the cells and so it ¡s also called cellular or internal respiration.
• During aerobic respiration, digested food (glucose), in the presence of oxygen, is completely broken down into carbon dioxide and water.
• The energy released during this process is stored in the form of ATP.

The overall equations is as under:

Question 43.
Explain: Respiration In plants.
Answer:
Respiration in plants:
1.  All plants need energy for their physiological processes. Like animals, plants also show gaseous exchange (or say respiration).
2. In plants, gaseous exchange takes place through stomata and the large inter-cellular spaces.
3. The gaseous exchange takes place through the process called diffusion. Diffusion mainly occurs through the surface of the plant leaves which is in direct contact with atmosphere.
4. The gases that are exchanged by diffusion are

• oxygen and
• carbon dioxide.

5. These gases may enter into plant cells or may move out of the plant cells into atmosphere. The direction of diffusion (i.e. movement of gases) depends upon

• environmental conditions and
• requirement of the plant.

6. In general, at night, when there is no photosynthesis, the plants eliminate maximum carbon dioxide (CO₂) whereas during daytime plants release oxygen (O₂).

Question 44.
Why do plant release oxygen during day time and carbon dioxide at night?
Answer:
Exchange of gases or say diffusion of gases is the process of respiration in the plants.

• Plants perform photosynthesis during day time. So, whatever carbon dioxide (CO₂) is generate due to respiration gets used in the photosynthesis. Hence, there is no carbon dioxide remaining to release. As a result, plants only release oxygen.
• Photosynthesis does not occur at night and hence plants release carbon dioxide.

Question 45.
How do aquatic animals respire?
Answer:
1. For respiration, aquatic animals use oxygen dissolved in water.
2. The amount of oxygen dissolved in water is quite low as compared to amount of oxygen present in air. As a result, the aquatic animals have to put more effort to obtain oxygen required for respiration. So, the rate of breathing in aquatic organisms is quite fast as compared to terrestrial organisms.
3. Fishes take in water through their mouths and force it past the gills where the dissolved oxygen is taken up by the blood.

Question 46.
What are the basic characteristics of respiratory surface in terrestrial animals?
Answer:
Important characteristics of the respiratory surface of terrestrial animals:

• Terrestrial animals have large surface area of their bodies which remains in contact with oxygen rich atmosphere.
• This surface is very thin, fine and delicate. The thin walls facilitates diffusion of respiratory gases O₂ and CO₂ and also H₂O in gaseous form.
• Usually, this surface lies within the body so that it can be properly protected. As a result there are passages that carry the oxygen to this surface from outside.
• There is also a mechanism for moving the air in and out of this area where oxygen is absorbed.

Question 47.
List out few organisms and their respiratory organs.
Answer:
Few organisms and their respiratory organs:

Question 48.
Explain the process of respiration in human beings along with the functions of the respiratory organs and a labelled diagram.
Answer:
Human respiratory system consists of nostril, nasal cavity, pharynx, Iaryngopharynx, trachea, bronchi, lungs, and diaphragm.
The process of respiration and functions of various respiratory organs of human beings are as follows:

1. Nose and nostrils:

• Human respiratory system starts with nose.
• Air is taken into the body from nostrils.

2. Nasal cavity:

• Nostril opens in the nasal cavity which is lined by fine nair and mucus.
• When air passes through the nasal cavity, the cavity traps the dust particles and microbes of the air and hence prevent them from entering the nose. Thus, filtered air moves further.

3. Throat:

• The air then passes through the throat and reaches the lungs.
• Cartilaginous rings are present in the throat. These rings ensure that the air passage remains open.

4. Lungs:

• Lungs are the chief respiratory organs.
•  In the lungs, air passage divides into smaller and smaller tubes which finally end in balloon like structures called alveoli (or alveolar sacs).
• The walls of alveoli are thin and covered by blood capillaries. This provides a large surface for exchange of gases.

Question 49.
Explain the mechanism of breathing.
Answer:
1. The function of human respiratory system is to breathe in O₂ and to breathe out CO₂.
2. The alternate process of inspiration (inhalation) and expiration (exhalation) is known as breathing.

(A) Inspiration:

• During this process, we lift our ribs and flatten our diaphragm. As a result, the chest cavity expands.
• Due to this, the oxygen rich air from the atmosphere rushes into the lungs and fills the expanded alveoli sacs.
• The blood then takes this oxygen in the alveolar blood vessels and transports it to all the cells of the body.

(B) Expiration:

• The blood brings carbon dioxide from all the body cells and release into alveoli. The diaphragm relaxes i.e. moves up and the air containing carbon dioxide is pushed out of the lungs into the atmosphere through nostrils.
• This process is known as exhalation or expiration.

Question 50.
What is a respiratory pigment? Explain human respiratory pigment. Also state its importance.
Answer:
1. When the body surface of an animal is large, only diffusion pressure is not sufficient to transport oxygen to all the parts of the body. So, a specialized pigment called respiratory pigment does this work.
2. In human body, hemoglobin (Hb) which has a very high affinity for oxygen works as the respiratory pigment.
3. Hemoglobin is present in the red blood cells. It carries oxygen from the respiratory surface of the lungs to the body cells.

Question 51.
How is carbon dioxide transported towards lungs in human beings?
Answer:
Carbon dioxide is a respiratory gas to be exhaled out of the body. CO₂ is more soluble in water than oxygen. So, majority of CO₂ is transported in the dissolved form in blood plasma.

Question 52.
How are the lungs designed to increase the surface area for exchange of gases?
Answer:
Lungs:

• Lungs are the chief respiratory organs.
•  In the lungs, air passage divides into smaller and smaller tubes which finally end in balloon like structures called alveoli (or alveolar sacs).
• The walls of alveoli are thin and covered by blood capillaries. This provides a large surface for exchange of gases.

Question 53.
Differentiate between breathing and respiration.
Answer:

Question 54.
State the role of blood in transportation. What are the basic requirements to run this transport system?
Answer:
Role of blood in transportation:

• Blood is a red coloured connective tissue in liquid form.
• Blood consists of fluid medium called plasma in which the cells remain in suspended form.
• Plasma does the work of transporting food, carbon dioxide, salts and nitrogenous wastes in dissolved form.
• The red blood cells transport oxygen.

Basic requirements to run this system:

• A pumping organ — To push blood around the body
• A network of tubes — To reach all the tissues
• A proper system —To ensure that this network can run properly and be repaired if damaged

Question 55.
Explain the structure of human heart along with a figure.
Answer:
1. Human heart is a muscular pumping organ to push blood around the body.
2. Its size is just as much as our fist.

Structure of human heart:

3. A human heart contains four chambers.
4. Both oxygen and carbon dioxide are transported by blood. So, in order to prevent the mixing of oxygen rich blood with blood containing carbon dioxide, there are four chambers in the heart.

The upper two chambers:

• These chambers are known as atria (singular : atrium).
• The left chamber is known as the left atrium while the right chamber is known as right atrium.
• The walls of atria are thin.

The lower two chambers:

• The lower chambers are known as ventricles.
• The left chamber is known as the left ventricle while the right chamber is known as the right ventricle.
• The walls of the ventricles are thick.
• All the four chambers are separated from each other by partitions called septa.

Valves:

• There is a bicuspid valve that allows the oxygenated blood to flow from left atrium to the left ventricle.
• Also, there is a tricuspid valve that allows deoxygenated blood to flow from right atrium to right ventricle.
• These valves prevent backward flow of blood from ventricles to the atria i.e. they are one-way valves.

Question 56.
Explain the flow of blood in human heart.
(Note : Although all the points of this answer are to be written as printed, but to understand the flow properly, read all the points marked as a’ together and then points marked as ‘b’ together.]
Answer:

Initially, all the tour chambers of the heart are in relaxed state. This state of the heart is known as the diastolic stage.

• A-1. During this stage, the deoxygenated blood from various organs (except lungs) passes through superior and inferior venacava and reaches the right atrium.
• B-1. At the same time, the oxygenated blood from the lungs passes through pulmonary veins and reaches left atrium.

Now, both the atria contract and the following process takes place:
A-2. The tricuspid valve opens and deoxygenated blood from the right atrium is poured Into the right ventricle.
B-2. The bicuspid valve opens and oxygenated blood form the left atrium is poured into the left ventricle.

Now both the ventricles contract and the following process takes place:

A-3. Deoxygenated blood from the right ventricle passes from pulmonary valve and pulmonary arteries and reaches lungs.

• In lungs, CO₂ is removed from blood and O₂ from air is added into it.

B-3. Oxygenated blood from the left ventricle passes through aortic valve, aorta and pulmonary arteries and gets distributed to all the parts of the body (except lungs).

• Thus, the blood flows from right ventricle towards lungs and that from left ventricle towards all the parts of the body, expect lungs.
• Since the blood circulates twice through the heart, it is called double circulation.
• The separation of both types of blood in the heart allows a highly efficient oxygen supply system to the body.

Question 57.
With the help of a schematic diagram, show the transport and exchange of oxygen and carbon dioxide.
Answer:

Schematic representation of transport and exchange of oxygen and carbon dioxide

Question 58.
Explain the evolution of heart starting from fishes to mammals. OR Explain the structure of hearts of lower animals (Note: For this question, write point (a) and (b) only).
Answer:
(a) 2-Chambered heart (Fishes):

• The heart of fishes have only two chambers.
• The blood is pumped to the grills. It is oxygenated there itself and passed directly to the rest of the body.
• Thus in fishes, the blood passes only once through the heart during one cycle of passage through the body.

(b) 3-Chambered heart (Amphibians and reptiles):

• In amphibians and many reptiles, there are three chambers in the heart. The oxygenated and deoxygenated blood gets mixed in the heart of such animals.
• Such organisms do not need high amount of energy like humans and so they can tolerate some Schematic representation of transport and exchange of oxygen and carbon dioxide

(c) 4-Chambered heart (Birds and mammals including humans):

• Birds and mammals have a four chambered heart.
• The separation of both types of blood i.e. oxygenated and deoxygenated in the heart allows a highly efficient oxygen apply system to the body.
• Such a system of separation is extremely useful for birds and mammals because they require very high energy for maintaining their body temperature.

Question 59.
Write a short note on blood vessels.
Answer:
Blood vessels:
Closed hollow tubes which transport blood from the heart to different organs and from different organs to the heart are called blood vessels. There are three types of blood vessels. They are:
(1) Arteries,
(2) Veins and
(3) Blood capillaries

(1) Arteries:

• Arteries carry blood from heart to all the organs of the body.
• For this, heart pumps blood into the arteries under high pressure.
• To sustain this pressure, walls of arteries are thick.
• After reaching the organs and tissues, arteries divide into many smaller vessels known as arterioles and finally into capillaries in order to make the blood reach to all the cells.

(2) Veins:

• Veins collect blood from different parts of the body and bring it back to the heart.
• Since blood is not under pressure while getting collected from the organs, the walls of veins are thin.
• Veins have valves to prevent backward flow of blood to ensure that the blood flows only in one direction.

(3) Capillaries:

• Capillaries are blood vessels having just a single-layered thick cell wall.
• Exchange of materials between blood and its surrounding talks place through these capillaries.
• These capillaries join to form venules and then veins.

Question 60.
What role do platelets play when we are Injured and bleeding? OR How does blood clot durIng an injury?
Answer:
1. If we get injured and our blood vessels rupture, the blood starts leaking form these vessels. This is called bleeding.
2. Due to leakage, the pressure of the blood will decrease. This will in turn decrease the efficiency of the pumping system i.e. the heart.
3. To avoid such a situation, blood contains platelet cells. The platelet cells keep on circulating around the body and plug the leaks in the vessels, if any. This clots the blood.

Question 61.
What is lymph? Write a short note on lymphatic system.
Answer:
1. Lymph is a colourless fluid which is also involved in transportation. It is also called tissue fluid. It consists of plasma and proteins.
2. Lymph does not contain red blood cells and so it is pale and colourless. Also, it contains less protein compared to blood.

Lymphatic system:

• The lymphatic system is a part of the immune system as well as circulatory system.
• Lymphatic system consists of lymph, lymph vessels, lymphatic capillaries and lymphatic nodes.

Flow of lymphatic system:

• Lymphatic flow begins in the areas around blood capillaries. In these areas, small amount of tissue fluid drains into lymphatic capillaries through its pores.
• The lymphatic capillaries then drain into larger lymph vessels that look like thin, transparent veins. Finally these veins open in large veins.

Functions of the lymphatic system:

• Lymph vessels carry digested and absorbed fats from intestine.
• It collects intercellular fluid through the medium of lymph vessels and returns ¡t to blood circulation.
• The system also protects against diseases.

Question 62.
Differentiate: Blood and lymph
Answer:

Question 63.
Differentiate: Atria and ventricles
Answer:

Question 64.
Differentiate: Arteries and veins
Answer:

Question 65.
State the need of transportation system in plants.
Answer:
Need of transportation in plants:
1. Different types of substances that are absorbed or synthesized in one part of the body are transported to the various parts of the body. This is known as transportation.
2. Under photosynthesis, plants convert solar energy into chemical energy by utilizing carbondioxide and water.
3. Plants also need other substances which they take up separately by the means of roots. Roots absorb these substances from soil.
4. If the distance between roots and leaves is small, the raw material and energy that roots and leaves possess can easily reach all the parts of the body through simple process called diffusion.
5. But If the distance is long, process of diffusion becomes inefficient in transportation. In such a situation, a proper transport system is required. However, since plants do not move from one place to another, their energy requirement is lesser compared to animals.

