## Selina Concise Physics Class 10 ICSE Solutions Refraction of Light at Plane Surfaces

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 4 Refraction of Light at Plane Surfaces. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 4 Refraction of Light at Plane Surfaces

Exercise 4(A)

Solution 1.

The change in the direction of the path of light, when it passes from one transparent medium to another transparent medium, is called refraction of light.

Solution 2.

Solution 3.

The ray of light which is incident normally on a plane glass slab passes undeviated. That is such a ray suffers no bending at the surface because here the angle of incidence is 0°. Thus if angle of incidence ∠i = 0°, then the angle of refraction ∠r = 0°. And the angle of deviation of the ray will also be 0°.

Solution 5.

The refraction of light (or change in the direction of path of light in other medium) occurs because light travels with different speeds in different media. When a ray of light passes from one medium to another, its direction (except for ∠i = 0°) changes because of change in its speed.

Solution 6.

Air is a rarer medium while water is denser than air with refractive index of 1.33. Therefore when light ray will travel from air to water it will bend towards the normal.

Solution 7.

Speed, intensity and wavelength

Solution 8.

The Snell’s laws of refraction are:

1. The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for the pair of the given media.

Solution 9.

The refractive index of second medium with respect to first medium is defined as the ratio of the sine of angle of incidence in the first medium to the sine of the angle of refraction in the second medium.
Refractive index of a medium is always greater than 1 (it cannot be less than 1) because the speed of light in any medium is always less than that in vacuum.

Solution 10.

Denser medium has a higher refractive index and therefore the speed of light in such medium is lower in comparison to the speed of light in a medium which has a lower refractive index.

Solution 11.

The speed of light decreases when it enters from a rarer medium to denser medium and increases when it enters from a denser medium to rarer medium.
Therefore, the speed of light increases when light ray passes from water to air and the speed of light decreases when light ray passes from water to glass.

Solution 12.

(a) Air (its refractive index is less than that of water)
(b) Glass (its refractive index is more than that of water)

Solution 13.

The refractive index of glass is 1.5 for white light means white light travels in air 1.5 times faster than in glass.

Solution 14.

Solution 16.

(a) The least for red colour and (b) the most for violet colour.

Solution 17.

1. Nature of a medium i.e. its optical density (e.g. μ= 1.5, μ= 1.33) – Smaller the speed of light in a medium relative to air, higher is the refractive index of that medium.
2. Physical condition such as temperature – with increase in temperature, the speed of light in medium increases, so the refractive index of medium decreases.

Solution 18.

Refractive index of a medium decreases with increase in wavelength of light.
Refractive index of a medium for violet light (least wavelength) is greater than that for red light (greatest wavelength).

Solution 19.

Refractive index of a medium decreases with the increase in temperature.
With increase in temperature, the speed of light in that medium increases; thus, the refractive index (= velocity of light in vacuum/velocity of light in medium) decreases.

Solution 21.

Solution 22.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

(i) The glass piece is not seen when the refractive index of liquid becomes equal to the refractive index of glass.
(ii) Light of a single colour is used because the refractive index of a medium (glass or liquid) is different for the light of different colours.

Solution 28.

When a ray of light from lighted candle fall on the surface of a thick plane glass mirror, a small part of light (nearly 4%) is reflected forming first image which is faint virtual image, while a large part of light (nearly 96%) is refracted inside the glass. This ray is now strongly reflected back by the silvered surface inside the glass. This ray is then partially refracted in air and this refracted ray forms another virtual image. This image is the brightest image because it is due to the light suffering a strong reflection at the silver surface.

Solution 29.

(a) When light travels from a rarer to a denser medium, its speed decreases
(b) When light travels from a denser to a rarer medium, its speed increases
(c) The refractive index of glass with respect to air is 3/2. The refractive index of air with respect to glass will be 2/3

Solution 1 (MCQ).

The ray of light bends towards the normal.
Reason: As the speed of light decreases in the denser medium, it bends towards the normal.

Solution 2 (MCQ).

(a) 0°
Reason: A ray of light which is incident normally (i.e. at angle of incidence = 0°) on the surface separating the two media, passes undeviated.

Solution 3 (MCQ).

Diamond
Reason: As the speed of light in diamond is the least, diamond has the highest refractive index.

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Exercise 4(B)

Solution 1.

Solution 2.

Solution 3.

The angle between the direction of incident ray and the emergent ray, is called the angle of deviation.

Solution 4.

Angle of deviation is the angle which the emergent ray makes with the direction of incident ray.

Solution 5.

In a prism the ray of light suffers refraction at two faces. The prism produces a deviation at the first surface and another deviation at the second surface. Thus a prism produces a deviation in the path of light.
The value of the angle of deviation (or the deviation produced by a prism) depends on the following four factors:
(a) the angle of incidence (i),
(b) the material of prism(i.e., on refractive index μ.),
(c) the angle of prism (A),
(d) The colour or wavelength (λ) of light used.

Solution 6.

As the angle of incidence increases, the angle of deviation decreases first and reaches to a minimum value (δm) for a certain angle of incidence. By further increasing the angle of incidence, the angle of deviation is found to increase.
Variation of angle of deviation (δ) with angle of incidence (i):

Solution 7.

False.
With the increase in the angle of incidence, the deviation produced by a prism first decreases and then increases.
A given prism deviates the violet light most and the red light least.

Solution 8.

Solution 9.

Changes in the angle of deviation as we increase

1. The wavelength of incident light
As we increase the wavelength, angle of deviation decreases.
2. The refracting angle of the prism
The angle of deviation increases with the increase in the angle of prism.

Solution 10.

The relation between the angle of incident (i), angle of emergence (e), angle of prism (A) and angle of deviation (δ) for a ray of light passing through an equilateral prism is δ = (i + e) – A

Solution 11.

(i) i2 = i1
(ii) Angle of deviation is minimum
Explanation: In minimum deviation position, the refracted ray inside the prism travels parallel to it if the prism is equilateral and the angle of incidence is equal to the angle of emergence.

Solution 12.

In case of an equilateral prism, when the prism is in the position of minimum deviation δ = δmin, the angle of incidence i1 is equal to the angle of emergence i2.

Solution 13.

Solution 14.

(i) Violet colour will deviate the most and (ii) Red colour will deviate the least.

Solution 15.

B made of flint glass. Because it has higher refractive index.

Solution 16.

The angle of deviation (δ) increases with the increase in the angle of prism (A).

Solution 17.

Let two rays OA and OL from a source O are incident on the prism. They are refracted along AB and LM from first face of the prism. These two rays again refract from the second face of the prism emerge out along BC and MN respectively such that they appear to come from I.

Solution 18.

(a) If the incident ray normal to prism then angle of incidence is 0o.
(b) In this case the angle of refraction from the first face r1= 0o.
(c) As the prism is equilateral so A=60o and r1=0o. So at the second face of the prism, the angle of incidence will be 60o.
(d) No the light will not suffer minimum deviation.

Solution 19.

Solution 1 (MCQ).

The light ray bends at both the surfaces of prism towards its base.

Solution 2 (MCQ).

The deviation produced by the prism does not depend on the size of prism.

Numericals

Solution 1.

Solution 2.

Exercise 4(C)

Solution 1.

Solution 2.

Consider a ray of light incident normally along OA. It passes straight along OAA’. Consider another ray from O (the object) incident at an angle i along OB. This ray gets refracted and passes along BC. On producing this ray BC backwards, it appears to come from the point I, and hence, AI represents the apparent depth, which is less than the real depth AO.

Solution 3.

The depth of the tank appears to be lesser than its real depth. This happens due to the refraction of light from a denser medium (water) to a rarer medium.

Solution 4.

Let any object B is at the bottom of a pond. Consider a light ray BC from the object that moves from water to air. After refraction from the water surface, the ray moves away from the normal N along the path CD. The produce of CD appears from the point B’ and a virtual image of the object at B appears at B’.

Solution 5.

A stick partially immersed in water in a glass container appears bent or raised as shown in figure above. This happens because the rays appear to come from P’ (which is the virtual image of the tip P of the stick) due to refraction from denser medium (water) to rarer medium (air) at the surface separating two media.

Solution 6.

Let the fish is looking from the point O. As the ray OP emerges out from water to air, it will bend away from the normal MN because air is a rarer medium in comparison of water. But if we extend ray OP then it will meet at Q due to which the plant AB will look taller than its actual height.

Solution 7.

(i) Part of the pencil which is immersed in water will look short and raised up.
(ii) The phenomena which is responsible for the above observation is refraction of light.
(iii) The required figure is

Solution 8.

The factors on which the magnitude of shift depends are:

1. The refractive index of the medium,
2. The thickness of the denser medium and
3. The colour (or wavelength) of incident light.

The shift increase with the increase in the refractive index of medium. It also increases with the increase in thickness of denser medium but the shift decreases with the increases in the wavelength of light used.

Solution 1 (MCQ).

Refraction of light
Hint: When a ray of light travels from denser to rarer medium, it moves away from the normal.

Solution 2 (MCQ).

The shift is maximum for violet light.
Hint: The shift is maximum for violet light because the refractive index of glass is the most for the violet light and apparent = (real depth)/(refractive index)

Numericals

Solution 1.

Solution 2.

Solution 3.

Exercise 4(D)

Solution 1.

Critical angle: The angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90° is called the critical angle.

Solution 2.

Solution 3.

Solution 4.

The critical angle for diamond is 24°. This implies that at an incident angle of 24° within the diamond the angle of refraction in the air will be 90°. And if incident angle will be more than this angle then the ray will suffer Total internal reflection without any refraction.

Solution 5.

When a ray is incident from a denser medium to a rarer medium at angle equal to critical angle (i = ic), the angle of refraction becomes 90°.

Solution 6.

The factors which affect the critical angle are:

1. The colour (or wavelength) of light, and
2. The temperature

Effect of colour of light: The critical angle for a pair of media is less for the violet light and more for the red light. Thus critical angle increases with increase in wavelength of light.
Effect of temperature: The critical angle increases with increase in temperature because on increasing the temperature of medium, its refractive index decreases.

Solution 7.

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

1. For red light the critical angle will be more than 45° and
2. For blue light the critical angle will be less than 45°.

Solution 8.

(a) Total internal reflection: It is the phenomenon when a ray of light travelling in a denser medium, is incident at the surface of a rarer medium such that the angle of incidence is greater than the critical angle for the pair of media, the ray is totally reflected back into the denser medium.
(b) The two necessary conditions for total internal reflection are:

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

(c) When incidence angle is more than critical angle i.e., in case of total internal reflection.

Solution 9.

(a) Total internal reflection occurs 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°.

Solution 10.

True

Solution 11.

(b) If refractive angle, r = 90°, the corresponding angle of incidence, i will be equal to critical angle.
(c) If the angle of incidenceexceeds the value of i obtained in part (b) (i.e., critical angle), total internal reflection will occur.

Solution 12.

Solution 13.

(a) Critical angle
Hint: The angle of incidence in the denser medium for which the angle of refraction in rarer medium is 90° is called the critical angle.

(b) 90°

(c) Total internal reflection.
Hint: When the angle of incidence is greater than the critical angle, the phenomenon of total internal reflection occurs due to which the ray of light is not refracted but is reflected back in the same medium.

(d) For the ray PR, the angle of incidence is less than the critical angle (i.e. ∠PQS ); hence, at the interface of two media as per the laws of reflection, ray PR suffers partial reflection and refraction.

Solution 14.

Solution 15.

Solution 16.

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. The light incident normally on any of its faces, suffers total internal reflection inside the prism.
Due to this behavior, a total reflecting prism is used to produce following three actions:

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

Solution 17.

As shown in diagram, a beam of light is incident on face AB of the prism normally so it passes undeviated and strikes the face AC where it makes an angle of 45° with the normal to AC. Because here the incident angle is more than critical angle so rays suffer total internal reflection and reflect at angle of 45°. The beam then strikes face BC, where it is incident normally and so passes undeviated. As a result the incident beam gets deviated through 90°.
Such a prism is used in periscope.

Solution 18.

(a) The angle of incidence at the face AC is 45° and angle of incidence at the face BC is 0°.
(b) The ray suffers total internal reflection at the face AC.

Solution 19.

Solution 20.

Solution 21.

Solution 22.

Solution 23.

A total reflecting prism can be used to turn a ray of light by 180°. The following diagram can make it further clear.

Solution 24.

When total internal reflection occurs from a prism, the entire incident light (100%) is reflected back into the denser medium. Whereas in ordinary reflection from a plane mirror, some light is refracted and absorbed so the reflection is partial.

Solution 25.

A total reflecting prism gives us an image much brighter than that obtained by using a plane mirror.

Solution 1 (MCQ).

42°

Solution 2 (MCQ).

Solution 3 (MCQ).

More Resources for Selina Concise Class 10 ICSE Solutions

## Selina Concise Physics Class 10 ICSE Solutions Force

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 1 Force. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 1 Force

Exercise 1(A)

Solution 1.

(a) When the body is free to move it produces translational motion.
(b) When the body is pivoted at a point, it produces rotational motion.

Solution 2.

The moment of force is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of force from the axis of rotation.
S.I. unit of moment of force is Newton metre (Nm).

Solution 3.

Moment of a force is a vector.

Solution 4.

Moment of force about a point depends on the following two factors:

1. The magnitude of the force applied and,
2. The distance of line of action of the force from the axis of rotation.

Solution 5.

When the body is pivoted at a point, the force applied on the body at a suitable point rotates the body about the axis passing through the pivoted point.
The direction of rotation can be changed by changing the point of application of force. The given figure shows the anticlockwise and clockwise moments produced in a disc pivoted at its centre by changing the point of application of force F from A to B.

Solution 6.

Moment of force about a given axis = force x perpendicular distance of force from the axis of rotation.

Solution 7.

Moment of force depends on the distance of line of action of the force from the axis of rotation. Decreasing the perpendicular distance from the axis reduces the moment of a given force.

Solution 9.

If the turning effect on the body is anticlockwise, moment of force is called anticlockwise moment and it is taken as positive while if the turning effect on the body is clockwise, moment of force is called clockwise moment and is taken negative.

Solution 10.

It is easier to open a door by applying the force at the free end of it because larger the perpendicular distance , less is the force needed to turn the body.

Solution 11.

The stone of hand flour grinder is provided with a handle near its rim so that it can be rotated easily about the iron pivot at its centre by a small force applied at the handle.

Solution 12.

It is easier to turn the steering wheel of a large diameter than that of a small diameter because less force is applied on steering of large diameter which is at a large distance from the centre of rim.

Solution 13.

A spanner (or wrench) has a long handle to produce larger turning moment so that nut can easily be turned with a less force.

Solution 14.

Solution 15.

Solution 16.

(a) Resultant force acting on the body = F-F=0 moment of forces = 0 i.e., no motion of the body
(b) The forces tend to rotate the body about the mid-point between two forces, Moment of forces= Fr

Solution 17.

At A and B, two equal and opposite forces each of magnitude F are applied. The two forces rotate the bar in anticlockwise direction.

Solution 18.

Two equal and opposite parallel forces not acting along the same line, form a couple. A couple is always needed to produce the rotation. For example, turning a key in a lock and turning a steering wheel.

Solution 19.

The moment of a couple is equal to the product of the either force and the perpendicular distance between the line of action of both the forces. S.I unit of moment of couple is Nm.

Solution 20.

At A and B, two equal and opposite forces each of magnitude F are applied. The two forces rotate the bar in anticlockwise direction. The perpendicular distance between two forces is AB which is called the couple arm.
Moment of force F at the end A
= F x OA(anticlockwise)
Moment of force F at the end B
= F x OB(anticlockwise)
Total moment of couple =F x OA + F x OB
= F x (OA +OB)= F x AB
= F x d(anticlockwise)
=Either force x perpendicular distance between the two forces (or couple arm)
Thus, Moment of couple = Force x Couple arm

Solution 21.

When a number of forces acting on a body produce no change in its state of rest or of motion, the body is said to be in equilibrium.

Solution 22.

(i) When a body remains in the state of rest under the influence of the applied forces, the body is in static equilibrium. For example a book lying on a table is in static equilibrium.
(ii) When a body remains in the same state of motion (translational or rotational), under the influence of the applied forces, the body is said to be in dynamic equilibrium. For example, a rain drop reaches the earth with a constant velocity is in dynamic equilibrium.

Solution 23.

For a body to be in equilibrium:

1. The resultant of all the forces acting on the body should be equal to zero.
2. The resultant moment of all the forces acting on the body about the point of rotation should be zero.

Solution 24.

According to the principle of moments, if the algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero, the body is in equilibrium. A physical balance (or beam balance) works on the principle of moments.

Solution 25.

Solution 1 (MCQ).

The moment of a force about a given axis depends on both on the force and its perpendicular distance from the axis.
Hint: Moment of force = Force x Perpendicular distance

Solution 2 (MCQ).

The body will have rotational as well as translational motion.

Numericals

Solution 1.

Moment of force= force x perpendicular distance of force from point O
Moment of force= F x r
5Nm= 10 x r
R= 5/10 =0.5 m

Solution 2.

Length, r=10cm =0.1m
F= 5N
Moment of force= F x r= 5 x 0.1 = 0.5 Nm

Solution 3.

Given , F= 2N
Diameter=2m
Perpendicular distance between B and O =1m
(i)Moment of force at point O
= F x r
= 2 x 1=2Nm (clockwise)
(ii)Moment of force at point A= F x r
= 2 x 2=4Nm (clockwise)

Solution 4.

Given AO=2m and OB=4m

(i) Moment of force F1(=5N) at A about the point O
=F1 x OA
=5 x 2= 10Nm (anticlockwise)

(ii) Moment of force F2(=3N) at B about the point O
= F2 x OB
=3 x 4=12 Nm(clockwise)

(iii) Total moment of forces about the mid-point O
= 12- 10=2Nm(clockwise)

Solution 5.

Given, AB=4m hence, OA=2m and OB =2m
Moment of force F(=10N) at A about the point O
= F x OA= 10 x 2= 20Nm (clockwise)
Moment of force F (=10N) at point B about the point O
= F x OB= 10 x 2 =20Nm (clockwise)
Total moment of forces about the mid-point O=
=20 +20= 40Nm(clockwise)

Solution 6.

(i) Perpendicular distance of point A from the force F=10 N at B is 0.5m , while it is zero from the force F=10N at A
Hence, moment of force about A is
= 10 N x 0.5m=5Nm(clockwise)

(ii) Perpendicular distance of point B from the force F=10 N at A is 0.5m, while it is zero from the force F=10N at B
Hence, moment of force about B is
= 10 N x 0.5m=5Nm(clockwise)

(iii) Perpendicular distance of point O from either of the forces F=10N is 0.25 m
Moment of force F(=10N) at A about O= 10N x 0.25m
=2.5Nm(clockwise)
And moment of force F(=10N) at B about O
=10N x 0.25m=2.5Nm(clockwise)
Hence, total moment of the two forces about O
=0.25 + 0.25=5Nm (clockwise)

Solution 7.

Solution 8.
Let the 50gf weight produce anticlockwise moment about the middle point of metre rule .i.e, at 50cm.
Let a weight of 100gf produce a clockwise moment about the middle point. Let its distance from the middle be d cm. Then,
according to principle of moments,
Anticlockwise moment = Clockwise moment
50gf x 50 cm=100gf x d

Solution 9.

(i) Weight mg (W) of rule produces an anti-clockwise moment about the knife edge O. In order to balance it, 20gf must be suspended at the end B to produce clockwise moment about the knife edge O.

Solution 10.

Anticlockwise moment= 40gf x 40 cm
Clockwise moment= 80gf x d cm
From the principle of moments,
Anticlockwise moment= Clockwise moment
40gf x 40 cm =80gf x d

Solution 11.

(i) Anticlockwise moment= 40gf x (50-10)cm
=40gf x 40cm=1600 gf x cm
Clockwise moment= 20gf x (90- 50) =20gf x 40cm
=800 gf x cm
Anticlockwise moment is not equal to clockwise moment. Hence the metre rule is not in equilibrium and it will turn anticlockwise.

(ii) To balance it, 40gf weight should be kept on right hand side so as to produce a clockwise moment about the middle point. Let its distance from the middle be d cm. Then,
clockwise moment= 20gf x 40cm + 40gf x d cm
From the principle of moments,
Anticlockwise moment= Clockwise moment
40 gf x 40 cm= 20gf x 40 + 40 x d cm
1600-800=40gf x dcm

Solution 12.

From the principle of moments,
Anticlockwise moment= Clockwise moment
20kgf x 2m =40kgf x d

Solution 13.

Solution 14.

(i) Total anticlockwise moment about O
= 150gf x 40 cm=6000gf cm

(ii) Total clockwise moment about O,
=250gf x 20 cm= 5000gf cm

(iii) The difference of anticlockwise and clockwise moment= 6000- 5000= 1000gf cm

(iv) From the principle of moments,
Anticlockwise moment= Clockwise moment
To balance it, 100gf weight should be kept on right hand side so as to produce a clockwise moment about the O. Let its distance from the point O be d cm. Then,
150gf x 40 cm=250gf x 20 cm +100gf x d
6000gf cm= 5000gf cm + 100gf x d
1000gf cm =100 gf x d

Solution 15.

