Class 10

Haryana Board HBSE 10th Class Maths Important Questions and Answers

HBSE 10th Class Maths Important Questions in English Medium

1. Real Numbers Class 10 Important Questions
2. Polynomials Class 10 Important Questions
3. Pair of Linear Equations in Two Variables Class 10 Important Questions
4. Quadratic Equations Class 10 Important Questions
5. Arithmetic Progressions Class 10 Important Questions
6. Triangles Class 10 Important Questions
7. Coordinate Geometry Class 10 Important Questions
8. Introduction to Trigonometry Class 10 Important Questions
9. Some Applications of Trigonometry Class 10 Important Questions
10. Circles Class 10 Important Questions
11. Constructions Class 10 Important Questions
12. Areas related to Circles Class 10 Important Questions
13. Surface Areas and Volumes Class 10 Important Questions
14. Statistics Class 10 Important Questions
15. Probability Class 10 Important Questions

HBSE 10th Class Maths Important Questions in Hindi Medium

• Chapter 1 वास्तविक संख्याएँ Important Questions
• Chapter 2 बहुपद Important Questions
• Chapter 3 दो चरों वाले रखिक समीकरण युग्म Important Questions
• Chapter 4 द्विघात समीकरण Important Questions
• Chapter 5 समांतर श्रेढ़ियाँ Important Questions
• Chapter 6 त्रिभुज Important Questions
• Chapter 7 निर्देशांक ज्यामिति Important Questions
• Chapter 8 त्रिकोणमिति का परिचय Important Questions
• Chapter 9 त्रिकोणमिति का अनुप्रयोग Important Questions
• Chapter 10 वृत्त Important Questions
• Chapter 11 रचनाएँ Important Questions
• Chapter 12 वृतों से संबंधित क्षेत्रफल Important Questions
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Important Questions
• Chapter 14 सांख्यिकी Important Questions
• Chapter 15 प्रायिकता Important Questions

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 Science Important Questions Chapter 7 Control and Coordination Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 7 Control and Coordination

Question 1.
What is the importance of control and co-ordination?
Answer:
1. Multicellular organisms are made-up of various organs and organ system.
2. It is a basic need of the body to control and co-ordinate these organs and organ systems, so that they can function properly and accomplish the voluntary as well as involuntary functions.

Question 2.
What is a stimulus?
Stimulus:
1. An event that encourages an action or creates sensation is called a stimulus. (Plural : Stimuli)
2. All living organisms — humans, plants and animals respond to changes occurring in their surroundings. These changes work as stimuli.
Example:

• On seeing sudden bright sunlight, our eyes gets closed. Here, the sudden bright light is stimulus whereas closing of eyes is response.
• If someone pours hot or cold water on us, we can feel it and also differentiate the temperature of water.
• A sunflower always faces the sun.
• Thus, heat, light, touch, weather changes, sound, etc. all are examples of external stimuli.
• If we are hungry, we feel like eating. Thus hunger, thirst, etc. are internal stimuli of the body.

Question 3.
What are receptors? Name few receptors found in human body, their location and function.
Answer:
Receptors:
1. A specialized structure in the human body that receives external stimuli is called a receptor.
2. These receptors are located in our sense organs. The main ones are —

Question 4.
What is a nerve cell? Explain its structure, function and working.
Answer:
Nerve cell:
Nerve cells or neurons are the building as well as functional units of the nervous system. They carry information from one part of the body to another.
Nerve cell has three components:
(i) Cell body,
(ii) Dendrites and
(iii) Axon
(i) Cell body: The cell body of a neuron is like a typical animal cell which contains cytoplasm and a nucleus.
A number of short or long fibers stretch out from the cell body. These fibers are called nerve fibers.
(ii) Dendrites: The short fibers on the cell body are known as dendrites.
(iii) Axon: The longest fiber on the cell body is called axon.

Movement of Impulse (working of nerve cell):
1. The end of the dendritic tip receives message from receptors. It then initiates a neuro-chemical reaction due to which an electric impulse known as nerve impulse is produced.
2. The dendrites pass the impulse to the cell body and then to axon upto the end of axon i.e. axon terminal bud.
3. Synapse — The gap between nerve ending of an axon of one neuron and dendrite of the next neuron is called synapse.
4. At the end of the axon terminal bud, the electrical impulse releases certain chemicals.
5. These chemicals cross the gap or synapse and start similar electrical impulse in the dendrite of the next neuron. This is how the impulse or message travels from one neuron to another in the body.
6. Finally with the help of similar synapse, the impulses are delivered from neurons to other cells such as muscle cells or glands.

Question 5.
What happens at the synapse between two neurons?
Answer:
Nerve cell: Nerve cells or neurons are the building as well as functional units of the nervous system. They carry information from one part of the body to another.
Nerve cell has three components:
(i) Cell body,
(ii) Dendrites and
(iii) Axon
(i) Cell body: The cell body of a neuron is like a typical animal cell which contains cytoplasm and a nucleus.
A number of short or long fibers stretch out from the cell body. These fibers are called nerve fibers.
(ii) Dendrites: The short fibers on the cell body are known as dendrites.
(iii) Axon: The longest fiber on the cell body is called axon.

Question 6.
What is a nervous tissue? What does It specialize in?
Answer:
1. Nervous tissue is made up of organized network of nerve cells or say neurons.
2. Nervous tissue specializes in conducting messages through electrical impulses from one part of the body to another.

Question 7.
Identify the parts of a neuron in a diagram: (1) where information is acquired, (2) through which information travels as an electrical impulse, and (3) where these impulse must be converted Into a chemical signal for onward transmission.
Answer:
Parts of a neuron: a — nucleus, b — cell body, c — dendrite, d — axon, e — nerve ending.

(1) Information is acquired at the end of the dendrite tip (c) of the nerve cell.
(2) Information travels as an electrical impulse through dendrite, cell impulse through dendrite, cell body and along the axon (d) to its end (e).
(3) At synapse, this impulse gets converted into a chemical signal for onward transmission.

Question 8.
Explain reflex action.
Answer:
Reflex action:

• Reflex action is an involuntary and instant response of the muscles or glands to a stimulus. It takes place without involving the brain.
• Reflex action is an automatic process. It is the simplest form of response in nervous system.

Example:
(1) We immediately pull-back our hands when we suddenly touch a hot vessel or when someone pricks us a pin.
→ In such cases, we react or say respond before the message can reach our brain. Such a response is called reflex action.
(2) Knee jerk response,
(3) Coughing,
(4) Blinking eyes,
(5) Yawning, movement of diaphragm,
(6) Contraction of pupil,
(7) Sneezing,
(8) Mouth watering or looking or smelling favourite food, etc. are other examples.

Question 9.
What role does reflex action plays when we suddenly see a bright light or when we cough?
Answer:
1. When we suddenly see a bright light, pupil of our eyes becomes small and it takes some time for the eyes to adjust to the bright light. In this case, reflex action protects the retina from damage that can be caused due to sudden and excessive light.
2. Coughing is another example in which reflex action does the job of clearing our wind pipe.

Question 10.
Explain reflex arc.
Answer:
Reflex arc:

• The pathway or say the route taken by the nerve impulses in a reflex action is known as the reflex arc.
• Reflex arcs allow rapid response.

Example:
1. A reflex action is an automatic response to a stimulus. So, when we suddenly touch a hot plate (a stimulus) we quickly, without thinking pull our hand back.
2. This type of very quick and automatic response is called reflex action.
3. The diagram shows the pathway taken by the nerve impulses in this reflex action.

• The heat is sensed by thermoreceptor in our hand.
• The receptor fingers an impulse in a sensory neuron which transmits the information to the spinal cord.
• From spinal cord, the impulse is passed to a relay neuron which in turn passes it to a motor neuron.
  Relay neurons are the communicating neurons between the sensory and motor neurons.
• The motor neuron transmits the impulse to the muscle of the arm. The muscle then contracts and pulls the hand away from the hot plate.
• The muscle of arm is an effector because it responds to the stimulus.
  This pathway along which the impulse travels is called the reflex arc.

Question 11.
Reflex actions have also evolved in animals. Explain.
Answer:
1. Generally, animals do not possess the complex neuron network needed for thinking. If at all they possess, the network is very basic. This means that animals cannot think as fast as humans.
2. However, they do need to respond to several sudden situations such as attack from other animals. To tackle such situations, the reflexes of animals are quite evolved.

Question 12.
How do Involuntary actions and reflex actions differ from each other?
Answer:

Question 13.
How does control and co-ordination take place in humans? Also, explain the functions of nervous system.
Answer:
There are two systems which co-ordinate different activities In humans:
(A) Nervous system and (B) Endocrine system (or Hormonal system)
These two systems work together to control and co-ordinate all our activities such as physical,emotional behaviour and thinking processes.

(A) Nervous system:

• It controls and co-ordinates all the parts of the body.
• The nervous system co-ordinates voluntary muscles, thus allowing a person to perform activities such as dancing, reading, writing, etc.
• The nervous system also co-ordinates certain involuntary functions like heart beat and breathing.
• The nervous system collects the information from the surrounding, interprets it and then responds accordingly.
• The nervous system also passes information from one system to other system.
• The units that make up the nervous system are called nerve cells or neurons. Thus, nerve cells are the functional units of a nervous system.

(B) Endocrine system (or Hormonal system):

• There are certain glands in the endocrine system which release chemical substances (chemical messengers) called hormones in the body.
• Generally, hormones regulate the slow activities of the body such as growth, metabolism, etc.
• The hormones are secreted in a small quantity and reach to the target organ by blood. Generally hormones do not travel long.

Question 14.
State the organization of human nervous system.
Answer:
Organs of Human Nervous System:

• The Central Nervous System of a human being is one of the important parts of the control and co-ordination system of the body.
• The Peripheral Nervous System does the task of establishing communication between the central nervous system and the other parts of the body.

Question 15.
Explain the structure and functioning of human brain along with a diagram.
Answer:
Human brain is divided into three major parts:
(i) Fore brain,
(ii) Mid brain and
(iii) Hind brain.

(I) Fore brain :

• Fore brain is the main thinking part of the brain.
• It has regions which receive sensory impulses from various receptors.
• Separate areas of the fore-brain are specialized for hearing, smell, sight, etc.
• There are other separate areas where this sensory information is interpreted by putting it together with information from other receptors as well as with information that is already stored in the brain. Finally, a decision is made by the brain considering all the information it gathers.

(ii) Midbrain:

• Mid brain connects the forebrain and hind brain.
•  It is the centre for visual and auditory reflexes.

(iii) Hind brain:

• Hind brain consists of three regions: (A) Cerebellum which lies on dorsal side and (B) Pons and (c) Medulla oblongata.
• It controls involuntary activities like breathing, heart-beat, blood pressure, peristaltic movements, etc.

Question 16.
How is the central nervous system protected? OR How is the brain protected from external injuries? OR How are the delicate organs of human nervous system protected? OR Give reason: The central nervous system is well protected.
Answer:
1. The central nervous system (CNS) consists of the brain and spinal cord.
2. The brain is protected by an extremely hard bony-box called cranium (skull). The skull envelopes the brain.
3. Inside the skull, the brain lies within the cerebro-spinal fluid-filled balloon-like structure which provides cushioning and works as a shock absorber.
4. The spinal cord is protected in a hard skeletal structure called a vertebral column or backbone.

