Class 10 ICSE Refraction through plane surface Selina Concise Physics Solution 4(C) & 4(D) -

ICSE CLASS 10 Chapter 4 selina solution concise physics

Exercise 4(C)

Question 1

How is the refractive index of a medium related to the real and apparent depths of an object in that medium?

Answer

The relation of refractive index μ with real and apparent depth is given as below,

$_{a} μ_{m} = Apparent Depth Real Depth$

Question 2

Prove that,

$Refractive index=Real depthApparent depth$

Answer

Consider a point object O kept at the bottom of a transparent medium ( such as water or glass ) separated from air by the surface PQ.

A ray of light OA, starting from the object O, is incident on the surface PQ normally, so it passes undeviated along the path AA’. Another ray OB, starting along the object O, strikes the boundary surface PQ at B and suffers refraction.

Since, the ray travels from denser to rarer medium so it bends away from normal N’BN drawn at the point of incidence B on the surface PQ and travels along BC in air.

When viewed by the eye , the ray BC appears to be coming from point I which is the virtual image of O, obtained on producing A’A and CB backwards.

Thus, any point seen from air will appear to be at I, which is lesser depth = AI than its actual depth AO.

Angle of incidence = ∠OBN’

Angle of refraction = ∠CBN.

Since, AO and BN’ are parallel and OB is transversal line, so

∠AOB = ∠OBN’ = i

Similarly, IA’ and BN are parallel and IC is the transversal line, so

∠BIA’ = ∠CBN = r

In right-angle triangle BAO,

$sin i = OB B A$

and,

In right-angle triangle IAB,

$sin r = I B B A$

For refraction from medium to air, by Snell’s law

$_{m} μ_{a} = sin r sin i = I B B A OB B A \Rightarrow_{m} μ_{a} = OB I B$ Hence, refractive index of medium with respect to air is,

$_{a} μ_{m} =^{m} μ ^{a} 1 = I B OB$

The object is viewed from a point vertically above the object O, since point B is very close to the point A. ∴ IB = IA and OB = OA

Hence,

$_{a} μ_{m} =^{m} μ ^{a} 1 = I A O A \Rightarrow Apparent depth Real depth$

Question 3

A tank of water is viewed normally from above.

(a) State how does the depth of tank appears to change.

(b) Draw a labelled ray diagram to explain your answer.

Answer

(a) When a tank is viewed normally from above from a rarer medium, the apparent depth of the tank is always less than its real depth.

As _aμ_w = 4 / 3, therefore, the depth of the pond appears three fourth of its real depth on seeing it from air in a nearly vertical direction.

(b) Below is the labelled ray diagram of a tank of water viewed normally from above:

A tank of water is viewed normally from above. State how does the depth of tank appears to change. Draw a labelled ray diagram to explain your answer. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

Question 4

Water in a pond appears to be only three-quarters of its actual depth.

(a) What property of light is responsible for this observation? Illustrate your answer with the help of a ray diagram.

(b) How is the refractive index of water calculated from its real and apparent depths?

Answer

(a) Water in a pond appears to be only three-quarters of its actual depth due to the refraction of light while travelling from one medium to another.

The light rays suffer refraction while travelling from denser medium (water or glass) to rarer medium (air), so it bends away from the normal. Below is the ray diagram for this:

Water in a pond appears to be only three-quarters of its actual depth. What property of light is responsible for this observation? Illustrate your answer with the help of a ray diagram. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

(b) Let us say an object (B) is at the bottom of a pond. Consider a ray of light BC from the object that moves from water to air.

The ray now moves away from the normal N along the path CD, after refraction from the water surface.

When we produce CD it appears from the point B’.

The virtual image of the object B appears at B’.

Hence, we get,

$Refractive index of water=Real depthApparent depth$

Question 5

Draw a ray diagram to show the appearance of a stick partially immersed in water. Explain your answer.

Answer

The above figure shows a stick placed obliquely in water. The portion OP of the stick under water when seen from air appears to be shortened and raised up as OP’. This is due to refraction of light from water to air.

The rays of light coming from tip P of the stick when pass from water to air, bend away from the normal and appear to be coming from point P’ which is the virtual image of the point P.