Solution 16.

Solution 17.

(i) From the principle of moments,
Clockwise moment= Anticlockwise moment
100g x (50-40) cm= mx(40-20) cm
100g x 10 cm = m x 20 cm = m =50 g

(ii) The rule will tilt on the side of mass m (anticlockwise), if the mass m is moved to the mark 10cm.

(iii) Anticlockwise moment if mass m is moved to the mark 10 cm= 50g x (40-10)cm =50 x 30=1500g cm
Clockwise moment=100g x (50-40) cm= 1000g cm
Resultant moment= 1500g cm -1000g cm= 500g cm (anticlockwise)

(iv) From the principle of moments,
Clockwise moment= Anticlockwise moment
To balance it, 50g weight should be kept on right hand side so as to produce a clockwise moment .Let its distance from fulcrum be d cm. Then,
100g x (50-40) cm + 50g x d =50g x (40-10)cm
1000g cm + 50g x d =1500 g cm
50 g x d= 500g cm
So, d =10 cm
By suspending the mass 50g at the mark 50 cm, it can be balanced.

Exercise 1(B)

Solution 1.

Centre of gravity is the point about which the algebraic sum of moments of weights of particles constituting the body is zero and the entire weight of the body is considered to act at this point.

Solution 2.

Yes, the centre of gravity can be situated outside the material of the body. For example, centre of gravity of ring.

Solution 3.

The position of centre of gravity of a body of given mass depends on its shape i.e., on the distribution of mass in it. For example: the centre of gravity of a uniform wire is at its mid-point. But if this wire is bent into the form of a circle, its centre of gravity will then be at the centre of circle.

Solution 4.

The position of centre of gravity of a
(a) rectangular lamina is at the point of intersection of its diagonals.
(b) cylinder is at the mid point on the axis of cylinder.

Solution 5.

(a) Centre of gravity of a triangular lamina is situated at the point of intersection of its medians.
(b) Centre of gravity of a circular lamina is situated at the centre of circular lamina.

Solution 6.

Centre of gravity of a uniform ring is situated at the centre of ring.

Solution 7.

Solution 8.

Take a triangular lamina. Make three fine holes at a, b, c near the edge of triangular lamina. Now suspend the given lamina along with a plumb line from hole ‘a’. Check that the lamina is free to oscillate about the point of suspension. When lamina has come to rest, draw straight line ad along the plumb line. Repeat the experiment by suspending the lamina through hole ‘b’ and then through hole ‘c’ for which we get straight lines be and cf respectively. It is noticed that the lines ad, be and cf intersect each other at a common point G which is the position of centre of gravity of triangular lamina i.e. the point of intersection of medians.

Solution 9.

(i) False. The position of centre of gravity of a body of given mass depends on its shape i.e., on the distribution of mass in it.
(ii) True.

Solution 10.

Solution 11.

Solution 1 (MCQ).

At its geometrical centre

Exercise 1(C)

Solution 1.

When a particle moves with a constant speed in a circular path, its motion is said to be the uniform circular motion. For example : Revolution of earth around sun is an example of uniform circular motion.

Solution 2.

Solution 3.

Yes, uniform circular motion has an accelerated motion with a constant speed.

Solution 4.

Motion of a cyclist on a circular track is an example of motion in which speed remains uniform, but the velocity changes.

Solution 5.

When the object moves in a circular path with uniform speed, it means that its magnitude of velocity does not change, only its direction changes continuously. Hence, it is considered as uniformly accelerated motion.

Solution 6.

 Uniform linear motion Uniform circular motion The body moves along a straight line. The body moves along a circular path. Speed and direction both remain constant. Speed is constant, but direction changes continuously. It is not an accelerated motion. It is an accelerated motion.

Solution 7.

Centripetal force is required for circular motion. It is always directed towards the centre of circle.

Solution 8.

Force acting on a body which is in circular motion is called centripetal force. It acts towards the centre of circular path.

Solution 9.

A planet moves around the sun in a nearly circular path for which the gravitational force of attraction on the planet by the sun provides the necessary centripetal force required for circular motion.

Solution 10.

(a) They act in opposite directions.
(b) No, centrifugal force is not the force of reaction of centripetal force.

Solution 11.

No, centrifugal force is a fictitious force.

Solution 12.

a. On standing outside the disc, we find that the pebble is moving on a circular path. On standing at the centre of the disc, we find that the pebble is stationary placed just in front of us.

Solution 13.

Force of tension in the thread provides the centripetal force.

Solution 15.

(a) False
(b) True
(c) True
(d) False

Solution 1 (MCQ).

Speed
Hint: Speed is scalar but velocity and acceleration are vector quantities. So, speed remains constant but velocity and acceleration change with the change in direction, and in circular motion the direction of motion changes at every point.

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## Selina Concise Physics Class 10 ICSE Solutions Machines

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 3 Machines. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 3 Machines

Exercise 3(A)

Solution 1.

(a) A machine is a device by which we can either overcome a large resistive force at some point by applying a small force at a convenient point and in a desired direction or by which we can obtain a gain in the speed.

(b) An ideal machine is a machine whose parts are weightless and frictionless so that which there is no dissipation of energy in any manner. Its efficiency is 100%, i.e. the work output is equal to work input.

Solution 2.

Machines are useful to us in the following ways:

1. In lifting a heavy load by applying a less effort.
2. In changing the point of application of effort to a convenient point.
3. In changing the direction of effort to a convenient direction.
4. For obtaining a gain in speed.

Solution 3.

(a) To multiply force: a jack is used to lift a car.
(b) To change the point of application of force: the wheel of a cycle is rotated with the help of a chain by applying the force on the pedal.
(c) To change the direction of force: a single fixed pulley is used to lift a bucket full of water from the well by applying the effort in the downward direction instead of applying it upwards when the bucket is lifted up without the use of pulley.
(d) To obtain gain in speed: when a pair of scissors is used to cut the cloth, its blades move longer on cloth while its handles move a little.

Solution 4.

The purpose of jack is to make the effort less than the load so that it works as a force multiplier.

Solution 6.

The ratio of the load to the effort is called mechanical advantage of the machine. It has no unit.

Solution 7.

The ratio of the velocity of effort to the velocity of the load is called the velocity ratio of machine. It has no unit.

Solution 8.

For an ideal machine mechanical advantage is numerically equal to the velocity ratio.

Solution 9.

It is the ratio of the useful work done by the machine to the work put into the machine by the effort.
In actual machine there is always some loss of energy due to friction and weight of moving parts, thus the output energy is always less than the input energy.

Solution 10.

(a) A machine acts as a force multiplier when the effort arm is longer than the load arm. The mechanical advantage of such machines is greater than 1.
(b) A machine acts a speed multiplier when the effort arm is shorter than the load arm. The mechanical advantage of such machines is less than 1.

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

Solution 11.

Solution 12.

Let a machine overcome a load L by the application of an effort E. In time t, let the displacement of effort be dE and the displacement of load be dL.
Work input = Effort X displacement of effort = E X dE
Work output = Load X displacement of load = L X dL

Solution 13.

The efficiency of such a machine is always less than 1, i.e. h<1. This is because there is always some loss in energy in form of friction etc.

Solution 14.

This is because the output work is always less than the input work, so the efficiency is always less than 1 because of energy loss due to friction.

Solution 15.

A lever is a rigid, straight or bent bar which is capable of turning about a fixed axis.
Principle: A lever works on the principle of moments. For an ideal lever, it is assumed that the lever is weightless and frictionless. In the equilibrium position of the lever, by the principle of moments,

Solution 16.

Solution 17.

The three classes of levers are:

1. Class I levers: In these types of levers, the fulcrum F is in between the effort E and the load L. Example: a seesaw, a pair of scissors, crowbar.
2. Class II levers: In these types of levers, the load L is in between the effort E and the fulcrum F. The effort arm is thus always longer than the load arm. Example: a nut cracker, a bottle opener.
3. Class III levers: In these types of levers, the effort E is in between the fulcrum F and the load L and the effort arm is always smaller than the load arm. Example: sugar tongs, forearm used for lifting a load.

Solution 18.

(a) More than one: shears used for cutting the thin metal sheets.
(b) Less than one: a pair of scissors whose blades are longer than its handles.

Solution 19.

When the mechanical advantage is less than 1, the levers are used to obtain gain in speed. This implies that the displacement of load is more as compared to the displacement of effort.

Solution 20.

A pair of scissors and a pair of pliers both belong to class I lever.
A pair of scissors has mechanical advantage less than 1.

Solution 21.

A pair of scissors used to cut a piece of cloth has blades longer than the handles so that the blades move longer on the cloth than the movement at the handles.
While shears used for cutting metals have short blades and long handles because as it enables us to overcome large resistive force by a small effort.

Solution 22.

(a) The weight W of the scale is greater than E.
It is because arm on the side of effort E is 30 cm and on the side of weight of scale is 10 cm. So, to balance the scale, weight W of scale should be more than effort E.

Solution 23.

Class II lever always have a mechanical advantage more than one.
Example: a nut cracker.
To increase its mechanical advantage we can increase the length of effort arm.

Solution 24.

Solution 25.

In these types of levers, the load L is in between the effort E and the fulcrum F. So, the effort arm is thus always longer than the load arm. Therefore M.A > 1.

Solution 26.

Solution 27.

Solution 29.

Solution 30.

In these types of levers, the effort is in between the fulcrum F and the load L and so the effort arm is always smaller than the load arm. Therefore M.A. < 1.

Solution 31.

With levers of class III, we do not get gain in force, but we get gain in speed, that is a longer displacement of load is obtained by a smaller displacement of effort.

Solution 32.

Solution 33.

(a) A bottle opener is a lever of the second order, as the load is in the middle, fulcrum at one end and effort at the other.

(b) Sugar tongs is a lever of the third order as the effort is in the middle, load at one end and fulcrum at the other end.

Solution 34.

Solution 35.

a. Class II
b. Class I
c. Class II
d. Class III

Solution 36.

(a) Class III.
Here, the fulcrum is the elbow of the human arm. Biceps exert the effort in the middle and load on the palm is at the other end.

(b) Class II.
Here, the fulcrum is at toes at one end, the load (i.e. weight of the body) is in the middle and effort by muscles is at the other end.

Solution 37.

Solution 38.

Class I lever in the action of nodding of the head: In this action, the spine acts as the fulcrum, load is at its front part, while effort is at its rear part.

Class II lever in raising the weight of the body on toes: The fulcrum is at toes at one end, the load is in the middle and effort by muscles is at the other end.

Class III lever in raising a load by forearm: The elbow joint acts as fulcrum at one end, biceps exerts the effort in the middle and a load on the palm is at the other end.

Solution 1 (MCQ).

M.A. x E = L

Solution 2 (MCQ).

Solution 3 (MCQ).

It can have a mechanical advantage greater than the velocity ratio.
Reason: If the mechanical advantage of a machine is greater than its velocity ratio, then it would mean that the efficiency of a machine is more than 100%, which is practically not possible.

Solution 4 (MCQ).

Effort is between fulcrum and load
Hint: Levers, for which the mechanical advantage is less than 1, always have the effort arm shorter than the load arm.

Solution 5 (MCQ).

M.A > 1
Hint: In class II levers, the load is in between the effort and fulcrum. Thus, the effort arm is always longer than the load arm and less effort is needed to overcome a large load. Hence, M.A > 1.

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Total length of rod=4 m = 400 cm

(a) 18kgf load is placed at 60 cm from the support.
W kgf weight is placed at 250 cm from the support.
By the principle of moments
18 x 60 = W x 250
W = 4.32 kgf

(b) Given W=5 kgf
18kgf load is placed at 60 cm from the support.
Let 5 kgf of weight is placed at d cm from the support.
By the principle of moments
18 x 60 = 5 x d
d = 216 cm from the support on the longer arm

(c) It belongs to class I lever.

Solution 9.

Solution 10.

Solution 11.

(a) The principle of moments: Moment of the load about the fulcrum=moment of the effort about the fulcrum
FB x Load = FA x Effort
(b) Sugar tongs the example of this class of lever.
(c) Given: FA=10 cm, AB = 500 cm, BF =500+10=510 cm.

Solution 12.

Exercise 3(B)

Solution 1.

Fixed pulley: A pulley which has its axis of rotation fixed in position, is called a fixed pulley.
Single fixed pulley is used in lifting a small load like water bucket from the well.

Solution 2.

The ideal mechanical advantage of a single fixed pulley is 1.
It cannot be used as force multiplier.

Solution 3.

There is no gain in mechanical advantage in the case of a single fixed pulley. A single fixed pulley is used only to change the direction of the force applied that is with its use, the effort can be applied in a more convenient direction. To raise a load directly upwards is difficult.

Solution 4.

The velocity ratio of a single fixed pulley is 1.

Solution 5.

The load rises upwards with the same distance x.

Solution 6.

Single movable pulley: A pulley, whose axis of rotation is not fixed in position, is called a single movable pulley.
Mechanical advantage in the ideal case is 2.

Solution 8.

The efficiency of a single movable pulley system is not 100% this is because

1. The friction of the pulley bearing is not zero ,
2. The weight of the pulley and string is not zero.

Solution 9.

The force should be in upward direction.
The direction of force applied can be changed without altering its mechanical advantage by using a single movable pulley along with a single fixed pulley to change the direction of applied force.
Diagram:

Solution 10.

The velocity ratio of a single movable pulley is always 2.

Solution 11.

The load is raised to a height of x/2.

Solution 12.

Ideal mechanical advantage of this system is 2. This can be achieved by assuming that string and the pulley are massless and there is no friction in the pulley bearings or at the axle or between the string and surface of the rim of the pulley.

Solution 13.

(b) The fixed pulley B is used to change the direction of effort to be applied from upward to downward.
(c) The effort E balances the tension T at the free end, so E=T
(d) The velocity ratio of this arrangement is 2.
(e) The mechanical advantage is 2 for this system (if efficiency is 100%).

Solution 14.

 Single fixed pulley Single movable pulley 1. It is fixed to a rigid support. 1. It is not fixed to a rigid support. 2. Its mechanical advantage is one. 2. Its mechanical advantage istwo. 3. Its velocity ratio is one. 3. Its velocity ratio is two. 4. The weight of pulley itself does not affect its mechanical advantage. 4. The weight of pulley itself reduces its mechanical advantage. 5. It is used to change the direction of effort 5. It is used as force multiplier.

Solution 15.

Solution 16.

Mechanical advantage = MA = L/E = 23
As one end of each string passing over a movable pulley is fixed, so the free end of string moves twice the distance moved by the movable pulley.
If load L moves up by a distance x, dL = x, effort moves by a distance 23x, dE = 23x

Solution 17.

Solution 18.

1. In a single fixed pulley, some effort is wasted in overcoming friction between the strings and the grooves of the pulley; so the effort needed is greater than the load and hence the mechanical advantage is less than the velocity ratio.
2. This is because of some effort is wasted in overcoming the friction between the strings and the grooves of the pulley.
3. This is because mechanical advantage is equal to the total number of pulleys in both the blocks.
4. The efficiency depends upon the mass of lower block; therefore efficiency is reduced due to the weight of the lower block of pulleys.

Solution 19.

(a) Multiply force: a movable pulley.
(b) Multiply speed: gear system or class III lever.
(c) Change the direction of force applied: single fixed pulley.

Solution 20.

1. The velocity ratio of a single fixed pulley is always more than 1.(false)
2. The velocity ratio of a single movable pulley is always 2.(true)
3. The velocity ratio of a combination of n movable pulleys with a fixed pulley is always 2n.(true)
4. The velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting the load. (true)

Solution 1 (MCQ).

It helps in applying effort in a convenient direction.
Explanation: A single fixed pulley though does not reduce the effort but helps in changing the direction of effort applied. As it is far easier to apply effort in downward direction, the single fixed pulley is widely used.

Solution 2 (MCQ).

The mechanical advantage of an ideal single movable pulley is 2.
Derivation: Consider the diagram given below:

Here the load L is balance by the tension in two segments of the string and the effort E balances the tension T at the free end, so
L = T + T = 2T and E = T
Assumption: Weight of the pulley is negligible.
We know that,

Solution 3 (MCQ).

Force multiplier
Explanation: The mechanical advantage of movable pulley is greater than 1. Thus, using a single movable pulley, the load can be lifted by applying an effort equal to half the load (in ideal situation), i.e. the single movable pulley acts as a force multiplier.

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

(d) (i) There is no friction in the pulley bearings, (ii) weight of lower pulleys is negligible and (iii) the effort is applied downwards.

Solution 8.

Solution 9.

Solution 10.

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## Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 2 Work, Energy and Power. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 2 Work, Energy and Power

Exercise 1(A)

Solution 1.

Work is said to be done only when the force applied on a body makes the body move. It is a scalar quantity.

Solution 2.

(i) When force is in direction of displacement, then work done , W = F x S
(ii) When force is at an angle θ to the direction of displacement, then work done, W= F S cos θ

Solution 3.

(a) When force is at an angle θ to the direction of displacement, then work done, W= F S cos θ
(b) (i) For zero work done, the angle between force and displacement should be 90o as cos 90= 0
W = FS cos90o= FS x 0 = 0
(ii) For maximum work done, the angle between force and displacement should be 0o as cos0= 1
Hence, W=FScos 0o=FS

Solution 4.

Two conditions when the work done is zero are:

1. When there is no displacement (S = 0) and,
2. When the displacement is normal to the direction of the force (θ = 90o).

Solution 5.

(i) If the displacement of the body is in the direction of force, then work done is positive.
Hence, W= F x S
For example: A coolie does work on the load when he raises it up against the force of gravity. The force exerted by coolie (=mg) and displacement, both are in upward direction.

(ii) If the displacement of the body is in the direction opposite to the force, then work done is negative.
Hence, W =- F x S
For example: When a body moves on a surface, the frictional forces between the body and the surface is in direction opposite to the motion of the body and so the work done by the force of friction is negative.

Solution 6.

Work is done against the force.

Solution 7.

When a body moves in a circular path, no work is done since the force on the body is directed towards the centre of circular path (the body is acted upon by the centripetal force), while the displacement at all instants is along the tangent to the circular path, i.e., normal to the direction of force.

Solution 8.

Work done by the force of gravity (which provides the centripetal force) is zero as the force of gravity acting on the satellite is normal to the displacement of the satellite.

Solution 9.

Work is done only in case of a boy climbing up a stair case.

Solution 10.

When a coolie carrying some load on his head moves, no work is done by him against the force of gravity because the displacement of load being horizontal, is normal to the direction of force of gravity.

Solution 11.

Force applied by the fielder on the ball is in opposite direction of displacement of ball. So, work done by the fielder on the ball is negative.

Solution 12.

When a coolie carries a load while moving on a ground, the displacement is in the horizontal direction while the force of gravity acts vertically downward. So the work done by the force of gravity is zero.

Solution 13.

S.I unit of work is Joule.
C.G.S unit of work is erg.
Relation between joule and erg :
1joule = 1N x 1m
But 1N = 105dyne
And 1m = 100 cm = 10cm
Hence, 1 joule = 105dyne x 102cm
= 107dyne x cm = 107erg
Thus, 1 Joule = 107 erg

Solution 14.

S. I unit of work is Joule.
1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 metre in its own direction.
Previous

Solution 15.

Relation between joule and erg :
1 joule = 1N x 1m
But 1N = 105dyne
And 1m = 100 cm= 10cm
Hence, 1 joule = 105dyne x 102cm
= 107dyne x cm = 107erg
Thus, 1 Joule = 107 erg

Solution 16.

Let a body of mass m fall down through a vertical height h either directly or through an inclined plane e.g. a hill, slope or staircase. The force of gravity on the body is F = mg acting vertically downwards and the displacement in the direction of force (i.e., vertical) is S=h. Therefore the work done by the force of gravity is
W = FS = mgh

Solution 17.

Let a boy of mass m climb up through a vertical height h either through staircase of using a lift. The force of gravity on the boy is F=mg acting vertically downwards and the displacement in the direction opposite to force (i.e., vertical) is S=-h. Therefore the work done by the force of gravity on the boy is
W= FS =-mgh
or,the work W=mgh is done by the boy against the force of gravity.

Solution 18.

The energy of a body is its capacity to do work. Its S.I unit is Joule (J).

Solution 19.

eV measures the energy of atomic particles.
1eV= 1.6 x 10-19J

Solution 20.

1 J = 0.24 calorie

Solution 21.

Calorie measures heat energy.
1calorie = 4.18 J

Solution 22.

1kWh is the energy spent (or work done) by a source of power 1kW in 1 h.
1kWh = 3.6 x 106J

Solution 23.

The rate of doing work is called power. The S.I. unit of power is watt (W).

Solution 24.

Power spent by a source depends on two factors:
(i) The amount of work done by the source, and
(ii) The time taken by the source to do the said work.
Example: If a coolie A takes 1 minute to lift a load to the roof of a bus, while another coolie B takes 2 minutes to lift the same load to the roof of the same bus, the work done by both the coolies is the same, but the power spent by the coolie A is twice the power spent by the coolie B because the coolie A does work at a faster rate.