Plants have two transport systems. They are:
(A) Xylem — For transporting water and minerals
(B) Phloem — For transporting food material produced by the plants

Question 66.
Explain transport of water In higher plants.
Answer:
Transport of water In higher plants:
1. Higher plants possess xylem for transporting water to all its parts.
2. Xylem contains tracheids and vessels.
3. Xylem tissues of all the organs of a plant are connected end to end with each other to form a network of conducting tubes.

Process of transport:
(a) Through diffusion:

• The cells of the roots are in direct contact with the soil and they take up ions from the soil.
• Due to this, a difference is created between the concentration of ions in the roots and that of ions in the soil.
• The ions present in the soil water are at higher concentration and so the water moves up from soil to the roots through osmosis.
• This water movement creates a water column under which water is steadily pushed upwards.

(b) Through conduction (transpiration):

• In higher plants transporting water till the highest point of the plant with this system is inefficient.
  As a result, higher plants take help of a process called transpiration.
• The transpiration or evaporation of water from the leaves create a suction which pulls up water from the xylem vessels.
• Thus the process of transpiration helps in the upward movement of water from roots to the leaves through the stem.
• Since the stomata are open during the daytime, the transpiration pull becomes a major driving force in the movement of water in the xylem.

Question 67.
Explain transpiration in brief.
Answer:
Transpiration:
1. The loss of water in the form of water vapour from the aerial parts of the plant is known as transpiration.
2. Plants absorb the water through the roots and transport it to all its parts.
3. This absorption is in fact driven by transpiration that occurs through the tiny pores on the leaves called stomata.
4. Transpiration creates a suction which pulls the water from the xylem cells of roots.
5. Thus, we can say transpiration plays an important role in making it possible for all the parts of the plant to receive water.
6. Transpiration also helps in regulating temperature.

Question 68.
What is translocation? Explain translocation of substances through the phloem tissue. OR How are food and other substances are transported In plants?
Answer:
Transportation of food and other substances:
1. The transportation of food and other substances i.e. the soluble products of photosynthesis is known as translocation.
2. Phloem is the chief tissue for conducting transportation.
3. Phloem transport the products of photosynthesis, some amino acids and other substances.
These products are transported from leaves to the other plant organs, while amino acids and the other products are transported towards the storage organs of roots, fruits, seeds and also to the growing parts of the plant.
4. Sieve tubes and companion cells are important components of the phloem. They can transport the food material in both directions i.e. upward and downward.
5. Sometimes translocation needs energy in the form of ATR in general, molecules like sucrose are transferred into phloem with the energy obtained from ATP. Here, the osmotic pressure in the cell increases, which pushes the food to the adjacent cell.
6. Interestingly in the spring season, the stored food in root or stem is transported to the buds to help them grow.

Question 69.
Differentiate: Xylem and phloem
Answer:

Question 70.
Differentiate: Transportation and translocation
Answer:

Question 71.
What is excretion? Discuss its need.
Answer:
Excretion:

• The harmful substances produced during several biochemical reactions are known as excretory substances.
• The biological process of removing these harmful nitrogen-based metabolic wastes in liquid form is known as excretion.
• Most excreted substances leave the body in the form of urine. Some of them are also lost in sweat and some in air that we breathe out.

Need for excretion:

• Body cells perform biochemical processes for sustaining life.
• During these processes, both useful as well as harmful toxic substances are produced.
• Accumulation of harmful substances in the body can harm the body and so they need to be removed from time to time.
• The process of removing excretory substances is simple in unicellular organisms.
• Unicellular organisms remove the excretory substances by simple diffusion from the body surface in the surrounding water.
• However, excretion is a complex process in multi-cellular organisms and so they possess special organs and even organ-system to perform excretion.

Question 72.
Explain excretory system of humans with a labeled diagram. OR Discuss the removal of urine from the body.
Answer:
Human excretory system is made up of the following organs:
(a) A pair of kidneys,
(b) A pair of ureter,
(c) A urinary bladder and
(d) A urethra

• Kidneys are located in abdomen on either side of abdomen. Kidney is bean shaped.
• Urine produced in the kidneys passes through the ureters into urinary bladder.
• The urinary bladder opens into urethra. Finally, urethra removes the urine outside the body.

Question 73.
Describe the structure of nephron, with labeled diagram.
Answer:
Nephron:

• Nephron is the main functional unit of kidney.
• Each human kidney possesses about 10 lakh nephrons.

Structure of nephron:
Bowman’s capsule:

• Each nephron has a double wailed cup-shaped bag at its upper end which is called Bowman’s capsule.
• The Bowman’s capsule contains a mass (bundle) of blood capillaries which is called glomerulus.

Tubule:

• The lower end of the Bowman’s capsule is called tubule.
• The part of tubule which is near the Bowman’s capsule is quite small and is called the neck.
• After the neck, the tubule becomes very narrow and coiled.
• This region of tubule consists of
  (A) A proximal convoluted tubule,
  (B) A Henle’s loop and
  (C) A distal convoluted tubule.
• The posterior end of nephron is known as collecting tube.
• Collecting tubule opens into renal pelvis.
• The renal pelvis opens into ureter.

Question 74.
Explain urine formation in human kidneys.
Answer:
1. The purpose of making urine is to filter out waste products from the blood.
2. Urine contains nitrogenous wastes such as urea or uric acid which are removed from blood in the kidneys.
3. The renal arteries bring the waste material and the blood from the body to the kidney.
4. This blood is filtered out from the blood capillaries into the owman’s capsule.
5. This filtration takes place under heavy pressure and is known as ultrafiltration.
6. Each kidney has large numbers of filtration units called nephrons packed close together. The filtrate passes through the tubular parts of nephron.
7. During this, useful substances such as glucose, amino acid, salts and a major amount of water are reabsorbed by blood capillaries that surround the nephron.
8. The amount of water re-absorbed depends on how much excess water does the body contains and how much of dissolved waste is yet to be excreted.
9. The remaining fluid contains excretory substances called urine.
10. The urine passes from ureter to urinary bladder and gets stored there.
11. When the bladder is filled with urine it expands and creates a pressure which then creates an urge to pass the urine out of the body via, urethra.

Question 75.
How is urine produced?
Answer:
1. The purpose of making urine is to filter out waste products from the blood.
2. Urine contains nitrogenous waste such as urea or uric acid which are removed from blood in the kidneys.
3. It is then no surprise that the basic filtration unit in the kidneys, like in the lungs, is a cluster of very thin-walled blood capillaries.
4. Each capillary cluster in the kidney is associated with the cup-shaped end of a coiled tube called Bowman’s capsule that collects the filtrate
5. Each kidney has large numbers of these filtration units called nephrons packed close together.
6. Some substances in the initial filtrate, such as glucose, amino acids, salts and a major amount of water, are selectively re-absorbed as the urine flows along the tube.
7. The amount of water re-absorbed depends on how much excess water there is in the body and on how much of dissolved waste there is to be excreted.
8. The urine forming in each kidney eventually enters a long tube, the ureter, which connects the kidneys with the urinary bladder.
9. Urine is stored in the urinary bladder until the pressure of the expanded bladder leads to the urge to pass it out through the urethra.
10. The bladder is muscular, so it is under nervous control, as we have discussed elsewhere.
11. As a result, we can usually control the urge to urinate.

Question 76.
Explain excretion in plants.
Answer:
Excretion in plants:
1. For plants, oxygen is one of the end products of photosynthesis and can also be considered as a waste product.
2. Plants emit this oxygen into atmosphere through diffusion process.
3. Plants get rid of excess water through transpiration.
4. For other waste products, many of the plant tissues contain dead cells in themselves.
5. Some of the material is also lost by falling leaves.
6. In the plant cell, vacuoles are the excretory organelle.
7.  Around the root system in the soil, plant excretes some of the waste products.
8. Peeling of the bark is also the example of plant excretion.
9. Resins and gums are excretions of plants.

Question 77.
Life on the earth depends on the sun. Give reason.
Answer:
1. All living beings need energy for their survival. They constantly need energy.
2. Food is the source of energy. This food directly or indirectly comes through the green plants.
3. All green plants trap solar energy or light energy and convert it in the form of food through the process of photosynthesis. Thus, the sun is the base of life on the earth.

Question 78.
Two green plants are kept separately in oxygen free containers, one In the dark and the other in continuous light. Which one will live longer? Why?
Answer:
1. The plant which is kept in dark will not be able to conduct photosynthesis. So, the container will soon be filled with the carbon dioxide released by the plant. Hence, the plant will die.
2. On the other hand, the plant kept in light would be able to carry out photosynthesis and thus convert the produced carbon dioxide into oxygen. Hence, this plant would live longer.

Question 79.
Leaves of a healthy potted plant were coated with Vaseline. Will this plant remain healthy for long? Give reasons for your answer.
Answer:
The plant covered with Vaseline will not remain healthy for long because Vaseline would make an impervious layer on the leaves. This will cause the following effects:
(a) Plant will not get oxygen for respiration.
(b) It will not get carbon dioxide for photosynthesis.
(c) Upward movement of water and minerals will be affected due to lack of transpiration.

Question 80.
Why mucus is secreted along with HCl in the stomach?
Answer:
Mucus makes a protective layer on the innermost layer of the stomach and protects it from the effect of HCl.

Question 81.
Why is small intestine in herbivores longer than in carnivores?
Answer:
1. Herbivores mainly eat green plants. So, cellulose found in green plants forms the largest part of an herbivore’s food.
2. Digesting cellulose takes quite a long time. Hence, herbivores have longer small intestine in which the food stays for a longer duration in order to get digested properly.
3. Carnivores do not eat plants and so their diet does not contain cellulose. So, the small intestine is small in them.

Question 82.
Why does absorption of digested food occur mainly in the small intestine?
Answer:
Maximum absorption of digested food occurs In small Intestine due to following reasons:
(a) Digestion is completed in small intestine.
(b) Inner lining of small intestine is provided with villi which increase the surface area to ensure better absorption.
(c) Wall of intestine is richly supplied with blood vessels which take the absorbed food to each and every cell of the body.

Question 83.
Hydrochloric acid Is an important constituent of gastric juice. Give reason.
Answer:
1. The stomach secretes gastric juices from its gastric glands for the chemical digestion of food.HCl is one of the constituents of gastric juice.
2.  Hydrochloric acid destroys the bacteria and other microorganisms that enter along with the food and thus prevents the decay of food in stomach.
3. HCl creates acidic medium in the stomach so that the gastric enzyme can act properly.

Question 84.
Name the following —
(a) The process in plants that links light energy with chemical energy
(b) Organisms that can prepare their own food
(c) The cell organelle where photosynthesis occurs
(d) Cells that surround a stomatal pore
(e) Organisms that cannot prepare their own food
(f) An enzyme secreted from gastric glands in stomach that acts on proteins.
Answer:
(a) Photosynthesis
(b) Autotrophs
(c) Chloroplasts
(d) Guard cells
(e) Heterotrophs
(f) Pepsin

Question 85.
What is a residual volume of air in lung? What is its importance
Answer:
1. During the breathe-in cycle, when air is taken in and let out, some amount of air remains in the lungs. This is known as a residual volume of air in lungs.
2. The residual volume of air in lungs is important to provide sufficient time for oxygen to be absorbed and for carbon dioxide to be released. This way the exchange of respiratory gases can be carried out continuously.

Question 86.
Name the energy currency ¡n the living organisms. When and where is It produced?
Answer:
1. ATP (Adenosine Triphosphate) is the energy currency in the living organisms. It Is produced at the end of respiration in mitochondria.
2. The energy released during respiration is used to make an ATP molecule from ADP and inorganic phosphate.

Question 87.
The rate of breathing in aquatic organisms much faster than in terrestrial organisms. Give reason.
Answer:
1. Aquatic organisms like fishes obtain oxygen from water present in dissolved state through their gills.
2. The amount of dissolved oxygen is quite low in water as compared to the amount of oxygen in the air. So, the breathing-rate of aquatic organisms is much higher.
3. Terrestrial organisms use the oxygen present in the atmosphere for respiration. Oxygen is available in plenty in the atmosphere and so terrestrial organisms do not need to breathe very fast.

Question 88.
Give the main points of difference between respiration in plants and respiration in animals.
Answer:

Question 89.
Differentiate: Inhalation and Exhalation
Answer:

Question 90.
The walls of ventricles are thicker than atria. Give reason.
Answer:
1. The human heart is a constantly pumping organ.
2. The blood flows in very high force in ventricles as compared to atria. Hence, the walls of ventricles are thicker than atria.

Question 91.
What is the need of transportation in human beings? Why circulatory system is a need of human being?
Answer:
1. The circulatory system is very important to transport various substances required for life processes.
2. Circulatory system is important for transporting respiratory gases between lungs and body cells.
3. It is important to transport nutrients (digested) from alimentary canal to body cells.
4. It is important to transport excretory substances towards kidneys.
5. It is also important to transport some of the enzymes and various hormones.

Question 92.
Describe the functions of various blood cells.
Answer:
Functions of various blood cells:
(1) Red blood cells: They contain hemoglobin. Hence, they transport O₂ from lungs to the body cells.
(2) White blood cells: They play an important role to provide immunity to the body. They protect the body from micro-organisms.
(3) Blood platelets: When any blood vessel gets cut, the platelets clot the blood and prevent bleeding.