Thus, the part PO of the stick appears to be P’O.

Hence, the immersed part of the stick appears to be raised and therefore, bent at the point O at the surface of water and the stick XOP appears as XOP’.

Question 6

A fish is looking at a 1.0 m high plant at the edge of a pond. Will the plant appear to the fish shorter or taller than its actual height? Draw a ray diagram to support your answer.

Answer

A fish is looking at a 1.0 m high plant at the edge of the pond. Will the plant appear to the fish shorter or taller than its actual height? Draw a ray diagram to support your answer. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

When a fish is looking at a 1.0 m high plant at the edge of the pond, it appears to be taller than its actual height due to the refraction of the light rays travelling from one medium to another.

When the fish is looking at the plant, as air is a rarer medium in comparison to water, the ray OP will bend away from the normal while emerging out from the water to air.

But, when we extend the ray it appears to be coming from a higher point.

Hence, the plant will look taller than its actual height.

Question 7

A student puts his pencil into an empty trough and observes the pencil from the position as indicated in the figure.

(i) What change will be observed in the appearance of the pencil when water is poured into the trough?

(ii) Name the phenomenon which accounts for the above-started observation.

(iii) Complete the diagram showing how the student’s eye sees the pencil through water.

Answer

(a) When a student puts his pencil into an empty trough, and later pours water in the same trough, the part of the pencil which is immersed in water will look short and raised up.

(b) The above observation is based on the phenomenon of refraction of light.

Question 8

An object placed in one medium when seen from the other medium, appears to be vertically shifted. Name two factors on which the magnitude of the shift depends and state how does it depend on them.

Answer

The shift by which the object appears to be raised depends on —

(i) The refractive index of the medium — Higher the refractive index of the medium, more is the shift.

(ii) The thickness of the denser medium — Thicker the medium, more is the shift.

(iii) The colour or wavelength of incident light — the shift decreases with the increase in the wavelength of light used.

Multiple Choice Type

Question 1

A small air bubble in a glass block when seen from above appears to be raised because of —

refraction of light ✓
reflection of light
reflection and refraction of light
none of the above

Question 2

An object in a denser medium when viewed from a rarer medium appears to be raised. The shift is maximum for—

red light
violet light ✓
yellow light
green light

Numericals

Question 1

A water pond appears to be 2.7 m deep. If the refractive index of water is 4/3, find the actual depth of the pond.

Answer

We know that,

Real depth = Apparent depth x Refractive index of water

Given,

Apparent depth = 2.7 m

$Refractive index of the water=43$

So, substituting the values in the formula we get,

$Real depth = 2.7 \times 34 \Rightarrow Real depth = 3.6 m$

Hence, actual depth = 3.6 m

Question 2

A coin is placed at the bottom of a beaker containing water (refractive index = 4/3) at a depth of 12 cm. By what height the coin appears to be raised when seen from vertically above?

Answer

As we know,

$Shift = Real depth \times (1-^{a} μ ^{m} 1)$

Given,

$Refractive index of the water=43$

and

Real depth = 12 cm

So, substituting the values in the formula we get,

$Shift = 12 \times (1- 43) Shift = 12 \times 41 Shift = 412 \Rightarrow Shift = 3$

Hence, the coin appears to be raised by a height of 3 cm when seen from vertically above.

Question 3

A postage stamp kept below a rectangular glass slab of refractive index 1.5 when viewed from vertically above it, appears to be raised by 7.0 mm. Calculate the thickness of the glass slab.

Answer

As we know,

$Shift = Real depth \times (1-^{a} μ ^{m} 1)$

Given,

Shift in the image = 7 mm or 0.7 cm

Refractive index of the glass block = 1.5

So, substituting the values in the formula we get,

$0.7 = Real depth \times (1- 1.51) 0.7 = Real depth \times 1.50.5 Real depth = 0.50.7 \times 1.5 \Rightarrow Real depth = 2.1$

Hence, the thickness of glass slab = 2.1 cm.
(Thickness of glass slab is same as the real depth of the postage stamp).

Exercise 4(D)

Question 1

Explain the term critical angle with the aid of a labelled diagram.