Solution 25.

 Work Power 1. Work done by a force is equal to the product of force and the displacement in the direction of force. 1. Power of a source is the rate of doing work by it. 2. Work done does not depend on time. 2. Power spent depends on the time in which work is done. 3. S.I unit of work is joule (J). 3. S.I unit of power is watt (W).

Solution 26.

 Energy Power 1. Energy of a body is its capacity to do work. 1. Power of a source is the energy spent by it in 1s. 2. Energy spent does not depend on time. 2. Power spent depends on the time in which energy is spent. 3. S.I unit of energy is joule (J). 3. S.I unit of power is watt (W).

Solution 27.

S.I unit of power is watt (W).
If 1 joule of work is done in 1 second, the power spent is said to be 1 watt.

Solution 28.

Horse power is another unit of power, largely used in mechanical engineering. It is related to the S.I unit watt as :
1 H.P =746 W

Solution 29.

Watt (W) is the unit of power, while watt hour (Wh) is the unit of work, since power x time = work.

Solution 30.

a. Energy is measured in kWh
b. Power is measure in kW
c. Energy is measured in Wh
d. Energy is meaused in eV
Concept insight: Energy has bigger units like kWh (kilowatt hour) and Wh (watt hour). Similarly bigger unit of power is kW (kilo watt).
The energy of atomic particles is very small, and hence, it is measured in eV (electron volt).

Solution 1 (MCQ).

746 W

Solution 2 (MCQ).

The unit kWh is the unit of energy.

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 12.

Solution 13.

(i) The work done by persons A and B is independent of time. Hence both A and B will do the same amount of work. Hence,

(ii) Power developed by the person A and B is calculated as follows:
A takes 20 s to climb the stairs while B takes 15 s, to do the same. Hence B does work at a much faster rate than A; more power is spent by B.

Solution 15.

Exercise 2(B)

Solution 1.

Two forms of mechanical energy are:

Solution 2.

Elastic potential energy is possessed by wound up watch spring.

Solution 3.

(a) Kinetic energy (K)
(b) Potential energy (U)
(c) Kinetic energy (K)
(d) Potential energy (U)
(e) Kinetic energy (K)
(f) Potential energy (U)

Solution 4.

Potential energy: The energy possessed by a body by virtue of its specific position (or changed configuration) is called the potential energy.
Different forms of P.E. are as listed below:

1. Gravitational potential energy: The potential energy possessed by a body due to its position relative to the centre of Earth is called its gravitational potential energy.
Example: A stone at a height has gravitational potential energy due to its raised height.
2. Elastic potential energy: The potential energy possessed by a body in the deformed state due to change in its configuration is called its elastic potential energy.
Example: A compressed spring has elastic potential energy due to its compressed state.

Solution 5.

Potential energy is possessed by the body even when it is not in motion. For example: a stone at a height has the gravitational potential energy due to its raised position.

Solution 6.

Gravitational potential energy is the potential energy possessed by a body due to its position relative to the centre of earth.
For a body placed at a height above the ground, the gravitational potential energy is measured by the amount of work done in lifting it up to that height against the force of gravity.
Let a body of mass m be lifted from the ground to a vertical height h. The least upward force F required to lift the body (without acceleration) must be equal to the force of gravity (=mg) on the body acting vertically downwards. The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
= mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energy U= mgh

Solution 7.

The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
=mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energy U= mgh

Solution 8.

A body in motion is said to possess the kinetic energy. The energy possessed by a body by virtue of its state of motion is called the kinetic energy.

Solution 9.

Solution 10.

According to the work-energy theorem, the work done by a force on a moving body is equal to the increase in its kinetic energy.

Solution 11.

Body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v and produces an acceleration a in moving a distance S.Then,
Work done by the force= force x displacement
W = F x S———(i)

Solution 12.

Both the masses have same momentum p. The kinetic energy, K is inversely proportional to mass of the body.
Hence light mass body has more kinetic energy because smaller the mass, larger is the kinetic energy.

Solution 13.

Kinetic energy is related to momentum and mass as
p = √2mK
As the kinetic energy of both bodies are same, momentum is directly proportional to square root of mass.
Now, mass of body B is greater than that of body A.
Hence, body B will have more momentum than body A.

Solution 15.
The three forms of kinetic energy are:

1. Translational kinetic energy- example: a freely falling body
2. Rotational kinetic energy-example: A spinning top.
3. Vibrational kinetic energy-example: atoms in a solid vibrating about their mean position.

Solution 16.

 Potential energy (U) Kinetic energy (K) 1. The energy possessed by a body by virtue of its specific position or changed configuration is called potential energy. 1.The energy possessed by a body by virtue of its state of motion is called the kinetic energy. 2.Two forms of potential energy are gravitational potential energy and elastic potential energy. 2. Forms of kinetic energy are translational, rotational and vibrational kinetic energy. 3.Example: A wound up watch spring has potential energy. 3. For example: a moving car has kinetic energy.

Solution 17.

(a) Motion.
(b) Position.

Solution 18.

When the string of a bow is pulled, some work is done which is stored in the deformed state of the bow in the form of its elastic potential energy. On releasing the string to shoot an arrow, the potential energy of the bow changes into the kinetic energy of the arrow which makes it move.

Solution 19.

The compressed spring has elastic potential energy due to its compressed state. When it is released, the potential energy of the spring changes into kinetic energy which does work on the ball if placed on it and changes into kinetic energy of the ball due to which it flies away.

Solution 20.

When water falls from a height, the potential energy stored in water at a height changes into the kinetic energy of water during the fall. On striking the ground, a part of the kinetic energy of water changes into the heat energy due to which the temperature of water rises.

Solution 21.

Yes, when force is normal to displacement, no transfer of energy takes place.

Solution 22.

Kinetic energy.

Solution 23.

The six different forms of energy are:

1. Solar energy
2. Heat energy
3. Light energy
4. Chemical or fuel energy
5. Hydro energy
6. Nuclear energy

Solution 24.

(a) Potential energy of wound up spring converts into kinetic energy.
(b) Chemical energy of petrol or diesel converts into mechanical energy (kinetic energy)
(c) Kinetic energy to potential energy
(d) Light energy changes into chemical energy
(e) Electrical energy changes into chemical energy
(f) Chemical energy changes into heat energy
(g) Chemical energy changes into heat and light energy
(h) Chemical energy changes into heat, light and sound energy

Solution 25.

(a) Electrical energy into sound energy
(b) Heat energy into mechanical energy
(c) Sound energy into electrical energy
(d) Electrical energy to mechanical energy
(e) Electrical energy into light energy
(f) Chemical energy to heat energy
(g) Light energy into electrical energy
(h) Chemical energy into heat energy
(i) Chemical energy into electrical energy
(j) Chemical energy to mechanical energy
(k) Electrical energy into heat energy
(l) Light energy into electrical energy
(m) Electrical energy into magnetic energy.

Solution 27.

No. This is because, whenever there is conversion of energy from one form to another apart of the energy is dissipated in the form of heat which is lost to surroundings.

Solution 1 (MCQ).

Potential energy
Hint: P.E. is the energy possessed by a body by virtue of its position.

Solution 2 (MCQ).

Chemical to electrical
Hint: When current is drawn from an electric cell, the chemical energy stored in it changes into electrical energy.

Numericals

Solution 1.

Solution 2.

Mass , m=1kg
Height, h=5m
Gravitational potential energy = mgh
=1 x 10 x 5=50J

Solution 3.

Gravitational potential energy=14700 J
Force of gravity = mg= 150 x 9.8N/kg= 1470N
Gravitational potential energy= mgh
14700 =1470 x h
h=10m

Solution 4.

Solution 5.

Mass =0.5 kg
Energy= 1 J
Gravitational potential energy= mgh
1=0.5 x10 x h
1=5h
Height, h= 0.2 m

Solution 6.

Force of gravity on boy=mg= 25 x 10 =250N
Increase in gravitational potential energy= Mg (h2-h1)
= 250 x (9-3)
=250 x6=1500 J

Solution 7.

Mass of water, m= 50kg
Height, h=15m
Gravitational potential energy= mgh
=50 x10 x 15
=7500 J

Solution 8.

Mass of man=50kg

(i) Work done by man =mgh2
=50 x 9.8 x10= 4900J

(ii) increase in his potential energy:
Height, h2= 10m
Reference point is ground, h1=0m
Gravitational potential energy= Mg (h2-h1)
= 50 x9.8x (10-0)
= 50 x 9.8 x10= 4900J

Solution 9.

F=150N

(a) Work done by the force in moving the block 5m along the slope =Force x displacement in the direction of force
=150 x 5=750 J.

(b) The potential energy gained by the block U =mgh where h =3m
=200 x 3=600 J
he potential energy gained by the block

(c) The difference i.e., 150 J energy is used in doing work against friction between the block and the slope, which will appear as heat energy.

Solution 10.

Solution 11.

If the speed is halved (keeping the mass same), the kinetic energy decreases, it becomes one-fourth (since kinetic energy is proportional to the square of velocity).

Solution 13.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

Solution 18.

Solution 19.

Solution 20.

Solution 21.

Solution 22.

Solution 23.

Exercise 2(C)

Solution 1.

According to the law of conservation of energy, energy can neither be created nor can it be destroyed. It only changes from one form to another.

Solution 2.

According to the law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy (i.e., the sum of kinetic energy K and potential energy U) remains constant i.e., K + U = constant when there are no frictional forces.

Mechanical energy is conserved only when there are no frictional forces for a given system (i.e. between body and air). Thus, conservation of mechanical energy is strictly valid only in vacuum, where friction due to air is absent.

Solution 3.

Motion of a simple pendulum and motion of a freely falling body.

Solution 4.

Kinetic energy of the body changes to potential energy when it is thrown vertically upwards and its velocity becomes zero.

Solution 5.

(a) Potential energy
(b) Potential energy and kinetic energy
(c) Kinetic energy

Solution 6.

Let a body of mass m be falling freely under gravity from a height h above the ground (i.e., from position A). Let us now calculate the sum of kinetic energy K and potential energy U at various positions, say at A (at height h above the ground), at B (when it has fallen through a distance x) and at C (on the ground).

(i) At the position A (at height h above the ground):
Initial velocity of body= 0 (since body is at rest at A)
Hence, kinetic energy K =0
Potential energy U = mgh
Hence total energy = K + U= 0 + mgh =mgh—–(i)

(ii) At the position B (when it has fallen a distance x):

Thus from equation (i), (ii) and (iii), we note that the total mechanical energy i.e., the sum of kinetic energy and potential energy always remain constant at each point of motion and is equal to initial potential energy at height h.

Solution 7.

When the bob swings from A to B, the kinetic energy decreases and the potential energy becomes maximum at B where it is momentarily at rest.

From B to A, the potential energy again changes into the kinetic energy and the process gets repeated again and again.
Thus while swinging, the bob has only the potential energy at the extreme position B or C and only the kinetic energy at the resting position A. At an intermediate position (between A and B or between A and C), the bob has both the kinetic energy and potential energy, and the sum of both the energies (i.e., the total mechanical energy) remains constant throughout the swing.

Solution 8.

(a) At position A, pendulum has maximum kinetic energy and its potential energy is zero at its resting position. Hence, K=mgh and U= 0.
(b) At B, kinetic energy decreases and potential energy increases. Hence, K= 0 and U=mgh
(c) At C also, kinetic energy K= 0 and potential energy U=mgh.

Solution 9.

a) Extreme position: Potential energy
b) Mean position: Kinetic energy
c) Between mean and extreme: Both kinetic energy and potential energy

Solution 10.

The gradual decrease of useful energy due to friction etc. is called the degradation of energy.
Examples:

1. When we cook food over a fire, the major part of heat energy from the fuel is radiated out in the atmosphere. This radiated energy is of no use to us.
2. When electrical appliances are run by electricity, the major part of electrical energy is wasted in the form of heat energy.

Solution 1 (MCQ).

Potential energy of the ball at the highest point is mgh.
Hint: At the highest point, the ball momentarily comes to rest and thus its kinetic energy becomes zero.

Solution 2 (MCQ).

The sum of its kinetic and potential energy remains constant throughout the motion.
Hint: In accordance with law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy remains constant.

Numericals

Solution 1.

Solution 2.

Solution 3.

(a)Potential energy of the ball =mgh
=2 x 10 x 5=100J
(b)Kinetic energy of the ball just before hitting the ground = Initial potential energy= mgh=2x10x5=100J
(c)Mechanical energy converts into heat and sound energy.

Solution 4.

Solution 5.

Solution 6.

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## Selina Concise Physics Class 10 ICSE Solutions Radioactivity

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 12 Radioactivity. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 12 Radioactivity

Exercise 12(A)

Solution 1.

Three constituent of an atom are:

1. Electrons: mass is 9.1 x 10-31 kg, charge is -1.6 x 10-19C
2. Neutron: mass is 1.6749 x 10-27 kg, charge is zero.
3. Protons: mass is 1.6726 x 10-27 kg, charge is +1.6 x 10-19 C

Solution 2.

Atomic number – the number of protons in the nucleus is called atomic number.
Mass number – the total number of nucleons in the nucleus is called mass number.

Solution 3.

• The nucleus at the centre of atom, whose size is of the order of 10-15 m to 10-14 m.
• The size of a nucleus is 10-5 to 10-4 times the size of an atom. It consists of protons and neutrons.
• If Z is the atomic number and A is the mass number of an atom, then the atom contains Z number of electrons; Z number of protons and A – Z number of neutrons.
• The atom is specified by the symbol ZXwhere X is the chemical symbol for the element.

Solution 4.

Atomic number Z = 11
Mass number A = 23
Number of neutrons A – Z = 12

Solution 5.

Isotopes: the atoms of the same element which have the same atomic number Z but differ in their mass number A are called isotopes.

Solution 6.

Isobars: the atoms of different elements which have the same mass number A, but differ in their atomic number Z are called isobars.

Solution 7.

Solution 8.

Radioactivity: Radioactivity is a nuclear phenomenon. It is the process of spontaneous emission of α or β and γ radiations from the nuclei of atoms during their decay.

Solution 9.

There will be no change in the nature of radioactivity. This is because radioactivity is a nuclear phenomenon.

Solution 10.

(a) Three types of radiations: Alpha, beta and gamma.

Solution 11.

(a) Gamma radiations have zero mass.
(b) Gamma radiations have the lowest ionizing power.
(c) Alpha particles have lowest penetrating power.
(d) Alpha particle has positive charge equal to 3.2 x 10-19C and rest mass equal to 4 times the mass of proton i.e. 6.68 x 10-27 kg.
(e) The gas is Helium.

(f) These radiations come from nucleus of the atom.

Solution 12.

• Radiations labeled A, B and C are α, β and γ respectively.
• Radiation labeled A is gamma radiation because they have no charge and hence under action of magnetic field they go undeflected.
• Radiation B is alpha radiation because its mass is large and it would be deflected less in comparison to beta radiation. The direction of deflection is given by Fleming’s left hand rule. Also directions of deflection of alpha and beta radiations are opposite as they have opposite charge.

Solution 13.

Solution 14.

(b) The radioactive substances are kept in thick lead containers with a very narrow opening, so as to stop radiations coming out from other directions because they may cause biological damage.

Solution 15.

This is because alpha and beta particles are charged particles, but gamma rays are neutral particles.

Solution 16.

No, it is not possible to deflect gamma radiation in a way similar to alpha and beta particles, using the electric or magnetic field because they are neutral and hence do not deflected under the action of electric or magnetic field.

Solution 17.

 Property α-particle β-particle γ-particle Nature Stream of positively charged particles, i.e. helium nuclei. Stream of negatively charged particles, i.e. energetic electrons. Highly energetic electromagnetic radiation. Charge Positive charge (Two times that of a proton) = + 3.2 x 10-19 C (or +2e) Negative charge = – 1.6 x 10-19 C (or -e) No charge Mass Four times the mass of proton i.e., 6.68 x 10-27 kg Equal to the mass of electron, i.e. 9.1 x 10-31 kg No mass (Rest mass is zero) Effect of electric field Less deflected More deflected than alpha particles but in direction opposite to those of α particles Unaffected

Solution 18.

Ionizing power of alpha radiation is maximum i.e., 10000 times of gamma radiation while beta particles have lesser ionizing power i.e., 100 times of gamma radiation and gamma radiation have least ionizing power.
Penetration power is least for alpha particle and maximum for gamma radiation.

Solution 19.

• Speed of α radiation is nearly 107 m/s.
• Speed of β radiation is about 90% of the speed of light or 2.7 x 108 m/s.
• Speed of γ radiation is 3 x 108 m/s in vacuum.

Solution 20.

• Alpha radiations are composed two protons and two neutrons.
• Beta particles are fast moving electrons.
• Gamma radiations are photons or electromagnetic waves like X rays.
• Alpha radiations have the least penetrating power.

Solution 21.

• Gamma radiation are produced when a nucleus is in a state of excitation (i.e., it has an excess of energy). This extra energy is released in the form of gamma radiation.
• Gamma radiations like light are not deflected by the electric and magnetic field.
• Gamma radiations have the same speed as that of light.

Solution 22.

It will become singly ionized helium He+.

Solution 23.

Any physical changes (such as change in pressure and temperature) or chemical changes (such as excessive heating, freezing, action of strong electric and magnetic fields, chemical treatment, oxidation etc.) do not alter the rate of decay of the radioactive substance. This clearly shows that the phenomenon of radioactivity cannot be due to the orbital electrons which could easily be affected by such changes. The radioactivity should therefore be the property of the nucleus. Thus radioactivity is a nuclear phenomenon.

Solution 24.

On emitting a β particle, the number of nucleons in the nucleus (i.e. protons and neutrons) remains same, but the number of neutrons is decreased by one and the number of protons is increased by one.

If a radioactive nucleus P with mass number A and atomic number Z emits a beta particle to form a daughter nucleus Q with mass number A and atomic number Z+1, then the change can be represented as follows:

(a) Atomic number ‘Z’ is not conserved. It is increased by 1.
(b) Mass number A is conserved.

Solution 25.

Solution 26.

(a) Atomic number decreases by 2.
(b) Atomic number increases by 1.
(c) Atomic number does not change.

Solution 27.

(a) After emitting an alpha particle the daughter element occupies two places to the left of the parent element in the periodic table.
Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of α-decay, then the α-decay can be represented as:

Thus, the resulting nucleus has an atomic number equal to (Z-2). Hence, it shifts two places to the left of the parent element in the periodic table.

(b) After emitting a -particle, the daughter element occupies one place to the right of the parent element in the periodic table.
Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of β-decay, then the β-decay can be represented as:

Thus, the resulting nucleus has an atomic number equal to (Z+1). Hence, it shifts one place to the right of the parent element in the periodic table.

(c) After emitting -radiation, the element occupies the same position in the periodic table.
Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of γ-decay, then the γ-decay can be represented as:

Thus, the resulting nucleus has atomic number equal to Z. Hence, it occupies the same position as the parent element in the periodic table.

Solution 28.

Solution 29.

(a) The composition of B – 82 protons and 126 neutrons.
(b) The composition of C – 83 protons and 125 neutrons.
(c) The mass number of nucleus A = no. of protons +no. of neurons = 84+128=212.
(d) Their will be no change in the composition of nucleus C.

Solution 30.

(a) The alpha particle was emitted.
(b) This is because the atomic number has decreased by 2 and mass number has decreased by 4.

Solution 31.

(a) This is allowed.
(b) This is not allowed because mass number is not conserved.

Solution 32.

Solution 33.

Solution 34.

The atomic number of P decreases by 2 and mass no. decreases by 4 due to the emission of one alpha particle and then increases by 1 due to the emission of each beta particle, so the atomic number of Q formed after the emission of one alpha and two beta particles is same as that of P. Hence P and Q are the isotopes.

Solution 36.

(a) The mass number (A) of an element is not changed when it emits beta and gamma radiations.
(b) The atomic number of a radioactive element is not changed when it emits gamma radiations.
(c) During the emission of a beta particle, the mass number remains same.

Solution 37.

Solution 38.

Radio isotopes: The isotopes of some elements with atomic number Z
Example: carbon (Z=6, A=14).
Radio isotopes are used in medical and scientific and industrial fields. Radio isotopes such as 92U232 are used as fuel for atomic energy reactors.

Solution 39.

Because they cannot penetrate the human skin.

Solution 40.

Gamma radiations have very high penetration power and can easily pass through the human body. Therefore they are used as radioactive tracers in medical science.

Solution 41.

When the number of neutrons exceeds much than the number of protons in a nuclei, it become unstable or radioactive.

Solution 42.

Solution 43.

Many diseases such as leukemia, cancer, etc., are cured by radiation therapy. Radiations from cobalt -60 are used to treat cancer by killing the cells in the malignant tumor of the patient.

Solution 44.

a  <  β  <  γ

An α-particle rapidly loses its energy as it moves through a medium and therefore its penetrating power is quite small. It can penetrate only through 3 – 8 cm in air. It can easily be stopped by a thin card sheet or a thick paper.