Question 93.
What is the importance of the structure of septum in human heart?
Answer:
1. Human heart (and also birds) require high amount of energy. They also need to maintain their body temperature constantly. So, they need constant energy supply and hence constant supply of oxygen.
2. The septum separates the right side and the left side of the heart. As a result, the oxygenated blood and deoxygenated blood remain in separate chambers and do not mix.
3. Such a separation made by septum enables a very efficient supply of oxygen to the body cells.

Very Short Answer Type Question:

Question 1.
What is the Importance of life processes?
Answer:
To maintain the body functions of the living organism and to reproduce.

Question 2.
Name important life processes to maintain life.
Answer:
(i) Nutrition
(ii) Respiration
(iii) Transportation
(iv) Excretion
(v) Control and Co-ordination
(vi) Movement and
(vii) Reproduction.

Question 3.
What happens when the guard cells lose water?
Answer:
They shrink and cause the pore to close

Question 4.
In which from is carbohydrate stored?
Answer:
In plants, the carbohydrates are stored in starch form, while in animals, the carbohydrates are stored in glycogen form.

Question 5.
Which is the most common source to provide energy to the living organisms?
Answer:
Carbohydrates are the most common source for providing energy to the living organisms.

Question 6.
Give the summary of photosynthesis in terms of a reaction.
Answer:
6CO₂ + 12H₂O + Chlorophyll + Sunlight → C₆H₁₂O₆ + 6O₂+ 6H₂O

Question 7.
Name the raw materials of photosynthesis.
Answer:
CO₂. Water, Sunlight and Chlorophyll (a plant pigment)

Question 8.
Give any two examples of plant parasites.
Answer:
Cascuta and loxanthus

Question 9.
Give any two examples of animal parasites.
Answer:
Liverfiuke and plasmodium

Question 10.
What is the source of oxygen liberated during pbtosynthesis?
Answer:
Water

Question 11.
Give examples of plant with variegated leaves.
Answer:
The examples of plant with variegated leaves are money plant and crotons.

Question 12.
What is the site of photosynthesis?
Answer:
Chloroplast found in plant leaf

Question 13.
Define Translocation.
Answer:
Transfer of products of photosynthesis and some other materials through phloem is called as translocation.

Question 14.
A few drops of Iodine solution were added to rice water. The solution turned blue-black in colour. What does this indicate about rice water?
Answer:
The blue-black colour of rice water confirms the presence of starch

Question 15.
Which is the first enzyme to mix with food in the digestive tract?
Answer:
Amylase

Question 16.
What Is the importance of assimilated food in human body?
Answer:
(i) As a fuel to get energy,
(ii) As a material for growth and repair of the body.

Question 17.
Define Ingestion.
Answer:
The process by which the organisms take in the food is called ingestion.

Question 18.
Define Egestion.
Answer:
The process of removal of undigested food from the body is called egestion.

Question 19.
Define Peristalsis.
Answer:
The muscles of alimentary canal contract in a designed rhythm to push the food forward, This is called peristalsis.

Question 20.
In nasal cavity, which substance traps the dust particles and microbes present in the air?
Answer:
Mucus

Question 21.
What is the reason for muscular cramps after hard work?
Answer:
During hard work (or exercise) muscles undergo partial breakdown of glucose and forms lactic acid. This lactic acid is accumulated in the muscles and causes muscular cramps.

Question 22.
Give two examples of animals who breathe through their cell membranes?
Answer:
Amoeba. planarian and paramecium.

Question 23.
What is the average breathing rate in an adult man?
Answer:
The average breathing rate In an adult man is 15 to 18 breathe per minute.

Question 24.
What is the name of the area through which the exchange of respiratory gases occurs in a woody stem?
Answer:
Woody stems of plants/trees have lenticels for exchanging respiratory gases.

Question 25.
Which valve regulates the flow of blood from left atrium to left ventricle?
Answer:
Bicuspid value

Question 26.
Define Transpiration.
Answer:
The loss of water in the form of vapour through the aerial parts of the plants is called transpiration.

Question 27.
Give the location and function of a companion cell.
Answer:
The companion cells are located just next to sieve tube cells n a phloem tissue. They have nucleus and many other important organelles. They control the function of the sieve tube cell.

Question 28.
Define Pulmonary cIrculation.
Answer:
In human circulatory system, the blood circulatory system which transports blood between the lungs and the heart Is called as pulmonary circulation.

Question 29.
Define Systemic circulation.
Answer:
In human circulatory system, the transportation pathway of blood between various organs (except lungs) and heart is known as systemic circulation.

Question 30.
What is the pulse rate of a healthy adult person?
Answer:
The pulse rate of a healthy adult person is 72 per minute in resting position.

Question 31.
What are the mediums of circulation In human body?
Answer:
Blood and lymph are the mediums of circulation in human body.

Question 32.
Arrange these body parts starting from the lower side of the body.
I. Urter, II. Urethra, Ill. Kidney, IV. Renal vein
Answer:
III, I. Il, IV

Question 33.
Define Osmoregulatlon,
Answer:
The process of regulating waste contents and ion concentration in the body is called osmooegulation.

Question 34.
Name two chemical waste products of the human body.
Answer:
The two chemical waste products of the human body are urea and uric acid.

Question 35.
What is Urethra?
Answer:
Urethra is a tube through which urine is passed out from the body.

Question 36.
Define Dialysis.
Answer:
The artificial procedure used for cleaning the blood of a person by separating the waste substance (urea) from it is called dialysis.

Fill in the Blanks

1. Plants store food in the form of ……………..
Answer:
Starch

2. Each stoma consists of minute pore surrounded by a pair of …………..
Answer:
Guard cells

3. Guard cells are present in ……………. of leaves.
Answer:
Stomata

4. In amoeba, the process of obtaining food is called ………….
Answer:
Phagocytosts

5. The digested food found in food vacuoles of amoeba is absorbed directly into cytoplasm through ………….
Answer:
Diffusion

6. ………….. is the last step in the respiration process of human beings.
Answer:
Release of carbon

7. The process of food ingestion in amoeba is called …………
Answer:
Phagocytosis

8. ……………. is the largest gland of the body.
Answer:
Liver

9. Pepsin digests………….
Answer:
Proteins

10. In human beings ……….. is a respiratory pigment.
Answer
Hemoglobin

11. When air is blown from mouth into a test-tube containing lime water, the lime water turns milky due to the presence of …………
Answer:
Carbon dioxide

12. During respiration exchange of gases take place in ……………..
Answer:
Alveoli of lungs

13. Lack of oxygen in muscles often leads to cramps among cricketers. This results due to…………..
Answer:
Non-conversIon of glucose to pyruvate

14. The scientific name of voice box of humans is……………..
Answer:
Larynx

15. Blood is a …… liquid tissue.
Answer:
Connective

16. The ………. blood from various organs of the body is received by the right atrium.
Answer:
De-oxygenated

17. Capillaries join together to form ………….
Answer:
Veins

18. The normal rate of heart beat is ………………..
Answer:
72/mm

19. The normal blood pressure is …………..
Answer:
120/80 mmHg

20. The instrument to measure blood pressure is called ……………..
Answer:
Sphygmomanometer

21. The extracellular fluid which always flows front body tissues to the heart is called ………………..
Answer:
Lymph

22. At night ……………. is important for water transportation in plants.
Answer:
Root pressure

23. In nephrons, the structure of tubule after the neck is ………………..
Answer:
narrow and coiled

24. The chief nitrogenous waste products in the human beings are and ………………..
Answer:
Urea; uric acid

25. Oxygen liberated during photosynthesis comes from………………..
Answer:
water

26. Life on the earth depends on based molecules.
Answer:
Carbon

27. Plants excretory product(s) other than O₂ and CO₂ is/are …………………
Answer:
Resins and Gums

True Or False

1. A plant which is not growing visibly is dead. — False
2. Most of the food sources on earth are oxygen based. — False
3. Bacteria can produce their own food. — True
4. The opening and closing of the stomatal pore depends upon oxygen in guard cells. — False
5. The green dots on plant leaves are chloroplasts. — True
6. In amoeba, the undigested food moves to the surface of the cell. — True
7. Lipase is present in pancreatic juice. — True
8. During respiration in humans glucose is oxidized. — True
9. Pepsin is a protein digesting enzyme — True
10. Exit of food from stomach into small intestine as well as from anus to outside the body is regulated by sphincter muscles. — True
11. The internal (cellular) energy reserve in autotrophs is glycogen. — False
12. During deficiency of oxygen in tissues of human beings, pyruvic acid is converted into lactic acid in the chloroplast. — False
13. The direction of diffusion is not fixed. It depends upon the environmental conditions and the requirements of plants. — True

14. Among plants, humans and fishes, plants use oxygen at the fastest rate whereas fishes at the lowest rate. — False
15. Plants have specific respiratory pigments to in take oxygen from the atmosphere. — False
16. In pisces, the heart does not pump oxygenated blood to different parts of the body. — True
17. The blood leaving the tissues becomes richer in oxygen. — False
18. The left atrium receives oxygenated blood from the lungs through pulmonary vein. — True
19. There is independent pathway for transportation in plants. — False
20. Movement of material in xylem occurs through physical fores whereas movement of material in phloem occurs by utilizing energy. — True
21. Complex multi-cellular organisms remove the excretory substances by diffusion. — False
22. Bowman’s capsule brings the waste material along with the blood to the kidneys. — False

Match the Following :

Question 1.

Answer: (A – R), (B – P), (C – Q)

Question 2.

Answer: (A – Q), (B – R), (C – P)

Question 3.

Answer: (A – R), (B – Q), (C – P)

Question 4.

Answer: (A – R), (B – Q), (C – S), (D – P)

Haryana State Board HBSE 10th Class Maths Notes Chapter 12 Areas Related to Circles Notes.

Haryana Board 10th Class Maths Notes Chapter 12 Areas Related to Circles

Introduction
In previous classes, we have learnt about methods of finding perimeters and areas of simple plane figures such as parallelogram, square, rectangle, rhombus and circle, In this chapter, we shall discuss review of the measurement of area and perimeter (circumference) of a circle, and leam about the concept of sector and a segment of a circle and their areas. We shall also learn about problems on finding the areas of some combinations of plane figures involving circles or parts of circles.

1. Circle: A circle is the locus of points which moves in such a way that its distance from a fixed point is constant.
2. Radius: A line segment joining the centre and a point on the circle is called its radius. The plural of radius is radii.
3. Diameter: The chord, which passes through the centre of the circle, is called a diameter of the circle.
Diameter = 2 × radius
4. Circumference: The length of the complete circle is called its circumference
OR
The perimeter of a circle is called its circumference.
5. Chord: A chord of a circle is a line segment joining any two points on the circle.
6. Arc: A piece of a circle between two points is called an are.
7. Sector of a circle: The region between an arc and the two radii, joining the centre to the endpoints of the arc is called a sector of the circle
8. Segment of the circle: The region between a chord and either of its arc is called a segment of the circular region or simply a segment of the circle.
9. Circular region: The region consisting of all points which are either on the circle or lies inside the circle is called the circular region or circular disc.
10. Semi-circular region: When two arcs are equal, that is, each is a semi-circle, then both segments and both sector become the same and each is known as the semi-circular region.
11. Concentric Circles: The circles which have same centre and different radii are called concentric circles.

Perimeter and area of a circle – A Review
Let us recall that a circle is the set of all those points in a plane each of which is at constant distance from a fixed point in that plane.

The fixed point is called the centre and given constant distance is known as the radius of the circle. The distance covered by travelling once around a circle is its perimeter usually called its circumference.

We know that the ratio of circumference of a circle to its diameter is constant and this constant ratio is denoted by π (a Greek letter, read as pie).
\(\frac{\text { Circumference }}{\text { Diameter }}=\pi\)
⇒ π = \(\frac{\mathrm{C}}{2 r}\) ⇒ C = 2πr
Where C is circumference and r is radius of circle.
Here π is an irrational number. For practical purposes, we generally take the value of π as \(\frac{22}{7}\) or 3.14, approximately.

(i) Area of a circle
Area of a circle = πr²
where, r is the radius of the circle.
(ii) Perimeter and Area of Semi-circle
(a) Perimeter of semi-circle = \(\frac{1}{2}\) × 2πr + 2r
(b) Area of semi-circle = \(\frac{1}{2}\)πr²
where r is the radius of a circle.

(iii) Perimeter and area of the quadrant of a circle

If r is the radius of the circle
(a) Perimeter of the quadrant
= \(\frac{1}{4}\) × 2πr + 2r
= \(\frac{\pi r}{2}+2 r=r\left(\frac{\pi}{2}+2\right)\)
(b) Area of quadrant = \(\frac{1}{4}\)πr²

(iv) Area enclosed by two concentric circles
If R and r are radii of two concentric circles then
Area enclosed by them = πR² – πr² = π(R² – r²)
= π(R + r)(R – r)
1. The distance covered by a rotating wheel in one revolution is equal to the circumference of the wheel.
2. The number of revolutions completed by a rotating wheel in one minute
\(=\frac{\text { distance covered in one minute }}{\text { circumference of wheel }}\)
3. When two circle touch internally distance between their centres = difference of their radii
4. When two circle touch externally, distance between their centres = sum of their radii.

Areas of a sector and segement of a circle
(a) Sector of the circle

The portion of circular region enclosed by two radii And corresponding arc is called a sector of the circle.
Let APB be in arc of a circle whose centre is O. Then region bounded by radii OA, OB and arc APB is called the sector of the circle. The sector OAPB is Major segment called the minor sector and OAQB is called the major sector. The angle of major sector is 360° – ∠AOB.