Question 17.
How does the nervous tissue cause an action?
Answer:
(a) Role of nervous tissue in causing the action:

• The neurons receive the stimuli from the surroundings with the help of sensory organs namely eyes, ears, nose, tongue and skin.
• Then the information is sent to the central nervous system (mostly to the brain).
• The brain analyzes the stimulus and makes the decision about what has to be done. According to this, the order is formulated. Now this decision order is sent to the related muscles through the neurons.
• Finally, the muscles act accordingly and thus the whole action response takes place.

(b) Role of muscle tissue in causing the action:

• When a nerve impulse reaches the muscle, the muscle fibre moves. This movement starts at the cell level.
• The muscle cells will move by changing their shape so that they can become short.
• The musde cells have special proteins that change their shape as well as their at angement in the cell in response to the nerve impulses. When this happen, the proteins get arranged in a different manner and hence muscle movement takes place.
• This movement of muscle is known as voluntary movement of muscle.

Question 18.
Write a note on movement of voluntary muscles.
Answer:
Role of muscle tissue in causing the action:

• When a nerve impulse reaches the muscle, the muscle fibre moves. This movement starts at the cell level.
• The muscle cells will move by changing their shape so that they can become short.
• The musde cells have special proteins that change their shape as well as their at angement in the cell in response to the nerve impulses. When this happen, the proteins get arranged in a different manner and hence muscle movement takes place.
• This movement of muscle is known as voluntary movement of muscle.

Question 19.
Explain plant movement In brief.
Answer:
Plant movement:

• Plants remain fixed at a place with their roots in the ground and so they cannot move from one place to another.
• However, movement of individual parts of a plant such as root, leaves, etc. is possible when they are subjected to some external stimuli like sunlight, gravitational force, water, touch, etc. Such movements of plants are called plant movements.
• The plant movement with respect to external stimuli falls into two main categories: (I) Tropism and (II) Nastism.

Question 20.
Explain “Plants show two different kinds of movements”. OR State and explain briefly the types of movements in plants.
Answer:
The movement of plants can be classified into two types. They are:
(1) Growth dependant movement:
→ When seeds are planted, they sprout as a plant and grow taller with time. This is directional movement.
(2) Growth independent movement.

Question 21.
How is co-ordination in plants different from co-ordination in animals? OR Plants co-ordinate slowly as compared to animals. OR Give scientific reason — “In general, the plants cannot respond to stimulus quickly”.
Answer:
1. The change in the environment to which the organisms respond and react is called stimulus (Plural: stimuli). Animals possess nervous system as well as hormones for co-ordinating their activities whereas plants have only hormones for responding to stimulus.
2. Animals have sense organs such as nose, ears and eyes which help them sense things and activities faster.
3. Plants too are capable to sense the effect of gravity, light, water, chemicals and touch but only by the means of their hormones.
4. Thus, plants identify all these environmental changes only with the help of their hormones. So, since plants are devoid of nervous system. they cannot respond to stimulus quickly.

Question 22.
How do some plants respond immediately on touching? Give an example. OR How do sensitive plants detect the touch and show movement?
Answer:
1. Generally, when we touch the plants (like chhui-mui or lajamani), they use electrical or chemical means to convey this information from cell to cell.
2. Here, the cells of plants change their shape by changing the amount of water in them. As a result, the cells may swell-up or shrink. Consequently. the shape of the cell gets changed.
Example:
When we touch the leaves of a chhui-mui (= Lajamani) plant i.e. a sensitive’ or ‘touch-me-not’ plant of mimosa family, it immediately begins to fold-up and droop.

Question 23.
What are tendrils? Explain. OR Explain the movement in the pea plant due to growth.
Answer:
1. In some plants, -a special thread like part arise from various parts of the plant, which are known as tendrils.
2. Such tendrils are highly sensitive to touch. When the thread structure comes in contact of any hard-support, it twists to it and starts growing in that specific direction.
3. Such type of plants are called climbers (t). For example, pea plant.
4. Because this growth is directional, it appears as if the plant is moving.

Question 24.
(a) What is meant by tropism? Explain with example.
(b) Mention the types of tropism. Explain each one in detail.
Answer:
(A) Tropism (Tropic movement) :

• It the growth movement (response) in a plant organ is due to the effect of an external and directional stimuli, then it is called tropism or tropic movement.
• It the plant shows growth in the direction of the stimulus, then it is called positive tropism whereas if the plant shows growth away from the stimulus, it is called negative tropism.
• Plants show movement with respect to five stimuli.

(B) Types of tropism:

Question 25.
What is the difference between photo nasty and thermonasty?
Answer:
Photonasty is a movement of the plant in response to light, whereas thermonasty is a movement of the plant in response to temperature. For example, flowers of sunflower and lotus open their petals in the morning when sunlight falls on them. Similarly, when temperature is increased, the crocus flower opens up.

Question 26.
What is chemical communication or hormonal co-ordination? Explain. OR What are hormones? Explain.
Answer:
1. One of the major limitations of nerve impulses is that they cannot reach each and every cell of animal body. Hence, there exists another means of communication which is called chemical communication or hormonal co-ordination.
2. Hormonal co-ordination takes place with the help of special chemical compounds called hormones.
The hormones are secreted by various glands of the endocrine system within the body.
3. Hormones are known as chemical co-ordinators or even chemical messengers.
4. Hormones are produced by a group of cells, get transported to a desired location and then act upon the cells of the target area simply by diffusion.
5. Although hormonal co-ordination takes place slowly, it can reach all the desired cells of the body.
6. Adrenaline, thyroxin, insulin, etc. are some of the hormones released in the body.

Question 27.
Describe plant hormones.
Plant hormone:
1. Plant hormones are special chemical compounds which are synthesized at one place/organ of the plants and migrate to the target organ to act. They play an important role in control and co ordination in plants, as well as in growth, development and responses to the environment.

Control and co-ordination in plants:

• Hormonal action is the main weapon to respond to various stimuli in the plants.
• Basically, there are two types of plant hormones. They are —

(A) Growth-promoting hormones:

(i) Auxins:

• It particularly acts on the areas of shoot-tip and root-ends, It helps the cells to grow longer.
• When light comes from one side of the plant. auxins diffuse towards the shady side of the shoot.
  The concentration of auxin stimulates the cells to grow longer on the shoot side, away from the light. Consequently, plants appears to bend towards light.

Gibberel lins:

• It helps in the growth of the cells and the stems.

Cytokinins:

• It is present in the higher concentration in the areas of the plants where rapid cell division occurs. For example, fruits and seeds.
• It promotes cell-division.

(B) Growth resisting / inhibitory hormones:

(i) Abscisic Acid (ABA):

•  It is responsible for wilting and falling of the leaves.
• It is responsible for the detachment of leaves, flowers and fruits.

Question 28.
Describe human endocrine system and give a list of glands in human body.
Answer:
1. Human nervous system and endocrine system work in the co-ordination with each other.
2. The endocrine system is the network of glands in the body. These glands release hormones.

Hormones:

• Basically, hormones are the chemical substances (chemical messengers).
• They play an important role in various metabolic processes.
• The hormones are secreted in a small quantity and reach to the target organ by blood. Generally, the target organ is nearby.

Glands: There are two types of glands.
They are — (1) Exocrine glands and (2) Endocrine glands.

Main endocrine glands of the human body

Question 29.
Give an idea about the hormones released by glands. Also state the main functions of these hormones.
Answer:

Question 30.
Write a short note on adrenal gland. OR Which hormone is called fight or flIght hormone? State its effect on the body.
Answer:
Adrenal gland:
1. The adrenal gland releases a hormone called adrenaline. Adrenaline hormone is also called fight or flight (run-away) hormone.
2. This hormone is secreted directly into the blood and carried to different parts of the body.

Effects of adrenaline:

• When adrenaline is released the heart beats become fast and so, more oxygen gets supplied to our heart.
• The muscles around the small arteries of heart and other organs contract due to which the digestive system and skin receive less blood. This diverts the blood to our skeletal muscles.
• Moreover, the breathing rate increases due to contractions of the diaphragm and the rib muscles.
• All these responses together enable the animal and human body to be ready for dealing with the situation of fight or flight i.e. fight or run-away immediately from the situation.

Question 31.
Write a note on thyroid gland and the importance of the hormone It releases.
Answer:
Thyroid gland:
1. Thyroid gland secretes a hormone called thyroxin.
2. Thyroxin controls metabolism rate of carbohydrate, protein and fat.
3. Deficiency of iodine can cause deficiency of thyroxin in the body.
4. Deficiency of iodine in our diet results in hypothyroidism and enlargement of thyroid gland. This results into a disease called goiter.
5. Neck of people suffering from goiter becomes swollen.
6. Goiter can be prevented by consuming iodized salt i.e. salt containing iodine.

Question 32.
It is highly recommended to consume iodized salt on a daily basis. Give reason.
Answer:
Adrenal gland:
1. The adrenal gland releases a hormone called adrenaline. Adrenaline hormone is also called fight or flight (run-away) hormone.
2. This hormone is secreted directly into the blood and carried to different parts of the body.

Effects of adrenaline:

• When adrenaline is released the heart beats become fast and so, more oxygen gets supplied to our heart.
• The muscles around the small arteries of heart and other organs contract due to which the digestive system and skin receive less blood. This diverts the blood to our skeletal muscles.
• Moreover, the breathing rate increases due to contractions of the diaphragm and the rib muscles.
• All these responses together enable the animal and human body to be ready for dealing with the situation of fight or flight i.e. fight or run-away immediately from the situation.

Question 33.
Write a short note on pancreas.
Answer:
Pancreas:
1. Pancreas is an endocrine gland as well as an exocrine gland.
2. As an endocrine gland. it secretes insulin and glucagon which maintain sugar level in the blood.
3. If insulin is not secreted in proper amounts, the sugar level in blood rises causing many harmful effects.
4. When the sugar level in blood rise, it gets detected by the cells of the pancreas. The cells respond to this rise by producing more insulin. As the blood sugar level falls, insulin secretion is reduced.

Question 34.
How is hormone secretion regulated?
Answer:
1. Hormone secretion is regulated by the feedback mechanism. In general, a particular gland secretes a specific hormone.
2. The hormone reaches to the target organ and performs its function. when that specific function completed, there is no need of that hormone. so, the message reaches to the brain and the brain stops that gland to secrete the hormone further. this is how hormone secretion is regulated.

Question 35.
Giving response to a stimulus is the basic characteristic of every living organism. give reason.
Answer:
1. The creation of living organisms is such that all of them experience the effects of changes in the atmosphere that surrounds them.
2. These effects work as stimuli for the organisms and they respond by making some changes in their bodies.
3. Slow or fast, but the organism will respond to stimuli such as heat, cold, sound, touch, etc.
4. Thus, response to stimuli is the basic characteristic of every living organisms.