The penetrating power of β-particles is more than that of the α-particles. They can pass through nearly 5 m in air, through thin card sheet, and even through thin aluminium foil, but a 5 mm thick aluminium sheet can stop them.

Whereas, the penetrating power of γ-rays is high. It is about 104 times that of α-particles and 102 times that of β-particles. They can pass through 500 m in air or through 30 cm thick sheet of iron. Thick sheet of lead is required to stop them.

Solution 45.

Two main sources of nuclear radiations are:

1. Radioactive fallout from nuclear plants and other sources.
2. Disposal of nuclear waste.

These radiations are harmful because when these radiations falls on the human body, they kill the human living tissues and cause radiation burns.

Solution 46.

The following safety measures must be taken in a nuclear power plant:

1. The nuclear reactor must be shielded with lead and steel walls so as to stop radiations from escaping out to the environment during its normal operation.
2. The nuclear reactor must be housed in an airtightbuilding of strong concrete structure which can withstand earthquakes, fires and explosion.
3. There must be back up cooling system for the reactor core, so that in case of failure of one system, the other cooling system could take its place and the core is saved from overheating and melting.

Solution 47.

The radioactive material after its use is known as nuclear waste.
It must be buried in the specially constructed deep underground stores made quite far from the populated area.

Solution 48.

Three safety precautions that we would take while handling the radioactive substances are:

3. Keep the radioactive substances in thick lead containers with a very narrow opening, so as to stop radiations coming out from other directions.

Solution 49.

Radioactive substance should not be touched by hands because these radiations are harmful; when radiation falls on the human body, they kill the human living tissues and cause radiation burns.

Solution 50.

Background radiation: These are the radioactive radiations to which we all are exposed even in the absence of an actual visible radioactive source.
There are two sources of background radiation:

1. Internal source: potassium, carbon and radium are present inside our body.

It is not possible for us to keep ourselves away from the background radiations.

Solution 1 (MCQ).

α or β
Hint: In a single radioactive decay, α and β particles are never emitted simultaneously. There will be either an α-emission or a α β-emission, which may be accompanied by γ emission.

Solution 2 (MCQ).

The nucleus of the atom.
Hint: Radioactivity is a nuclear phenomenon. Hence, electrons come out from the nucleus. Electron is created as a result of decay of one neutron into a proton inside the nucleus and it is not possible for the electron to stay inside the nucleus; thus, it is spontaneously emitted.

Solution 3 (MCQ).

(a) α – particles

An α – particle rapidly loses its energy as it moves through a medium and therefore its penetrating power is quite small. It can penetrate only through 3 – 8 cm in air. It can easily be stopped by a thin card sheet or a thick paper.

Solution 4 (MCQ).

(b) β-particles

β-particles are negatively charged, so they get deflected by the electric and magnetic fields. The deflection of β-particle is more than that of a-particle since a β-particle is lighter than the α-particle. Whereas, gamma radiations are not deflected by the electric and magnetic fields since they are not charged particles.

Exercise 12(B)

Solution 1.
Energy released by combining of nuclei of an atom or by decay of an unstable radioactive nucleus during a nuclear reaction i.e., during fusion or fission is known as nuclear energy.

Solution 1 (MCQ).
(d) neutron
A neutron is used in nuclear fission for bombardment.

Solution 1 (Num).
1 a.m.u. = 1.66 × 10-27 kg
→ 0.2 a.m.u. = 0.2 × 1.66 × 10-27 kg
Δm = 0.332 Δ 10-27 kg

Solution 2.
Einstein’s mass-energy equivalence relation : E = Δmc2
Where E is the energy released due to the loss in the mass Δm and c is the speed of light.

Solution 2 (MCQ).
(d) 10K
To make the fusion possible, a high temperature of approximately 107 K and high pressure is required.

Solution 2 (Num).
Given that Δm = 0.0265 a.m.u.
1 a.m.u. liberates 931.5 MeV of energy. Thus, energy liberated equivalent to 0.0265 a.m.u. is
= 0.0265 a.m.u. × 931.5 MeV
= 24.7 meV

Solution 3.
(a) The mass of atomic particles is expressed in terms of atomic mass unit (a.m.u.). 1 a.m.u. of mass is equivalent to 931 MeV of energy.
(b) Mass of proton = 1.00727 a.m.u.
Mass of neutron = 1.00865 a.m.u.
Mass of electron = 0.00055 a.m.u.

Solution 4.
Nuclear fission is the process in which a heavy nucleus is splits into two light nuclei nearly of the same size by bombarding it with slow neutrons.

Solution 5.
(a) $$_{ 92 }^{ 235 }{ U }$$ and $$_{ 92 }^{ 235 }{ U }$$
(b) Experimentally it is found that isotope of $$_{ 92 }^{ 235 }{ U }$$ is more easily fissionable because the fission of
is $$_{ 92 }^{ 235 }{ U }$$ possible by sloe neutron unlike $$_{ 92 }^{ 238 }{ U }$$ where fission is possible only by the fast neutrons.
(c) Slow and fast both.

Solution 6.
Nearly 190 MeV of energy is released due to fission of one nucleus of $$_{ 92 }^{ 235 }{ U }$$ The cause of emission of this energy is the loss in mass i.e., the sum of masses of product nuclei is less than the sum of mass of the parent nucleus and neutron.

Solution 7.

Solution 8.
A chain reaction is a series of nuclear fissions whereby the neutrons produced in each fission cause additional fissions, releasing enormous amount of energy.
It is controlled by absorbing some of the neutrons emitted in the fission process by means of moderators like graphite, heavy water, etc. then the energy obtained in fission can be utilized for the constructive purposes

Solution 9.
(i) It is used in a nuclear bomb.
(ii) It is used in a nuclear reactor where the rate of release of energy is slow and controlled which is used to generate electric power.

Solution 10.

 Radioactive decay Nuclear Fission It is a self process. It does not occur by itself. Neutrons are bombarded on a heavy nucleus. The nucleus emits either the a or b particles with the emission of energy in form of g rays which is not very large. A tremendous amount of energy is released when a heavy nucleus is bombarded with neutrons and the nucleus splits in two nearly equal fragments. The rate of radioactive decay cannot be controlled. The rate of nuclear fission can be controlled.

Solution 11.
Nuclear fission is the process in which a heavy nucleus is splits into two light nuclei nearly of the same size by bombarding it with slow neutrons.
When uranium with Z = 92 is bombarded with neutron, it splits into two fragments namely barium (Z = 56) and krypton (Z = 36) and a large amount of energy is released which appears due to decrease in the mass.

Nuclear fusion is also known as thermo-nuclear reaction. This is because nuclear fusion takes place at very high temperature.

Solution 12.
When two nuclei approach each other, due to their positive charge, the electrostatic force of repulsion between them becomes too strong that they do not fuse. Thus, nuclear fusion is not possible at ordinary temperature and ordinary pressure.
Hence to make the fusion possible, a high temperature of approximately 107 K and high pressure is required. At such a high temperature, due to thermal agitations both nuclei acquire sufficient kinetic energy so as to overcome the force of repulsion between them when they approach each other, and so they get fused.

Solution 13.

(b) In all three deuterium nuclei fuse to form a helium nucleus with a release of 21·6 MeV energy.
(c) When two deuterium nuclei ($$_{ 1 }^{ 2 }{ H }$$) fuse, nucleus of helium isotope ($$_{ 3 }^{ 2 }{ He }$$ ) is formed and 3·3 MeVenergy is released. This helium isotope again gets fused with one deuterium nucleus to form a helium nucleus ($$_{ 4 }^{ 2 }{ He }$$
) and 18·3 MeV of energy is released in this process.

Solution 14.
(a)

Solution 15.
(a) Nuclear fusion
(b) Nuclear fission

Solution 16.
Both fission and fusion create release of neutrons and large amount of energy.
Nuclear fission: A heavy nucleus splits in two nearly equal light fragments when bombarded with neutrons. It is possible at very ordinary temperature and pressureNuclear fusion: Two light nuclei combine to form a heavy nucleus at very high temperature and high pressure. Possible only at a very high temperature (≈107 K) and a very high pressure.

Solution 17.

Solution 18.
The source of energy in the Sun and stars is the nucleus fusion of light nuclei such as hydrogen present in them in their inner part. This takes place at a very high temperature and high pressure due to which helium nucleus is formed with the release of high amount of energy.

Solution 19.
(a) Nuclear fission
(b) Nuclear fusion

More Resources for Selina Concise Class 10 ICSE Solutions

## Selina Concise Physics Class 10 ICSE Solutions Calorimetry

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 11 Calorimetry. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 11 Calorimetry

Exercise 11(A)

Solution 1.

The kinetic energy due to random motion of the molecules of a substance is known as its heat energy.

Solution 2.

S.I. unit of heat is joule (symbol J).

Solution 3.

One calorie of heat is the heat energy required to raise the temperature of 1 g of water from 14.5oC to 15.5 oC.
1 calorie = 4.186 J

Solution 4.

One kilo-calorie of heat is the heat energy required to raise the temperature of 1 kg of water from 14.5oC to 15.5oC.

Solution 5.

The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature.
S.I. unit kelvin (K).

Solution 6.

 Heat Temperature The kinetic energy due to random motion of the molecules of a substance is known as its heat energy. The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature. S.I. unit joule (J). S.I. unit kelvin (K). It is measured by the principle of calorimetry. It is measured by a thermometer.

Solution 7.

The measurement of the quantity of heat is called calorimetry.

Solution 8.

The heat capacity of a body is the amount of heat energy required to raise its temperature by 1oC or 1K.
S.I. unit is joule per kelvin (JK-1).

Solution 9.

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through by 1oC (or 1K).
S.I. unit is joule per kilogram per kelvin (Jkg-1K-1).

Solution 10.

Heat capacity = Mass x specific heat capacity

Solution 11.

• Heat capacity of the body is the amount of heat required to raise the temperature of (whole) body by 1oC whereas specific heat capacity is the amount of heat required to raise the temperature of unit mass of the body by 1oC.
• Heat capacity of a substance depends upon the material and mass of the body. Specific heat capacity of a substance does not depend on the mass of the body.
• S.I. unit of heat capacity is JK-1 and S.I. unit of specific heat capacity is Jkg-1K-1.

Solution 12.

Water has the highest specific heat capacity.

Solution 13.

Specific heat capacity of water=4200 J kg-1 K-1.

Solution 14.

1. The heat capacity of a body is 50JK-1 means to increase the temperature of this body by 1K we have to supply 50 joules of energy.
2. The specific heat capacity of copper is 0.4Jg-1K-1 means to increase the temperature of one gram of copper by 1K we have to supply 0.4 joules of energy.

Solution 16.

The quantity of heat energy absorbed by a body depends on three factors :

1. Mass of the body – The amount of heat energy required is directly proportional to the mass of the substance.
2. Nature of material of the body – The amount of heat energy required depends on the nature on the substance and it is expressed in terms of its specific heat capacity c.
3. Rise in temperature of the body – The amount of heat energy required is directly proportional to the rise in temperature.

Solution 17.

The expression for the heat energy Q
Q = mcΔt (in joule)

Solution 18.

Heat capacity of liquid A is less than that of B.
As the substance with low heat capacity shows greater rise in temperature.

Solution 19.

Solution 20.

The principle of method of mixture:
Heat energy lost by the hot body = Heat energy gained by the cold body.
This principle is based on law of conservation of energy.

Solution 21.

Solution 22.

In the absence of water, if on a cold winter night the atmospheric temperature falls below 0oC, the water in the fine capillaries of plant will freeze, so the veins will burst due to the increase in the volume of water on freezing. As a result, plants will die and the crop will be destroyed. In order to save the crop on such cold nights, farmers fill their fields with water because water has high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0oC.

Solution 23.

The specific heat capacity of water is very high. It is about five times as high as that of sand. Hence the heat energy required for the same rise in temperature by a certain mass of water will be nearly five times than that required by the same mass of sand. Similarly, a certain mass of water will give out nearly five times more heat energy than that given by sand of the same mass for the same fall in temperature. As such, sand gets heated or cooled more rapidly as compared to water under the similar conditions. Thus a large difference in temperature is developed between the land and the sea due to which land and sea breezes are formed. These breezes make the climate near the sea shore moderate.

Solution 24.

The reason is that water does not cool quickly due to its large specific heat capacity, so a hot water bottle provides heat energy for fomentation for a long time.

Solution 25.

By allowing water to flow in pipes around the heated parts of a machine, heat energy from such part is removed. Water in pipes extracts more heat from surrounding without much rise in its temperature because of its large specific heat capacity. So, Water is used as an effective coolant.

Solution 26.

2. To avoid freezing of wine and juice bottles.

Solution 28.

A calorimeter is a cylindrical vessel which is used to measure the amount of heat gained or lost by a body when it is mixed with other body.
It is made up of thin copper sheet because:

1. Copper is a good conductor of heat, so the vessel soon acquires the temperature of its contents.
2. Copper has low specific heat capacity so the heat capacity of calorimeter is low and the amount of heat energy taken by the calorimeter from its contents to acquire the temperature of its contents is negligible.

Solution 29.

By making the base of a cooking pan thick, its thermal capacity becomes large and it imparts sufficient heat energy at a low temperature to the food for its proper cooking. Further it keeps the food warm for a long time, after cooking.

Solution 1 (MCQ).

JK-1

Solution 2 (MCQ).

J kg-1K-1

Solution 3 (MCQ).

4200 J kg-1K-1

Numericals

Solution 1.

The size of 1 degree on the Kelvin scale is the same as the size of 1 degree on the Celsius scale. Thus, the difference (or change) in temperature is the same on both the Celsius and Kelvin scales.
Therefore, the corresponding rise in temperature on the Kelvin scale will be 15K.

Solution 2.

Solution 3.

1. We know that heat energy needed to raise the temperature by 15o is = heat capacity x change in temperature.
Heat energy required= 966 J K-1 x 15 K = 14490 J.
2. We know that specific heat capacity is = heat capacity/ mass of substance
So specific heat capacity is = 966 / 2=483 J kg-1 K-1.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

Solution 12.

Solution 13.

Exercise 11(B)

Solution 1.

(a) The process of change from one state to another at a constant temperature is called the change of phase of substance.
(b) There is no change in temperature during the change of phase.
(c) Yes, the substance absorbs or liberates heat during the change of phase.

Solution 2.

Melting: The change from solid to liquid phase on heating at a constant temperature is called melting.
Melting point: The constant temperature at which a solid changes to liquid is called the melting point.

Solution 3.

1. Average kinetic energy of molecules changes.
2. Average potential energy of molecules changes.

Solution 4.

(a) Average kinetic energy does not change.
(b) Average potential energy increases.
Explanation: When a substance is heated at constant temperature (i.e. during its phase change state), the heat supplied makes the vibrating molecules gain potential energy to overcome the intermolecular force of attraction and move about freely. This means that the substance changes its form.

However, this heat does not increase the kinetic energy of the molecules, and hence, no rise in temperature takes place during the change in phase of a substance.
This heat supplied to the substance is known as latent heat and is utilized in changing the state of matter without any rise in temperature.

Solution 5.

The melting point of ice decreases by the presence of impurity in it.
Use: In making the freezing mixture by adding salt to ice. This freezing mixture is used in preparation of ice creams.

Solution 6.

The melting point of ice decreases by the increase in pressure. The melting point of ice decrease by 0.0072oC for every one atmosphere rise in pressure.

Solution 7.

(a) AB part shows rise in temperature of solid from 0 to T1oC, BC part shows melting at temperature T1oC, CD part shows rise in temperature of liquid from T1oC to T3oC , DE part shows the boiling at temperature T3oC.
(b) T1oC.
(c) T3oC.

Solution 8.

Solution 9.

Boiling: The change from liquid to gaseous phase on heating at a constant temperature is called boiling.
Boiling point: The particular temperature at which vaporization occurs is called the boiling point of liquid.

Solution 10.

Volume of water wills increases when it boils at 100oC.

Solution 11.

The boiling point of water increases on adding salt.

Solution 12.

The boiling point of a liquid increases on increasing the pressure.

Solution 14.

In a pressure cooker, the water boils at about 120oC to 125oC.

Solution 15.

This is because at high altitudes atmospheric pressure is low; therefore boiling point of water decreases and so it does not provide the required heat energy for cooking.

Solution 16.

(a) When ice melts, its volume decreases.
(b) Decrease in pressure over ice increases its meltingpoint.
(c) Increase in pressure increases the boiling point of water.
(d)A pressure cooker is based on the principle that boiling point of water increases with the increase in pressure.
(e) The boiling point of water is defined as the constant temperature at which water changes to steam.
(f) water can be made to boil at 115°C by increasing pressure over its surface.

Solution 17.

Latent heat: The heat energy exchanged in change of phase is not externally manifested by any rise or fall in temperature, it is considered to be hidden in the substance and is called the latent heat.

Solution 18.

The quantity of heat required to convert unit mass of ice into liquid water at 00C (melting point) is called the specific latent heat of fusion of ice.
Its S.I. unit is Jkg-1.

Solution 19.

Specific latent heat of ice: 336000 J kg-1.

Solution 20.

It means 1 g of ice at 0oC absorbs 360 J of heat energy to convert into water at 0oC.

Solution 21.

Latent heat is absorbed by ice.

Solution 22.

1 g of water at 0oC has more heat than 1 g of ice at 0oC. This is because ice at 0oC absorbs 360 J of heat energy to convert into water at 0oC.

Solution 23.

(a) 1 g ice at 0oC requires more heat because ice would require additional heat energy equal to latent heat of melting.
(b) 1 g ice at 0oC first absorbs 336 J heat to convert into 1 g water at 0oC.

Solution 24.

This is because 1 g of ice at 0oC takes 336 J of heat energy from the mouth to melt at 0oC. Thus mouth loses an additional 336 J of heat energy for 1 g of ice at 0oC than for 1g of water at 0oC. Therefore cooling produced by 1 g of ice at 0oC is more than for 1g of water at 0oC.

Solution 25.

This is because 1 g of ice at 0oC takes 336 J of heat energy from the bottle to melt into water at 0oC. Thus bottle loses an additional 336 J of heat energy for 1 g of ice at 0oC than for 1 g iced water at 0oC. Therefore bottled soft drinks get cooled, more quickly by the ice cubes than by iced water.

Solution 26.

The reason is that after the hail storm, the ice absorbs the heat energy required for melting from the surrounding, so the temperature of the surrounding falls further down and we feel colder.

Solution 27.

The reason is that the heat energy required for melting the frozen lake is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it became very cold.

Solution 29.

1. The reason is that the specific latent heat of fusion of ice is sufficiently high, so when the water of lake freezes, a large quantity of heat has to be released and hence the surrounding temperature becomes pleasantly warm.
2. Heat supplied to a substance during its change of state, does not cause any rise in its temperature because this is latent heat of phase change which is required to change the phase only.

Solution 1 (MCQ).

J kg-1

Solution 2 (MCQ).

80cal g-1

Numericals

Solution 1.

Mass of ice=10g = 0.01kg
Amount of heat energy absorbed, Q=5460J
Specific latent heat of fusion of ice=?
Specific heat capacity of water = 4200Jkg-1K-1
Amount of heat energy required by 10g (0.01kg) of water at 0oC to raise its temperature by 50oC= 0.01X4200X50=2100J.
Let Specific latent heat of fusion of ice=L Jg-1.
Then,
Q = mL + mcΔT
5460 J = 10 x L + 2100J
L = 336Jg-1.

Solution 2.

Mass of water m = 5.0 g
specific heat capacity of water c = 4.2 J g-1 K-1
specific latent heat of fusion of iceL =336 J g-1
Amount of heat energy released when 5.0 g of water at 20oC changes into water at 0oC = 5 x 4.2 x 20 = 420 J.
Amount of heat energy released when 5.0g of water at 0oC changes into ice at 0oC=5x336J=1680J.
Total amount of heat released =1680 J + 420 J = 2100 J.

Solution 3.

Solution 4.

Solution 5.

Mass of ice m1 =17 g
Mass of water m2 =40 g.
Change in temperature =34-0=34K
Specific heat capacity of water is 4.2Jg-1K-1.
Assuming there is no loss of heat, heat energy gained by ice (latent heat of ice), Q= heat energy released by water
Q = 40 x 34 x 4.2 = 5712 J.

Solution 6.

Let whole of the ice melts and let the final temperature of the mixture be ToC.
Amount of heat energy gained by 10g of ice at -10oC to raise its temperature to 0oC= 10x10x2.1=210J
Amount of heat energy gained by 10g of ice at 0oC to convert into water at 0oC=10×336=3360 J
Amount of heat energy gained by 10g of water (obtained from ice) at 0oC to raise its temperature to ToC = 10×4.2x(T-0)=42T
Amount of heat energy released by 10g of water at 10oC to lower its temperature to ToC = 10×4.2x(10-T)=420-42T
Heat energy gained = Heat energy lost
210 + 3360 + 42T = 420-42T
T = -37.5oC
This cannot be true because water cannot exist at this temperature.
So whole of the ice does not melt. Let m gm of ice melts. The final temperature of the mixture becomes 0oC.
So, amount of heat energy gained by 10g of ice at -10oC to raise its temperature to 0oC= 10x10x2.1=210J
Amount of heat energy gained by m gm of ice at 0oC to convert into water at 0oC=mx336=336m J
Amount of heat energy released by 10g of water at 10oC to lower its temperature to 0oC = 10×4.2x(10-0)=420
Heat energy gained = Heat energy lost
210 + 336m = 420
m = 0.625 gm

Solution 7.