(b) Segment of a circle

The circular region enclosed between a chord and corresponding arc is called a segment of a circle.
Let AB be a chord of the circle with centre O. Then shaded region APB is a minor segment of the circle and AQB is a major segment of the circle.

Length of an arc and area of a sector of a circle

1. Write the length of an arc of a sector of sector with radius and angle with degree measure θ.
Let OAPB be a sector of a circle with centre O and radius r such that ∠AOB = θ. If θ then AB is minor are of the circle.
If θ is increases then length of arc AB increases. When an are subtends an angle 180°, i.e., ∠AOB = 180°.
Then length of arc = Circumference of semicircle
= πr
If the arc subtends an angle of at the centre.
Then its arc length (l)
= \(\frac{\theta}{180}\) × πr ……(1)
(l) = \(\frac{\theta}{360}\) × 2πr
⇒ l = \(\frac{\theta}{360}\) (Circumference of the circle)
When an arc subtends an angle 180°, at the centre the corresponding sector is semi-circular region of area = \(\frac{\pi r^2}{2}\)
If the arc subtends an angle at the centre then area of corresponding sector (A),

Area of a segment of a circle
Let r be the radius of the circle with centre O. Let AB be a chord make a minor segment APB.
Let ∠AOB Draw OC ⊥ AB.
Area of minor segment APB = Area of sector OAPB – Area of ΔAOB
= \(\frac{\theta}{360^{\circ}} \pi r^2\) – \(\frac{1}{2}\)AB × OC
= \(\frac{\theta \pi r^2}{360^{\circ}}-\frac{1}{2}\) × 2AC × OC …… (1)
In right ΔACO, we have
\(\sin \frac{\theta}{2}=\frac{\mathrm{AC}}{\mathrm{OA}}\)

AC = OA \(\frac{\theta}{2}\)sin = r sin\(\frac{\theta}{2}\) and

Areas of combinations of plane figures
In this section we will learn to calculate the areas of some plane figures & designs which are combinations of more than one plane figures such that flowerbeds, drain covers, window design, designes on table covers etc. We illustrate the process of calculating areas of these figures through some examples.

Haryana State Board HBSE 10th Class Maths Notes Chapter 11 Constructions Notes.

Haryana Board 10th Class Maths Notes Chapter 11 Constructions

Introduction
In class IX, we have learnt about some constructions namely drawing the perpendicular bisector of a line segment, bisecting an angle, some constructions of triangles also gave their justifications In this chapter we shall study some more constructions by using the knowledge of the earlier constructions e.q. division of a line segment drawing a triangle similar to a given triangle and drawing of tangents to a circle.

Division of a line Segment
First of all, we will know about some basic terms.
1. Construction: The process or art of constructing, OR the act of devising and forming.
2. Similar: Two geometrical figures are similar if they are of the same shape but not necessarily of the same size.
3. Bisect: To divide into two equal parts.
4. Arc: The part of a curve between two given points on the curve.
5. Altitude: A line through one vertex of a triangle and perpendicular to the opposite side.
6. Tangent: It is a straight line which touches the circle at one point only
7. Point of Contact: The point at which the tangent touches the circle is called the point of contact
8. Concentric circles: Two circles are known as concentric circles, if they have some centre and different radii.
9. Corresponding : (i) Similar in character, form or function.
(ii) Able to be matched, joined or interlocked.

Construction 1:
To divide a line segment in a given ratio
Given a line segment AB, we want to divide it in the ratio m : n here m and n are positive integers. We take m = 2, n = 3.
Steps of Construction:
1. Draw any ray AX, making an acute angle with AB.
2. Along AX mark (2 + 3) = 5 points A₁, A₂, A₃, A₄ and A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
3. Join A₅B.
4. From the point A₂ draw A₂C || A₅B, meeting AB at C. Then AC : BC = 2 : 3.

Justification: In ΔACA₂ and ΔABA₅ we have CA₂ || BA₅
\(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{AA}_2}{\mathrm{~A}_2 \mathrm{~A}_5}\)
[By Basic proportionality theorem]
\(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{2}{3}\)
Hence, AC : BC = 2 : 3.

Alternative Method:
Steps of Construction:
1. Draw a line segment AB.
2. Draw a ray AX making an acute angle with AB.
3. Draw a ray BY (On opposite side of AX) parallel to AX making ∠ABY = ∠BAX.
4. Along AX mark the points A₁, A₂ and along By mark the points B₁, B₂, B₃ such that AA₁ = A₁A₂ = BB₁ = B₁B₂ = B₂B₃.

5. Join A₂B₃ intersecting AB at point C. Then
AC : BC = 2 : 3.
Justification: Here, AX || BY
∠CAX = ∠ABY (Alternate interior ∠S)
∠ACA₂ = ∠BCB₃ (Vertically opposite ∠S)
ΔCAA₂ ~ ΔCBB₃ (By AA similarity criterian)
⇒ \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{AA}_2}{\mathrm{BB}_3}\)
⇒ \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{2}{3}\)
Hence, AC : BC = 2 : 3.

Construction 2:
To construct a triangle similar to a given triangle as per given scale factor.
Scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle e.g.
1. Scale factor \(\frac{2}{3}\) means the sides of the constructed triangle is \(\frac{2}{3}\) of the sides of given triangle.
2. Scale factor \(\frac{5}{4}\) means the sides of the constructed triangle is \(\frac{5}{4}\) of sides of given triangle.
Let us take the following examples for understanding the construction involved.

Construction of Tangents to a Circle
We study in circles (Chapter-10) that tangent to circle is a line that intersects the circle at only one point. The point is called the point of contact. The tangent at any point of a circle is perpendicular to the radius through the point of contact.
(a) Construction of Tangents to a Circle from a point outside the circle when its centre is known:
Steps of Construction:
1. Draw a Circle with centre O and given radius.
2. Mark a point P outside the circle.
3. Join OP and draw perpendicular bisector of PO meeting PO at M.
4. Draw a circle with M as centre and radius equal to PM = OM. intersecting the given circle at points Q and R.

5. Join PQ and PR.
Then PQ and PR are the required tangents.

(b) Construction of Tangents to a circle from a point outside the circle when the centre of circle is not known:
Steps of Construction:
1. Taking three points A, B, C on the circle and join AB and BC.
2. Draw perpendicular bisectors of AB and BC which intersect each other at point O.
3. Then O is the required centre of the given circle.

4. Mark a point P outside the circle.
5. Join PO and draw perpendicular bisector of PO meeting PO at M.
6. M as the centre, PM = MO as the radius draw another circle which intersects the previous circle at the points Q and R.
7. Join PQ and PR.
Then PQ and PR are the required tangents.

Haryana State Board HBSE 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 1.
How many elements are known to man? Why did he classify them?
Answer:
1. Today, man has discovered 118 elements.
2. All the elements have different properties. These elements are either used individually or in combination with other elements to form an endless variety of substances.
3. Classifying the elements help us to understand their properties and produce various products.

Question 2.
Explain Dobereiner’s classification of elements or Dobereiner’s triads.
Answer
1. In 1817, German chemist Johann Wolfgang Dobereiner started classifying elements on the basis of their chemical properties.
2. He discovered that there exists several ‘triads’ or ‘groups of three elements each’ that appeared to have similar chemical properties.
3. Law of Triads: If three elements are arranged in the increasing order of their atomic masses, the atomic mass of the intermediate (i.e. second) element would be almost equal to the average of atomic masses of first and third elements. This is known as Law of Triads. It was given by Dobereiner.

Based on this principle he identified the following three triads:

• Lithium, Sodium and Potassium,
• Chlorine, Bromine and Iodine,
• Calcium, Strontium and Barium.

Example:

• Atomic mass of Lithium (Li) is 6.9 u, Sodium (Na) is 23.0 u and that of potassium (K) is 39.0 u.
• Here, the atomic mass of sodium is the average of lithium and potassium i.e. = ((6.9 + 39)/2) = 23. Hence, these three elements form a triad.

Question 3.
The elements in the Dobereiner’s table were arranged in such a manner that the atomic mass of the intermediate (i.e. second) element would be almost equal to the average of atomic masses of first and third elements. Demonstrate this with the help of Dobereiner’s triads.
Answer:
1. Table given below shows Dobereiner’s triacis.
2. We can see that in all these three triads, the atomic mass of the second element is equal or nearly equal to the average of atomic mass of first and third element.

Question 4.
State the limitations of Doberelner’s classification of elements.
Answer:
Limitation of Dobereiner’s classification :

• Under Dobereiner’s classification, overall only a limited number of elements could be classified into triads.
• After arranging the elements in triads, it was found that there were certain other elements which could not be classified by Dobereiner’s method.

Question 5.
Explain Newlands’ method of classifying elements (or Newlands’ Law of Octaves)
Answer:
Newlands’ Law of Octaves :
1. When elements are arranged in the increasing order of their atomic masses, properties of every 8th element are found to be similar to the properties of the first element.
2. In 1866, a scientist named John Newlands arranged the elements in the increasing order of their atomic masses.
3. Newlands was the first person to arrange elements in their increasing order of atomic masses.
4. During this arrangement, he found properties of every 8th element to be similar i.e. property of 1st and then 8th element would be similar. Similarly property of 2nd and 9th element would be similar and so on.
5. He called this periodicity pattern as the Law of Octaves (octaves = eight). The law can be compared with the octaves or say 8 notes found in music where in the 8th note is similar to the 1st note.

Question 6.
Discuss limitations of Newlands’ Law of Octaves. OR How can you say that Newlands’ table worked well with lighter elements only?
Answer:
Limitation of Newlands’ Law of Octaves:
(1) The law of octaves was applicable only upto calcium. After calcium, every 8th element did not possess properties similar to that of 1st
(2) Newlands thought that there were only 56 elements in nature. He also thought that no more elements would be discovered in the future. However, later, several new elements were discovered that could not be arranged in the table as per Newlands’ law.
(3) In order to fit elements any how into his table, Newlands adjusted two elements in the slot even if the properties of elements did not match with other elements.

Question 7.
How did Mendeleev drill to arrange the elements into his Periodic Table? OR How did Mendeleev construct his Periodic Table?
Answer:
When in 1869, Russian chemist Dmitri Ivanovich Mendeleev started his work of classifying elements, 63 elements were known to man.
Mendeleev started by examining the relationship of
(a) ‘Atomic mass of an element with
(b) ‘Physical property of the element’ and
(c) Chemical property of the element.

(i) Classification on the basis of chemical properties:

• For classifying the elements on the basis of chemical properties, Mendeleev started studying the compounds that a particular element formed with oxygen as well as hydrogen.
•  A compound formed when an element combines with oxygen is called oxide whereas with hydrogen is called hydride. (For example, CaO, NaH, K₂H, etc.) The formulae of such oxides and hydrides were taken as ‘one of the basis of classification’.
• Mendeleev selected hydrogen and oxygen because they are very reactive and form compounds with most elements.

(ii) Classification on the basis of atomic mass:

• Mendeleev took 63 cards (1 card per element) and on each card he wrote the ‘atomic mass’ as well as ‘properties’ of a particular element.
• Mendeleev arranged the cards in the increasing order of atomic masses of elements starting from atomic number 1 to 63. Then he put the cards that showed similar properties into single group. This way he arranged elements into 8 groups. Each of these 8 groups were further divided in two sub-blocks, A and B.
• Hence, while arranging the elements in the increasing order of their atomic masses, he also arranged them in groups on the basis of their chemical properties.
• Through this classification, he concluded that “The properties of elements are periodic function (i.e. periodic in nature) of their atomic mass.” This law came to be known as Mendeleev’s Periodic Law.

Mendeleev’s observation:

1. Mendeleev observed that most of the elements got a place in his periodic table.
2. The elements got arranged in the order of increasing atomic masses.
3. Elements having similar physical and chemical properties occur periodically.

Question 8.
What observations did Mendeieev make while classifying the elements into his Periodic Table?
Answer:
Mendeleev’s observation:

1. Mendeleev observed that most of the elements got a place in his periodic table.
2. The elements got arranged in the order of increasing atomic masses.
3. Elements having similar physical and chemical properties occur periodically.

Question 9.
Which criteria did Mendeleev use for developing his periodic table?
Answer:
Criteria used by Mendeieev for developing periodic table:

• The properties of elements are the periodic function of their atomic masses. Hence, arranging elements in the increasing order of their atomic masses.
• Elements with similar properties are arranged in the same group.
• The formula of oxides and hydrides formed by an element.

Question 10.
Discuss the anomalies (irregularities) of Mendeleev’s Periodic Table.
Answer:
Anomalies (irregularities) of Mendeleev’s Periodic Table:
(1) Sequence of few elements was inverted:

•  In Mendeleev’s table, the elements were arranged in increasing order of their atomic masses. However, at few instances Mendeleev did not follow this arrangement.
• At few places he placed elements with slightly higher atomic mass before elements with lesser atomic mass. This means he inverted his sequence. He did this so that elements with similar properties could be grouped together.

(2) Gaps were kept at few places:

• Mendeleev had left some blank spaces in his Periodic Table. He was very sure that there exists some elements that would fit these spaces.
• Mendeleev named those blank places with a Sanskrit numeral prefix called ‘Eka

Example:
‘Eka’ means one place below something. So, Eka-aluminium means one space (left blank) below aluminium. In this manner scandium (Sc), gallium (Ga) and germanium (Ge) discovered later had properties similar to Eka-boron, Eka-aluminium and Eka-silicon respectively.