Question 36.
Label the parts of a neuron in the given figure.

Answer:
(a) Dendrite
(b) Cell body
(c) Axon
(d) Nerve ending

Question 37.
We respond very quick to reflexes but very slow to the act of thinking. Give reason.
Answer:
1. Thinking is a complex activity and so it involves a complicated interaction of several nerve impulses that too from many neurons.
2. Thinking center is located in forebrain.
3. When this centre receives signals through sensory nerves from various parts of the body, it analyzes the signals and formulates the instructions. The instructions then transmit from brain to the target organ. This process takes some time and hence the response is slow.
4. In reflex action, initially the brain is not involved in decision making. So, the instruction is generally given by spinal cord before the message could reach the brain for analysis and decision making purpose. As a result, the decision making is quite quick.
5. Hence, we respond very quick to reflexes but very slow to the act of thinking.

Question 38.
Label the parts (a), (b), (c) and (d) and show the direction of flow of electrical signals in Figure.
Answer:
(a) Sensory neuron
(b) Spinal cord (CNG)
(c) Motor neuron
(d) Effector (muscle in arm)

Question 39.
Name the plant hormones responsible for the ………………
(a) elongation of cells
(b) growth of stem
(c) promotion of cell division
(d) falling of senescent leaves
Answer:
(a) Auxiri – elongation of cells
(b) Gibberellin – growth of stem
(c) Cytokinin – promotion of cell division
(d) Abscisic acid – falling of senescent leaves

Question 40.
Out figure (a), (b) and (c), which appears more accurate and why?
Answer:
Figure (a) is more appropriate.
Reason: In a plant, shoots are negatively geotropic and hence grow upwards and roots are positively geotropic and so grow downwards.

Question 41.
Why do multicellular organisms use hormonal system in addition to the nervous system? Multicellular organisms use hormonal system in addition to the nervous system due to following two reasons:
Answer:
1. The nervous system transmit electrical impulses. The electrical impulse can reach only those cells that are connected to the PNS.
2. Secondly, after the generation of the electrical impulses, the cell takes some time to reset or to normalize.
3. As a result, multicellular organisms use hormonal system in addition to the nervous system.

Question 42.
State the characteristics of animal hormones.
Answer:
Characteristics of animal hormones:
1. Hormones are chemical compounds secreted from specific endocrine glands.
2. Hormones are chemical co-ordinators and are also called chemical messengers.
3. Generally, they control slow activities of the body such as growth and metabolism.
4. They are synthesized at places away from where they need to act. They then travel to the target cells, tissues or organs and act by simple diffusion.
5. They are transported through blood.
6. They are secreted in very small quantities.

Question 43.
Give an idea about hormones related with growth in human.
Answer:
Mainly two hormones are related with growth in humans. They are —
(1) Growth hormone:

• It is secreted by the pituitary gland.
• Growth hormone regulates growth and development of the body.

(2) Thyroxln:

• It is secreted by thyroid gland.
• Thyroxin regulates carbohydrate, protein and fat metabolism in the body. This is necessary for balanced growth of body.

Question 44.
Which disorders can we face If hormonal balance gets disturbed?
Answer:
If hormones are secreted in inappropriate quantities, we may face the following disorders:
(1) Gigantlsm: It is caused due to excess secretion of growth hormone. Under this disorder, an individual becomes extremely tall.
(2) Dwarfism: Deficiency of growth hormone in childhood leads to dwarfism. Under this disorder, the height of an individual does not grow much and he remains dwarf-sized.

(3) Goiter: It is caused by the deficiency of iodine in our diet.

• Iodine deficiency affects the synthesis of thyroxin and causes goiter.
• The thyroid gland gets enlarged which results in enlargement of the neck region.

(4) Diabetes: If insulin hormone is not secreted in sufficient amounts, the sugar level rises in the blood. This causes diabetes.

Question 45.
A person suffering from diabetes is given the injections of insulin. Give reason.
Answer:
1. In human bodies, pancreas release a hormone called insulin which regulates the level of sugar in the blood.
2. When insulin is not secreted in proper amounts, the sugar level in the blood rises and the person may suffer from a disease called diabetes.
3. Diabetes can be harmful for heart, brain, eyes, kidneys, etc.
4. Hence, in order to save the diabetic person from such harmful effects, he is given the injections of insulin which regulates his sugar level in the blood.

Question 46.
Give differences between: Plant hormones and animal hormones
Answer:

Question 47.
Label the endocrine glands shown In the figure. Also state their locations.

Answer:
The endocrine glands and its location are as follows:

Question 48.
Differentiate between tropic movement and nastic movement.
Answer:

Question 49.
Differentiate between response in plants and response in animals.
Answer:

Question 50.
Differentiate between movement of leaves of sensitive plants and movement of shoot towards light.
Answer:

Question 51.
Differentiate between central nervous system and Autonomous nervous system.
Answer:

Question 52.
Differentiate between cerebrum and cerebellum.
Answer:

Question 53.
Smita’s father was complaining about frequent urination, pain in legs and a frequent weight loss to Smite’s mother and she discussed the things with her daughter, Smita she when returned from her school. Listening to this, Smite said to her mother that her father should visit a doctor, who told her mother that her husband is having an elevated level of blood glucose. He should take care of his diet and should exercise regularly to maintain his normal glucose level. On the basis of given text, answer the following questions.

Questions:
(1) Name the disease he is suffering from and name the hormone, whose deficiency causes it.
(2) Identify the gland that secretes it and mention the function of this hormone.
(3) Explain, how the time and amount of secretion of this hormone is regulated in human system?
(4) What values were shown by Smita and her father’?
Answers:
(1) Smita’s father is suflenng from diabetes. It Is caused by the deficiency of insulin hormone in the body.
(2) Pancreas secretes insulin. Insulin helps in regulating the sugar level in the blood.
(3) When the level of sugar increases in the blood, the body responds by producing more insulin. When blood sugar level falls, secretion of insulin gets reduced.
(4) The values shown by Smita are empathy and alertness in decision making. The values shown by her father are neglecting his health.

Question 54.
On a beautiful sunny day, Pritesh was in a beautiful mood f ride. His father was equally keen and both of them rode on his motorcycle. As soon as they reached the main cross road, they saw a tragic accident of a motorcycle with a car. The rider was not wearing helmet. So, on getting hit, the rider of the motorcycle fell and was again hit by the road divider. Crowd gathered and the rider was immediately shifted to the nearby hospital through an ambulance. Prltesh’s father looked at Pritesh with utmost anger and merciful state at the same time because Pritesh too was not wearing the helmet and he too could meet such an accident.

Questions:
(1) Which part of brain is involved when you are learning to drive a vehicle?
(2) Injury to which part of brain leads to instant death of an individual? Why?
(3) Pritesh became quite fearful. Which gland do you think became more active when he saw the accident? Which hormone was poured in blood? What changes he might have felt in his body?
(4) Which values did Pritesh’s father displayed?
Answers:
(1) Cerebrum and cerebellum are involved while learning to ride.
(2) Injury to medulla oblongata in the brain leads to instant death. This is because all the involuntary activities are controlled by It.
(3) Seeing such an event, the adrenal gland of Pritesh must have become active. As a result. adrenaline hormone must have secreted in blood. Under such situation, Pritesh perspired and his heartbeats and breathing rate increased.
(4) Pritesh’s father showed the value of empathy and respecting one’s life because it is extremely precious. Moreover, he also showed the value of concern of a father for the son.

Question 55.
After studying chapter Control and Co-ordination, you learnt that probably your maid is suffering from one of the diseases discussed in that chapter. The front portion of her neck was bulged out abnormally. On discussing with her you also found that she faced difficulty in breathing and even swallowing food. Now, look at the questions given here and answer them.

Questions:
(1) From which disease do you think your maid is suffering?
(2) What is the main cause of this disease?
(3) How can you help her at an immediate basis?
(4) What values did you display here?
Answers:
(1) The symptoms show that she is suffering from goiter.
(2) Goiter is mainly caused due to deficiency of iodine.
(3) On an immediate basis we can provide her with iodine supplements as well as recommend her to use only iodized salt in her food.
(4) The values of alertness, caring and sound decision making are displayed here.

Question 56.
There are three plants A, B and C. The flowers of plant A open their petals in bright light during the day but close them when it gets dark at night. On the other hand, the flowers of plants B open their petals at night but close them during the day when there is bright light. The leaves of plant C fold up when touched with fingers or any other solid object.

(a) Name the phenomenon shown by the flowers of (j) plant A, and (ii) plant B.
(b) Name one flower each which behaves like the flower of (i) plant A, and (ii) plant B.
(c) Name the phenomenon exhibited by the leaves of plant C.
(d) Name a plant whose leaves behave like those of plant C.
(e) Which plant(s) exhibit phenomenon based on growth movement?
Answer:
(a) (i) Photonasty (ii) Photonasty
(b) (i) Dandelion flower (ii) Moonflower
(c) Thigmonasty
(d) Sensitive plant (Mimosa pudica)
(e) A and B

Very Short Answer Type Question

Question 1.
Which two body systems are important in control and co-ordination?
Answer:
Hormonal system (= Endocrine system) and Nervous system.

Question 2.
Name main components of a neuron.
Answer:
Dendrites, Cell body and Axon.

Question 3.
What Is the function of dendrite?
Answer:
It carries nerve impulses which has been received from the axis of the previous neuron towards the cell-body of a neuron.

Question 4.
What is the role of axon?
Answer:
It carries nerve impulses from the cell body towards next neuron’s dendrite or tissue.

Question 5.
Define stimuli.
Answer:
An event that encourages the action or creates sensation is called stimulus. (Plural = Stimuli)

Question 6.
Define Response.
Answer:
The reaction of living organisms against the stimulus induced by the change in the external environment is called response.

Question 7.
Define Co-ordination.
Answer:
Different organs of the body jointly function systematically against the stimuli and react to express proper response against the stimuli. This process is called coordination.

Question 8.
What is a spinal cord? What ¡s its function?
Answer:
Spinal cord is a cylindrical structure which is the extension of medulla oblongata. Spinal cord is a part of Central Nervous System (CNS).

Question 9.
State the function of spinal cord.
Answer:
(i) To conduct sensory information from peripheral nervous system to the brain.
(ii) To conduct motor information from brain to various effectors.

Question 10.
Reflex arc of some animals is much evolved. Why?
Answer:
Because thinking process of the brain of these animals is not so fast. So, the evolved reflex arc serves as an efficient way of performing various functions.

Question 11.
How do we feel that we have eaten enough?
Answer:
The centre of hunger and satiety are located in a separate part of the fore-brain. They give the feeling of being hungry or satisfied.

Question 12.
Name the most important part of the human brain.
Answer:
Cerebrum

Question 13.
State one function each of cerebellum and Pons.
Answer:
Cerebellum maintains balance of body and posture. Pons regulates respiration.

Question 14.
Give the functions of medulla oblongata.
Answer:
Medulla oblongata controls involuntary functions such as breathing, heart beats, digestion, blood pressure, pristaltic movements of alimentary canal, etc, It also controls reflexes such as coughing, sneezing, swallowing, secretion of saliva and vomiting.