Answer

When a ray of light passes from a denser to a rarer medium, at a certain angle of incidence i = C, the angle of refraction becomes 90°.

This angle C is called the critical angle.

Thus, critical angle is defined as the angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90°.

Question 2

How is the critical angle related to the refractive index of a medium?

Answer

Knowing the refractive index of the denser medium with respect to the rarer medium, we can calculate the critical angle C for that pair of medium with the help of the relation

$_{a} μ_{g} = sin C 1 = cosec C$

Question 3

State the approximate value of the critical angle for

(a) glass-air surface

(b) water-air surface.

Answer

(a) As we know,

$_{a} μ_{g} = sin C 1 = cosec C$

Refractive index is

$_{a} μ_{g} = 23$

Substituting the values in the formula we get,

$23 = sin C 1 sin C = 23 1 sin C = 32$

As,

$sin 42° = 32$

Hence, critical angle for glass air surface is 42°.

(b) As we know,

$_{a} μ_{w} = sin C 1 = cosec C$

Refractive index is

$_{a} μ_{w} = 34$

Substituting the values in the formula we get,

$34 = sin C 1 sin C = 34 1 sin C = 43 sin 49° = 43$

Hence, critical angle for water air surface is 49°.

Question 4

What is the meant by the statement ‘the critical angle for diamond is 24°’?

Answer

The statement ‘the critical angle for diamond is 24°’ implies that at an incident angle of 24°, the angle of refraction in the air will be 90° within the diamond.

If the incident angle is more than the critical angle then the ray will suffer total internal reflection without any refraction.

Question 5

A light ray is incident from a denser medium on the boundary separating it from a rarer medium at an angle of incidence equal to the critical angle. What is the angle of refraction for the ray?

Answer

The angle of refraction becomes equal to 90° when a light ray is incident from a denser medium on the boundary separating it from a rarer medium at an angle of incidence equal to the critical angle. Below diagram shows the path of the ray when angle of incidence is equal to the critical angle:

A light ray is incident from a denser medium on the boundary separating it from a rarer medium at an angle of incidence equal to the critical angle. What is the angle of refraction for the ray? Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

Question 6

Name two factors which affect the critical angle for a given pair of media. State how do the factors affect it.

Answer

The critical angle depends on two factors:

(a) The colour or wavelength of light — The refractive index of a transparent medium decreases with the increase of wavelength of light. (It is most for violet light and least for red light), therefore the critical angle for a pair of media is least for the violet light and most for the red light, i.e. the critical angle increases with the increase in wavelength of light.

(b) The temperature — On increasing the temperature of medium, its refractive index decreases. So, the critical angle for the pair of medium increases with increase in temperature.

Question 7

The critical angle for glass-air is 45° for the light of yellow colour. State whether it will be less than, equal to, or more than 45° for

(i) red light,

(ii) blue light ?

Answer

As the wavelength of light decreases (or increases) refractive index becomes more (or less) and critical angle becomes less (or more).

(i) As the wavelength for red light is more than that for yellow colour, the refractive index becomes less and the critical angle will be more than 45°.

(ii) As the wavelength for blue light is less than that for yellow colour, the refractive index becomes more and the critical angle will be less than 45°.

Question 8

Which colour of light has higher critical angle ? Red light or green light.

Answer

As we know, the relation between critical angle and refractive index is

$sin c = μ 1$

where, c is critical angle and μ is refractive index.

As μ for red light is less as compared to that for green light and μ and c are inversely related.

Hence, the critical angle of red light is higher than that of green light.

Question 9

(a) What is total internal reflection?

(b) State two conditions necessary for total internal reflection to occur.

Answer

(a) When a ray of light travelling in a denser medium, is incident at the surface of a rarer medium at the angle of incidence greater than the critical angle for the pair of media, the ray is totally reflected back into the denser medium. This phenomenon is known as total internal reflection.

(b) The two necessary conditions for total internal reflection are —

The light must travel from a denser medium to a rarer medium.
The angle of incidence must be greater than the critical angle for the pair of media.

Critical Angle Diagram:

Total Internal Reflection Diagram:

Question 10

Fill in the blanks to complete the following sentences —

(a) Total internal reflection occurs only when a ray of light passes from a ………. medium to a ………. medium.