Question 11.
Discuss the achievements of Mendeleev’s Periodic Table.
Answer:
Achievements of Mendeleev’s Periodic Table:
(1) The gaps that Mendeleev had left in his table got properly filled when new elements were discovered.
(2) Chemists not only accepted Mendeleev’s table but also conferred him as the ‘originator of the concept’ on which Modern Periodic Table is based.
(3) Noble gas elements such as helium (He), neon (Ne), etc. were not known during Mendeleev’s time. So, those elements were not present in his table. But, Mendeleev’s table was so precisely designed that when these gases were discovered they were easily placed in a new separate group without disturbing the existing order.

Question 12.
Elements in Mendeleev’s Periodic Table were arranged in increasing order of their atomic masses.
Then why did Mendeleev place cobalt Co (atomic mass 58.93) before element nickel Ni (atomic mass 58.71) when In fact atomic mass of nickel was lesser than cobalt?
Answer:
1. Although mass of cobalt (58.93) was slightly more than Nickel (58.71), even then cobalt was placed before Nickel. Mendeleev did this so that he could maintain the properties of element placed in one group.
2. The properties of Co, Rh and Ir are similar and that of Ni, Pd and Pt are similar. Hence, Mendeleev placed Co before Ni so that elements Co, Rh and Ir having similar properties could be arranged in the same group.

Question 13.
Compare the properties of Eka-aluminium and gallium to sho the brilliant prediction Mendeleev’s table had.
Answer:
1. During Mendeleev’s time, gallium was not discovered. But, Mendeleev proposed properties of elements that could be filled in blank spaces that he left in his table. One such element was eka-aluminium.
2. The properties given by Mendeleev for Eka-aluminium and the actual element gallium that later replaced Eka-aluminium were quite similar. This can be seen in the table below.

Question 14.
Why did Mendeleev leave gaps in the Periodic Table?
Answer:
Gaps were kept at few places:

• Mendeleev had left some blank spaces in his Periodic Table. He was very sure that there exists some elements that would fit these spaces.
• Mendeleev named those blank places with a Sanskrit numeral prefix called ‘Eka

Example:
‘Eka’ means one place below something. So, Eka-aluminium means one space (left blank) below aluminium. In this manner scandium (Sc), gallium (Ga) and germanium (Ge) discovered later had properties similar to Eka-boron, Eka-aluminium and Eka-silicon respectively.

Question 15.
Discuss limitations of Mendeleev’s classification.
Answer:
Limitation of Mendeleev’s classification:
(1) Position of hydrogen:
Mendeleev had placed hydrogen (H) element in group 1 i.e. group of alkalis. He did so because like alkali metals, hydrogen combines with halogens and oxygen and sulphur to form compounds having similar formulae.

Contrary to this hydrogen also showed properties similar to elements of group 17 i.e. group of halogens. Thus, hydrogen could be placed in group 17 as well. Due to this confusion hydrogen could not be assigned a fixed place in Mendeleev’s Periodic Table.

(2) Position of isotopes:

• Isotopes are atoms of the same element having similar chemical properties but different atomic masses.
• Several elements have isotopes. For example, hydrogen has three isotopes and all have separate masses. Even then all these three were placed in one single position as hydrogen element H.

(3) Wrong order of some elements:

Mendeleev had arranged the elements on the basis of increasing order of atomic masses however, he broke this rule for certain elements. For example. he placed cobalt (Co) having larger mass before nickel (Ni) having lesser mass. This made it difficult to predict how many elements can be discovered between such two elements.

Question 16.
State the three limitations of Mendeleev’s Periodic Table.
Answer:
(1) The position of hydrogen could not be correctly assigned.
(2) The table could not assign proper position for the isotopes of various elements.
(3) Some elements were not arranged on the basis of their increasing atomic mass. This posed a question as to how many elements could still take the place between two elements.

Question 17.
What is a periodic table?
Answer:
Periodic table:

• The Periodic Table is a chart in which all the elements known to us are arranged in a systematic manner.
• A Periodic Table is divided into rows (periods) and columns (groups).
• Elements having similar properties are placed in the same group (i.e. vertical column).

(For Information only: Why is Periodic Table called so?

• During classification of elements it was found that the elements show similar physical and chemical properties after a fix interval or say ‘period’ i.e. the elements are periodic in their nature.
• The table was then prepared based on the periodic nature or periodicity of elements and so the table is called Periodic Table.)

Question 18.
How was Periodic Law of Mendeleev Improved Into Modern Periodic Law?
Answer:
1. Mendeleev had arranged the elements in his table on the basis of increasing atomic masses. In 1913, Henry Moseley showed that, rather than ‘atomic mass’, ‘atomic number’ is a ‘better fundamental property’. Henry said, atomic number and not atomic mass determine the chemical properties of elements.
2. Looking to the facts, the Periodic Law given by Mendeleev was modified into Modern Periodic Law which states “Properties of elements are a periodic function of their atomic number”.

Question 19.
State Mendeleev’s Periodic Law and Modern Periodic Law.
Answer:
(a) Mendeleev’s Periodic Law:
Properties of elements are the periodic function of their atomic masses.

(b) Modern Periodic Law:
Properties of elements are a periodic function of their atomic numbers.

Question 20.
On what basis were the elements arranged in the Modern Periodic Table? Why?
Answer:
1. The elements in the Modern Periodic Table were arranged on the basis of Modern Periodic Law which states that elements are a periodic function of their atomic numbers.
2. Atomic number (Z) tells us the number of protons in the nucleus of the atom of an element.
3. This number i.e. the atomic number goes on increasing as one moves from one element to the next.
4. Thus, in Modern Periodic Table the elements were arranged in the increasing order of their atomic number Le. Z.

Question 21.
Discuss the arrangement of elements In groups and periods in Modern Periodic Table.
Answer:
The Modern Periodic Table has 18 vertical columns known as ‘groups’ and 7 horizontal rows known as ‘periods
(a) Placement of an element in a group:
(i) Elements are placed in a particular group on the basis of the ‘valence electrons’ i.e. elements having same number of valence electrons will be placed same group.

Example:

• The electronic configuration of fluorine (f) is 2, 7 and that of chlorine (Cl) is 2, 8, 7. This means both these elements have valency = 1. Hence, both these elements are placed in same group i.e. group no. 17.
• Note that as we go down in a group, the number of shells increase. Fluorine has only 2 shells whereas chlorine has 3.

(b) Placement of an element In a period:
(i) Within a horizontal period, as one moves from left to right, the ‘elements have same number of shells’ but, ‘different valence electrons
(ii) Moreover, on moving left to right, the number of electrons of valence shell increase by 1 unit because the elements are arranged in the increasing order of their atomic number by 1 unit.
(iii) Thus, elements having same number of shells are placed in the same period. For example, Na, Mg, Al, Si, etc. have 3 shells and hence are placed In the 3rd period.

Question 22.
How are electrons arranged in various shells? How do we determine the number of elements in each period on the basis of number of elements in a shell?
Answer:
Arrangement of electrons in a shell:
1. The maximum number of electrons that can be accommodated in a shell depends on the formula 2n² (where n = the number of given shells from the nucieusý’
2. Letter ‘K’ denotes the first shell, ‘L’ second, M’ third and so on. Thus for shell ‘K, n = 1, for ‘L, n = 2 and for ‘M’, n = 3.
3. By knowing the values of ‘n’ for K, L and M we can derive the number of electrons that each of these shells can hold. It is:

• K shell = 2(n)² = 2(1)² = 2 electrons
• L shell = 2(n)² = 2(2)² 2(4) = 8 electrons
• M shell = 2(n)² = 2(3)² = 2(9) = 18 electrons

4. Based on this calculation we say that the 1st period has 2 elements and period has 8 elements.
5. Although M shell can hold 18 electrons but third period can hold only 8 electrons. Therefore, 3 period can hold only 8 elements.

Question 23.
What do you mean by periodic properties?
Answer:
1. The properties which are determined by the electronic configuration of elements or which depend on the electronic configuration of elements are known as periodic properties.
2. Also, the properties which show a recurring gradation within the same group or along a period are known as periodic properties.
Example: Valency, atomic radii (atomic size), metallic property, electronegativity, etc. exhibit periodic properties.

Question 24.
Bring out the trends of change In valency, atomic size and metallic and non-metallic properties in groups and periods.
Answer:

Question 25.
What is valency? How is it calculated?
Answer:
1. Relative ability of an element to combine with other element is known as a valency. OR Valency is the combining capacity of an atom of an element to acquire noble gas configuration.
2. Valency depends on the number of valence electrons that an atom of an element has.
3. The valency of an element is

• Either equal to the number of electrons in the valence shell OR
• Equal to eight minus the number of electrons in the valence shell.

Question 26.
Explain the trend of valency within a period and a group. OR How does the valency vary in a period on going from left to right and while going down In a group?
Answer:
1. As you move in the period from left to right, the valency first increases from 1 to 4. Then it decreases from 4 to 0.
2. All the elements in a given group possess equal number of electrons. Hence, valency does not change within a group.

Question 27.
Explain atomic size (atomic radius).
Answer:
Atomic size:

• The radius of the atom is called the atomic size of the atom.
• Atomic size (or radius) can be visualized as the distance between the centre of the nucleus and the outermost shell of an isolated atom.
• Atomic radius is expressed in angstrom (A°), or centimeter (cm) or picometer (pm). 1 pm = 10 12m.
• For example, the atomic radius of hydrogen atom is 37 pm.

Question 28.
Why does atomic radii increase as we go down in a group, while decrease as we move from left to right in a period? OR Explain the trends in atomic size of elements In a period and a group.
Answer:
Situation in a group :
1. In a group as we move from top to down, new orbits get added in the elements.
2. For example, in the first group, Lithium (Li) has 2 orbits, Sodium (Na) has 3 orbits, Potassium (K) has 4 orbits and so on.
3. Since the orbits increase as we move down, naturally their radii also increase.

Situation in a period :

• In a period, as we move form left to right, no new orbits are added unlike in group.
• Secondly, in the period, the positive electric charge of nucleus attracts more electrons and so the radii of atoms decrease.
• Thus, as we move from left to right in a period, the atomic radii decreases.

Question 29.
Explain the trends of the metallic character in a period and a group.
Answer:
Trend of the metallic character in a group:
On moving down in a group, the effective nuclear charge experienced by valence electrons decreases because the outermost electrons are far away from the nucleus. Therefore, tendency of the element to lose electrons increases and electrons can be lost easily. Hence, the metallic character increases on moving down in a group.

Trend of the metallic character In a period:

• On moving from left to right in a period the effective nuclear charge experienced by valence electrons increases.
• Therefore, the tendency to lose electrons will decrease.
• Thus metallic character decreases in a period on moving from left to right.

Question 30.
Explain the trends of the non-metallic character in a period and a group.
Answer:
Trend of the non-metallic character in a group:
On moving down within a group, the non-metallic character decreases.

Trend of the non-metallic character in a period:

• On moving from left to right in a period, the nuclear charge experienced by valence electrons increases. So, the tendency to attract electrons increase.
• As a result, the non-metallic character increases in a period on moving from left to right.

Question 31.
What are metalloids or semi-metallic elements? Give example.
Answer:
1. Elements which possess properties of both metals and non-metals are known as metalloids or semi-metallic elements.
2. In the Modern Periodic Table, a zig-zag line separates metals and non-metals. The border line elements such as boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te) and polonium (Po) on this zig-zag line are known as metalloids or semi-metals.

Question 32.
Give difference between ‘Elements of a group’ and ‘Elements of a period’
Answer:

Question 33.
Possibility of groups instead of triads arose. Explain.
Answer:
1. The method of classifying elements on the basis of triads was given by Dobereiner.
2. Under this method, Dobereiner stated that if three elements having similar chemical properties are arranged in their increasing order of atomic masses, the atomic mass of the intermediate (second) element would be similar to the average of atomic masses of first and third element.
3. However, under this classification, overall only a limited number of elements could be classified and so this method failed later on.
4. It was also discovered that more elements could be added to these triads.
5. For example, fluorine could be added to the triad of chlorine, bromine and iodine. Similarly magnesium could be added to the triad of calcium, strontium and barium.
6. Hence, scientists thought that rather than classifying elements only in triads, perhaps they could be arranged in larger groups.

Question 34.
Newlands’ law of arranging elements came to be known as ‘Law of Octaves’. Give reason.
Answer:
Newlands’ Law of Octaves :
1. When elements are arranged in the increasing order of their atomic masses, properties of every 8th element are found to be similar to the properties of the first element.
2. In 1866, a scientist named John Newlands arranged the elements in the increasing order of their atomic masses.
3. Newlands was the first person to arrange elements in their increasing order of atomic masses.
4. During this arrangement, he found properties of every 8th element to be similar i.e. property of 1st and then 8th element would be similar. Similarly property of 2nd and 9th element would be similar and so on.
5. He called this periodicity pattern as the Law of Octaves (octaves = eight). The law can be compared with the octaves or say 8 notes found in music where in the 8th note is similar to the l note.

Question 35.
Differentiate between groups and periods of Modern Periodic Table.
Answer:

Question 36.
Two elements X and Y belong to groups 1 and 3 respectively In the same period. Compare these elements with respect to their —
Answer:
(a) Metallic character (b) Size of atoms (c) Formulae of their oxides and chlorides
(a) As one moves in a period from left to right, the metallic character decreases. Hence, the metallic character decreases while moving from X to Y.
(b) As one moves in a period from left to right, the atomic radius decreases. Hence, atoms of element Y are smaller than that of element X.
(c) Oxides of element X and Y: X₂O, X₂O₃, Chlorides of element X and Y: XCl, YCl₃

Question 37.
Atomic number of magnesium is 12. What information can you get from this? (Hint: Keep In mind the Modern Periodic Table.)
Answer:
1. The atomic number of magnesium is 12. So, its electronic configuration is (2, 8, 2).
2. From this, we can say that it is positioned in the third period arid second group in the Periodic Table.
3. There are 3 orbits (2, 8, 2) and there are 2 electrons in its outermost orbit.
4. It has 2 electrons in the last orbit and so it will lose these 2 electrons and hence become mg²⁺ ion. Thus, its vaiency is +2.