Question 15.
What does CNS stand for?
Answer:
CNS is the abbreviation for Central Nervous System.

Question 16.
Which is the most important part of the human brain?
Answer:
Cerebrum. It is also the largest part of the brain.

Question 17.
State two examples of movement that we make for protecting ourselves.
Answer:
(1) When we suddenly see bright light, we compress our pupils.
(2) We pull back our hand when we touch a hot object.

Question 18.
What are the components of peripheral nervous system?
Answer:
Cranial nerves and spinal nerves

Question 19.
Why thinking Is considered a complex activity?
Answer:
Thinking involves a highly complex interaction of several nerve impulses from many neurons. Hence,……..

Question 20.
How does a muscle cell move?
Answer:
Muscle cells move by changing their shape i.e. either by shortening or elongating.

Question 21.
What are phytohormones?
Answer:
The chemical substances which help in control and co-ordination in plants are known as phytohormones.

Question 22.
Give examples of phytohormones.
Answer:
Auxins, Gibberellins, Cytokinins, Abscisic acid (ABA), Ethylene, etc.

Question 23.
What is photoperlodism?
Answer:
The developmental response of plants to the relative lengths of light and dark periods is known as photoperiodism.

Question 24.
What are the main types of a plant movement?
Answer:
(i) Growth dependant movement and
(ii) Growth independent movement

Question 25.
Name the plant which shows thigmonasty.
Answer:
Mimosa plant

Question 26.
Give an example of movement of a plant part which is caused by the loss of water.
Answer:
Thigmonastic movement as seen in mimosa plant.

Question 27.
What does a root do in response to gravity? What is this phenomenon known as?
Answer:
In response to gravity, the root grows towards the earth. This phenomenon is known as geotropism.

Question 28.
What does a stem do in response to light? What is this phenomenon known as?
Answer:
Stem bends towards the light. This phenomena is called phototropism. For stems, it is also called positive phototropism.

Question 29.
What does a stem do In response to gravity? What is this phenomenon known as?
Answer:
Stem always grows away from the earth i.e. away from gravity. This is called negative geotropism.

Question 30.
What does a root do in response to light? What Is this phenomenon known as?
Answer:
The roots of the plant bend away from the light.
This is called negative phototropism.

Question 31.
Name one hormone secreted by the pituitary gland.
Answer:
Thyroid Stimulating Hormone (TSH).

Question 32.
Where are hormones synthesized (made) in the human body?
Answer:
Hormones are made by specialized glands or tissues throughout the body. Most but not all hormones are part of endocrine system.

Question 33.
Which gland secretes the growth hormone ?
Answer:
Pituitary gland

Question 34.
Name the disease caused by the deficiency of insulin hormone in the body.
Answer:
Diabetes

Question 35.
Why are hormones called chemical messengers?
Answer:
Hormones carry information in the form of chemicals which regulate the biological processes of body. Hence,………..

Question 36.
Why both, chemical signal and electrical impulses are needed in higher animals?
Answer:
In higher animals, electrical impulses bring immediate response to only those cells that are connected by nervous tissue while chemical
signals reach to each arid every cells of body. Hence, …..

Question 37.
Which mechanism is considered as a second way of control and co-ordination in our body?
Answer:
Secretion of hormones by the endocrine system is considered as a second way of control and co-ordination in our body.

Question 38.
Where is growth hormone auxin synthesized and where is it diffused?
Answer:
Auxin is synthesized at the shoot tip and is diffused at that side of the shoot which is in shade.

Question 39.
State the site of secretion as well as the function of growth hormone releasing factor.
Answer:
Site of secretion: Hypothalamus; Function: Stimulates the pituitary gland to release growth hormone.

Fill in the Blanks

1. ………… is the unit of human nervous system.
Answer:
neuron

2. The word ‘reflex’ means …………
Answer:
Sudden action

3. In a neuron, conversion of electrical signal to a chemical signal occurs at/in …………
Answer:
axonal end

4. The highest co-ordinating part …………
Answer:
brain

5. Aging process in the plants is phytochrome
Answer:
Cytokinins

6. …………  causes seed dormancy in plants.
Answer:
ABA

7. The main function of abscisic acid in plants is to …………
Answer:
inhibit growth

8. The substance that triggers the fall of mature leaves and fruits from plants is due to (hormone).
Answer:
abscisic acid

9. The plant movement with respect to external stimuli falls into two main categories which are ……… and ………..
Answer:
Tropism; Nastism

10. In comparison to animals, plants are devoid of …………  for coordinating activities.
Answer:
Nervous system

11. Growth of pollen tube is an example of …………
Answer:
Chemotroplsm

12. The auxin hormone acts particularly on and of …………  the plant body.
Answer:
Shoot; root

13. ………… is also known as a master gland.
Answer:
Pituitary gland

14. Iodine is necessary for the synthesis of …………  hormone.
Answer:
Thyroxln

15. Dwarfism results due to …………
Answer:
Less secretion of growth hormone

16. The hormone which increases the fertility in males is called …………
Answer:
testosterone

True Or False

1. Except unicellular organisms, all respond to stimuli. — False
2. We can feel touch due to thigmo receptors. — True
3. Both neurons and hormones can be considered as messengers. — True
4. Generally, the nervous system co-ordinates fast activities whereas endocrine system co ordinates slow activities. — True
5. Electrical impulses will reach only those cells which are connected by nervous tissue. — True
6. Mid brain is the centre for visual and auditory reflexes. — True
7. A nerve impulse makes the muscle move. — True
8. Generally, hormones travel far off in the body to deliver the message. — False
9. Plants show two types of movement. — True
10. Cytokinin is present in the higher concentration in the areas of the plants where rapid cell division occurs. — True
11. Pancreas maintains carbohydrate, protein and fat metabolism in the body. — False
12. Pancreas is an endocrine gland as well as an exocrine gland. — True

Match the Following 

Question 1.

Answer:
(A q), (B – s), (C – p), (D – r)

Question 2.

Answer:
(A -q), (B – r), (C – p)

Question 3.

Answer:
(A – q), (B – s), (C – r), (D – p)

Question 4.

Answer:
(A – r), (B – t), (C – p), (D – q) (E – s)

Haryana State Board HBSE 10th Class Maths Notes Chapter 2 बहुपद Notes.

Haryana Board 10th Class Maths Notes Chapter 2 बहुपद

→ बहुपद की पात- घर x के बहुपद p (x) में x को उच्चतम घात (power) यहुपद की घात कहलाती है; जैसे बहुपद 5x³ – 4x² + x – \(\sqrt{2}\) चर x में घात 3 का बहुपद है।

→ घातों एक, दो और तीन के बहुपद क्रमशः रखिक बहुपद, द्विघात बहुपद एवं त्रिवात बहुपद कहलाते हैं।

→ एक द्वियात बहुपद ax² + bx + c जहाँ a, b, c वास्तविक संख्याएँ हैं और a ≠ 0 है के रूप का होता है।

→ यदि p (x), x में कोई बहुपद है और कोई वास्तधिक संख्या है, तो p(x) में x को k से प्रतिस्थापित करने पर प्राप्त वास्तविक संख्या p(x) का x = k पर मान कहलाती है और इसे P (k) से निरूपित करते हैं।

→ कोई वास्तविक संख्या k, बहुपद p(x) का शून्यक कहलाती है यदि p(k) = 0 हो।

→ किसी रैखिक बहुपद ax + b का शून्यक \(\frac{-b}{a}\) होता है।

→ एक बहुपद p (x) के शून्यक उन बिंदुओं के x-निर्देशांक होते हैं जहाँ y = p(x) का ग्राफ x-अक्ष को प्रतिच्छेद करता है।

→ एक द्विघात बहुपद के अधिक-से-अधिक दो शून्यक और एक त्रिघात बहुपद के अधिक-से-अधिकं तीन शून्यक हो सकते हैं।

→ यदि द्विघात बहुपद ax² + bx + c के शून्यक α और β हो, तो उनका योगफल व गुणनफल निग्न होगा-
α + β = \(-\frac{b}{a}\), αβ = \(\frac{c}{a}\)

→ यदि किसी द्विधात बहुपद के दो शून्यक दिए गए हों तो बहुपद होगा-
बहुपद = x² – (शून्यकों का योग) x + शून्यकों का गुणनफल

→ यदि α, β, γ त्रिपात बहुपद ax³ + bx² + cx + d = 0 के शून्यक हों, तो उनका योगफल व गुणनफल निम्न होगा-
α + β + γ = \(\frac{-b}{a}\)
αβ + βγ + γα = \(\frac{c}{a}\)
और αβγ = \(\frac{-d}{a}\)

→ यदि α, β व γ एक विधात बहुपद के शून्यक हों तो बहुपद = x³ – (α + β + γ) x² + (αβ + βγ + γα) x – αβγ

→ विभाजन एल्गोरिथ्म के अनुसार दिए गए बहुपद p (x) और शून्येतर बहुपद g(x) के लिए दो ऐसे बहुपदों q(x) तथा r(x) का अस्तित्व है कि P(x) = g (x) q(x) + r(x),
जहाँ r(x) = 0 है या बात r(x) < घात g(x) है।

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 9 Some Applications of Trigonometry Important Questions and Answers.

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

Short/Long Answer Type Questions

Question 1.
From a point on a bridge across a river the angles depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 4m from the banks. Find the width of river.
Solution :
Let A and B represent points on the bank on opposite sides of the river so that AB is the width of river. Pis a on the bridg at a height of 4 m is DP = 4m. We are required to determine the width of the river, which is the length of side AB of ΔAPB
∴ AB = AD + BD

Angles of depression of the banks on opposite sides of the river are
⇒ ∠APR = 30°
⇒ ∠PAD = 30°
[alternate interior angles]
and ∠BPQ = 45°
∠PBD = 45°
In right triangle APD, we have
tan 30° = \(\frac {PD}{AD}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac {4}{AD}\)
⇒ AD = 4\(\sqrt{3}\) …………..(1)
In right triangle BPD, we have
tan 45° = \(\frac {PD}{BD}\)
⇒ 1 = \(\frac {4}{BD}\)
⇒ BD = 4 ………….(2)
Adding equ. (1) and (2), we get
AD + BD = 4\(\sqrt{3}\) + 4
= 4 × 1.732 + 4
⇒ AB = 6.928 + 4 = 10.928 m
Hence, width of river = 10.928 m.

Question 2.
As observed from the top of a 100 m high lighthouse from the sea level, the angles of depression of two ships are 30 and 45°. If one ship is exactly behind the other on same side of lighthouse, find the distance between two ships.
Solution :
Let the AB be the lighthouse of height of 100 m. Let D and C be position of two ships. The angle of depression of two ships are 30° and 45°

⇒ ∠DAP = 30°
⇒ ∠ADB = 30°
[Alternate interior angles]
and ∠CAP = 45°
⇒ ∠ACB = 45°
In right triangle ADB, we have
tan 30° = \(\frac {AB}{BD}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac {100}{BD}\)
⇒ BD = 100\(\sqrt{3}\)
⇒ 10 × 1.732 = 173.2 m
In right triangle ACB, we have
tan 45° = \(\frac {AB}{BC}\)
⇒ 1 = \(\frac {100}{BC}\)
⇒ BC = 100 m
⇒ DC = BD – BC
173.2 – 100 = 73.2 m
Hence distance between two ships
= DC = 73.2 m

Question 3.
The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of depression from the top of the tower of the foot of the hill is 30°. If tower is 50 m high, find the height of hill.
Solution :
Let AB be the hill of height hm and CD be the tower of height 50 m.