Solution 8.

Solution 9.

Since the whole block does not melt and only 2 kg of it melts, so the final temperature would be 0 oC.
Amount of heat energy gained by 2 kg of ice at 0oC to convert into water at 0oC=2 x 336000 = 672000 J
Let amount of water poured = m kg.
Initial temperature of water = 100oC.
Final temperature of water = 0oC.
Amount of heat energy lost by m kg of water at 100oC to reach temperature 0oC =m x 4200 x 100 = 420000m J
We know that heat energy gained = heat energy lost.
672000J = mX420000J
m = 672000/420000=1.6kg

Solution 10.

Amount of heat energy gained by 100 g of ice at -10C to raise its temperature to 0C = 100 × 2.1 × 10 = 2100 J
Amount of heat energy gained by 100 g of ice at 0C to convert into water at 0C = 100 × 336 = 33600 J
Amount of heat energy gained when temperature of 100 g of water at 0C rises to 100C = 100 × 4.2 × 100 = 42000 J
Total amount of heat energy gained is = 2100 + 33600 + 42000 = 77700 J = 7.77 × 104 J

Solution 11.

Amount of heat energy gained by 1kg of ice at -10oC to raise its temperature to 0oC= 1 x 2100 x 10 = 21000 J
Amount of heat energy gained by 1kg of ice at 0oC to convert into water at 0oC = L
Amount of heat energy gained when temperature of 1kg of water at 0oC rises to 100oC= 1 x 4200 x 100 = 420000 J
Total amount of heat energy gained = 21000 + 420000 + L = 441000 + L.
Given that total amount of heat gained is = 777000J.
So,
441000 + L = 777000.
L = 777000 – 441000.
L = 336000JKg-1

Solution 12.

Mass of ice, mice = 200 g

Time for ice to melt, t1 = 1 min = 60 s

Mass of water, mw = 200 g

Temperature change of water, ΔT = 20 °C

Rate of heat exchange is constant. So, power required for converting ice to water is same as the power required to increase the temperature of water.

Exercise 11(C)

Solution 1.

The earth receives heat radiations from sun which reach us after passing through its atmosphere. The earth’s atmosphere is transparent for the visible and thermal radiations of short wavelengths coming from the sun. The earth’s surface and objects on it thus become warm in the day time. After the sunset, the earth’s surface and the objects on it radiate the infrared radiations of long wavelengths. A part of these radiations are reflected back by the clouds and a part of it is absorbed by the green house gases like carbon dioxide, methane, water vapours and chlorofluorocarbons. Thus the clouds and green house gases prevents a large fraction of radiations given out by the earth’s surface, from escaping into the space. This phenomenon is called greenhouse effect.

Solution 2.

Carbon dioxide, methane, water vapours and chlorofluorocarbons.

Solution 3.

Solution 4.

Coal, petroleum, natural gas.

Solution 5.

From sun, we receive 1366W m-2 energy at the top of our earth’s atmosphere, out of which only 235Wm-2 energy reaches near the earth’s surface. The earth and ocean surface absorbs 168 W m-2 energy and only 67 W m-2 energy remains in the lower atmosphere. With this much energy received on earth surface, its actual surface temperature would have been around -180C which is quite uncomfortable for human living. Fortunately the greenhouse gases present in the earth’s atmosphere contribute in trapping the heat energy within the atmosphere and they produce an average warming effect of about 330C to keep the effective temperature around 150C. So, greenhouse effect helps in keeping the temperature of earth’s surface suitable for living of human beings.

Solution 6.

Three reasons for increase of greenhouse gases:

1. The burning of fuels, deforestation and industrial production
2. Increase of population.
3. Imbalance of carbon dioxide cycle

Solution 7.

The effect of enhancement of greenhouse effect are:

1. The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place.
2. It has affected the blooming season of the different plants.
3. The climate changes have shown the immediate effect on simple organism and plants.
4. It has affected the world’s ecology.
5. It has increased the heat stroke deaths.

Solution 8.

Global warming means the increase in the average effective temperature of earth’s surface due to an increase in the amount of greenhouse gases in its atmosphere.

Solution 9.

The effect of enhancement of greenhouse effect are:

1. The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place.
2. It has affected the blooming season of the different plants.
3. The climate changes have shown the immediate effect on simple organism and plants.
4. It has affected the world’s ecology.
5. It has increased the heat stroke deaths.

Solution 10.

Due to rise in sea level the building and roads in the coastal areas will get flooded and they could suffer damage from hurricanes and tropical storms.

Solution 11.

Due to global warming, many new diseases have emerged because bacteria can survive better in increased temperature and they can multiply faster. It is extending the distribution of mosquitoes due to increase in humanity levels and their frequent growth in warmer atmosphere. This has resulted in increase of many new diseases. The deaths due to heat stroke have certainly increased.

Solution 12.

Due to global warming, many new diseases have emerged because bacteria can survive better in increased temperature and they can multiply faster. It is extending the distribution of mosquitoes due to increase in humanity levels and their frequent growth in warmer atmosphere. This has resulted in increase of many new diseases. The deaths due to heat stroke have certainly increased.

Solution 13.

At the present rate of increase of green house effect, it is expected that nearly 30% of the plant species will extinct by the year 2050 and up to 70% by the end of the year 2100. In the near future, warming of nearly 3oC will result in poor yield in farms in low latitude regions. This will increase the rise of malnutrition.

Solution 14.

At the present rate of increase in green house effect, is estimated that nearly 30% of the plant and animal species will extinct by the year 2050 and upto 70 % by the end of year 2100. This will disrupt ecosystem. The animals from the equatorial region will shift to higher latitude in search of cold regions. The absorption of carbon dioxide by ocean will cause acidification due to which marine species will migrate.

Solution 16.

The tax calculated on the basis of: carbon emission from industry, number of employee hour and turnover of the factory is called carbon tax.
This tax shall be paid by industries. This will encourage the industries to use the energy efficient techniques.

Solution 1 (MCQ).

-18oC

Solution 2 (MCQ).

The increase in sea levels.
Explanation: Due to global warming, the average temperature of the Earth has increased and has lead to the melting of ice around both the poles. This melting of ice has lead to an increase in the level of water in sea.

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## Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism

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Selina ICSE Solutions for Class 10 Physics Chapter 10 Electro-magnetism

Exercise 10(A)

Solution 1.

Experiment:
In Fig , AB is a wire lying in the north- south direction and connected to a battery through a rheostat and a tapping key. A compass needle is placed just below the wire. It is observed that

1. When the key is open i.e., no current passes through the wire, the needle shows no deflection and it points in the N-S direction (i.e. along the earth’s magnetic field). In this position, the needle is parallel to the wire as shown in Fig. (a).
2. When the key is pressed, a current passes in the wire in the direction from A to B (i.e. From south to north) and the north pole(N) of the needle deflects towards the west [Fig. (b)].
3. When the direction of current in the wire is reversed by reversing the connections at the terminals of the battery, North Pole (N) of the needle deflects towards the east [Fig. (c)].
4. If the compass needle is placed just above the wire, the North Pole (N) deflects towards east when the direction of current in wire is from A to B [Fig. (d)], but the needle deflects towards west as in fig (e), if the direction of current in wire is from B to A.

The above observations of the experiment suggest that a current carrying wire produces a magnetic field around it.

Solution 2.

Solution 3.

(a) On decreasing the current the magnetic field lines become rarer.
(b) The direction of magnetic field lines will get reversed.

Solution 4.

Right hand thumb rule determines the direction of magnetic field around a current carrying wire.
It states that if we hold the current carrying conductor in right hand such that the thumb points in the direction of flow of current, then the fingers encircle the wire in the direction of the magnetic fields lines.

Solution 5.

(a) The direction of magnetic field at a point just underneath is towards east.
(b) Right hand thumb rule.

Solution 6.

A current carrying conductor produces a magnetic field around it and the magnetic needle in this magnetic field experience a torque due to which it deflects to align itself in the direction of magnetic field.

Solution 7.

Solution 8.

Face of the coil exhibit North polarity.

Solution 9.

(i) Along the axis of coil inwards.
(ii) Along the axis of coil outwards.

Solution 10.

Solution 11.

Right hand thumb rule: If we hold the current carrying conductor in right hand such that the thumb points in the direction of flow of current, then the fingers encircle the wire in the direction of the magnetic fields lines.

Solution 12.

(a) A – North pole, B – South pole.
(b) The magnet will be repelled because the end of the solenoid near the north pole of magnet becomes the north pole as current at this face is anticlockwise and the two like poles repel.

Solution 13.

(a) A – North pole, B – South pole.
(b) The north pole of compass needle will deflect towards west.
Reason: The end A of the coil behaves like north pole which repels north pole of compass needle towards west.

Solution 14.

The magnetic field due to a solenoid can be made stronger by using:

1. By increasing the number of turns of winding in the solenoid.
2. By increasing the current through the solenoid.

Solution 15.

A current carrying freely suspended solenoid at rest behaves like a bar magnet. It is because a current carrying solenoid behaves like a bar magnet.

Solution 16.

The needle of the compass will rest in in the direction of magnetic field due to solenoid at that point.

Solution 17.

Magnetic fielddue to a solenoid carrying current increases if a soft iron bar is introduced inside the solenoid.

Solution 18.

(A) When current flows in a wire, it creates magnetic field around it.
(B) On reserving the direction of current in a wire, the magnetic field produced by it gets reversed.
(C) A current carrying solenoid behaves like a bar magnet
(D) A current carrying solenoid when freely suspended, it always rest in north-south direction.

Solution 19.

Solution 20.

(a) X-north pole, Y –south pole.
(b) By reducing resistance of circuit by mean of rheostat to increase current.

Solution 21.

(a) Solenoid is a cylindrical coil of diameter less than its length.
(b) The device so obtained is electromagnet.
(c) It is used in electric relay.

Solution 22.

Solution 23.

An electromagnet is a temporary strong magnet made from a piece of soft iron when current flows in the coil wound around it. It is an artificial magnet.

The strength of magnetic field of an electromagnet depends on:

1. Number of turns: The strength of magnetic increases on increasing the number of turns of winding in the solenoid.
2. Current: The strength of magnetic field increases on increasing the current through the solenoid.

Solution 24.

At A-south pole and at B-north pole.

Solution 25.

The strength of an electromagnet can be increased by following ways:

1. By increasing the number of turns of winding in the solenoid.
2. By increasing the current through the solenoid.

Solution 27.

1. An electromagnet can produce a strong magnetic field.
2. The strength of the magnetic field of an electromagnet can easily be changed by changing the current in its solenoid.

Solution 28.

 Electromagnet Permanent magnet It is made up of soft iron It is made up of steel. The magnetic field strength can be changed. The magnetic field strength cannot be changed. The electromagnets of very strong field can be made. The permanent magnets are not so strong.

Solution 29.

The soft iron bar acquires the magnetic properties only when an electric current flows through the solenoid and loses the magnetic properties as the current is switched off. That’s why soft iron is used as the core of the electromagnet in an electric bell.

Solution 30.

If an a.c. source is used in place of battery, the core of electromagnet will get magnetized, but the polarity at its ends will change. Since attraction of armature does not depend on the polarity of electromagnet, so the bell will still ring on pressing the switch.

Solution 31.

Solution 32.

The material used for making the armature of an electric bell is soft iron which can induce magnetism rapidly.

Solution 1 (MCQ).

The presence of magnetic field at a point can be detected by a compass needle.
Note: In the presence of a magnetic field, the needle of compass rests only in the direction of magnetic field and in the absence of any magnetic field, the needle of compass can rest in any direction. In the earth’s magnetic field alone, the needle rests along north-south direction.

Solution 2 (MCQ).

By reversing the direction of current in a wire, the magnetic field produced by it gets reversed in direction.
Hint: On reversing the direction of current in a wire, the polarity of the faces of the wire also reverses. Thus, the direction of magnetic field produced by it also gets reversed.

Exercise 10(B)

Solution 1.

The magnitude of force on a current carrying conductor placed in a magnetic field depends on:

1. On strength of magnetic field B.
2. On current I in the conductor.
3. On length of conductor.

Magnitude of force on a current carrying conductor placed in a magnetic field depends directly on these three factors.

Solution 2.

(a) When current in the conductor is in the direction of magnetic field force will be zero.
(b) When current in the conductor is normal to the magnetic field.

Solution 3.

Direction of force is also reversed.

Solution 4.

Fleming’s left hand rule: Stretch the forefinger, middle finger and the thumb of your left hand mutually perpendicular to each other. If the forefinger indicates the direction of magnetic field and the middle finger indicates the direction of current, then the thumb will indicate the direction of motion of conductor.

Solution 6.

Unit is: Newton/ampere x meter (or NA-1m-1).

Solution 7.

(a) The coil will experience a torque due to which it will rotate.
(b) The coil will come to rest when their plane become normal to the magnetic field.
(c) (i) When plane of a oil is parallel to the magnetic field,
(ii) When plane of coil is normal to the magnetic field.
(d) The instrument which makes use of the principle stated above is d.c. motor.

Solution 8.

(a) The coil begins to rotate in anticlockwise direction.
(b) This is because, after half rotation, the arms AB and CD get interchanged, so the direction of torque on coil reverses. To keep the coil rotating in same direction, commutator is needed to change the direction of current in the coil after each half rotation of coil.

Solution 9.

Electric motor: An electric motor is a device which converts the electrical energy into the mechanical energy.
Principle: An electric motor (dc motor) works on the principle that when an electric current is passed through a conductor placed normally in a magnetic field, a force acts on the conductor as a result of which the conductor begins to move and mechanical energy is obtained.

Solution 10.

Solution 11.

Electrical energy converts into mechanical energy.

Solution 12.

The speed of rotation of an electric motor can be increased by:

1. Increasing the strength of current.
2. Increasing the number of turns in the coil.

Solution 13.

Electric motor is used in electrical gadgets like fan, washing machine, juicer, mixer, grinder etc.

Solution 1 (MCQ).

In an electric motor, the energy transformation is from electrical to mechanical.
Note: An electric motor is a device which converts electrical energy into mechanical energy.

Exercise 10(C)

Solution 1.

(a) Electromagnetic induction: whenever there is change in number of magnetic field lines associated with conductor, an electromotive force is developed between the ends of the conductor which lasts as long as the change is taking place.

1. When the magnet is stationary there is no deflection in galvanometer. The pointer read zero. [Fig. (a)]
2. When the magnetwith north pole facing the solenoid is moved towards the solenoid, the galvanometer shows a deflection towards the right showing that a current flows in the solenoid in the direction as shown in [Fig (b)]
3. As the motion of magnet stops, the pointer of the galvanometer comes to the zero position [Fig (c)]. This shows that the current in the solenoid flows as long as the magnet is moving.
4. If the magnet is moved away from the solenoid, the current again flows in the solenoid, but now in a direction opposite to that shown in [Fig. (b)] and therefore the pointer of the galvanometer deflects towards left[ Fig. (d)].
5. If the magnet is moved away rapidly i.e. with more velocity, the extent of deflection in the galvanometer increases although the direction of deflection remains the same.It shows that more current flows now.
6. If the polarity of the magnet is reversed and then the magnet is brought towards the solenoid, the current in solenoid flows in the direction opposite to that shown in Fig (b) and so the pointer of galvanometer deflect towards left [Fig. (e)].

Solution 2.

Faraday’s formulated two laws of electromagnetic induction:

1. Whenever there is a change in the magnetic flux linked with a coil, an e.m.f. is induced. The induced e.m.f. lasts so long as there is a change in the magnetic flux linked with the coil.
2. The magnitude of the e.m.f. induced is directly proportional to the rate of change of the magnetic flux linked with the coil. If the rate of change of magnetic flux remains uniform, a steady e.m.f. is induced.

Solution 3.

Magnitude of induced e.m.f depend upon:

1. The change in the magnetic flux.
2. The time in which the magnetic flux changes.

Solution 4.

(a) Mechanical energy changes to the electrical energy.
(b) Phenomenon is called electromagnetic induction.

Solution 5.

(a) When there is a relative motion between the coil and the magnet, the magnetic flux linked with the coil changes. If the north pole of the magnet is moved towards the coil, the magnetic flux through the coil increases as shown in above figure. Due to change in the magnetic flux linked with the coil, an e.m.f. is induced in the coil. This e.m.f. causes a current to flow in the coil if the circuit of the coil is closed.
(b) The source of energy associated with the current obtained in part (a) is mechanical energy.

Solution 6.

1. When the magnet is stationary there is no deflection in galvanometer. The pointer read zero. [Fig (a)]
2. When the magnetwith north pole facing the solenoid is moved towards the solenoid, the galvanometer shows a deflection towards the right showing that a current flows in the solenoid in the direction as shown in Fig (b)
3. As the motion of magnet stops, the pointer of the galvanometer to the zero position [Fig (c)]. This shows that the current in the solenoid flows as long as the magnet is moving.
4. If the magnet is moved away from the solenoid , the current again flows in the solenoid , but now in the direction opposite to that shown in [Fig. (b)] and therefore the pointer of the galvanometer deflects towards left [Fig. (d)].
5. If the magnet is moved away rapidly i.e. with more velocity, the extent of deflection in the galvanometer increases although the direction of deflection remains the same.It shows that more current flows now.
6. If the polarity of the magnet is reversed and then the magnet is brought towards the solenoid, the current in solenoid flows in the direction opposite to that shown in Fig (b) and so the pointer of galvanometer deflect towards left [Fig. (e)].

(b) Magnitude of induced e.m.f depend upon:

1. The change in the magnetic flux.
2. The time in which the magnetic flux changes.

(c) The direction of induced e.m.f depends on whether there is an increase or decrease in the magnetic flux.

Solution 7.

The current induced in a closed circuit only if there is change in number of magnetic field lines linked with the circuit.

Solution 8.

1. Yes.
2. Yes.
3. Yes.
4. No.

Solution 9.

Fleming’s right hand rule determines the direction of current induced in the conductor.

Solution 10.

Fleming’s right hand rule: Stretch the forefinger, middle finger and the thumb of your right hand mutually perpendicular to each other. If the forefinger indicates the direction of magnetic field and the thumb will indicates the direction of motion of conductor, then the middle finger indicates the direction of induced current.

Solution 11.

Lenz’s law: It states that the direction of induced e.m.f. (or induced current) is such that it always tends to oppose the cause which produces it.

Solution 12.

When a coil has a large number of turns, then magnitude of induced e.m.f. in the coil become more and then by Lenz’s law it will oppose more.

Solution 13.

So that the mechanical energy spent in producing the change, is transformed into the electrical energy in form of induced current.

Solution 14.

Lenz’s law implies the law of conservation of energy. It shows that the mechanical energy spent in doing work, against the opposing force experienced by the moving magnet, is transformed into the electrical energy due to which current flows in the solenoid.

Solution 15.

The pointer of galvanometer deflects. The deflection last so long as the coil moves.

(a) Deflection becomes twice.
(b) Deflection becomes thrice

Solution 16.

1. The pointer of galvanometer deflects towards left. The deflection lasts so long as the coil moves.
2. (a) Deflection becomes twice (b) Deflection becomes thrice.

Solution 17.

(a)

1. When switch is closed suddenly, the galvanometer needle deflects for a moment.
2. If switch is kept closed then galvanometer needle returns to zero.
3. If switch is opened again then galvanometer needle deflects again but in opposite to the direction of deflection in case (a).

(b) This can be explained by Faraday’s law which states that whenever there is change in the magnetic flux linked with a coil, an e.m.f. is induced. The induced e.m.f. lasts as long as there is a change in the magnetic flux linked with the coil.

Solution 18.

An A.C. generator works on the principle of ‘electromagnetic induction’.
Statement: Whenever a coil is rotated in a magnetic field, the magnetic flux linked with the coil changes, and therefore, an EMF is induced between the ends of the coil. Thus, a generator acts like a source of current if an external circuit containing load is connected between the ends of its coil.

Solution 19.

The number of rotations of the coil in one second or the speed of rotation of the coil.

Solution 20.

Solution 21.

(a)In an a.c generator, if the speed at which the coil rotates is doubled, the frequency is also doubled.
(b) Maximum output voltage is also doubled.

Solution 22.

Two ways in an a.c generator to produce a higher e.m.f. are:

1. By increasing the speed of rotation of the coil.
2. By increasing the number of turns of coil.

Solution 23.

Mechanical energy changes into the electrical energy.

Solution 24.