Question 38.
Elements of which groups of the Periodic Table can easily lose electrons and elements of which groups of the Periodic Table can easily gain electrons?
Answer:
1. Elements of group IA, IIA, IIIA and that of group IB. IIB and IIIB will easily lose electrons.
2. Elements of group IVA, IVA, VIIIA and VIIIB will neither lose nor gain electrons.
3. Elements of group VA, VIA, VIIA and that of group VB, VIB and VIIB will easily gain electrons.

Question 39.
The position of three elements A, B and C in the Modern Periodic Table is shown below:

Giving reason for the following statements in one or two sentences.
(a) Element A is a metal.
(b) Atom of element C has a larger size than atom of element B.
(c) Element B has a valency of 3.
Answer:
(a) Element A has 1 valence electron. So it can lose this electron and become electropositive. Hence, we can consider element A as metal.
(b) Element B belongs to 2nd period and so has 2 shells whereas element C belongs to 3rd period and so it has 3 shells. As a result, atoms of element C are bigger than atoms of element B.
(c) Element B belongs to group 13 and so its valency is 3.

Question 40.
The atomic number of elements A,B, C, D and E are given below:

From the above table, answer the following questions:
(a) Which two elements are chemically similar? (b) Which is an inert gas? (c) Which elements belong to 31 d period of Periodic Table? (d) Which element among these is a non-metal?
Answer:
(a) Element C and D have similar electronic configuration, i.e. 2, 8, 2 and 2, 2. Hence, these two elements are similar.
(b) Electronic configuration of element B is 2, 8. Hence, it has 8 electrons in its outermost shell. So, element B is an inert gas.
(c) Atomic number of C is 12 and so its electronic configuration is (2, 8, 2) i.e. it has 3 shells. Hence, element C belongs to 3rd period of the Modern Periodic Table.
(d) Element A (2, 5) is non-metal.

Question 41.
What is meant by periodicity? Why are the properties of elements in the same group similar?
Answer:
1. The properties of the elements n the Modern Periodic Table depend on regular changes that are seen in the electronic configuration of elements arranged in groups and periods. This law is known as periodicity.
2. Valency decides the properties that an element possesses.
3. The electronic configuration of the valence orbit of all the elements in a group is same.
4. Thus, in the same group, properties of an element remains same.

Question 42.
The electronic configuration of an element ‘X’ is 2, 8, 8, 2. To which (a) period and (b) group of the Modern Periodic Table does ‘X’ belong? State its valency. Justify your answer in each case.
Answer:
(a) The electronic configuration of element X states that the element has four shells namely K, L, M and N. Hence, element X belongs to 4th period.
2. There are 2 electrons in the outermost i.e. N shell and so element X belongs to group 2.

Question 43.
Atomic number is considered to be a more appropriate parameter than atomic mass for classification of elements in a periodic table. Why?
Answer:
1. Atomic mass does not properly tell us the properties of elements. Moreover, the elements cannot be arranged in the increasing order of atomic mass because of certain anomalies.
2. The properties of elements depend upon the number of electrons present in the valence shell. The number of electrons can be known by its atomic number. In this regard, an element can be classified in a better way through atomic number.
3. Hence, atomic number is considered to be a more appropriate parameter than atomic mass for classification of elements in the Periodic Table.

Question 44.
From the part of a Periodic Table, answer the following questions:

(a) Atomic number of oxygen is 8. What would be the atomic number of Fluorine?
(b) Out of ‘X’ and ‘Q’ which element has larger atomic size. Give reason for your answer.
(c) Out of ‘Y’ and ‘Z’ which element has smaller atomic size. Give reason for your answer.
Answer:
(a) In the Modern Periodic Table, elements are arranged on the basis of increasing order of their atomic numbers. Fluorine comes after oxygen and so atomic number of fluorine is 9.
(b) As one moves from, left to right in a period, the atomic size of the element goes on decreasing. Hence, atomic size of element Q is smaller than that of element X.
(c) As one moves down from one period to another, the numbers of shells increase. Element Y lies above element Z. So, element Y will have lesser shells than Z and hence Y’s atomic size will be smaller.

Question 45.
The atomic number of an element ‘X’ is 20. Write —
(a) Its valency
(b) Whether it Is a metal or non-metal
(c) The formula & compound formed when the element ‘W reacts with an element ‘Y’ of atomic number 8.
Answer:
(a) Since the atomic number of X is 20, its electronic configuration will be 2, 8, 8, 2. Hence, its valency is 2.
(b) Its valency is 2, so it belongs to group 2 i.e. group of metals. Moreover, since its valency is 2. it will lose 2 electrons which means it shows the property of metal.
(c) Atomic number of Y is 8. So its electronic configuration is (2, 6). In other words, valency of Y is 2. Moreover valency of X is also 2. So, the compound formed by sharing 2 electrons will be compound ‘XV’.

Question 46.
An element X belongs to 13th group of periodic table. State its valency. What will be the formula of its sulphate?
Answer:
Element X belonging to group 13th has 3 valence electrons and so its valency is 3. Sulphate of X = X₂(SO₄)₃.

Question 47.
The elements Li, Na and K, each having one valence electron, are in period 2,3 and 4 respectively of Modern Periodic Table.
(a) In which group of the periodic table should they be?
(b) Which one of them is least reactive?
(c) Which one of them has the largest atomic radius? Give reason to justify your answer in each case.
Answer:
(a) All these elements have 1 valence electron and so they belong to Group 1.
(b) Li is the least reactive. Since Li is 2nd period it has only 2 shells. So, its outermost orbit is very near the nucleus. As a result, is difficult to remove the electron. This makes it least reactive of the three elements.
(c) Element K lies in 4th perioded has 4 shells. Hence, it has largest atomic radius of the three.

Question 48.
Sudha madam asked students to place three elements namely lithium, sodium and potassium in one group or say triad. Now, answer the following questions.
(a) How could she place all the three elements in one group?
(b) What is the name of this group? State the properties of the elements of this group.
Answer:
(a) Sudha madam placed the three elements in the same group because all three elements have similar properties.
(b) The name of this group is ‘alkali metal group’. Properties of the elements of this group are-

• All these elements are metals.
• Valency of each element is 1.
• These metals readily react with water and form alkalis and hydrogen gas.

Question 49.
In the Science class this morning, Prakash sir taught students how Newlands classified the elements known to him. He then gave the students an assignment to role-play Newland. Prakash sir gave them some elements along with their atomic weights which are as follows.

The students were to answer the following questions. Study them and draft your answer.
(a) Which two elements will have similar properties on the basis of Newland’s law of octaves? Justify the reason for your selection.
Answer:
Newlands had arranged the elements in the increasing order of their atomic weights. So, first we arrange the elements.

As per Newland’s law of Octaves, if elements are arranged in the increasing order of the atomic weights then every 8th element will have properties similar to the just one. Hence, element ‘a’ having atomic mass 2 will have properties similar to element ‘d’ having atomic mass 23. Similarly, element ‘b’ and ‘g’ will can be grouped together.

Question 50.
Introduction to the groups of elements of the Modern Periodic Table

Introduction:

118 elements are known to us. Majority of these elements are metals. These metals are not very reactive. They are malleable and ductile and good conductors of heat and electricity. The remaining elements are non-metals and semi-metals (metalloids).

Brief explanation of each group of the Modern Periodic Table is as follows:

(1) Group 1 elements (Alkali metals): Elements of this group are known as ‘Alkali metals’. There are in total 6 alkali metals from Lithium to Francium. Alkali metals are very soft and highly reactive metals. They are so reactive that they have to be stored in substances like oil because if kept open they will start reacting with atmospheric gases.

If they are not found freely then how do we obtain them? Well, chemists extract them from their compounds. Can explode when exposed to air The metals are malleable, ductile and good conductors. They have only 1 valence electron. They are always eager to lose this 1 electron and join with other electron to form a positive ion or say cation.

(2) Group 2 elements (Alkaline earth metals): Elements of this group are known as ‘alkaline earth metals’. There are in total 6 elements from Beryllium to Radium. Alkaline metals are also reactive metals but not as reactive as alkali metals. They have 2 electrons in their outer shell. So, they tend to lose 2 electrons to form positive ions.

(3) Group 3 to Group 12 (Transition metals (elements)): Elements of this group are known as ‘Transition metals’. They are moderately reactive. They are malleable and ductile. They have high melting and boiling points

(4) Group 13 to 16: Group 13 to 16 contains a mix of metals, metalloids and non-metals.

Metalloids: Metalloids are elements that have few properties of metals and few of non-metals.
Metalloids are the smallest class of elements (the other two classes of elements are metals and nonmetals). There are just seven rnetalloids namely boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb). tellurium (Te) and polonium (PO).

(5) Group 17 (Halogens): Halogens are non-metals. Examples of halogens are Fluorine (F), chlorine (Cl), Bromine (Br), Iodine (I), etc. The term “halogen” means salt-former” and compounds containing halogens are called “salts”.All halogens have 7 electrons in their last shell. So, they gain 1 electron They are highly reactive with alkali metals (group 1) arid alkaline metals (group 2).

(6) Group 18 (Noble gases): The 18th group (extreme right group) consists of non-metal elements known as noble gases or inert gases. The valence (outermost) shells of all noble gases are completely filled with electrons (2 for Helium,8 for all others). Hence, the noble gases are very stable and extremely less reactive.There are 6 noble gases that occur naturally. They are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn).

(7) Inner-transItion metals: Inner-transition metals are the elements listed below the Modern Periodic Table. These are 30 rare earth metals.

Why are they placed below the table and not In the table?
If you look at the Modern Periodic Table, you will notice that after the element Lanthanum (atomic number 57), the next element that appears is Hafnium (atomic number 72). This means that the elements with atomic number 58 to 71 are missing. Similarly, after actinium (atomic number 89) the next element that occurs is Rutherfordium (atomic number 104).

The missing elements having atomic number 58 to 71 are placed below the table and are called lanthanides because they exhibit properties similar to element Lanthanum (atomic number 57). Similarly, the missing elements having atomic number 90 to 103 are placed below the table and are called actinides because they exhibit properties similar to element actinium (atomic number 89)

The inner-transition metals were discovered quite recently. Scientists and chemists are yet to determine their exact characteristics and properties. Hence, it was decided to keep these elements separate below the Modern Periodic Table.

(8) The case of hydrogen — an important point: Hydrogen is placed in group 1. However, its position is a matter of discussion. The reason for it follows.
Hydrogen shows similarity to group 1 elements i.e. alkalis as well as to group 17 elements i.e. halogens.

(a) Hydrogen’s similarity with alkalis: Hydrogen has only 1 electron in its valence shell. All the electrons that have 1 electron in their outer shells are placed in group 1 For example, lithium (2, 1), sodium (2, 8, 1) and so on. This means, hydrogen’s electronic configuration resembles that of alkali metals and hence it is placed in group 1. Hence, both alkalis and hydrogen have the tendency to lose 1 electron and become positive ion. The way alkali metals react with halogens (i.e. elements of group 17), oxygen and sulphur, hydrogen also reacts in a similar way. Hence, hydrogen is placed in the grotip of alkalies.

(b) Hydrogen’s simIlarity with halogens: The hydrogen also shows properties to halogens i.e. elements of group 17. All the elements of halogen group are 1 electron short of attaining their octet configuration. For example, Fluorine (2, 7),chlorine (2, 8, 7), etc. Hydrogen with electronic configuration of (1) is also 1 short of attaining octet configuration. So, the way the halogens require 1 electron to complete their configuration and become stable, hydrogen also require 1 electron to complete its configuration.

Hence, both halogens and hydrogen have the tendency to accept 1 electron and become negative ion. Halogens exist as diatornic molecules i.e. the molecule of halogens is made up of 2 atoms. For example, fluorine exists as, F₂, chlorine as C₁₂ and so on. Similarly, hydrogen also exists as a diatomic molecule i.e. as H₂. Just like halogens, hydrogen also combines with metals and non-metals to form covalent compounds. Considering the above points we conclude that hydrogen also resembles halogens.

(c) Conclusion: Since hydrogen shows properties similar to alkanes (group 1 elements) and halogens (group 2 elements). its position in the periodic table is till date contradictory and hence not fixed. However, we place hydrogen in group I simply for the purpose of convenience (technically, it can be placed in halogen group as well.

Very Short Answer Type Question :

Question 1.
What is the need to classify elements?
Answer:
We can learn a lot about the properties and characteristics of elements when they are systematically classified. Hence, there arose a need to classify them.

Question 2.
What were elements first classified into?
Answer:
When elements were classified for the first time they were classified as metals and non-metals.

Question 3.
Who was Dobereiner?
Answer:
John Wolfgang Dobereiner was a German chemist who in 1817 tried to arrange the elements with similar properties into groups.

Question 4.
What are Dobereiners Triads?
Answer:
Dobereiner identified some elements that could be grouped into groups of three on the basis of the atomic mass of elements. Such groups were called Dobereiner’s Triads.