Angle of elevation of the top of a hill from the foot of a tower is 60° i.e. ∠ACB = 60° and angle of depression from top of the tower of the foot of the hill is 30 i.e. ∠BDE = 30°.
⇒ ∠DBC = 30°
[Alternate interior angles]
In right triangle BCD, we have
tan 30° = \(\frac {CD}{BC}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac {50}{BC}\)
⇒ BC = 50\(\sqrt{3}\) ……..(1)
In right triangle ABC, we have
tan 60° = \(\frac {AB}{BC}\)
⇒ \(\sqrt{3}\) = \(\frac {h}{BC}\)
⇒ h = \(\sqrt{3}\)BC ………..(2)
substituting, the value of BC in equ. (2), we get
h = \(\sqrt{3}\) × 50\(\sqrt{3}\)
= 50 × 3 = 150 m
Hence height of hill = 150 m

Question 4.
A vertical tower stands on a horizontal plane and is sur mounted by a vertical flag staff of height 6 m. At a point on the plane, the angle of elevation of the bottom and top of the flag staff are 30° and 45° respectvely. Find the height of the tower. Take \(\sqrt{3}\) = 1.73]
Solution :
Let AB be tower of height h m and BC be vertical flag staff of height 6m. At point D on the plane angle of elevation of the bottom and top of the flag staff are 30° and 45° respectively i.e. ∠ADB = 30° and ∠ADC = 45°
In right triangle ABD, we have

⇒ h = \(\frac{6 \times(1.73+1)}{(\sqrt{3})^2-(1)^2}\)
⇒ h = \(\frac{6 \times 2.73}{3-1}\)
⇒ h = \(\frac{6 \times 2.73}{2}\)
⇒ h = 3 × 2.73
⇒ h = 8.19 m
Hence, height of the tower h = 8.19 m

Question 5.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 48 meters high. Find the height of the building.
Solution :
Let AB be tower of the height 48 m and CD be building of height h m. Angle of elevation of the top of a building from the foot of the tower is 30° Le. ∠CBD = 30° and angle of elevation of top of the tower from the foot of the building is 60° L.E. ∠ACB = 60°,

Question 6.
An aeroplane when flying at a height of 4000 m. from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of two planes from the same point on the ground are 60° and 45° repectively. Find the vertical distance between the aeroplanes at that instant.
Solution :
Let A and B be position of two aeroplanes. When A be the vertical above B and AC = 4000 m.
Let the angles of elevation of two nlanes from the point A be 60 and 45° respectively i.e. ∠ADC = 60° and ∠BDC = 45°

Hence, vertical distance between two aeroplanes = 1690.53 m.

Question 7.
Two men on either side of the cliff 80 m high observes the angles of the elevation of the top of the cliff to be 30° and 60° respectively. Find the distance between the two men.
Solution :
Let the CD be cliff of height 80 m. Let A abd B be position of eyes of two men. The angles of elevation of eyes of two men from top of the cliff be 30° and 60° respectively i.e. ∠CAD = 30° and ∠CBD = 60°.

Hence, distance between two men = 184.80 m

Question 8.
At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in lake, at A is 60°. Find the distance of cloud from A.
Solution :
Let BR be surface of lake and A be point of observation P be position of cloud and C be the position of reflection of cloud in the lake. Draw AQ ⊥ PR. Angle of elevation of cloud from point A is 30° i.e. ∠PAQ = 30° and angle of depression of reflection of cloud in the lake is 60° i.e. ∠CAQ = 60°
Let PQ be h m, then PR = (h + 20) m and
RC = PR = (h + 20) m

substituting the value of h in equ. (1), we get
AQ = \(\sqrt{3}\) × 20 = 20\(\sqrt{3}\)m
In right ΔAPQ, we have
⇒ AP² = AQ² + PQ²
⇒ AP² = (20\(\sqrt{3}\))² + (20)²
⇒ AP² = 1200 + 400
⇒ AP² = 1600
⇒ AP = \(\sqrt{1600}\)
⇒ AP = 40
Hence, distance of cloud from point A = 40 m.

Question 9.
Amit standing on a horizontal plane, find a bird flying at a distance of 200m. From him at an elevation of 30°. Deepak standing on the roof of a 50m hight building, find the angle of elevation of the same bird to be 45°. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.
Solution :
Let A and B be two positions of Amit and Deepak. Let P be position of bird. Bird flying at a distance from A 200 m i.e. AP = 200 m. Angle of elevation of P from Amit is 30° i.e. ∠PAR = 30° and Angle of elevation of P from Deepak is 45° i.e.
∠PBQ = 45°

In right ΔAPR, we have
sin 30° = \(\frac {PR}{AP}\)
⇒ \(\frac {1}{2}\) = \(\frac{P Q+Q R}{200}\)
⇒ PQ + QR = \(\frac {200}{2}\)
⇒ PQ + 20 = 100 [∴ QR = BS]
⇒ PQ = 100 – 50 = 50 m
In right ΔPBQ, we have
sin 45° = \(\frac {PQ}{PB}\)
⇒ \(\frac{1}{\sqrt{2}}\) = \(\frac {50}{PB}\)
⇒ PB = 50\(\sqrt{2}\) m
Hence, distance of the bird from Deepak is 50\(\sqrt{2}\) m.

Question 10.
A vertical tower stands on a horizontal plane and is sur mounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and top of the flag staff are λ and β respectively. Prove that the height of the tower is \(\frac {h tan α}{tan β – tan α}\)
Solution :
Let AB be the vertical tower of height H and AD be flag staff of height h. The angle of elevation of bottom and top of flag staff are λ and β respectively i.e.
⇒ ∠ACB = α and
∠DCB = β

Hence proved.

Fill in the Blanks

Question 1.
……… is the height to which something is raised above a point of reference.
Solution :
Elevation

Question 2.
………. is the depth to which something is lowered below point of reference.
Solution :
depression

Question 3.
Numerically the angle of elevation is… to the angle of depression.
Solution :
equal

Question 4.
The angle of elevation and angle of depression both are measured with the ………. .
Solution :
horizontal

Question 5.
If two lines are ………, then the alternate interior angles formed are equal.
Solution :
parallel

Question 6.
Two angles having a sum of ………… are called complementary angles.
Solution :
90°

Question 7.
As the ……….. of an object increases, the angles of elevation of its top with the horizontal level also increases.
Solution :
height

Multiple Choice Questions

Choose the correct answer for each of the following:

Question 1.
A 20 m long ladder rests against a wall. If the feet of the ladder is 10 m away from the wall, then angle of the elevation is:
(a) 30°
(b) 60°
(c) 90°
(d) 45°
Solution :

Question 2.
When the sun is 30° above the horizontal, the length of the shadow casts by a 50 m high building is :
(a) 50\(\sqrt{3}\) m
(b) 25 m
(c) 25\(\sqrt{3}\) m
(d) \(\frac{50}{\sqrt{3}}\)m

Solution :
∵ Tan 30° = \(\frac {BC}{AB}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac {50}{AB}\)
⇒ Length of shadow AB = 50\(\sqrt{3}\) m
Hence correct choice is (a)

Question 3.
The angle of elevation of a tower from a distance 100 m from its foot is 30°. Height of tower is :
(a) 50\(\sqrt{3}\)
(b) 100\(\sqrt{3}\)
(c) \(\frac{100}{\sqrt{3}}\)
(d) \(\frac{50}{\sqrt{3}}\)
Solution :
In ΔABC

Hence, height of tower = \(\frac{100}{\sqrt{3}}\)m
Hence correct choice is (c).

Question 4.
A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is :
(a) 45°
(b) 30°
(c) 60°
(d) 90°
Solution :
In ΔABC

sin θ = sin 30°
⇒ θ = 30°
Hence angle of elevation 30°
So correct choice is (b).

Question 5.
If the elevation of the Sun changed from 30° to 60°, then difference between the length of shadows of a pole of height 15 m is :
(a) 7.5 m
(b) 15 m
(c) 103 m
(d) 573 m
Solution :
In ΔBCD

⇒ x + 5\(\sqrt{3}\) = 15\(\sqrt{3}\)
⇒ x = 15\(\sqrt{3}\) – 5\(\sqrt{3}\)
⇒ x = 10\(\sqrt{3}\) m
Hence required difference x = 10\(\sqrt{3}\) m
So correct choice is (c)

Question 6.
From a window 15 above the ground, the angle of elevation of the top of a tower is α and angle of depression of the foot of the tower is β. Then height of the tower is :
(tan α = \(\frac {5}{3}\), tan β = \(\frac {2}{3}\))
(a) 50 m
(b) 52.5 m
(c) 55 m
(d) 53.5 m
Solution :
Tan α = \(\frac {5}{3}\), Tan β = \(\frac {2}{3}\)
In ΔАВС Tan β = \(\frac{2}{3}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{15}{\mathrm{BC}}\)

= 37.5 m
∴ Height of tower = CD + DE
= 15 + 37.5 m
= 52.5 m
So correct choice is (b).

Question 7.
A, B, C are three collinear points on the ground such that B lies between A and Cand AB = 10 m. If the angles of elevation of the top of a vertical tower at C are respectively 30° and 60° as seen from A and B, then the height of tower is :
(a) 5\(\sqrt{3}\) m
(b) 5 m
(c) \(\frac{10}{\sqrt{3}}\)
(d) \(\frac{20}{\sqrt{3}}\)
Solution :
Let CD = x, and BC = y
In ΔBCD,

So correct choice is (a).

Question 8.
A person of height 2 m wants to get a fruit which is on a pole of height \(\frac {10}{3}\)m. If he stands at a distance of \(\frac {4}{3}\)m from the foot of the pole, then the angle at which he should throw the stone, so that it hits the fruits, is :
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Solution :
From the figure

= Tan 45°
∴ θ = 45°
So correct choice is (c)

Haryana State Board HBSE 10th Class Science Solutions Chapter 9 Heredity and Evolution Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 9 Heredity and Evolution

HBSE 10th Class Science Heredity and Evolution Textbook Questions and Answers

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as …………………
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer:
(c) TtWW

Question 2.
An example of homologous organs is ……………………
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above
Answer:
(d) all of the above

Question 3.
In evolutionary terms, we have more in common with ………………
(a) a Chinese school-boy
(b) a chimpanzee
(c) a spider
(d) a bacterium!
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light – coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:

Reason:
1. No, the given information is not enough to consider the dominance or recessiveness of eye colour.

2. However, we can observe that generally in a population, the proportion of the people having light coloured eyes is much less than the people having dark coloured eyes. Based on this observation, we may consider that the light coloured eyes are a recessive trait and dark coloured eyes may be dominant.