Two dissimilarities between D.C. motor and A.C. generator:

 A.C. Generator D.C. Motor 1. A generator is a device which converts mechanical energy into electrical energy. 1. A D.C. motor is a device which converts electrical energy into mechanical energy. 2. A generator works on the principle of electromagnetic induction. 2. A D.C. works on the principle of force acting on a current carrying conductor placed in a magnetic field.

Similarity: Both in A.C generator and D.C motor, a coil rotates in a magnetic field between the pole pieces of a powerful electromagnet.

Solution 25.

The voltage of a.c. can be stepped up by the use of step-up transformer at the power generating station before transmitting it over long distances. It reduces the loss of electrical energy as heat in the transmission line wires. On the other hand, if d.c. is generated at the power generating station, its voltage cannot be increased for transmission, and so due to passage of high current in the transmission line wires, there will be a huge loss of electrical energy as heat in the line wires.

Solution 26.

The purpose of the transformer is to step up or step down the a.c. voltage.
No, a transformer cannot be used with a direct current source.

Solution 28.

Solution 29.

Solution 30.

The device is step up transformer.
It works on the principle of electromagnetic induction.

Solution 31.

Step-up transformer: The step-up transformer is used to change a low voltage alternating e.m.f. to a high voltage alternating e.m.f. of same frequency.

Working: When the terminals of primary coil are connected to the source of alternating e.m.f., a varying current flows through it which also produces a varying magnetic field in the core of the transformer. Thus, the magnetic field lines linked with the secondary coil vary and induce an e.m.f. in the secondary coil. The induced e.m.f. varies in the same manner as the applied e.m.f. in the primary coil varies, and thus, has the same frequency as that of the applied e.m.f.
The magnitude of e.m.f. induced in the secondary coil depends on the ‘turns ratio’ and the magnitude of the applied e.m.f.
For a transformer,

Two characteristics of the primary coil as compared to its secondary coil:

1. The number of turns in the primary coil is less than the number of turns in the secondary coil.
2. A thicker wire is used in the primary coil as compared to that in the secondary coil.

Solution 32.

Working: In a step down transformer, the number of turns in secondary coil are less than the number of turns in the primary coil i.e., turns ratio NS/NP<1.
As Es/Ep = NS/NP.
So Es/Eps is less than Ep.

Two uses of step down transformer are:

1. With electric bells
2. At the power sub-stations to step-down the voltage before its distribution to the customers.

Solution 34.

A is the primary coil. B is the secondary coil.
We have drawn laminated core in the diagram.
The material of this part is soft iron.
This transformer a step-down transformer because the number of turns in primary coil is much greater than that in the secondary coil.

Solution 35.

(a) Soft iron core is used. The core is made up from the thin laminated sheets of soft iron of T and U shape, placed alternately one above the other and insulated from each other by paint or varnish coating over them.

Solution 36.

The secondary windings of a transformer in which the voltage is stepped down are usually made up of thicker than the primary because more current flows in the secondary coil. The use of thicker wire reduces its resistance and therefore the loss of energy as heat in the coil.

Solution 37.

To reduce the energy losses due to eddy currents.

Solution 38.

1. In a step-up transformer, the number of turns in the primary is less than the number of turns in the secondary.
2. The transformer is used in alternating current circuits.
3. In a transformer, the frequency of A.C. voltage remain same.

Solution 39.

Solution 40.

The energy loss in a transformer is called ‘copper loss’.

Copper losses: Primary and secondary coils of a transformer are generally made of copper wire. These copper wires have resistance. When current flows through these wires, a part of the energy is lost in the form of heat. This energy lost through the windings of the transformer is known as copper loss.

This loss can be minimized by using thick wires for the windings. Use of thick wire reduces its resistance and therefore reduces the loss of energy as heat in the coil.

Solution 41.

 Step up transformer Step down transformer It increases the a.c. voltage and decrease the current. It decreases the a.c. voltage and increase the current. The wire of primary coil is thicker than that in the secondary coil. The wire in the secondary coil is thicker than that in the primary coil.

Solution 42.

Soft iron is used in all.

Solution 1 (MCQ).

Fleming’s right hand rule
Statement: According to Fleming’s right hand rule, if we stretch the thumb, middle finger and forefinger of our right hand mutually perpendicular to each other such that the forefinger indicates the direction of magnetic field and thumb indicates the direction of motion of conductor, then the middle finger will indicate the direction of induced current.

Solution 2 (MCQ).

N> NP
Hint: Since a step-up transformer is used to change a low voltage alternating e.m.f. to a high voltage alternating e.m.f. of same frequency, the number of turns in the secondary coil is more than the number of turns in the primary coil, i.e. N> NP.

Numericals

Solution 2.

Solution 3.

Solution 4.

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## Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 9 Electrical Power and Household Circuits. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 9 Electrical Power and Household Circuits

Solution 1.

The electric power is generated at 11 KV, 50Hz at the power generating station.

Solution 2.

At a power generating station, the electric power is generated at 11 kV. From here, the alternating voltage is transmitted to the grid sub-station and stepped up to 132 kV using a step-up transformer. It is then transmitted to the main sub-station where the voltage is stepped down to 33 kV using a step-down transformer and is then transmitted to the intermediate sub-station. At the intermediate sub-station, the voltage is stepped down to 11 kV using a step-down transformer and is transmitted to the city sub-station, where the voltage is further stepped down to 220 V and is supplied to our houses.

Solution 3.

Electric power from the generating station is transmitted at 11 kV because voltage higher than this causes insulation difficulties, while the voltage lower than this involves high current and loss of energy in form of heat (I2Rt).

Solution 4.

At 220 V of voltage and 50 Hz of frequency, the a.c. is supplied to our houses.

Solution 5.

(a) Step-up transformer
(b) Step-down transformer

Solution 6.

(a) The three connecting wires used in a household circuit are:

1. Live (or phase) wire (L),
2. Neutral wire (N), and
3. Earth wire (E).

(b) Among them neutral and earth wires are at the same potential.
(c) The switch is connected in the live wire.

Solution 7.

Before the electric line is connected to the meter in a house, a fuse of rating (≈ 50 A) is connected in the live wire at the pole or just before the meter. This fuse is called the pole fuse.
Its current rating is ≈ 50 A.

Solution 8.

1. After the company fuse, the cable is connected to a kWh meter and from this meter; connections are made to the distribution board through a main fuse and a main switch.
2. Main fuse is connected in the live wire and in case of high current it gets burnt and cut the connections to save appliances.
3. Main switch is connected in the live and neutral wires. It is used to cut the connections of the live as well as the neutral wires simultaneously from the main supply.

Solution 9.

The electric meter in a house measures the electrical energy consumed in kWh.
Its value in S.I. unit is 1kWh = 3.6 x 106J.

Solution 10.

The main fuse in a house circuit is connected on the distribution board, in live wire before the main switch.

Solution 12.

Advantages of ring system over tree system

1. In a ring system the wiring is cheaper than tree system.
2. In ring system the sockets and plugs of same size can be used while in a tree system sockets and plugs are of different size.
3. In ring system, each appliance has a separate fuse due to which if there is a fault and the fuse of one appliance burns it does not affect other appliances; while in a tree system when fuse in one distribution line blows, it disconnects all the appliances connected to that distribution circuit.

Solution 13.

Solution 14.

All the electrical appliances in a building should be connected in parallel at the mains, each with a separate switch and a separate fuse connected in the live wire so that the switching on or off in a room has no effect on other lamps in the same building.

Solution 16.

In set A, the bulbs are connected in series. Thus, when the fuse of one bulb blows off, the circuit gets broken and current does not flow through the other bulbs also.
In set B, the bulbs are connected in parallel. Thus, each bulb gets connected to its voltage rating (= 220 V) and even when the fuse of one bulb blows off, others remain unaffected and continue to glow.

Solution 1 (MCQ).

The main fuse is connected in live wire.
Hint: The main fuse is connected in live wire so that if the current exceeds its rating, the fuse melts and breaks the circuit; thus, preventing the excessive current from flowing into the circuit.

Solution 2 (MCQ).

Electrical appliances in a house are connected in parallel.
Hint: On connecting the electrical appliances in parallel, each appliance works independently without being affected whether the other appliance is switched on or off.

Solution 3 (MCQ).

Energy
Hint: The electric meter in a house records the amount of electrical energy consumed in a house.

Exercise 9(B)

Solution 1.

An electric fuse is a safety device, which is used to limit the current in an electric circuit. The use of fuse safeguards the circuit and appliances connected in that circuit from being damaged.
An alloy of lead and tin is used as a material of fuse because it has low melting point and high resistivity.

Solution 2.

‘Fuse’ is used to protect electric circuits from overloading and short circuiting. It works on heating effect of current.

Solution 3.

(a) A fuse is a short piece of wire of material of high resistance and low melting point.
(b) A fuse wire is made of an alloy of lead and tin. If the current in a circuit rises too high, the fuse wiremelts
(c) A fuse is connected in series with the live wire.
(d) Higher the current rating, Thicker is the fuse wire.

Solution 4.

The fuse wire is fitted in a porcelain casing because porcelain is an insulator of electricity.

Solution 5.

Solution 6.

Solution 7.

The fuse wire is always connected in the live wire of the circuit because if the fuse is put in the neutral wire, then due to excessive flow of current when the fuse burns, current stops flowing in the circuit, but the appliance remains connected to the high potential point of the supply through the live wire. Now if a person touches the appliance, he may get a shock as the person will come in contact with the live wire through the appliance.

Solution 8.

The 20 A fuse wire will be thicker so that its resistance be low.

Solution 9.

It means that the line to which this fuse is connected has a current carrying capacity of 5 A.

Solution 10.

The safe limit of current which can flow through the electrical appliance is I = P/V = 5000/200 = 25 A; which is greater than 8 A. So, such fuse cannot be used.

Solution 11.

Solution 12.

A switch is an on-off device for current in a circuit (or in an appliance). The switch should always be connected in the live wire so that the appliance could be connected to the high potential point through the live wire. In this position the circuit is complete as the neutral wire provides the return path for the current. When the appliance does not work i.e., in off position of the switch, the circuit is incomplete and no current reaches the appliance.

On the other hand, if switch is connected in the neutral wire, then in ‘off’ position, no current passes through the bulb. But the appliance remains connected to the high potential terminal through the live wire.

Thus, if the switch is connected in the neutral wire, it can be quite deceptive and even dangerous for the user.
Precaution while handling a switch: A switch should not be touched with wet hands.

Solution 13.

A switch should not be touched with wet hands. If water reaches the live wire, it forms a conducting layer between the hand and the live wire of the switch through which the current passes to the hand and the person may get a fatal shock.

Solution 14.

Let a switch S1 be fitted at the bottom and a switch S2 at the top of the staircase. Fig. (a) shows the off position of the bulb.

The bulb can now be switched on independently by either the switch S1 or the switch S2. If the switch S1 is operated, the connection ‘ab’ is changed to ‘bc’, which completes the circuit and the bulb lights up [Fig. (b)].

Similarly, on operating the switch S2, the connection ‘bc’ changes to ‘ba’, which again completes the circuit [Fig. (c)].

Similarly if the bulb is in on position as shown in Fig. (b) or (c), one can switch off the bulb either from the switch S1 or the switch S2.

Solution 15.

All electrical appliances are provided with a cable having a plug at one end to connect the appliance to the electric supply.
In this three way pin plug, the top pin is for earthing (E), the live pin (L) in on the left and the neutral pin (N) is on the right.

Solution 16.

The three pins in the plug are labelled as

Here E signifies the earth pin,
L is for live wire, and
N is for neutral wire.

1. The earth pin is made long so that the earth connection is made first. This ensures the safety of the user because if the appliance is defective, the fuse will blow off. The earth pin is thicker so that even by mistake it cannot be inserted into the hole for the live or neutral connection of the socket.
2. The pins are splitted at the end to provide spring action so that they fit in the socket holes tightly.

Solution 17.

Solution 18.

(a) 1 – Earth, 2 – Neutral, 3 – Live
(b) Terminal 1 is connected to the outer metallic case of the appliance.
(c) The fuse is connected to live wire joined to 3 so that in case of excessive flow of current fuse melts first and breaks down the circuit to protect appliances.

Solution 19.

Local earthing is made near kWh meter. In this process a 2 – 3 metre deep hole is dug in the ground. A copper rod placed inside a hollow insulating pipe, is put in the hole. A thick copper plate of dimensionsis 50 cm x 50 cm welded to the lower end of the copper rod and it is buried in the ground. The plate is surrounded by a mixture of charcoal and salt to make a good earth connection.
To keep the ground damp, water is poured through the pipe from time to time. This forms a conducting layer between the plate and the ground. The upper end of the copper rod is joined to the earth connection at the kWh meter.

Solution 20.

If the live wire of a faulty appliance comes in to direct contact with the metallic case due to some reason then the appliance acquires the high potential of live wire. This may results in shock if any person touches the body of appliance. But if the appliance is earthed then as soon as the live wire comes in to contact with the metallic case, high current flows through the case to the earth. The fuse connected to the appliance will also blows off, so the appliance get disconnected.

Solution 21.

(a) The fuse must be connected in the live wire only. If the fuse is in the neutral wire, then although the fuse burns due to the flow of heavy current, but the appliance remains at the supply voltage so that on touching the appliance current flows through the appliance to the person touching it.
(b) Metallic case of the appliance should be earthed.

Solution 22.

The paint provides an insulating layer on the metal body of the appliance. To make earth connection therefore, the paint must be removed from the body part where connection is to be made.

Solution 23.

1. According to new international convention
2. Live wire is brown in colour.
3. Neutral is light blue and
4. Earth wire is yellow or green in colour.

Solution 25.

(a) The three wires are: Live wire, Earth wire and Neutral wire.
(b) The heating element of geyser should be connected to live wire and neutral wire.
(c) The metal case should be connected to earth wire.
(d) The switch and fuse should be connected to live wire.

Solution 26.

One may get an electric shock from an electrical gadget in the following two cases:

1. If the fuse is put in the neutral wire instead of live wire and due to fault, if an excessive current flows in the circuit, the fuse burns, current stops flowing in the circuit but the appliance remains connected to the high potential point of the supply through the live wire. In this situation, if a person touches the faulty appliance, he may get an electric shock as the person will come in contact with the live wire through the appliance.
Preventive measure: The fuse must always be connected in the live wire.
2. When the live wire of a faulty appliance comes in direct contact with its metallic case due to break of insulation after constant use (or otherwise), the appliance acquires the high potential of the live wire. A person touching it will get a shock because current flows through his body to earth.
Preventive measure: Proper ‘earthing’ of the electric appliance should be done.

Solution 27.

Power circuit carries high power and costly devices. If there is some unwanted power signal (noise) in the wire it can damage the device. To reduce this effect earth is necessary.
Lighting circuit carries low power (current).So, we ignore the earth terminal.

Solution 28.

A high tension wire has a low resistance and large surface area.

Solution 29.

To carry larger current, the resistance of the wire should be low, so its area of cross section should be large. Therefore 15 A current rated wire will be thicker.

Solution 30.

(a) Switches 2 and 3.
(b) The lamps are connected in series.

Solution 31.
(a)

(b)

 Wire no. Wire name Colour (Old convention) Colour (New convention) 1 Neutral wire Black Light blue 2 Earth wire Green Green or yellow 3 Live wire Red Brown

(c) The bulbs are joined in parallel.

Solution 1 (MCQ).

5 A
Hint: The electric wiring for light and fan circuit uses a thin fuse of low current rating (= 5 A) because the line wire has a current carrying capacity of 5 A.

Solution 2 (MCQ).

A switch must be connected in live wire.
Explanation: A switch must be connected in live wire, so that when it is in ‘off’ position, the circuit is incomplete and no current reaches the appliance through the live wire.

More Resources for Selina Concise Class 10 ICSE Solutions

## Selina Concise Physics Class 10 ICSE Solutions Current Electricity

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 8 Current Electricity. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 8 Current Electricity

Exercise 8(A)

Solution 1.

Current is defined as the rate of flow of charge.
I=Q/t
Its S.I. unit is Ampere.

Solution 2.

Electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point. Its unit is the volt.

Solution 3.

The potential difference between two points is equal to the work done in moving a unit positive charge from one point to the other.
It’s S.I. unit is Volt.

Solution 4.

One volt is the potential difference between two points in an electric circuit when 1 joule of work is done to move charge of 1 coulomb from one point to other.

Solution 7.

In a metal, the charges responsible for the flow of current are the free electrons. The direction of flow of current is conventionally taken opposite to the direction of motion of electrons.

Solution 8.

It states that electric current flowing through a metallic wire is directly proportional to the potential difference V across its ends provided its temperature remains the same. This is called Ohm’s law.
V = IR

Solution 11.

Solution 12.

Ohmic Resistor: An ohmic resistor is a resistor that obeys Ohm’s law. For example: all metallic conductors (such as silver, aluminium, copper, iron etc.)

Solution 13.

Solution 14.

1. Ohmic resistor obeys ohm’s law i.e., V/I is constant for all values of V or I; whereas Non-ohmic resistor does not obey ohm’s law i.e., V/I is not same for all values of V or I.
2. In Ohmic resistor, V-I graph is linear in nature whereas in non-ohmic resistor, V-I graph is non-linear in nature.

Solution 16.

In the above graph, T1 > T2. The straight line A is steeper than the line B, which leads us to conclude that the resistance of conductor is more at high temperature Tthan at low temperature T2. Thus, we can say that resistance of a conductor increases with the increase in temperature.

Solution 17.

Solution 18.

Resistance of a wire is directly proportional to the length of the wire.
R ∝ l
The resistance of a conductor depends on the number of collisions which the electrons suffer with the fixed positive ions while moving from one end to the other end of the conductor. Obviously the number of collisions will be more in a longer conductor as compared to a shorter conductor. Therefore, a longer conductor offers more resistance.

Solution 19.

With the increase in temperature of conductor, both the random motion of electrons and the amplitude of vibration of fixed positive ions increase. As a result, the number of collisions increases. Hence, the resistance of a conductor increases with the increase in its temperature.
The resistance of filament of a bulb is more when it is glowing (i.e., when it is at a high temperature) as compared to when it is not glowing (i.e., when it is cold).

Solution 20.

Iron wire will have more resistance than copper wire of the same length and same radius because resistivity of iron is more than that of copper.

Solution 21.

1. Resistance of a wire is directly proportional to the length of the wire means with the increase in length resistance also increases.
R ∝ l
2. Resistance of a wire is inversely proportional to the area of cross-section of the wire. If area of cross-section of the wire is more, then resistance will be less and vice versa.
R ∝ 1/A
3. Resistance increases with the increase in temperature since with increase in temperature the number of collisions increases.
4. Resistance depends on the nature of conductor because different substances have different concentration of free electrons.Substances such as silver, copper etc. offer less resistance and are called good conductors; but substances such as rubber, glass etc. offer very high resistance and are called insulators.

Solution 22.

The resistivity of a material is the resistance of a wire of that material of unit length and unit area of cross-section.
Its S.I. unit is ohm metre.

Solution 23.

Solution 24.

Metal < Semiconductor < Insulator

Solution 26.

Manganin

Solution 28.

‘Copper or Aluminium’ is used as a material for making connection wires because the resistivity of these materials is very small, and thus, wires made of these materials possess negligible resistance.

Solution 30.

Manganin is used for making the standard resistor because its resistivity is quite large and the effect of change in temperature on their resistance is negligible.

Solution 31.

Generally fuse wire is made of an alloy of lead and tin because its resistivity is high and melting point is low.

Solution 32.

A wire made of tungsten is used for filament of electric bulb because it has a high melting point and high resistivity.

A nichrome wire is used as a heating element for a room heater because the resistivity of nichrome is high and increase in its value with increase in temperature is high.

Solution 33.

A superconductor is a substance of zero resistance at a very low temperature. Example: Mercury at 4.2 K.

Solution 34.

Superconductor

Solution 1 (MCQ).

Nichrome is an ohmic resistance.
Hint: Substances that obey Ohm’s law are called Ohmic resistors.

Solution 2 (MCQ).

For carbon, resistance decreases with increase in temperature.
Hint: For semiconductors such as carbon and silicon, the resistance and resistivity decreases with the increase in temperature.

Numericals

Solution 1.

Solution 9.

Solution 11.

Solution 12.

Exercise 8(B)

Solution 1.

e.m.f.: When no current is drawn from a cell, the potential difference between the terminals of the cell is called its electro-motive force (or e.m.f.).
Terminal voltage: When current is drawn from a cell, the potential difference between the electrodes of the cell is called its terminal voltage.
Internal Resistance: The resistance offered by the electrolyte inside the cell to the flow of electric current through it is called the internal resistance of the cell.

Solution 2.

 e.m.f. of cell Terminal voltage of cell 1. It is measured by the amount of  work done in moving a unit positive charge in the complete circuit inside and outside the cell. 1. It is measured by the amount of work done in moving a unit positive charge in the circuit outside the cell. 2. It is the characteristic of the cell i.e., it does not depend on the amount of current drawn from the cell 2. It depends on the amount of current drawn from the cell. More the current is drawn from the cell, less is the terminal voltage. 3. It is equal to the terminal voltage when cell is not in use, while greater than the terminal voltage when cell is in use. 3. It is equal to the emf of cell when cell is not in use, while less than the emf when cell is in use.