Question 5.
State the principle of Dobereiner’s Triads.
Answer:
When three elements are arranged in the order of increasing atomic masses, the atomic mass of middle element is roughly the average of the atomic masses of other two elements.

Question 6.
The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called as? Give one example of such a set of elements.
Answer:
Such an arrangement of elements is called Dobereiner’s triads.

Example:
Lithium (ll): atomic mass = 6.9, Sodium (Na): atomic mass = 23, Potassium (K): atomic mass = 39
Atomic mass of Na =\(\frac{39+6.9}{2}\)= 23

Question 7.
Can Na, Si, Cl form Dobereiner’s triad?
Answer:
Although the atomic mass of Si Is approximately the average of atomic masses of Na and Cl. still these three elements cannot form Dobereiner’s triads because the properties of these elements are different.

Question 8.
What can be considered as Newland’s biggest discovery?
Answer:
Newland’s thought that on arranging elements on the basis of increasing atomic masses, every 8th element has properties similar to the 1st can be considered his biggest discovery.

Question 9.
Fill the table given below with first seven elements as arranged by Newland’s.

Answer:

Question 10.
State Newland’s Law of Octaves.
Answer:
When elements are arranged in the order of increasing atomic masses, the properties of the eight elements (starting from a given element) are repetitions of the properties of the first element.

Question 11.
As per Patel Sir, Newland Law of Octaves was applicable only to tighter elements. Did he pass a correct statement? Why?
Answer:
Yes, Reason: Newland’s Law of Octaves is applicable only to lighter elements having mass upto element calcium. After calcium, the 1st and 8th element do not have similar properties.

Question 12.
Define Mendeleev’s Periodic Law.
Answer:
The properties of elements are a periodic function of their atomic masses,

Question 13.
What was the base of classification adopted by Mendeleev?
Answer:
(a) Arranging elements on the basis of increasing atomic masses.
(b) Grouping elements that have similar properties.

Question 14.
Why did Mendeleev leave some gaps in his periodic table?
Answer:
Mendeleev predicted that the gaps will be occupied by elements that would be discovered in some time future.

Question 15.
Magnesium and calcium are metals. Is this enough to classity both these elements into same group? Why?
Answer:
No, Reason: Elements should be put into same group on the basis of some fundamental properties that are common to such elements.

Question 16.
Mendeleev had knowledge about how many elements?
Answer:
63

Question 17.
What did Mendeleev focus on while considering chemical properties as one of the bases of classification?
Answer:
While considering chemical properties as one of the bases of classification, Mendeleev concentrated on compounds that the elements formed by combining with oxygen and hydrogen.

Question 18.
Why did Mendeleev consider compounds that elements formed with hydrogen and oxygen as a base of classification?
Answer:
Hydrogen and oxygen are very reactive and they form compounds with most elements. This provides a great help in classifying the elements. Hence,……….

Question 19.
Why were noble gases discovered very late?
Answer:
Elements are discovered on the basis of their reactivity with oxygen. Noble gases are inert i.e, not reactive. Hence, it was quite difficult to imagine existence of such elements.

Question 20.
Why hydrogen dId not find a suitable position in Mendeleev’s Periodic Table?
Answer:
Hydrogen is an element which shows properties similar to the elements of group 1 as well as to the elements of group 17 ie. halogens. Hence, till date where should hydrogen be placed is not correctly determined even in the Modern Periodic Table.

Question 21.
What are isotopes?
Answer:
Atoms of an element having same atomic number but different atomic masses are called isotopes.

Question 22.
Define valency.
Answer:
The combining capacity of an atom of an element to acquire noble gas configuration is called valency.

Question 23.
State Modern Periodic Law.
Answer:
Properties of elements are a periodic function of their atomic number.

Question 24.
What are groups in a Modern Periodic Table?
Answer:
The 18 vertical columns of the Modern Periodic Table are called groups.

Question 25.
What are periods?
Answer:
The 7 horizontal rows of the Modern Periodic Table are called periods.

Question 26.
Look at the table given below.

To which group do these elements belong? Wily?
Answer:
All these elements belong to Group 1 because every element has one valence electron in its outermost shell.

Question 27.
Loot at the table below.
Answer:

State the electronic configuration of each element. To which period will these elements belong? Why?
Answer:
1. Mg – 2, 8, 2, Si – 2, 8, 4, S – 2, 8, 6 Ar – 2, 8, 8
2. Electrons of all these elements are arranged in 3 shells and so all lie in 3rd period.

Question 28.
An element Q lies in group 2 whereas element R lies in group 15 of the periodic table. Now fill the table.

Answer:

Question 29.
Why all the elements in spite of belonging to one same period have different properties?
Answer:
Within a period the number of valence electrons is different for each element. So, the tendency to lose or gain electrons for each element varies. Hence …………

Question 30.
Define atomic size (radius).
Answer:
The distance between the centre of the nucleus and the outermost shell of an isolated atom is called atomic radius or atomic size.

Question 31.
What are metalloids?
Answer:
Elements that have some properties of metals and some of non-metals are known as metal bids.

Question 32.
State the number of valence electrons and valency of most of the elements of group 18.
Answer:
Generally, the elements of group 18 have 8 valence electrons and their valency is O (zero).

Question 33.
An element Q lies in group 2 of the periodic table. State the formula of its chloride and oxide.
Answer:
Chloride — XCl₂; Oxide XC.

Question 34.
State the formula of the compound formed when an element X of group 14 combines with element Y of group 16.
Answer:
Electronic configuration of X : 2, 8, 4
Electronic configuration of Y : 2, 8, 6
∴ X will combine with 2 atoms of Y.
∴ The formula of compound formed will be XY₂.

Question 35.
Write the positions of metallic, non-metallic and metalloid elements in the Modern Periodic
Table.
Answer:
In the Modern Periodic Table, metallic elements are on the left side while non-metallic elements are on the right side. Metalloids or semi-elements are in the middle of the Periodic Table.

Question 36.
The construction of Modern Periodic Table owes to research of which scientist?
Answer:
(A) Sir Ramsay
(B) Moseley
(C) Lord Rayleigh
(D) Mendeleev

True or False

1. ………… elements are known to us and out of them are available naturally.
Answer:
118; 98

2. Elements with similar properties were arranged in groups for the first time in the year …………
Answer:
1817

3. It was ………… who found that properties of elements are a periodic function of their atomic masses.
Answer:
Newlands

4. As per the ‘Law of Octaves’, lithium shows properties similar to …………
Answer:
Sodium

5. Newland’s Periodic Table ended at element number named ………… named …………
Answer:
56; Thorium

6. The main credit for classifying elements goes to …………  (write full name).
Answer:
Dmitri Ivanovich Mendeleev

7. While considering chemical properties as one of the parameter for classifying the elements, Mendeleev focused on compounds that the elements formed with ……………
Answer:
Hydrogen and oxygen

8. Element X has atomic number 20. It should find its place in ………… period.
Answer:
Fourth

9. Element that took place of Eka-silicon was …………
Answer:
Germanium

10. Atomic mass of Eka-aluminium: 68; Atomic mass of Eka-gallium……………
Answer:
69.7

11. Hydrogen’s properties resemble ………… to ……….. and (Name of groups).
Answer:
Alkalis; halogens

12. In 1913 …………….. showed that the atomic number of an element is a more fundamental property than its atomic mass.
Answer:
Henry Moseley

13. Atomic number is denoted as ……………….
Answer:
‘Z’

14. The Modern Periodic Table takes care of limitations of Mendeleev’s Periodic Table. (state the number)
Answer:
Three

15. With respect to group number, hydrogen can be placed in
Answer:
Group number 1 and also 7.

16. The number of shells that can be accommodated in a shell can be calculated using the formula ………..
Answer:
2n²

14. Atomic radius is measured in ………….
Answer:
Pico meter (pm)

18. 1 pm = ………………………
Answer:
10^-12m

19. Noble gases are also known as ……………..
Answer:
Inert gases

20. Metallic elements have …………….. electrons n their outermost orbit.
Answer:
1 to 3

21. If you move from right to left in a period, the atomic radius will ……………..
Answer:
Increase

22. Boron, silicon and germanium are examples of
Answer:
Metalloids (or semi-metals)

23. Oxides of elements tend to be basic.
Answer:
Metallic

24. Non-metals are electro-
Answer:
Negative

True or False

1. Triads are written in the increasing order of their atomic number. — False
2. Look at the table below.

Do these elements satisfy even to be a triad? (Yes = True, No = False) — True (Yes)
3. Dobereiner could identify only three triads and so his system of classification was not useful. — True
4. Newlands’ law failed after calcium — True
5. Newlands thought that after 63rd element no more elements will be discovered — False
6. Newlands put nickel and carbon in same slot — False
7. Newlands’ law worked well for heavier elements — False
8. Mendeleev knew about 7 more elements as compared to Newlands. — True
9. Nickel (atomic mass 58.93) appeared before cobalt (atomic mass 58.71) in Mendeleev’s periodic table. So as to group them together on the basis of similar properties. — False
10. Mendeleev’s Table did not consist of noble gases. — True
11. Halogens have valency of 7. — True
12. The position of hydrogen in the Modern Periodic Table is till date a matter of argument. — True
13. An element having atomic number 25.5 can be easily placed between Mn and Fe. — False
14. Each period signifies a new electronic shell getting filled. — True
15. As we go down in a group, the distance between the outermost electron and the nucleus decreases. — False
16. A zigzag line separates metals from non-metals within the Modern Periodic Table. — True
17. As the effective nucleus charge acting on the valence shell electrons increases across a period, the tendency to lose electrons decrease. — True
18. Metals are electro-positive. — True

Match the Following:

Question 1.

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

Question 2.

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

Haryana State Board HBSE 10th Class Maths Notes Chapter 10 Circles Notes.

Haryana Board 10th Class Maths Notes Chapter 10 Circles

Introduction
We have studied in class IX that a circle is a collection of all the points in a plane which are at a constant distance from a fixed point. The constant distance is called the radius and fixed point is known as the centre. We have also studied various terms related to a circle like chord, sector, segment, arc etc.

In this chapter we shall learn about tangents to a circle and some of their properties. Let us now examine the different situations that can arise when a line and a circle are given in a plane.

Tangent of a circle
Consider a circle with centre O and a line P in the plane of the circle. There are three types of possibilities arise as shown in the figures below:

In the figure (i), the line P and the circle have no common point hence, the line P is known as a non intersecting line with respect to the circle.
In the figure (ii), the line P intersects the circle in two distinct points A and B. It is called a secant of the circle.
In the figure (iii), the line P intersects the circle in one and only one point A and is said to be a tangent to the circle. The point A at which the tangent line meets the circle is called the Point of contact.

Normal: The line containing the radius through the point of contact is also sometimes called the normal to the circle at the point.
Supplementary angles: Two angles having sum 180° are called supplementary angles.
Co-interior angles: Interior angles on the same side of the transversal are called co-interior angles or consecutive interior angles.
Concentric circles: Circles having the same centre are called concentric circles.
Parallelogram: A quadrilateral having each pair of opposite sides equal and parallel is called parallelogram.

Rhombus : A parallelogram having all the sides equal is called a rhombus.
Circumscribed circle: The circumscribed circle or circumcircle of a polygon is a circle passing through all the vertices of the polygon. The centre of this circle is called circumcentre and its radius is called the circumradius.

Inscribed circle: Inscribed circle or incircle of a polygon is the largest circle that can be contained in the polygon and it touches each side of the polygon at a point. Hence each side of the polygon is a tangent to the incircle. The centre of this circle is called incentre and its radius is called inradius.

A line which intersects the circle at only one point is known as the tangent to the circle. In the given figure PQR is a tangent to the circle and point Q is the point of contact.

The word tangent to a circle has been derived from the latin word “Tangere”, which means ‘to touch’ and was introduced by the Denish mathematician Thomas Fineke in 1583.

The tangent to a circle is a special case of the secant, when the two end points of its chord coincide.

Some Properties of tangent to a circle
Theorem 10.1:
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: A circle with centre O and a tangent AB to the circle at a point P.
To Prove: OP ⊥ AB.
Construction: Take any point R, other than P on the tangent AB. Join OR. Suppose OR meets the circle at Q.
Proof :
OP = OQ (Radii of the same circle)
But OP < OQ + QR
OP < OR

Thus, OP is shorter than any other line segment joining O to any point of AB, other than P.
We know that among all line segment joining the point O to a point on AB, the shortest one is perpendicular to AB.
Hence, OP ⊥ AB. Proved

Number of Tangents from a point on a circle
To get an idea of the number of tangents from a joint on a circle, let us perform the following activity:

From figure (III) PR₁ and PR₂ are two tangents drawn from a point P lying outside the circle. These tangents touch the circle at R₁ and R₂ respectively. So, R₁ and R₂ are called points of contact of tangenta PR₁ and PR₂.
(i) There is no tangent to a circle passing through a point lying inside the circle.
(ii) There is one and only one tangent to a circle passing through a point lying on the circle.
(iii) There are exactly two tangents to a circle through a point lying outside the circle.

Length of Tangent: The length of segment of the tangent from the external point P and point of contact with the circle is called the length of the tangent from the point P to the circle.
From figure (III), PR₁ and PR₂ are the length of tangents from P to the circle.

Theorem 10.2:
The lengths of tangents drawn from an external point to a circle are equal.
Given: AP and AQ are two tangents from a point A to a circle C (O, r).
To Prove: AP = AQ
Construction: Join OA, OP and OQ.
Proof: AP is a tangent at P and OP is the radius through P.