Question 5.
How are the areas of study – evolution and classification – interlinked?
Answer:
1. In classification, the groups of organisms are based on the similarities and differences of characteristics of the organisms.

2. The organisms that have more similarities belong to nearer groups while the organisms having lesser similarities i.e. having more differences belong to distant groups.

3. In evolution, a study of identifying hierarchies of characteristics between the species is done. So, in such studies small groups of species with recent common ancestors are formed. Thus, the areas of study – evolution and classification are interlinked.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
1. In dogs, the coat colour is either black or white. The character of black coloured coat is indicated by ‘B’ and white coloured coat is indicated by ‘b’.

2. First of all the homozygous/pure parents with opposite characters are fertilized. Thus, applying Mendelian’s law, in the above case, the black coloured coat is a dominant character. So, black coat is dominant trait in dog.

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
1. The remains of dead organisms buried under the earth for millions of years are known as fossils.
2. Fossils are impressions of dead plants or animals that lived in the past.
3. When plants or animals die, the micro-organisms get decomposed in the presence of moisture and oxygen.
4. However, sometimes due to environmental conditions, their bodies do not decompose completely.
5. Such body parts of the plants and animals become fossil and can be available on digging the earth.
6. Fossils can be in the form of imprints, burrow of a worm, or even an actual bone.

Example:

• If the dead leaf gets caught in the mud, leaf will not decompose completely.
• The mud around the leaf will set around it as a mould which will then slowly become hard to form a rock and retain the impression of the leaf. Thus fossil of a leaf is formed.
• By studying these fossils, scientists learn how organisms evolved over time.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
1. It is believed that life must have developed from simple inorganic molecules which were present on earth soon after it was formed.
2. It was also speculated that the conditions on earth at that time could have given rise to more complex organic molecules that were necessary for life.
3. It is believed that the first primitive organisms would have arisen from further chemical synthesis.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:
(1) First, in sexual reproduction, there occurs a combination of genetic material of two different parents i.e. a male and a female. Hence, each new generation is a combination of two different parental DNAs.

(2) Second, during the formation of gametes i.e. formation of sperms and ovum, the cell undergoes cell division phase called meiosis.

• During meiosis, the genetic material gets distributed randomly. Moreover, all the gametes are not always similar and so there are more chances of variation during fertilization, in other words, there are so many sperm cells but only one fertilizes with the ovum. Naturally, this will create variation.
• Owing to these two reasons, there occur more variations in sexual reproduction as compared to asexual.
• The combination of different types of gametes which takes place is the root of variations. This also gives rise to new type of characters too. This also favours evolution.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
1. During gametogenesis (production of male and female sex cells), the meiosis type of cells division occurs.
2. During, meiosis the number of chromosomes becomes half. The DNA gets distributed to half in each gamete.
3. During fertilization, both the types of sex cells get fused and the quantity of DNA remains maintained. Thus, equal genetic contribution of male and female parents is ensured in the progeny.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes, this is true.

Reason:

• If the variation achieved by an organism is helpful to adapt to the surrounding environment then it increases the probability of the survival of that organism.
• The survived organisms then produces more offspring with such genetic variations and thus the species survives.

HBSE 10th Class Science Heredity and Evolution InText Activity Questions and Answers

Textbook Page no – 143

Question 1.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait B which exists in 60% of population of an asexually reproducing species must have arisen earlier than trait A.

Reason:

• In asexually reproducing species, DNA copy error takes place, in very small amount and so new traits do not occur very fast.
• Since, the percentage of trait B is very large as compared to trait A. This suggests that trait B must have arisen quite before to trait A.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
1. Variations in a species are created through two reasons,

• Due to inaccuracies in DNA copying or
• During sexual reproduction.

2. Different individuals gain different types of advantages with these variations. Several variations enable the individual to adapt to the environmental conditions. This increases the chances of their survival.

3. The individuals that survive then reproduce offspring and the existence of species remains continued.

Textbook Page no – 147

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
1. Mendel conducted cross fertilization between pure tall plants (TT) and pure short (tt) plant. This resulted in all (Tt) plants in F₁ generation.

2. This shows that single copy of T is enough to make the plant tall. The trait that gets expressed over the other is called the dominant trait and the other recessive.

3. In Mendel’s trial, the trait which gets expressed in 75% individuals in F₂ generation after self-fertilization is dominant whereas the trait which appears in 25% individual is recessive.

Question 2.
How do Mendel’s experiments show that traits are inherited independently?
Answer:
1. Mendel performed the experiments of heredity on pea plant. He studied the heredity of two characters simultaneously and performed the dihybrid experiment.

2. On the basis of this experiment we can say that even though the F₁ progeny plants were dominant in both the characters, the F₂ progeny showed the new combinations. In F₂ progeny plants some were tall and wrinkled, while some were short and round.

3. The experiment clearly shows that the factors that control the shape, size and height of the seeds are inherited independently and that is the main reason why new combinations could be seen in F₂ generation.

Question 3.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
No, the given information is not enough to determine which trait is dominant.

Reason:

• The daughter (or any child) receives two copies of genes, one from each parent.
• The trait of blood group depends on the genotype i.e. the types of genes which get combined. Hence, the provided information is insufficient and so we cannot tell which one is a dominant blood group.

Question 4.
How is the sex of the child determined in human beings?
Answer:
1. In human beings, sex of the child to be born is determined by the sex chromosome of the father.

2. The father or say male contains XY chromosomes and so during spermatogenesis, the male produces two types of sperms. 50% of these sperms have ‘X’ sex chromosome and 50% sperms have ‘Y’ sex chromosome.

3. If the sperm having ‘X’ chromosome fertilizes the ovum, a female child (= a girl) is born and if the sperm having ‘Y’ chromosome fertilizes the ovum, the male child (= a boy) is born.

Textbook Page no – 150

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:
A particular trait may increase in a population through two main ways. They are –

• Natural selection – It directs evolution with a survival advantage.
• Genetic drift – It leads to accumulation of different changes.

Question 2.
Why are traits acquired during the life-time of an individual not inherited?
Answer:
Change in non-reproductive tissues cannot be passed on to the DNA of the germ cells. Hence the experiences of an individual during its lifetime can neither be passed on to its progeny nor can such experiences result in evolution.

Example:

• If we breed a group of mice, all their progeny will have tails.
• Suppose we remove the tails of each generation of these mice by surgery then it does not mean that the progeny is tailless.
• Artificially removing the tail does not bring any change in the genes of the germ cells of the mice.
• So, the cut-tail that a generation of mice experienced cannot become a trait to be passed on to the next generation.

Question 3.
Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics?
Answer:
1. A small population generally does not show much variation from generation to generation.
2. A very small population of surviving tigers indicates that very less amount of variation will occur in the genetic characteristics. This means they will not be able to adapt to the major environmental changes if any, in the future.
3. The tigers will not survive and the species will become extinct.

Textbook Page no – 151

Question 1.
What factors could lead to the rise of a new species?
Answer:
Factors that can lead to the rise of a new species are:

• Accumulated variations favourable to natural environment,
• Geographic isolation of a population,
• Gene flow,
• Genetic drift and
• Natural selection

Question 2.
Will geographical isolation be a major factor in the speciation of a self-pollinating plant species? Why or why not?
Answer:
No, geographical isolation will not be a major factor in the speciation of a self-pollinating plant species.

Reason:

• Only a single parent is involved in self-pollination. So, there is no gene flow between two geographically isolated populations.
• If it would be a cross-pollination species, the geographical isolation would be a major factor as it would lead to faster accumulation of variation (genetic drift) in the two geographically separated populations.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
No, geographical isolation will not be a major factor to the speciation that reproduces asexually.

Reason:
Asexually reproducing organisms show very little variation over generations. These little variations are not sufficient enough to raise a new species having such variations.

Textbook Page no – 156

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
1. One of the ways to determine how close two species are with respect to evolution is to obtain evidence from their homologous organs.

2. For example, the basic design of internal structure of bones of forelimbs of a frog, lizard, bird, bat and man is same, even though these organs perform different functions.

3. This indicates that all these forelimbs have evolved from a common ancestral animal, which had a same basic internal structure.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
No, the wings of a butterfly and the wings of a bat cannot be considered homologous organs.

Reason:
Although the wings of a butterfly and the wings of a bat perform similar function, but the design, structure and components of their wings are very different. Hence, the wings of these two can be considered as analogous organs and not homologous.

Question 3.
What are fossils? What do they tell us about the process of evolution?
Answer:
1. The remains of dead organisms buried under the earth for millions of years are known as fossils.
2. Fossils are impressions of dead plants or animals that lived in the past.
3. When plants or animals die, the micro-organisms get decomposed in the presence of moisture and oxygen.
4. However, sometimes due to environmental conditions, their bodies do not decompose completely.
5. Such body parts of the plants and animals become fossil and can be available on digging the earth.
6. Fossils can be in the form of imprints, burrow of a worm, or even an actual bone.

Example:

• If the dead leaf gets caught in the mud, leaf will not decompose completely.
• The mud around the leaf will set around it as a mould which will then slowly become hard to form a rock and retain the impression of the leaf. Thus fossil of a leaf is formed.
• By studying these fossils, scientists learn how organisms evolved over time.

Textbook Page no – 156

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer:
1. Across the earth, there is a great diversity in humans and so we look so different from each other in terms of size, colour and looks.

2. Initially, human race was identified on the basis of skin colour and the humans were known as yellow, black, white or brown. However, in the recent years, it has been proved that all the human beings have evolved from a single species called the Homo sapiens.

3. Hence, in spite of having different skin colour, looks and size, we all belong to the same species.

Question 2.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design? Why or why not?
Answer:
1. In evolutionary terms, we can say that chimpanzees have a better developed body design.
2. The reason for this is that chimpanzees are highly complex organisms compared to remaining three.

Activities:

Activity -1

Aim: To study inheritance in the type of earlobes.

Background:

• The lowest part of the ear pinna (external ear) is known as an earlobe.
• The earlobe may be either closely attached to the (lateral) side of the head, or not-attached i.e. might be free.

Procedure:

• Observe the ears of all the students in the class. Prepare a list of students having free or attached earlobes and
• Find out about the earlobes of the parents of each student in the class. Correlate the earlobe type of each student with that of their parents. Based on this evidence, suggest a possible rule for the inheritance of earlobe types.
• Students can observe the earlobes of other students. They can also ask about the type of earlobes of their parents.
• Make a chart as given below. Write 1F’ for the free earlobes and A’ for the attached earlobes.

Observation:
We can observe that more number of students have inherited the trait of ‘free earlobes’ i.e. trait F as compared to attached earlobes i.e. A.

Conclusion:
From studies it has been proved that free earlobe is a dominant trait whereas an attached earlobe is a recessive trait.

Activity 2.

In Fig. 9.3 of textbook, what experiment would we do to confirm that the F₂ generation did in fact have a 1:2:1 ratio of TT, it and tt trait combinations?