Solution 3.

Internal resistance of a cell depends upon the following factors:

1. The surface area of the electrodes: Larger the surface area of the electrodes, less is the internal resistance.
2. The distance between the electrodes: More the distance between the electrodes, greater is the internal resistance.

Solution 4.

Solution 5.

(a) Terminal voltage is less than the emf : Terminal Voltage < e.m.f.
(b) e.m.f. is equal to the terminal voltage when no current is drawn.

Solution 6.

When the electric cell is in a closed circuit the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. So, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

(a) series
(b) parallel
(c) parallel
(d) series

Solution 11.

For the same change in I, change in V is less for the straight line A than for the straight line B (i.e., the straight line A is less steeper than B), so the straight line A represents small resistance, while the straight line B represents more resistance. In parallel combination, the resistance decreases while in series combination, the resistance increases. So A represents the parallel combination.

Solution 1 (MCQ).

In series combination of resistances, current is same in each resistance.
Hint: In a series combination, the current has a single path for its flow. Hence, the same current passes through each resistor.

Solution 2 (MCQ).

In parallel combination of resistances, P.D. is same across each resistance.
Hint: In parallel combination, the ends of each resistor are connected to the ends of the same source of potential. Thus, the potential difference across each resistance is same and is equal to the potential difference across the terminals of the source (or battery).

Solution 3 (MCQ).

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

Solution 12.

Solution 13.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

Solution 18.

Solution 19.

Solution 20.

Solution 21.

Solution 23.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

Solution 28.

Solution 29.

Solution 30.

Exercise 8(C)

Solution 1.

Solution 2.

Solution 3.

Solution 4.

The S.I. unit of electrical energy is joule.
1Wh = 3600 J

Solution 5.

The power of an appliance is 100 W. It means that 100 J of electrical energy is consumed by the appliance in 1 second.

Solution 6.

The S.I. unit of electrical power is Watt.

Solution 7.

(i) The household unit of electricity is kilowatt-hour (kWh).
One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.
(ii) The voltage of the electricity that is generally supplied to a house is 220 Volt.

Solution 8.

(i) Electrical power is measured in kW and
(ii) Electrical energy is measured in kWh.

Solution 9.

One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.
Its value in SI unit is 1kWh = 3.6 x 106J

Solution 10.

Kilowatt is the unit of electrical power whereas kilowatt-hour is the unit of electrical energy.

Solution 11.

Solution 12.

An electrical appliance such as electric bulb, geyser etc. is rated with power (P) and voltage (V) which is known as its power rating. For example: If an electric bulb is rated as 50W-220V, it means that when the bulb is lighted on a 220 V supply, it consumes 50 W electrical power.

Solution 13.

It means that if the bulb is lighted on a 250 V supply, it consumes 100 W electrical power (which means 100J of electrical energy is converted in the filament of bulb into the light and heat energy in 1 second).

Solution 14.

Solution 15.

Solution 16.

When current is passed in a wire, the heat produced in it depends on the three factors:

1. on the amount of current passing through the wire,
2. on the resistance of wire and
3. on the time for which current is passed in the wire.
• Dependence of heat produced on the current in wire: The amount of heat H produced in the wire is directly proportional to the square of current I passing through the wire,  i.e., H ∝ I2
• Dependence of heat produced on the resistance of wire: The amount of heat H produced in the wire is directly proportional to the resistance R of the wire, i.e., H ∝ R
• Dependence of heat produced on the time: The amount of heat H produced in the wire is directly proportional to the time t for which current is passed in the wire, i.e., H ∝ t

Solution 1 (MCQ).

Solution 2 (MCQ).

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

When one lamp is connected across the mains, it draws 0.25 A current, while if two lamps are connected in series across the mains, current through each bulb becomes

(i.e., current is halved), hence heating (= I2Rt) in each bulb becomes one-fourth, so each bulb appears less bright.

Solution 11.

Solution 12.

Solution 13.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

Solution 18.

Solution 19.

Solution 20.

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## Selina Concise Physics Class 10 ICSE Solutions Sound

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 7 Sound. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 7 Sound

Exercise 7(A)

Solution 2.

(a) Amplitude: The maximum displacement of the particle of medium on either side of its mean position is called the amplitude of wave. Its S.I. unit is metre (m).
(b) Frequency: The number of vibrations made by a particle of the medium in one second is called the frequency of the waves.
It is also defined as the number of waves passing through a point in one second. Its S.I. unit is hertz (Hz).
(c) Wavelength: The distance travelled by the wave in one time period of vibration of particle of medium is called its wavelength. Its S.I. unit is metre (m).
(d) Wave velocity: The distance travelled by a wave in one second is called its wave velocity. Its S.I. unit is metre per second (ms-1).

Solution 3.

(i) Wavelength (or speed) of the wave changes, when it passes from one medium to another medium.
(ii) Frequency of a wave does not change when it passes from one medium to another medium.

Solution 4.

Two factors on which the speed of a wave travelling in a medium depends are:

1. Density: The speed of sound is inversely proportional to the square root of density of the gas.
2. Temperature: The speed of sound increases with the increase in temperature.

Solution 5.

The light waves can travel in vacuum while sound waves need a material medium for propagation.
The light waves are electromagnetic waves while sound waves are the mechanical waves.

Solution 6.

If a person stands at some distance from a wall or a hillside and produces a sharp sound, he hears two distinct sounds: one is original sound heard almost instantaneously and the other one is heard after reflection from the wall or hillside, which is called echo.

The condition for the echo: An echo is heard only if the distance between the person producing the sound and the rigid obstacle is long enough to allow the reflected sound to reach the person at least 0.1 second after the original sound is heard.

Solution 7.

t = 2d/V = 2 x 12/340 = 24/340 < 0.1 seconds so the man will not be able to hear the echo. This is because the sensation of sound persists in our ears for about 0.1 second after the exciting stimulus ceases to act.

Solution 8.

The applications of echo:

1. Dolphins detect their enemy and obstacles by emitting the ultrasonic waves and hearing their echo.
2. In medical science, the echo method of ultrasonic waves is used for imaging the human organs such as the liver, gall bladder, uterus, womb etc. This is called ultrasonography.

Solution 9.

Sound is produced from a place at a known distance say, d at least 50 m from the reflecting surface. The time interval t in which the echo reaches the place from where the sound was produced, is noted by a stop watch having the least count 0.01 s. then the speed of sound is calculated by using the following relation

Solution 10.

Bats, dolphin and fisherman detect their enemies or obstacles or position of fish by emitting/sending the ultrasonic waves and hearing/detecting the echo.

Solution 11.

Bats can produce and detect the sound of very high frequency up to about 1000kHz. The sounds produced by flying bats get reflected back from any obstacle in front of it. By hearing the echoes, bats come to know even in the dark where the obstacles are. So they can fly safely without colliding with the obstacles.

Solution 12.

The process of detecting obstacles with the help of echo is called sound ranging. It’s used by the animals like bats, dolphin to detect their enemies.

Solution 13.

The ultrasonic waves are used for the sound ranging. Ultrasonic waves have a frequency more than 20,000 Hz but the range of audibility of human ear is 20Hz to 20,000 Hz

Solution 14.

Sonar is sound navigation and ranging. Ultrasonic waves are sent in all directions from the ship and they are received on their return after reflection from the obstacles. They use the method of echo.

Solution 15.

In medical science, echo method of ultrasonic waves is used for the imaging of human organs such as liver, gall bladder, uterus, womb; which is called ultrasonography.

Solution 1 (MCQ).

17 m.

Explanation: An echo is heard distinctly if it reaches the ear at least 0.1 s after the original sound.
If d is the distance between the observer and the obstacle and V is the speed of sound, then the total distance travelled by the sound to reach the obstacle and then to come back is 2d and the time taken is,
t = Total distance travelled/Speed of sound = 2d/V
or, d = V t/2
Putting t = 0.1 s and V = 340 m/s in air at ordinary temperature, we get:
d = (340 x 0.1)/2 = 17 m
Thus, to hear an echo distinctly, the minimum distance between the source and the reflector in air is 17 m.

Solution 2 (MCQ).

Ultrasonic waves

Numericals

Solution 1.

(i) Frequency or the number of waves produced per second
= Velocity/Wavelength
= 24 / 20 x 10-2
= 120
(ii) Time = 1/ frequency = 1/ 120= 8.3 x 10-3 seconds

Solution 2.

Velocity = 2D/Time
350 = 2 x D/ 0.1
D = 350 x 0.1 / 2 = 17.5 m

Solution 3.

Velocity = 2D/Time
1400 = 2 x D/ 0.1
D = 1400 x 0.1/ 2 = 70 m

Solution 4.

(a) Velocity = 2D/Time
Time = 2 x 25 / 350 = 0.143 seconds
(b) Yes, because the reflected sound reaches the man 0.1 second after the original sound is heard and the original sound persists only for 0.1 second.

Solution 5.

Velocity = 2D/Time
3 x 108 = 2 x 45 x 1000/ Time
Time = 90000/ 3 x 10= 3 x 10-4 second

Solution 6.

Velocity = 2 x D/Time
Time after which an echo is heard = 2 D/Velocity = 2 x 48 / 320 = 0.3 seconds

Solution 7.

2 D = velocity x time
D = (velocity x time) / 2 = 1450 x 4 / 2 = 2900 m = 2.9 km

Solution 8.

5 vibrations by pendulum in 1 sec
So 8 vibrations in 8/5 seconds = 1.6 sec
Velocity = 2 x D/ time
340 = 2 x D/ 1.6
D = 340 x 1.6 / 2 = 272 m

Solution 9.

The distance of first cliff from the person, 2 x D1 = velocity x time
D1 = 320 x 4 / 2 = 640 m
Distance of the second cliff from the person, D2 = 320 x 6 / 2 = 960 m
Distance between cliffs = D1 + D2 = 640 + 960 = 1600 m

Solution 10.

Distance of hill from the man
D1 = velocity x time/ 2 =v x 5 / 2—————– (eqn 1)
Now, D1 – 310 = v x 3 / 2————————— (eqn 2)
By subtracting eqn 2 form eqn 1 ,we get310 = v x (5/2-3/2)
So, v = 310m/s

Solution 11.

Depth of the sea = velocity x time/2 = 1400 x 1.5 / 2 = 1050 m

Exercise 7(B)

Solution 1.

The vibrations of a body in the absence of any external force on it are called the free vibrations. Eg.: When we strike the keys of a piano, various strings are set into vibration at their natural frequencies.

Solution 2.

When each body capable of vibrating is set to vibrate freely and it vibrates with a frequency f. It is the natural frequency of vibration of the body.
The natural frequency of vibration of a body depends on the shape and size of the body.

Solution 3.

Solution 4.

The free vibrations of a body occur only in vacuum because the presence of medium offer some resistance due to which the amplitude of the vibration does not remain constant, but it continuously decreases.

Solution 5.

The frequency of sound emitted due to vibration in an air column depends on the length of the air column.

Solution 6.

The frequency of the note produced in the air column can be increased by decreasing the length of the air column.

Solution 7.

The frequency of vibration of the stretched string can be increased by increasing the tension in the string, by decreasing the length of the string.

Solution 8.

A stringed instrument is provided with the provision for adjusting the tension of the string. By varying the tension, we can get the desired frequency.

Solution 9.

a) (i) Diagram is showing the principal note.
b) (iii)Diagram has frequency four times that of the first.
d) Ratio is 1:2

Solution 10.

Strings of different thickness are provided on a stringed instrument to produce different frequency sound waves because the natural frequency of vibration of a stretched string is inversely proportional to the radius (thickness) of the string.

Solution 11.

The frequency of vibrations of the blade can be lowered by increasing the length of the blade or by sticking a small weight on the blade at its free end.

Solution 12.

The presence of the medium offers some resistance to motion, so the vibrating body continuously loses energy due to which the amplitude of the vibration continuously decreases.

Solution 13.

The periodic vibrations of a body of decreasing amplitude in the presence of resistive force are called the damped vibrations.
The amplitude of the free vibrations remains constant and vibrations continue forever. But, the amplitude of damped vibrations decreases with time and ultimately the vibrations ceases.
For eg, When a slim branch of a tree is pulled and then released, it makes damped vibrations.
A tuning fork vibrating in air excute damped vibrations.

Solution 14.

1. Damped vibrations
2. Example: When a slim branch of a tree is pulled and then released, it makes damped vibrations.
3. The amplitude of vibrations gradually decreases due to the frictional (or resistive) force which the surrounding medium exerts on the body vibrating in it. As a result, the vibrating body continuously loses energy in doing work against the force of friction causing a decrease in its amplitude.
4. After sometime, the vibrating body loses all of its energy and stops vibrating.

Solution 15.

The tuning fork vibrates with the damped oscillations.

Solution 16.

Solution 17.

The vibrations of a body which take place under the influence of an external periodic force acting on it, are called the forced vibrations. For example: when guitar is played, the artist forces the strings of the guitar to execute forced vibrations.

Solution 18.

The vibrations of a body in the absence of any resistive force are called the free vibrations. The vibrations of a body in the presence of an external force are called forced vibrations.

In free vibrations, the frequency of vibration depends on the shape and size of the body. In forced vibrations, the frequency is equal to the frequency of the force applied.

Solution 19.

Resonance is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body rapidly begins to vibrate with an increased amplitude. This phenomenon is known as resonance.
Mount two identical tuning forks A and B of same frequency upon two separate sound boxes such that their open ends face each other as shown.

If the prong A is struck on a rubber pad, it starts vibrating. On putting A on its sound box, tuning fork B also starts vibrating and a loud sound is heard. The vibrations produced in B are due to resonance.

Solution 20.

Condition for resonance:
Resonance occurs when the frequency of the applied force is exactly equal to the natural frequency of the vibrating body.

Solution 21.

forced,equal to the

Solution 22.

Solution 23.

At resonance, the body vibrates with large amplitude thus conveying more energy to the ears so a loud sound is heard.

Solution 24.

a) The vibrating tuning fork A produces the forced vibrations in the air column of its sound box. These vibrations are of large amplitude because of the large surface area of air in the sound box. They are communicated to the sound box of the fork B. The air column of B starts vibrating with the frequency of the fork A. Since the frequency of these vibrations is same as the natural frequency of the fork B, the fork B picks up these vibrations and starts vibrating due to resonance.
b) On putting the tuning fork A to vibrate, the other tuning fork B will also start vibrating. The vibrations produced in the second tuning fork B are due to resonance.

Solution 25.

(a) Set the pendulum A into vibration by displacing it to one side, normal to its length. It is observed that pendulum D also starts vibrating initially with a small amplitude and ultimately it acquires the same amplitude as the pendulum A initially had. When the amplitude of the pendulum D becomes maximum, the amplitude of the pendulum A becomes minimum since the total energy is constant. After some time the amplitude of the pendulum D will decreases and amplitude of A increases. The exchange of energy takes place only between the pendulums A and D because their natural frequencies are same. The pendulums B and C also vibrate, but with very small amplitudes.
(b) The vibrations produced in pendulum A are communicated as forced vibrations to the other pendulums B, C and D through XY. The pendulums B and C remain in the state of forced vibrations, while the pendulum D comes in the state of resonance.

Solution 26.

The phenomenon responsible for producing a loud audible sound is named resonance. The vibrating tuning fork causes the forced vibrations in the air column. For a certain length of air column, a loud sound is heard. This happens when the frequency of the air column becomes equal to the frequency of the tuning fork.

Solution 27.

(a) No loud sound is heard with the tubes A and C, but a loud sound is heard with the tube B.
(b) Resonance occurs with the air column in tube B whereas no resonance occurs in the air column of tubes A and C. The frequency of vibrations of air column in tube B is same as the frequency of vibrations of air column in tube D because the length of the air column in tube D is 20-18 = 2cm and that in tube B is 20-14 = 6 cm (3 times). On the other hand, the frequency of vibrations of air column in tubes A and C is not equal to the frequency vibrations of air column in tube B.
(c) When the frequency of vibrations of air column is equal to the frequency of the vibrating tuning fork, resonance occurs.

Solution 28.

When a troop crosses a suspension bridge, the soldiers are asked to break steps. The reason is that when soldiers march in steps, all the separate periodic forces exerted by them are in same phase and therefore forced vibrations of a particular frequency are produced in the bridge. Now, if the natural frequency of the bridge happens to be equal to the frequency of the steps, the bridge will vibrate with large amplitude due to resonance and suspension bridge could crumble

Solution 29.

The sound box is constructed such that the column of the air inside it, has a natural frequency which is the same as that of the strings stretched on it, so that when the strings are made to vibrate, the air column inside the box is set into forced vibrations. Since the sound box has a large area, it sets a large volume of air into vibration, the frequency of which is same as that of the string. So, due to resonance a loud sound is produced.

Solution 30.

When we tune a radio receiver, we merely adjust the values of the electronic components to produce vibrations of frequency equal to that of the radio waves which we want to receive. When the two frequencies match, due to resonance the energy of the signal of that particular frequency is received from the incoming waves. The signal received is then amplified in the receiver set.
The phenomenon involved is resonance. It is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body rapidly begins to vibrate with an increased amplitude. This phenomenon is known as resonance.

Solution 1 (MCQ).

It executes free vibrations.
Hint: The periodic vibrations of a body of constant amplitude in the absence of any external force on it are called free vibrations.

Solution 2 (MCQ).

Forced vibrations
Hint: The vibrations of a body which take place under the influence of external periodic force acting on it are called the forced vibrations.

Solution 3 (MCQ).

A tuning fork of frequency 256 Hz will resonate with another tuning fork of frequency 256 Hz.
Hint: Resonance occurs when the frequency of an externally applied periodic force on the body is equal to its natural frequency.

Exercise 7(C)

Solution 1.

The following three characteristics of sound are:

1. Loudness
2. Pitch or shrillness
3. Quality or timber.

Solution 2.

(a) Amplitude – The louder sound corresponds to the wave of large amplitude.
(b) Loudness is directly proportional to the square of amplitude.

Solution 3.

Loudness will be four times because loudness is directly proportional to the square of amplitude.

Solution 4.

(a) Ratio of loudness will be 1:9
(b) The ratio of frequency will be 1:1

Solution 5.

Solution 6.

The unit of loudness is phon.

Solution 7.

Because the board provides comparatively a large area and forces a large volume of air to vibrate and thereby increases the sound energy reaching our ears.

Solution 8.

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point. Its unit is microwatt per metre2.

Solution 9.

Relationship between loudness L and intensity I is given as:
L = K log I, where K is a constant of proportionality.

Solution 10.

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point.
The loudness of a sound depends on the energy conveyed by the sound wave near the eardrum of the listener. Loudness, being a sensation, also depends on the sensitivity of the ears of the listener. Thus the loudness of sound of a given intensity may differ from listener to listener. Further, two sounds of the same intensity but of different frequencies may differ in loudness even to the same listener because of the sensitivity of ears is different for different frequencies. So, loudness is a subjective quantity while intensity being a measurable quantity is an objective quantity for the sound wave.

Solution 11.

The loudness of the sound heard depends on:

1. Loudness is proportional to the square of the amplitude.
2. Loudness is inversely proportional to the square of distance.
3. Loudness depends on the surface area of the vibrating body.

Solution 12.

Decibel is the unit used to measure the sound level

Solution 13.

Upto 120 dB

Solution 14.

The disturbance produced in the environment due to undesirable loud and harsh sound of level above 120 dB from the various sources such as loudspeaker, moving vehicles etc. is called noise pollution.

Solution 15.

Pitch of sound is determined by its wavelength or the frequency. Two notes of the same amplitude and sounded on the same instrument will differ in pitch when their vibrations are of different wavelengths or frequencies.

Solution 16.

Pitch

Solution 17.

Pitch is the characteristic of sound which enables us to distinguish different frequencies sound. Pitch is the characteristic of sound by which an acute note can be distinguished from a grave or flat note.

Solution 18.

Solution 19.

As the water level in a bottle kept under a water tap rises, the length of air column decreases, so the frequency of sound produced increases i.e., sound becomes shriller and shriller. Thus by hearing sound from a distance, one can get the idea of water level in the bottle.

Solution 20.

Trumpet. Because its frequency is highest.

Solution 21.

(a) increases
(b) one-fourth

Solution 23.

Quality or timber of sound.

Solution 24.

The two sounds of same loudness and same pitch produced by different instruments differ due to their different waveforms.
The waveforms depend on the number of the subsidiary notes and their relative amplitude along with the principal note.
Diagram below shows the wave patterns of two sounds of same loudness and same pitch but emitted by two different instruments. They produce different sensation to ears because they differ in waveforms: one is a sine wave, while the other is a triangular wave.

Solution 25.

Since the guitars are identical, they will have a similar waveform and so the similar quality.

Solution 26.

Different instruments emit different subsidiary notes. A note played on one instrument has a large number of subsidiary notes while the same note when played on other instrument contains only few subsidiary notes. So they have different waveforms.