Similarly AQ is a tangent at Q and OQ is the radius through Q.
∴ OQ ⊥ AQ
In the right ΔOPA and ΔOQA, we have
OP = OQ [equal radii of the same circle]
AO = AO [common]
∠OPA = ∠OQA [each is 90°]
∴ ΔOPA ≅ ΔOQA [By RHS congruence]
⇒ AP = AQ [By CPCT Proved]
Hence, AP = AQ. Proved

Haryana State Board HBSE 10th Class Maths Notes Chapter 9 Some Applications of Trigonometry Notes.

Haryana Board 10th Class Maths Notes Chapter 9 Some Applications of Trigonometry

Introduction
We have studied about trigonometric ratios in earlier classes. In this chapter, we shall use the trigonometric ratios, to solve the problems regarding the heights and distances of various objects. Two common terms used in this chapter are angle of elevation and angle of depression.
1. Elevation: The height to which something is raised above a point of reference.
2. Depression: The depth to which something is lowered below a point of reference.
3. Line of sight: The line drawn from the eye of an observer to the point in the object viewed by the observer.
4. Complementary angles: Two angles having a sum of 90°, are called complementary angles.

5. Speed = Distance/Time
6. Alternate interior angle: The pair of angles on opposite sides of the transversal but inside the two lines are called alternate interior angles.
If the two lines are parallel, then the alternate interior angles formed are equal.

Heights and Distances
Angle of Elevation: If a person is looking up at an object, the acute angle measured from the horizontal level to the line of sight when the object being viewed is called the angle of elevation. Here

O is the point of observation, P is the position of the object, OP is the line of sight. OA is horizontal level and α is the angle of elevation.

Angle of Depression: The angle between the line of sight and horizontal level through the eye of the observer, when the object being viewed is below the horizontal level is called angle of depression. Here position of observer is at O. OM is the horizontal level through O. OP is the line of sight. β is the angle of depression when the object at P is observed from O.

Remarks:
(a) Numerically the angle of elevation is equal to the angle of depression.
(b) The angle of elevation and angle of depression both are measured with the horizontal.
(c) The angle of elevation or depression increases as the observer moves towards the object.

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

Haryana Board 10th Class Maths Important Questions Chapter 10 Circles

Short / Long Answer Type Questions

Question 1.
Two concetric circles of raddi a and b (a > b) are given. Find the length of chord of the larger circle which touches the smaller circle.
Solution :
Let two concetric circles of centre O with raddi a and b (a > b)
AO = OB = a and OM = b
OM ⊥ AB (By theorem 10.1)

In right triangle AMO, we have
AO² = AM² + OM²
(By Pythagoras theorem)
⇒ a² = AM² + b²
⇒ AM² = a² – b²
⇒ AM = \(\sqrt{a^2-b^2}\) units
we know that perpendicular drawn from centre to chord bisect the chord
∴ AM = MB
AB = 2 × AM
= 2 × \(\sqrt{a^2-b^2}\) units
Hence, length of chord of larger circle
= 2\(\sqrt{a^2-b^2}\) units

Question 2.
How many tangents can be drawn on the circle of radius 5 cm from a point lying outside the circle at distance 9 cm from the centre.
Solution :
Draw a circle of radius 5 cm with centre O. Let P be the point outside 9 cm from the centre O.

From external point P, we can draw two tangents PA and PB only.

Question 3.
In the given figure, two circles touch each other at the point C. Prove that common tangent to the circles at C, bisects the common tangents at P and Q.

Solution :
PT and CT are tangents at circle with centre A from external point T
∴ PT = CT …… (1) [by theorem 10.2]
Similarly
QT = CT ……… (2)
From equ. (1) and (2)
PT = QT
Now, PQ = PT + QT
⇒ PQ = 2 PT
⇒ PT = QT = \(\frac {1}{2}\)PQ
Hence, common tangent CT bisects the common tangents at P and Q. Hence proved

Question 4.
In the given figure ΔABC is circumscribing a circle, the length of BC is ….. cm.

Solution :
BQ = BP = 3 cm
[by theorem 10.2]
AR = AP = 4 cm
[by theorem 10.2]
Now,
CR = AC – AR
= 11 – 4 = 7
CR = QC
⇒ QC = 7 cm
BC = BQ + QC
= 3 + 7 = 10 cm

Question 5.
In the given figure, PQ and PR are tangents to the circle with centre O such that ∠QPR = 50°, then find ∠OQR.

Solution :
Join QR
PQ = PR by theorem 10.2]
⇒ ∠PQR = ∠PRQ [Angles Opp. to equal sides are equal] … (1)
In ΔPQR, we have
∠PQR + ∠PRO + ∠QPR = 180°
⇒ ∠PQR + ∠PRO + 50° = 180° [using equ. (1)]
⇒ 2∠PQR = 180° – 50°
⇒ ∠PQR = \(\frac {130°}{2}\)
⇒ ∠PQR = 65°
Now, OQ ⊥ PQ
[By theorem 10.1]
⇒ ∠PRO = 90°
⇒ ∠OQR = ∠PQO – ∠PQR
= 90° – 65° = 25°
Hence, ∠OQR = 25°

Question 6.
In the given figure PQ is a chord of length 6 cm of the circle of radius 6 cm. TP and TQ are tangents to the circle at points P and Q respectively. Find the ∠PTQ.

Solution :
Here PQ = 6 cm
PO = OQ = 6 cm
[equal raddi]
∴ PO = OQ = PQ
⇒ POQ is a equilateral triangle
∠POQ = 60°
Now, OP ⊥ PT and OQ ⊥ TQ
[By theorem 10.1]
In quadrilateral POQT, we have
∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360°
⇒ 60° + 90° + ∠PTQ + 90° = 360°
⇒ 240° + ∠PTQ = 360°
⇒ ∠PTQ = 360° – 240°
⇒ ∠PTQ = 120°

Question 7.
In the figure, AB and CD are common tangents to two circles of unequal raddi. Prove that AB = CD.

Solution :
Produced AB and CD to meet at P
Now, PA = PC
(By theorem 10.2) …. (1)

PB = PD
(By theorem 10.2) … (2)
Subtracting equ. (2) from equ. (1), we get
PA – PB = PC – PD
⇒ AB = CD Hence proved

Question 8.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q.

If PA = 12 cm, QC = DQ = 3 cm, then find PC + PD.
Solution :
AC = CQ
(By theorem 10.2) ……. (1)
BD = DQ
(By theorem 10.2) …….(2)
But CQ = DQ …….(3)
By equ. (1). (2) and (3), we get
AC = BD ……….(4)
Now,
AP = PB
(by theorem 10.2)
⇒ CP + AC = PD + BD
⇒ CP + AC = PD + AC
[using equ. (4)]
⇒ CP = PD
CP = AP – AC = 12 – CQ [∴ CQ = AC]
⇒ CP = 12 – 3 = 9 cm
[CQ = DQ = 3 cm]
∴ DP = CP = 9 cm
PC + PD = 9 + 9 = 18 cm

Question 9.
In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ΔABC is 54 cm², then find the lengths of sides AB and AC.

Solution :
Join OA, OB and OC
Draw OE ⊥ AC and OF ⊥ AB
Area of ΔABC = 54 cm² (given)
⇒ ar (ΔBOC) + ar (ΔAOC) + ar (ΔAOB) = 54
⇒ \(\frac {1}{2}\)BC × OD + \(\frac {1}{2}\)AC × OE + \(\frac {1}{2}\)AB × OF = 54
⇒ \(\frac {1}{2}\)[BC × 3 + AC × 3 + AB × 3] = 54
⇒ \(\frac {1}{2}\)[BC + AC + AB] = 54
⇒ AB + BC + AC = \(\frac{54 \times 2}{3}\)
⇒ AB + BC + AC = 36 cm ……….(1)
AF = AE [by theorem 10.2]
Let AF = AE = x cm
BD = BF and CD = CE [by theorem 10.2]
⇒ BF = 6 cm and CE = 9 cm

Now AB + BC + AC = 36
⇒ AF + BF + BD + CD + CE + AE = 36
⇒ x + 6 + 6 + 6 + 9 + 9 + 1 = 36
⇒ 2x + 30 = 36
⇒ 2x = 6
⇒ x = 3
∴ AB = 6 + 3 = 9 cm, AC = 9 + 3 = 12 cm.

Question 10.
Prove that the parallelogram circular scribing a circle is rhombus.
Solution :
Let ABCD be a parallelogram
∴ AB = CD and AD = BC
we know that length of tangents drawn to a circle from an exterior point are equal in length.
AP = AS …(1)
PB = BQ …(2)
CR = CQ …(3)
DR = DS ……(4)

Adding equations (1), (2), (3) and (4), we get
AP + PB + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AD + BC
⇒ AB+ AB = BC + BC
⇒ 2 AB = 2 BC
⇒ AB = BC
since, in a parallelogram ABCD adjacent sides AB and BC are equal.
So, ABCD is a rhombus.

Fill in the Blanks

Question 1.
Tangent is perpendicular to the …….. through the point of contact.
Solution :
radius

Question 2.
Only …….. tangents can be drawn to a circle from an external point.
Solution :
two

Question 3.
Lengths of tangents from an external point to a circle are ……..
Solution :
equal

Question 4.
The line containing the radius through the point of contact is also sometimes called the …….. to the circle at the point.
Solution :
normal

Question 5.
Circles having the same …….. are called concentric circles.
Solution :
centre

Question 6.
The word ……. to circle has been deribed from the latin word “……..”
Solution :
tangent

Question 7.
The point at which the tangent line meets the circles is called the ……..
Solution :
point of contact

Multiple Choice Questions

Choose the correct answer each of the following :

Question 1.
In the adjoining figure, if PA and PB are tangents to the circle with centre such that ∠APB = 50°, then ∠OAB is equal to :

(a) 25°
(b) 30°
(c) 40°
(d) 50°
Solution :
From figure
∠1 + ∠P = 180°

[Opposite angles of cyclic ∠1]
⇒ ∠1 + 50° = 180°
⇒ ∠1 = 180° – 50° = 130°
In ΔOAB,
OA = OB (radius)
⇒ ∠2 = ∠3 (Isosceles Δprop)
∠2 + ∠3 + ∠AOB = 180°
(Angle sum property of Δ)
⇒ ∠2 + ∠2 + 130° = 180°
⇒ 2∠2 = 180° – 130° = 50°
⇒ ∠2 = ∠OAB = \(\frac {50°}{2}\) = 25°
Hence correct choice is (a).

Question 2.
In the adjoining figure, if AP = 4 cm, CR = 5 cm and BQ = 6 cm the perimeter of ΔABC is:

(a) 15 cm
(b) 30 cm
(c) 25 cm
(d) 20 cm
Solution :
From Fig. BQ = BP = 6 cm
CQ = CR = 5 cm
PA = AR = 4 cm
The perimeter of ΔABC = BQ + CQ + CR + AR + AP + BP
= 60 + 5 + 5 + 4 + 4 + 6
= 30 cm
Hence correct choice is (b).

Question 3.
In the adjoining figure, if ∠AOB = 125°, then ∠COD is equal to :
(a) 62.5°
(b) 45°
(c) 35°
(d) 55°

Solution :
From fig. ∠AOB + ∠COD = 180°
125° + ∠COD = 180°
⇒ ∠COD = 180° – 125°
= 55°
Hence correct choice is (d).

Question 4.
In the adjoining figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to :
(a) 4 cm
(b) 2 cm
(c) 2\(\sqrt{3}\) cm
(d) 4\(\sqrt{3}\) cm

Solution :
∠OAT = 90° (Angle between radius and tangent)
Cos 30° = \(\frac {AT}{OT}\)

\(\frac{\sqrt{3}}{2}\) = \(\frac {AT}{OT}\)
⇒ AT = 2\(\sqrt{3}\)cm
Hence correct choice is (c).

Question 5.
In the given figure QR is a common tangent to the given circle, which touch externally at P. If QP = 3.8 cm, the length of QR is :
(a) 38 cm
(b) 7.6 cm
(c) 5 cm
(d) 1.9 cm

Solution :
From Fig QP = PT = PR = 3.8 cm
So, QR = QP + PR
= 3.8 + 3.8
= 7.6 cm
Hence correct choice is (b).

Question 6.
In the given figure, if ∠APO = 40°. Then ∠AOB is :
(a) 100°
(b) 80°
(c) 50°
(d) 40°

Solution :
Given ∠APO = 40°
and ∠OAP = 90°
(radius is ⊥ to tangent)
∴ In ΔPAQ
∠AOP = 180 – (40 + 90)
= 50°
∴ ∠AOB = 2 × ∠AOP
= 2 × 50°
= 100°
So correct choice is (a)

Question 7.
In the adjoining figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(a) 20°
(b) 40°
(c) 35°
(d) 45°

Solution :
∠DQR = 90°
(Angle between radius and tangent)
∠BQD = ∠DQR – ∠BQR
= 90° – 70°
= 20°
Similarly ∠AQD = 20
∠AQB = 20° + 20°
So correct choice is (b).

Question 8.
If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm then length of each tangent is equal to :
(a) \(\frac {3}{2}\)\(\sqrt{3}\) cm
(b) 6 cm
(c) 3 cm
(d) 3\(\sqrt{3}\) cm
Solution :
∠OBA = 90°
(Angle between radius and tangent)
In right ΔOBA
Tan 30° = \(\frac {OB}{AB}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac {3}{AB}\)

AB = 3\(\sqrt{3}\)
∴ AC = AB = 3\(\sqrt{3}\) (tangents from same external point)
So correct choice is (d).