Observation:
Although the experiment phenotypically shows the ratio of 3 : 1 ratio between tall and dwarf, genetically the ratio is 1 : 2 : 1.
This can be proved by the following experiment. We can raise F₃ generation by allowing self-pollination of F₃ generation plants.

Then we may calculate ratio for —

• Tall plants which produced only tall plants. These plants will be pure dominant (TT).
• Tall plants which produced both tall and dwarf pea plants. These plants are hybrid (Tt).
• Dwarf plants produced only dwarf plants. This means they are pure recessive (tt).

Observing the data carefully, one can find the appearance of 1 : 2 : 1 ratio of TT Tt and tt trait combination in F₃ generation (also known as genotypic ratio).

Haryana State Board HBSE 10th Class Maths Notes Chapter 1 वास्तविक संख्याएँ Notes.

Haryana Board 10th Class Maths Notes Chapter 1 वास्तविक संख्याएँ

→ यूक्लिड विभाजन प्रमेयिका- दो धनात्मक पूर्णाक a और b दिए होने पर, ऐसी अद्वितीय पूर्ण संख्याएँ q और r विद्यमान होती हैं कि a = bq + r, 0 ≤ r < b है।

→ यूक्लिड विभाजन एल्गोरिम द्वारा- दो धनात्मक पूर्णाकों, c और d(c > d) का HCF ज्ञात करने के लिए नीचे दिए हुएं चरणों का अनुसरण किया जाता है-
चरण I-c और के लिए यूक्लिड विभाजन प्रमेयिका से हम ऐसे q और r ज्ञात करते हैं कि c = dq + r, 0 ≤ r < d हो।
चरण II-यदि r = 0 है, तो d पूर्णाको c और d का HCF है। यदि r ≠ 0 है, तो और के लिए, यूक्लिड विभाजन प्रमेविका का प्रयोग पुनः कीजिए।
चरण III-इस प्रक्रिया को तब तक जारी रखिए, जब तक शेषफल 0न प्राप्त हो जाए। इसी स्थिति में, प्राप्त भाजक ही वांछित HCF है।

→ अंकगणित की आधारभूत प्रमेय- प्रत्येक भाज्य संख्या को अभाज्य संख्याओं के एक गुणनफल के रूप में व्यक्त (गुणनखडित) किया जा सकता है तथा यह गुणनखंडन अभाज्य गुणनखंडों के आने वाले क्रम के बिना अद्वितीय होता है।

→ यदि p कोई अभाज्य संख्या है और p, a² को विभाजित करता है तो p, a को भी विभाजित करेगा, जहाँ a एक धनात्मक पूर्णाक है।

→ किन्हीं दो धनात्मक पूर्णांकों a और b के लिए HCF(a, b) × LCM(a, b) = a × b होता है।

→ परिमेय और अपरिमेय संख्याएँ मिलकर वास्तविक संख्याएँ बनाती हैं।

→ \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\) तथा व्यापक रूप में \(\sqrt{p}\) अपरिमेय संख्याएँ हैं, जहाँ पर p एक अभाज्य संख्या है।

→ यदि x एक परिमेय संख्या हो जिसका दशमलब प्रसार सांत हो, तो हम x को \(\frac{p}{q}\) के रूप में व्यक्त कर सकते हैं, जहाँ p औरq सहअभाज्य होते हैं तथा q का अभाज्य गुणनखंडन 2ⁿ5^m के रूप का होता है, जहाँ n, m प्रणेतर पूर्णाक होते हैं।

→ माना x = \(\frac{p}{q}\) एक ऐसी परिमेय संख्या है कि q का अभाज्य गुणनखंडन 2ⁿ5^m के रूप का है, (जहाँ n, m ऋणेतर पूर्णाक है) तो x का दशमलव प्रसार सांत होगा।

→ माना x = \(\frac{p}{q}\) एक ऐसी परिमेय संख्या है कि q का अभाज्य गुणनखंडन 2ⁿ5^m के रूप का नहीं है, (जहाँ n, m ऋणेतर पूर्णाक है) तो x का दशमलव प्रसार असांत आवर्ती होगा।

→ किन्हीं तीन संख्याओं p, q तथा r के लिए LCM होगा-

→ किन्हीं तीन संख्याओं P, q तथा r के लिए HCF होगा-

→ एक परिमेय संख्या और एक अपरिमेय संख्या का योग या अंतर एक अपरिमेय संख्या होती है।

→ एक शून्येतर परिमेय संख्या और एक अपरिमेय संख्या का गुणनफल या भागफल एक अपरिमेय संख्या होती है।

Haryana Board HBSE 10th Class Maths Notes

HBSE 10th Class Maths Notes in English Medium

• Chapter 1 Real Numbers Notes
• Chapter 2 Polynomials Notes
• Chapter 3 Pair of Linear Equations in Two Variables Notes
• Chapter 4 Quadratic Equations Notes
• Chapter 5 Arithmetic Progressions Notes
• Chapter 6 Triangles Notes
• Chapter 7 Coordinate Geometry Notes
• Chapter 8 Introduction to Trigonometry Notes
• Chapter 9 Some Applications of Trigonometry Notes
• Chapter 10 Circles Notes
• Chapter 11 Constructions Notes
• Chapter 12 Areas Related to Circles Notes
• Chapter 13 Surface Areas and Volumes Notes
• Chapter 14 Statistics Notes
• Chapter 15 Probability Notes

HBSE 10th Class Maths Notes in Hindi Medium

• Chapter 1 वास्तविक संख्याएँ Notes
• Chapter 2 बहुपद Notes
• Chapter 3 दो चरों वाले रखिक समीकरण का युग्म Notes
• Chapter 4 द्विघात समीकरण Notes
• Chapter 5 समांतर श्रेढ़ियाँ Notes
• Chapter 6 त्रिभुज Notes
• Chapter 7 निर्देशांक ज्यामिति Notes
• Chapter 8 त्रिकोणमिति का परिचय Notes
• Chapter 9 त्रिकोणमिति का अनुप्रयोग Notes
• Chapter 10 वृत्त Notes
• Chapter 12 वृतों से संबंधित क्षेत्रफल Notes
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Notes
• Chapter 14 सांख्यिकी Notes
• Chapter 15 प्रायिकता Notes

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

Haryana Board 10th Class Science Solutions Chapter 10 Light Reflection and Refraction

HBSE 10th Class Science Light Reflection and Refraction Textbook Questions and Answers

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay

Question 2.
The Image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At Infinity
(d) Between the optical centre of the lens and its principal focus.
Answer:
(b) At twice the focal length

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be …….
(a) both concave
(b) both convex
(c) the mirror Is concave and the lens is convex
(d) the mirror is convex, but the lens is concave
Answer:
(a) both concave

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be …………
(a) plane
(b) concave
(c) convex
(d) either plane or convex
Answer:
(d) either plane or convex

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm
Answer:
(c) A convex lens of focal length 5 cm

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer:
1. When the object is placed between the pole and the principal focus of the concave mirror, we get an erect, virtual and enlarged image.
2. We are given that the focal length of concave mirror = 15 cm
3. So, the object should be placed in front of the concave mirror at a distance less than 15 cm.

Question 8.
Name the type of mirror used in the following situations.
(a) Headlights of a car
(b) Side/rear-view mirror of a vehicle
(c) Solar furnace
Support your answer with reason.
Answer:
(a) Concave mirrors are used as reflectors in headlights of cars. When a bulb is located at the focus of the concave mirror, the light rays after reflection from the mirror travel over a large distance as a parallel beam of high intensity.

(b) A convex mirror is used as a side/rear-view mirror of a vehicle because of the following reasons:

• A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
• A convex mirror has a wider field of view than a plane mirror of the same size.

(c) Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces. Concave mirrors are able to raise the temperature drastically.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
As shown in figure, when the lower half of the convex lens is covered with a black paper, it still forms the complete image of the object, but the intensity of the image is reduced.

The rays coming from the object get refracted by the upper half of the lens and meet at the other side of the lens to form the image.

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the Image formed.
Answer:
The final position, nature and size of the image A’B’ are:
Here, Object size h = +5 cm
Object distance u = – 25 cm
Focal length of lens f = + 10 cm
( ∵ Converging, i.e., Convex lens)
Image distance y =?
Image height h’ =?

Image distance y is positive. This shows that the image formed is real and on the other side of the lens, at 16.67 cm from the lens as shown in the figure. AB : Object, A’B’: Image
Now, Magnification m = \(\frac{h^{\prime}}{h}=\frac{v}{u}\)

The negative (minus) sign shows that the image is inverted, real, diminished (3.3 cm). Figure 10.60 show the positive, size and nature of the image formed.

Question 11.
A concave lens of focal length 15 cm forms an Image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
Scale: 1 cm = 5 cm
Focal length, f = – 15 cm [f is – ve for a concave lens]
Image distance, v = – 10 cm
[Concave lens forms virtual image on same side as the object, so v is – ve]

Object distance, u = – 30 cm.
The negative sign shows that the object is placed on the left side of the lens.

Question 12.
An object Is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Object distance, u = – 10 cm
Focal length, f = + 15 cm [f is +ve for a convex mirror]

Image distance, v = + 6 cm.
As v is +ve, so a virtual, erect image is formed at a distance 6 cm behind the mirror.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
Magnification of a mirror \(m=\frac{h^{\prime}}{h}=\frac{-v}{u}\)
Here, for a plane mirror, m = + 1, So, h’ = h and y = – u

• m = 1 indicates that both ¡mage and object are of same size.
• Positive sign of m indicates that a virtual image is formed behind the mirror.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Object size, h = + 5cm
Object distance, u = – 20 cm
Radius of curvature, R = + 3.0 cm [R is +ve for a convex mirror]
Focal length, \(f=\frac{R}{2}\) = + 15 cm
From mirror formula,

A virtual, erect image of height 2.2 cm is formed behind the mirror at a distance of 8.6 cm from the mirror.

Question 15.
An object of size 7.0 cm Is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharply focused image can be obtained? Find the size and the nature of the Image.
Answer:
Object size, h = + 7.0 cm
Focal length, t = – 18 cm
Image size, h’ = ?

Object distance, u = – 27 cm
image distance,v = ?
The screen should be placed at a distance of 54 cm on the object side of the mirror to obtain sharp image.

The image is real, inverted and enlarged in size.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Here, P = – 2.0 D f =\(\frac{1}{P}=\frac{1}{-2.0 D}\) = – 0.5 m.
Since the power of lens is negative, the lens is concave.

Question 17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Here P = +1.5D
\(f=\frac{1}{P}=\frac{1}{+1.5 D}=+\frac{10}{15} m\) = + 66.67cm.

• As the focal length is positive, the prescribed lens is converging.
• Opposite side at F, real and inverted, highly diminished (point sized)

HBSE 10th Class Science Light Reflection and Refraction InText Activity Questions and Answers

Textbook Page no – 168

Question 1.
Define the principal focus of a concave mirror.
Answer:
The point (F) on the principle axis where the beam of light parallel to the principle axis either actually converges to or diverges from, after reflection from a mirror is called the principle focus (F).