(b) Critical angle is the angle of ………. in denser medium for which the angle of ………. in rarer medium is ……….

Answer

(a) Total internal reflection occurs only when a ray of light passes from a denser medium to a rarer medium.

(b) Critical angle is the angle of incidence in denser medium for which the angle of refraction in rarer medium is 90°.

Question 11

State whether the following statement is true or false —

If the angle of incidence is greater than the critical angle, light is not refracted at all, when it falls on the surface from a denser medium to a rarer medium.

Answer

True

Question 12

The refractive index of air with respect to glass is expressed as

$_{g} μ_{a} = sin r sin i$

(a) Write down a similar expression for _aμ_g in terms of the angles i and r.

(b) If angle r = 90°, what is the corresponding angle i called?

Answer

(a) The expression for _aμ_g = $sin i sin r$

(b) When refractive angle r = 90° then, the corresponding angle of incidence i will be equal to critical angle.

(c) The phenomenon of total internal reflection occurs when the angle of incidence exceeds the value of i obtained in part (b).

Question 13

Figure below shows two rays A and B travelling from water to air. If the critical angle for water-air surface is 48°, complete the ray diagram showing the refracted rays for each. State conditions when the ray will suffer total internal reflection.

Figure shows two rays A and B travelling from water to air. If the critical angle for water-air surface is 48°, complete the ray diagram showing the refracted rays for each. State conditions when the ray will suffer total internal reflection. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

Answer

Completed ray diagram with refracted rays labelled is shown below:

Completed ray diagram with refracted rays labelled for two rays A and B travelling from water to air with conditions when the ray will suffer total internal reflection. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

Ray A is incident normally at water air boundary so its angle of incidence is 0°. Hence, the refracted ray shown as AA’ in the diagram passes undeviated.

For Ray B, the angle of incidence is equal to the critical angle for water air boundary so it is partly reflected as BC’ and partly refracted as BD’.

Two necessary conditions for the total internal reflection are —

The light must travel from a denser to a rarer medium.
The angle of incidence must be greater than the critical angle for the pair of medium.

Question 14

Figure shows a point source P inside a water container. Three rays A, B and C starting from the source P are shown up to the water surface.

(a) Show in the diagram, the path of these rays after striking the water surface. The critical angle for water-air surface is 48°.

(b) Name the phenomenon which the rays A, B and C exhibit.

Answer

(a) Below diagram shows the path of the rays after striking the water surface:

(b) Rays A and B show the phenomenon of ‘refraction of light’ and ray C shows the phenomenon of ‘total internal reflection’.

Question 15

In the figure, PQ and PR are the two light rays emerging from an object P. The ray PQ is refracted as QS.

(a) State the special name given to the angle of incidence ∠PQN of the ray PQ.

(b) What is the angle of refraction for the refracted ray QS?

(d) The ray PR suffers partial reflection and refraction on the water-air surface. Give reason.

(e) Draw in the diagram the refracted ray for the incident ray PR and hence show the position of image of the object P by the letter P’ when seen vertically from above.

Answer

(a) The angle of incidence ∠PQN is known as the critical angle.

(b) For the refracted ray QS, the angle of refraction is 90°.

(d) For the ray PR, the angle of incidence is less than ∠PQN ( i.e the critical angle ). So according to the laws of reflection, ray PR suffers partial reflection and refraction.

(e) Below diagram shows the refracted ray and the position of image of the object P when seen vertically from above:

Question 16

The refractive index of glass is 1.5. From a point P inside a glass slab, draw rays PA, PB and PC incident on the glass-air surface at an angle of incidence 30°, 42° and 60° respectively.

(a) In the diagram show the approximate direction of these rays as they emerge out of the slab.

(b) What is the angle of refraction for the ray PB?