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
R = 20cm.
Focal length (f) = \(\frac{\mathrm{R}}{2}=\frac{20}{2}\) = 10 cm

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
A convex mirror is preferred as a rear-view mirror because —

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

Textbook Page no – 171

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
The radius of curvature and focal length of a convex mirror are positive.
∴ R = + 32cm and f = \(\frac{\mathrm{R}}{2}\) = + 16cm

Question 2.
A concave mirror produces three times magnified (enlarged) real Image of an object placed at 10 cm in front of It. Where is the image located?
Answer:
As the image is real, so magnification m must be negative.
\(m=\frac{h^{\prime}}{h}=-\frac{v}{u}=-3\)
∴ v = 3 x (-10) = – 30 cm.
Thus the image is real and it is located at a distance of 30 cm from the mirror in the front of mirror.

Textbook Page no – 176

Question 1.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
1. Light travels faster in rarer medium air and slower in denser medium water.
2. Since the ray of light travelling in air enters obliquely into water, it slows down and bends towards the normal.

Question 2.
Light enters from air to glass having refractive Index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 x 108 mr1.
Answer:
1. Refractive index of glass, ng = 1.50
Speed of light in vacuum, c = 3 x 10⁸ ms^-1

2. Absolute refractive index of glass, ng = ((c/v))
∴ Speed of light in glass \(v=\frac{c}{n_g}=\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{1.5}=2 \times 10^8 \mathrm{~ms}^{-1}\)

Question 3.
Find out, from table 10.3 of the text book, the medium having highest optical density. Also find the medium with lowest optical density.
Answer:
1. From table 10.3, diamond has highest refractive index (= 2.42), so it has largest optical density.
2. Air has lowest refractive index (= 1.0003), so it has lowest optical density.

Question 4.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given In table 10.3 of the text book.
Answer:
For kerosene, n = 1. 44,   For turpentine oil, n = 1. 47,   For water, n = 1. 33
As water has lowest refractive index, so light travels fastest in this optically rarer medium compared to kerosene and turpentine oil.

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
It indicates that the ratio of speed of light in air to that in diamond is 2.42.

Textbook Page no – 184.

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre (1 D) is the power of a lens whose focal length is 1 meter.

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed ¡n front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
1. Here, image distance y = + 50 cm (∴ y is positive for real images)
2. As mentioned, the real image is of the same size as the object.

Thus, object (needle) is placed at 50 cm from the lens.

Power of lens:

∴ Focal length (f) = 25 cm = 0.25 m
Now, \(P=\frac{1}{f}=\frac{1}{+25}=+4 D\)
Thus, power of lens = + 4D

Question 3.
Find the power of a concave lens of focal length 2 m.
Answer:
Since the focal length of a concave lens is negative; f = – 2m.
Power \(P=\frac{1}{f}=\frac{1}{-2 m}=-0.5 \mathrm{D}\)

Activities

Activity 1.

Aim: To study the formation of images by both sides of a large shining spoon.

Apparatus required: A large shining spoon.

Procedure:

• Take a large shining spoon. Try to view face in its curved surface.
• Do you get the image? Is it smaller or larger?
• Move the spoon slowly away from your face. Observe the Image. How does it change?
• Reverse the spoon and repeat the activity. How does the image look like now?
• Compare the characteristics of the image on the two surfaces.

Observation and Conclusion:

• Initially, we can see our tace inside the curved portion of the spoon. The image of our face looks erect and enlarged. As we move the spoon at a farther distance, the image goes on becoming smaller and then gets inverted. The inner portion of the spoon behaves as a concave mirror.
• When we see the spoon from outer surface, it behaves as a convex mirror. The image formed is erect and small Irrespective of the distance of face from the spoon.

Activity 2.

Aim: To find the approximate focal length of a concave mirror.

Apparatus required: Concave mirror, a sheet of paper and meter scale

Caution: Do not look at the sun directly or even Into a mirror reflecting sunlight. It may damage your eyes.

Procedure:

• Hold a concave mirror in your hand and direct its reflecting surface towards the sun.
• Direct the light reflected by the mirror onto a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually unto you find on the paper sheet a bright, sharp spot of light

Question 1.
Hold the mirror and the paper in the same position for a few minutes. What do you observe? Why?
Answer:
Observation
Initially, the paper turns blackish and emits smoke while burning slowly. The paper then catches fire.

Conclusion:

• The point at which the sun-rays get converged to produce tire is the principal focus (F) of the concave mirror.
• The distance of this image of sun at point F gives the approximate value of focal length (t) of the concave mirror.

Activity 3.

Aim: To locate the image formed by a concave mirror for different positions of the object.

Procedure:
1. Take a concave mirror. Find out its approximate focal length. Let f = 10 cm.

2. Mark a line on a table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.

3. Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror. These lines will correspond to the position of the points P, F and C, respectively.

4. Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the
mirror till you obtain a sharp bright image of the candle flame on it.

5. Observe the image carefully. Note down its nature, position arid relative size with respect to the object size.

6. Repeat the experiment by placing the candle —
(a) just beyond C, (b) at C, (C) between F and C, (d) at F, and (e) between P and F

7. In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Then, look for its virtual image in the mirror itself.

8. Note down and tabulate your observations.

Observation:
Table 1. Nature, size and position of Images formed by a concave mirror for different positions of the object

Activity 4.

Students should do this activity on their own. Make use of activity 3 for drawing the diagrams.

Activity 5.

Aim: To locate the image formed by a convex mirror for different positions of the object.

Take a convex mirror. Hold it in one hand. Hold a pencil in the upright position in the other hand.

Question 1.
Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged?
Answer:
The image is erect, diminished and virtual.

Question 2.
Move the pencil away from the mirror slowly. Does the image become smaller or larger?
Answer:
The image formed is smaller and erect.

Question 3.
Repeat this activity carefully. State whether the image will move closer to or farther away from the focus as the object is moved away from the mirror?
Answer:
As the object is moved away from the mirror, the image moves closer to the focus of the mirror.

Observation:

Activity 6.

Aim: To demonstrate that convex mirror has a wide field of view.

Question 1.
Observe the Image of a distant object, say a distant tree, in a plane mirror. Could you see a full-length Image?
Answer:
No, we cannot see the full-length image of the object in plane mirror.

Question 2.
Try with plane mirrors of different sizes. Did you see the entire object In the Image?
Answer:
When we try with plane mirrors of different sizes, we find that the entire object in the image is seen only when the size of the plane mirror is at least half the size of the object.

Question 3.
Repeat this activity with a concave mirror. Did the mirror show full length Image of the object?
Answer:
The full length image is not formed even in concave mirror.

Question 4.
Now try using a convex mirror. Did you succeed? Explain your observations with reason.
Answer:
A full length image of the object is obtained in a small convex mirror.

Reasons for these observations:

• In a plane mirror, size of image is always equal to the size of the object.
• In case of a concave mirror, when the object is between P and F only then we get an enlarged, virtual
  and erect image behind the mirror. But here, since the object is far away, its full length image cannot be seen.
• In a convex mirror, the image is always virtual, erect and smaller than the object.

Activity 7.

Aim: To demonstrate refraction of light.

1. Place a coin at the bottom of a bucket filled with water.
2. With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
3. Repeat the activity. Why did you not succeed in doing in one go?
4. Ask your friends to do this. Compare your experience with theirs.

Observation and conclusion:

• Due to refraction of light from water, the coin appears to be slightly above its actual position.
• This visual illusion makes it difficult to identify exact position of the coin and pick it. However, after trying a while we can pick it up.

Activity 8.

1. Place a large shallow bowl on a table and put a coin in it.
2. Move away slowly from the bowl. Stop when the coin just disappears from your sight.
3. Ask a friend to pour water gently into the bowl without disturbing the coin.
4. Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?

Observation and Conclusion: On pouring water into the bowl, the coin becomes visible again. Moreover, due to refraction of light, the coin appears slightly raised above its actual position. This happens due to refraction of light.

Activity 9.

1. Draw a thick straight line in ink, over a sheet of white paper placed on a table.
2. Place a glass slab over the line in such a way that one of its edges makes an angle with the line.

Question 1.
Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges?
Answer:
We see the line first in the air medium and then in the glass. So, the line under the glass slab appears bent at the edges-due to refraction.

Question 2.
Next, place the glass slab such that it ¡s normal to the line. What do you observe now? Does the part of the line under the glass slab appear bent?
Answer:
Now, the line does not appear bent. The reason for this is that the light rays incident normally at the glass and air do not undergo refraction.

Question 3.
Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?
Answer:
Yes, part of the line under the glass slab appears to be raised. This is due to the refraction of light rays when they travel from glass to air.

Activity 10.

1. Fix a sheet of white paper on a drawing board using drawing pins.
2. Place a rectangular glåss slab over the sheet in the middle.
3. Draw the outline of the slab with a pencil. Let us name the outline as ABCO.
4. Take four identical pins.
5. Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
6. Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
7. Remove the pins and the slab.
8. Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
9. Join O and O’. Also produce EF up to P, as shown by a dotted line in Figure.

Activity 11.

Aim: To show the converging nature of convex lens.
Caution: Do not look at the Sun directly or through a lens while doing this Activity or otherwise. You may damage your eyes if you do so.

Procedure:

• Hold a convex lens in your hand. Direct it towards the Sun.
• Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun.
• Hold the paper and the lens in the same position for a while. Keep observing the paper. What happens? Why? Recall your experience in Activity 2.

Observation and conclusion:

• The paper begins to burn producing smoke. Later, it catches fire.
• We conclude that the light rays from the sun gets converged when they pass through the lens and form a bright spot on the paper. This spot shows the focus of the lens.

Activity 12.

Aim: To locate the position and examine the nature of the image formed by a convex lens for different positions of the object.

1. Take a convex lens. Find its approximate focal length in the way described in Activity 11.

2. Draw five parallel straight lines, using chalk on a long table such that the distance between the successive lines is equal to the focal length of the lens.

3. Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.

4. The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F2 and 2F₂, respectively.

5. Place a burning candle, far beyond 2F₁ to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.

6. Note down the ‘nature, position and relative size of the image.

7. Repeat this activity by placing the object just behind 2F₁, between F₁ and 2F₁ at F₁ at F₁, between F₁ and O. Note down and tabulate your observations.

Observation and conclusion:

The nature, position and size of the images formed by a convex lens depend on the position of the object in relation, to the points O, F₁ and 2F₁. After nothing the various characteristics of the image, the observations are mentioned in the table below.

Activity 13.

Aim: To locate the position and examine the nature of the image formed by a concave lens for different positions of the object.

1. Take a concave lens. Place it on a lens stand.
2. Place a burning candle on one side of the lens.
3. Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not, observe the image directly through the lens.
4. Note down the nature, relative size and approximate position of the image.
5. Move the candle away from the lens. Note the change In the size of the image. What happens to the size of the image when the candle is placed too far away from the lens.

Observation and Conclusion:

• On observing the image directly through the lens, we can see that the image of the candle is virtual, erect, smaller than the candle and closer to the lens.
• On moving the candle away from the lens, the size of the image decreases.
• When the candle is placed very far from the lens, the image becomes highly diminished and appears point-like.

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.