(Take sin 42° = 2 / 3)

Answer

(a) Diagram with rays PA, PB and PC is shown below:

The refractive index of glass is 1.5. From a point P inside a glass slab, draw rays PA, PB and PC incident on the glass-air surface at an angle of 30°, 42° and 60° respectively. In the diagram show the approximate direction of these rays as they emerge out of the slab. What is the angle of refraction for the ray PB? Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

(b) Given,

μ_g = 1.5

As we know,

$sin i_{c} = μ 1$

Substituting the values in the formula we get,

$sin i_{c} = 1.51 \Rightarrow sin i_{c} = 0.667 \Rightarrow i_{c} = 41.8$

Hence, we can round off i_c = 42°

Applying,

$sin i sin r =_{a} μ_{g} sin r =_{a} μ_{g} \times sin i sin r =_{a} μ_{g} \times sin 42°$

Given,

$sin 42° = 32$

and

$_{a} μ_{g} = 23$

Substituting the values in the formula we get,

$sin r = 23 \times 32 \Rightarrow sin r = 1$

As sin r = 1 , so angle of refraction = 90°.

Question 17

A ray of light enters a glass slab ABDC as shown in figure and strikes at the centre O of the circular part AC of the slab. The critical angle of glass is 42°. Complete the path of the ray till it emerges out from the slab. Mark the angles in the diagram wherever necessary.

Answer

Completed diagram showing the path of the ray till it emerges out from the slab is shown below:

Question 18

What is a total reflecting prism? State three actions that it can produce. Draw a diagram to show one such action of the total reflecting prism.

Answer

A prism having an angle of 90° between its two refracting surfaces and the other two angles each equal to 45° is called a total reflecting prism. In such a prism, the light incident normally on any of its faces suffers total internal reflection inside the prism.

Due to this property, a total reflecting prism is used for the following three purposes —

To deviate a ray of light through 90°.
To deviate a ray of light through 180°.
To erect the inverted image without producing deviation in its path.

Below diagram shows the deviation of a ray of light through 90° by a total reflecting prism:

What is a total reflecting prism? State three actions that it can produce. Draw a diagram to show one such action of the total reflecting prism. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

Question 19

Show with the help of a diagram how a total reflecting prism can be used to turn a ray of light through 90°. Name one instrument in which such a prism is used.

Answer

Below diagram shows the deviation of a ray of light through 90° by a total reflecting prism:

Show with the help of a diagram how a total reflecting prism can be used to turn a ray of light through 90°. Name one instrument in which such a prism is used. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

ABC is a total reflecting prism. A beam of light is incident normally at the face AB. It passes undeviated into the prism and strikes at the face AC making an angle of incidence equal to 45°.

For glass air interface, the critical angle is equal to 42°, therefore the beam of light suffers total internal reflection at the face AC because the angle of incidence is greater than the critical angle.

The reflected beam inside the prism then strikes the face BC, where it is incident normally and therefore passes undeviated. Hence, the incident beam gets deviated through 90°.

This action of prism is used in a periscope where a total reflecting prism is preferred over a plane mirror.

Question 20

A ray of light XY passes through a right-angled isosceles prism as shown in the adjacent diagram.

A ray of light XY passes through a right-angled isosceles prism as shown in the diagram. What is the angle through which the incident ray deviates and emerges out of the prism? Name the instrument where this action of prism is put into use. Which prism surface will behave as a mirror? Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

(a) What is the angle through which the incident ray deviates and emerges out of the prism?

(b) Name the instrument where this action of prism is put into use.

Answer

(a) The angle through which the incident ray deviates and emerges out of the prism is the angle of deviation.

The angle of deviation = the angle which the emergent ray makes with the incident ray XY = 90°.

(b) The instrument where this action of prism is put into use is the periscope.

Question 21

Draw a diagram of a right angled isosceles prism which is used to make an inverted image erect.

Answer

Below diagram shows the usage of a right angled isosceles prism which is used to make an inverted image erect:

Draw a diagram of a right angled isosceles prism which is used to make an inverted image erect. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

This action of prism is used in a slide.

Question 22

In the figure, a ray of light PQ is incident normally on the hypotenuse of an isosceles right angled prism ABC.

(a) Complete the path of the ray PQ till it emerges from the prism. Mark in the diagram the angle wherever necessary.

(b) What is the angle of deviation of the ray PQ?

Answer

(a) Below diagram shows the path of the ray PQ till it emerges from the prism with all the angles labelled:

(b) The angle of deviation of the ray PQ is 180°.

Question 23

In the figure given below, a ray of light PQ is incident normally on the face AB of an equilateral glass prism. Complete the ray diagram showing its emergence into air after passing through the prism. Take critical angle for glass = 42°.

In the figure, a ray of light PQ is incident normally on the face AB of an equilateral glass prism. Complete the ray diagram showing its emergence into air after passing through the prism. Take critical angle for glass = 42°. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

(a) Write the angles of incidence at the faces AB and AC of the prism.

(b) Name the phenomenon which the ray of light suffers at the face AB, AC and BC of the prism.

Answer

(a) The angle of incidence at the faces AB, i = 0°.
The angle of incidence at the faces AC, i = 60°.

(b) The phenomenon which the ray of light suffers at:
face AB — refraction
face AC — total internal reflection
face BC — refraction

Question 24

Copy the diagram given below and complete the path of the light ray till it emerges out of the prism. The critical angle of glass is 42°. In your diagram mark the angles wherever necessary.

Answer

Below is the completed diagram showing the path of the light ray till it emerges out of the prism with all angles marked:

Completed diagram showing the path of the light ray till it emerges out of the prism with all angles marked. The critical angle of glass is 42°. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

Question 25

Draw a neat labelled ray diagram to show total internal reflection of a ray of light incident normally on one face of a 30°, 90°, 60° prism.

Answer

Below is the labelled ray diagram showing total internal reflection of a ray of light incident normally on one face of a 30°, 90°, 60° prism

Draw a neat labelled ray diagram to show total internal reflection of a ray of light incident normally on one face of a 30°, 90°, 60° prism. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

Question 26

What device other than a plane mirror, can be used to turn a ray of light through 180°? Draw a diagram in support of your answer. Name an instrument in which this device is used.

Answer

A total reflecting prism is used to turn a ray of light through 180°.

This device is used in a binocular and camera to invert the image without the loss of intensity.

Question 27

Mention one difference between the reflection of light from a plane mirror and total internal reflection of light from a prism.

Answer

Plane mirror	Prism
In reflection of light from a plane mirror, only a part of the light is reflected while rest is refracted and absorbed. So, the reflection is partial.	In total internal reflection of light from a prism, the entire incident light is reflected back into the denser medium. So there is no loss of light energy.

Question 28

State one advantage of using a total reflecting prism as a reflector in place of a plane mirror.

Answer

The advantage of using a total reflecting prism as a reflector in place of a plane mirror, is that the image is much brighter and the brightness remains unchanged even after the long use of the total reflecting device.

In the case of a plane mirror, the image is less bright and the brightness gradually decreases as the silvering on mirror becomes old and rough.

Question 29

Two isosceles right-angled glass prisms P and Q are placed near each other as shown in figure. Complete the path of the light ray entering the prism P till it emerges out of the prism Q.

Answer

Below diagram shows the path of the light ray entering the prism P till it emerges out of the prism Q:

Two isosceles right-angled glass prisms P and Q are placed near each other as shown in figure. This ray diagram shows the path of the light ray entering the prism P till it emerges out of the prism Q. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

Question 30

Complete the path of ray PQ through the glass prism ABC as shown in figure till it emerges out of the prism. Given the critical angle of glass is 42°.

Complete the path of ray PQ through the glass prism ABC as shown in figure till it emerges out of the prism. Refraction of light at plane surfaces, Concise Physics Class 10 Solutions.

Multiple Choice Type

Question 1

The critical angle for the glass-air interface is:

24°
48°
42° ✓
45°

Question 2

A total reflecting right-angled isosceles prism can be used to deviate a ray of light through:

30°
60°
75°
90° ✓

Question 3

A total reflecting equilateral prism can be used to deviate a ray of light through:

30°
60° ✓
75°
90°

ICSE CLASS 10 Chapter 4 selina solution concise physics

Exercise 4(C)

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Multiple Choice Type

Question 1

Question 2

Numericals

Question 1

Question 2

Question 3

Exercise 4(D)

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Multiple Choice Type

Question 1

Question 2

Question 3

CLICK here for 4A & 4 B