Solution 27.

It is because the vibrations produced by the vocal chord of each person have a characteristic waveform which is different for different persons.

Solution 28.

1. Frequency
2. Amplitude
3. Waveform

Solution 29.

1. Loudness
2. Quality or timbre
3. Pitch

Solution 30.

Solution 31.

1. IV
2. I
3. II

Solution 33.

(i) b, since amplitude is largest
(ii) a, since frequency is lowest

Solution 34.

Musical note is pleasant, smooth and agreeable to the ear while noise is harsh, discordant and displeasing to the ear.
In musical note, waveform is regular while in noise waveform is irregular.

Solution 1 (MCQ).

By reducing the amplitude of the sound wave, its loudness decreases.
Hint: Loudness of sound is proportional to the square of the amplitude.

Solution 2 (MCQ).

Waveforms
Explanation: The waveform of a sound depends on the number of the subsidiary notes and their relative amplitude along with the principal note. The resultant vibration obtained by the superposition of all these vibrations gives the waveform of sound.

Solution 3 (MCQ).

B is shrill, A is grave
Explanation: Shrillness or pitch of a sound is directly proportional to the frequency of the sound wave. Greater the frequency, shriller will be the note.

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## Selina Concise Physics Class 10 ICSE Solutions Spectrum

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 6 Spectrum. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 6 Spectrum

Exercise 6(A)

Solution 1.

The deviation produced by the prism depends on the following four factors:

1. The angle of incidence – As the angle of incidence increases, first the angle of deviation decreases and reaches to a minimum value for a certain angle of incidence. By further increasing the angle of incidence, the angle of deviation is found to increase.
2. The material of prism (i.e., on refractive index) – For a given angle of incidence, the prism with a higher refractive index produces a greater deviation than the prism which has a lower refractive index.
3. Angle of prism- Angle of deviation increases with the increase in the angle of prism.
4. The colour or wavelength of light used- Angle of deviation increases with the decrease in wavelength of light.

Solution 2.

The deviation caused by a prism increases with the decrease in the wavelength of light incident on it.

Solution 3.

Speed of light increases with increase in the wavelength.

Solution 4.

Red colour travels fastest and Blue colour travels slowest in glass.

Solution 5.

Colour of light is related to its wavelength.

Solution 6.

(i) 4000 Å to 8000 Å
(ii) 400 nm to 800 nm

Solution 7.

(i) For blue light, approximate wavelength = 4800 Å
(ii) For red light, approximate wavelength = 8000 Å

Solution 8.

Seven prominent colours of the white light spectrum in order of their increasing frequencies:
Red, Orange, Yellow, Green, Blue, Indigo, Violet

Solution 10.

Green, Yellow orange and red have wavelength longer than blue light.

Solution 11.

A glass prism deviates the violet light most and the red light least.

Solution 12.

(a) In vacuum, both have the same speeds.
(b) In glass, red light has a greater speed.

Solution 13.

The phenomenon of splitting of white light by a prism into its constituent colours is known as dispersion of light.

Solution 14.

When white light is incident on the first surface of a prism and enters in glass, light of different colours due to different speeds in glass, is refracted or deviated through different angles. Thus the dispersion of white light into its constituent colours takes place at the first surface of prism. Thus the cause of dispersion is the change in speed of light with wavelength or frequency.

Solution 15.

When white light is incident on the first surface of a prism and enters in glass, light of different colours due to different speeds in glass, is refracted or deviated through different angles. Thus the dispersion of white light into its constituent colours takes place at the first surface of prism.

On the second surface, only refraction takes place and different colours are deviated through different angles. As a result, the colours get further separated on refraction at the second surface (violet being deviated the most and red the least).

Solution 16.

The colour band obtained on a screen on passing white light through a prism is called the spectrum.

Solution 17.

(a) Violet, Indigo, Blue, Green, Yellow, Orange, Red.
(b) No, different colours have different widths in the spectrum.
(c) (i) Violet colour is deviated the most. (ii) Red colour is deviated the least.

Solution 18.

Solution 19.

• Constituent colours of white light are seen on the screen after dispersion through the prism.
• When a slit is introduced in between the prism and screen to pass only the light of green colour, only green light is observed on the screen.
• From the observation, we conclude that prism itself produces no colour.

Solution 20.

• If a monochromatic beam of light undergoes minimum deviation through an equi-angular prism, then the beam passes parallel to the base of prism.
• White light splits into its constituent colours i.e., spectrum is formed.
• We conclude that white light is polychromatic.

Solution 1 (MCQ).

Both deviation and dispersion.
Hint: When a white light ray falls on the first surface of a prism, light rays of different colours due to their different speeds in glass get refracted (or deviated) through different angles. Thus, the dispersion of white light into its constituent colours takes place at the first surface of prism.

Solution 2 (MCQ).

The colour of the extreme end opposite to the base of the prism is red.
Hint: The angle of deviation decreases with the increase in wavelength of light for a given angle of incidence. Since the red light has greatest wavelength, it gets deviated the least and is seen on the extreme end opposite to the base of prism.

Numericals

Solution 1.

Solution 2.

Exercise 6(B)

Solution 1.

(a) Five radiations in the order of their increasing frequencies are:
Infrared waves, Visible light, Ultraviolet, X-rays and Gamma rays.
(b) Gamma rays have the highest penetrating power.

Solution 2.

(a) Gamma rays, X-rays, infrared rays, micro waves, radio waves.
(b) Microwave is used for satellite communication.

Solution 3.

(a) Gamma ray.
(b) Gamma rays have strong penetrating power.

Solution 5.

(a) X-rays are used in the study of crystals.
(b) It is also used to detect fracture in bones.

Solution 8.

4000 Å to 8000 Å

Solution 9.

(i) Infrared
(ii) Ultraviolet

Solution 10.

The part of spectrum beyond the red and the violet ends is called the invisible spectrum as our eyes do not respond to the spectrum beyond the red and the violet extremes.

Solution 11.

Solution 12.

(i) Ultraviolet rays-wavelength range 100Å to 4000Å

(ii) Visible light-wavelength range 4000Å to 8000Å

(iii) Infrared radiations-wavelength range 8000Å to 107Å

Solution 13.

(i) Infrared radiations are longer than 8 x 10-7m.
(ii) ultraviolet radiations are shorter than 4 x 10-7 m.

Solution 14.

Solution 15.

(i) Microwaves are used for satellite communication.
(ii) Ultraviolet radiations are used for detecting the purity of gems, eggs, ghee etc.
(iii) Infrared radiations are used in remote control of television and other gadgets.
(iv) Gamma rays are used in medical science to kill cancer cells.

Solution 16.

Solution 17.

(a) A- Gamma rays, B-infrared radiations
(b) Ratio of speeds of these waves in vacuum is 1:1 as all electromagnetic waves travel with the speed of light in vacuum.

Solution 18.

All heated bodies such as a heated iron ball, flame, fire etc., are the sources of infrared radiations.
The electric arc and sparks give ultraviolet radiations.

Solution 19.

Infrared radiations are the electromagnetic waves of wavelength in the range of 8000Å to 107Å.

Detection: If a thermometer with a blackened bulb is moved from the violet end towards the red end, it is observed that there is a slow rise in temperature, but when it is moved beyond the red region, a rapid rise in temperature is noticed. It means that the portion of spectrum beyond the red end has certain radiations which produce a strong heating effect, but they are not visible. These radiations are called the infrared radiations.

Use: The infrared radiations are used for therapeutic purposes by doctors.

Solution 20.

The electromagnetic radiations of wavelength from 100Å to 4000Å are called the ultraviolet radiations.

Detection: If the different radiations from the red part of the spectrum to the violet end and beyond it, are made incident on the silver-chloride solution, it is observed that from the red to the violet end, the solution remains unaffected. However just beyond the violet end, it first turns violet and finally it becomes dark brown. Thus there exist certain radiations beyond the violet end of the spectrum, which are chemically more active than visible light, called ultraviolet radiations.

Use: Ultraviolet radiations are used for sterilizing purposes.

Solution 21.

• Ultraviolet radiations travel in a straight line with a speed of 3 x 108 m in air (or vacuum).
• They obey the laws of reflection and refraction.
• They affect the photographic plate.

Solution 22.

• Ultraviolet radiations produce fluorescence on striking a zinc sulphide screen.
• They cause health hazards like cancer on the body.

Solution 23.

• Infrared radiations travel in straight line as light does, with a speed equal to 3 x 108m/s in vacuum.
• They obey the laws of reflection and refraction.
• They do not cause fluorescence on zinc sulphide screen.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

1. Infrared radiations are used in photography in fog because they are not much scattered by the atmosphere, so they can penetrate appreciably through it.
2. Infrared radiations are used as signals during the war as they are not visible and they are not absorbed much in the medium.
3. Infrared lamps are used in dark rooms for developing photographs since they do not affect the photographic film chemically, but they provide some visibility.
4. Infrared spectrum can be obtained only with the help of a rock-salt prism since the rock-salt prism does not absorb infrared radiations whereas a glass prism absorbs them.
5. A quartz prism is used to obtain the spectrum of the ultraviolet radiations as they are not absorbed by quartz, whereas ordinary glass absorbs the ultraviolet light.
6. Ultraviolet bulbs have a quartz envelope instead of glass as they are not absorbed by quartz, whereas ordinary glad absorbs the ultraviolet light.

Solution 1 (MCQ).

Gamma rays

Solution 2 (MCQ).

Carbon arc-lamp

Solution 3 (MCQ).

Hint: Infrared radiations produce strong heating effect.

Numericals

Solution 1.

(a) Frequency =500MHz =500 x 106Hz
Wavelength= 60 cm=0.6 m
Velocity of wave= frequency x wavelength
= 500x 106 x 0.6=3 x 108m/s
(b) Electromagnetic wave is travelling through air.

Solution 2.

Exercise 6(C)

Solution 1.

When white light from sun enters the earth’s atmosphere, the light gets scattered i.e., the light spreads in all directions by the dust particles, free water molecules and the molecules of the gases present in the atmosphere. This phenomenon is called scattering of light.

Solution 2.

The intensity of scattered light is found to be inversely proportional to the fourth power of wavelength of light. This relation holds when the size of air molecules is much smaller than the wavelength of the light incident.

Solution 3.

Violet colour is scattered the most and red the least as the intensity of scattered light is found to be inversely proportional to the fourth power of wavelength of light.

Solution 6.

Since the wavelength of red light is the longest in the visible light, the light of red colour is scattered the least by the air molecules of the atmosphere and therefore the light of red colour can penetrate to a longer distance. Thus red light can be seen from the farthest distance as compared to other colours of same intensity. Hence it is used for danger signal so that the signal may be visible from the far distance.

Solution 7.

On the moon, since there is no atmosphere, therefore there is no scattering of sun light incident on the moon surface. Hence to an observer on the surface of moon (space), no light reaches the eye of the observer except the light directly from the sun. Thus the sky will have no colour and will appear black to an observer on the moon surface.

Solution 8.

Scattering property of light is responsible for the blue colour of the sky as the blue colour is scattered the most due to its short wavelength.

Solution 9.

As the light travels through the atmosphere, it gets scattered in different directions by the air molecules present in its path. The blue light due to its short wavelength is scattered more as compared to the red light of long wavelength. Thus the light reaching our eye directly from sun is rich in red colour, while the light reaching our eye from all other directions is the scattered blue light. Therefore, the sky in direction other than in the direction of sun is seen blue.

Solution 10.

At the time of sunrise and sunset, the light from sun has to travel the longest distance of atmosphere to reach the observer. The light travelling from the sun loses blue light of short wavelength due to scattering, while the red light of long wavelength is scattered a little, so is not lost much. Thus blue light is almost absent in sunlight reaching the observer, while it is rich in red colour.

Solution 11.

At noon, the sun is above our head, so we get light rays directly from the sun without much scattering of any particular colour. Further, light has to travel less depth of atmosphere; hence the sky is seen white.

Solution 12.

The clouds are nearer the earth surface and they contain dust particles and aggregates of water molecules of sizes bigger than the wavelength of visible light. Therefore, the dust particles and water molecules present in clouds scatter all colours of incident white light from sun to the same extent and hence when the scattered light reaches our eye, the clouds are seen white.

Solution 1 (MCQ).

Blue colour
Hint: When light of certain frequency falls on that atom or molecule, this atom or molecule responds to the light, whenever the size of the atom or molecule comparable to the wavelength of light. The sizes of nitrogen and oxygen molecules in atmosphere are comparable to the wavelength of blue light. These molecules act as scattering centers for scattering of blue light. This is also the reason that we see the sky as blue.

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## Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 5 Refraction through Lens. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 5 Refraction through Lens

Exercise 5(A)

Solution 1.

A lens is a transparent refracting medium bounded by two curved surfaces which are generally spherical.

Solution 2.

Solution 3.

Convex lens:

1. It converge the incident rays towards the principal axis.
2. It has a real focus.

Concave lens:

1. It diverges the incident rays away from the principal axis.
2. It has a virtual focus.

Solution 4.

Equiconvex lens is converging.

Solution 5.

Concave lens will show the divergent action on a light beam.

Solution 6.

As shown in the figure the convex lens has two glass prisms and one glass block. One of the glass prisms is situated above the glass block and one below the block.

Solution 7.

As shown in the figure the concave lens has two glass prisms and one glass block. One of the glass prisms is situated above the glass block and one below the block.

Solution 8.

If a parallel beam of light is incident on a convex lens then the upper part of the lens bends the incident ray downwards. The lower part bens the ray upwards while the central part passes the ray undeviated.

But in case of a concave lens the upper part of the lens bends the incident ray upwards and lower part bends the ray downwards while the central part passes the ray undeviated.

Solution 9.

It is the line joining the centers of curvature of the two surfaces of the lens.

Solution 10.

It is point on the principal axis of the lens such that a ray of light passing through this point emerges parallel to its direction of incidence.
It is marked by letter O in the figure. The optical centre is thus the centre of the lens.

Solution 11.

Solution 12.

A lens is called an equiconvex or equiconcave when radii of curvature of the two surfaces of lens are equal.

Solution 13.

A light ray can pass through a lens from either direction. Therefore, a lens has two principal foci.

For a convex lens, the first focal point is a point F1 on the principal axis of the lens such that the rays of light starting from it or passing through it, after refraction through lens, become parallel to the principal axis of the lens.

The second focal point for a convex lens is a point F2 on the principal axis such that the rays of light incident parallel to the principal axis, after refraction from the lens, pass through it.

Solution 14.

A light ray can pass through a lens from either direction. Therefore, a lens has two principal foci.

For a concave lens, the first focal point is a point F1 on the principal axis of the lens such that the incident rays of light appearing to meet at it, after refraction from the lens become parallel to the principal axis of the lens.

The second focal point for a concave lens is a point F2 on the principal axis of the lens such that the rays of light incident parallel to the principal axis, after refraction from the lens, appear to be diverging from this point.

Solution 15.

Solution 16.

Solution 17.

Solution 18.

Solution 19.

Solution 20.

Solution 21.

The distance from the optical centre O of the lens to its second focal point is called the focal length of the lens.

Solution 22.

A plane passing through the focal point and normal to the principal axis of the lens is called the first focal plane.

Solution 23.

(i) If a lens has both its focal length equal medium is same on either side of lens.
(ii)If a ray passes undeviated through the lens it is incident at the optical centre of the lens.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

Solution 28.

Solution 29.

(a) If half part of a convex lens is covered, the focal length does not change, but the intensity of image decreases.
(b) A convex lens is placed in water. Its focal length will increase.
(c) The focal length of a thin convex lens is more than that of a thick convex lens.

Solution 1 (MCQ).

First focus

Solution 2 (MCQ).

Its second focus

Exercise 5(B)

Solution 1.

1. A ray of light incident at the optical centre O of the lens passes undeviated through the lens.
2. A ray of light incident parallel to the principal axis of the lens, after refraction passes through the second focus F2 (in a convex lens) or appears to come from the second focus F2 (in a concave lens).
3. A ray of light passing through the first focus F1 (in a convex lens) or directed towards the first focus F1 (in a concave lens), emerges parallel to the principal axis after refraction.

Solution 2.

Solution 3.

 Real image Virtual image 1. A real image is formed due to actual intersection of refracted (or reflected) rays. 1. A virtual image is formed when the refracted (or reflected) rays meet if they are produced backwards. 2. A real image can be obtained on a screen. 2. A virtual image can not be obtained on a screen. 3. A real image is inverted with respect to the object. 3. A virtual image is erect with respect to the object.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

(ii) The position of the images will be more than twice the focal length of lens.
(iii) The image will be magnified, real and inverted.
(iv) As the object move towards F1 the image will shift away from F2 and it is magnified. At Fthe image will form at infinity and it is highly magnified. Between F1 and optical centre, the image will form on the same side of object and will be magnified.

Solution 8.

Solution 9.

Solution 10.

Solution 13.

Let the candle is placed beyond 2F1 and its diminished image which is real and inverted is formed between F2 and 2F2.

Here the candle is AB and its real and inverted image is formed between F2 and 2F2.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

The object is placed between focal point F1 and convex lens and its image is formed at the same side of the lens which is enlarged.
So this lens can be used as a magnifying lens.

Solution 18.

The sun is at infinity so convex lens forms its image at second focal point which is real and very much diminished in size.

While using the convex lens as burning glass, the rays of light from the sun (at infinity) are brought to focus on a piece of paper kept at the second focal plane of the lens. Due to sufficient heat of the sun rays, the paper burns. Hence this lens is termed as ‘burning glass’.

Solution 19.

(a) This is convex lens.
(b) The nature of the image is real.

Solution 20.

(a) Convex lens.
(b) Virtual.

Solution 21.

(a) Concave lens
(b) Image is diminished

Solution 23.

Image formed by a concave lens is virtual and diminished.

Solution 24.

The virtual image formed by a convex lens will be magnified and upright.

Solution 25.

(a) at focus,
(b) at 2F,
(c) between F and 2F,
(d) between optical centre and focus.

Solution 26.

 Type of lens Position of object Nature of image Size of image Convex Between optic centre and focus Virtual and upright Magnified Convex At focus Real and inverted Very much magnified Concave At infinity Virtual and upright Highly diminished Concave At any distance Virtual and upright Diminished

Solution 27.

1. When the object is situated at infinity, the position of image is at F2, it is very much diminished in size and it is real and inverted.
2. When the object (AB) is situated beyond 2F1, the position of image (A’B’) is between F2 and 2F2, it is diminished in size and real and inverted.
3. When the object (AB) is situated at 2F1, the position of image (A’B’) is at 2F2, it is of same size as the object and real and inverted.
4. When the object (AB) is situated between 2F1and F1, the position of image (A’B’) is beyond 2F2, it is magnified in size and real and inverted.
5. When the object (AB) is situated at F1, the position of image is at infinity; it is very much magnified in size and real and inverted.
6. When the object (AB) is situated between lens and F1, the position of image (CD) is on the same side, behind the object; it is magnified in size and virtual and upright.

Solution 28.

1. When object (AB) is situated at infinity then parallel rays from object appears to fall on concave lens. Due to which image forms at focus. This image is highly diminished in size and virtual and upright.
2. When object (AB) is situated at any point between infinity and optical centre of the lens then image forms between focus and optical centre. This image is diminished in size and virtual and upright.

Solution 29.

(a) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and diminished.
(b) An object is placed at a distance 2f from a convex lens of focal length f. The image formed is equal to that of the object.
(c) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and magnified.

Solution 30.

(a) False
(b) False
(c) False
(d) True
(e) False

Solution 1 (MCQ).

The focal length of the convex lens is 10 cm.
Hint: As the object distance = image distance, the object must be kept at 2f.
Therefore, 2f = 20 cm or f = 10 cm.

Solution 2 (MCQ).

Virtual and enlarged.
Explanation: When the object is kept between optical centre and focus of a convex lens, the image is formed on the same side, behind the object. The image thus formed is virtual, enlarged and erect.

Solution 3 (MCQ).

Virtual, upright and diminished
Hint: Concave lens forms virtual, upright and diminished image for all positions of the object.

Exercise 5(C)

Solution 1.

1. The axis along which the distances are measured is called as the principal axis. These distances are measured from the optical centre of the lens.
2. All the distances which are measured along the direction of the incident ray of the light are taken positive, while the distances opposite to the direction of the incident ray are taken as negative.
3. All the lengths that are measured above the principal axis are taken positive, while the length below the principal axis is considered negative.
4. The focal length of the convex lens is taken positive and that of concave lens is negative.

Solution 1 (MCQ).

Magnification is -0.5. The negative sign of magnification indicates that the image is real while 0.5 indicates that the image is diminished. A convex lens only forms a real and diminished image of an object. Hence, the correct answer is option (d).

Solution 2.

(i) The positive focal length of a lens indicates that it is a convex lens.
(ii) The negative focal length of a lens indicates that it is a concave lens.

Solution 3.

Lens formula:

• The distance of the object from the optical centre is called the object distance (u).
• The distance of the image from the optical centre is called the image distance (v).
• The distance of the principal focus from the optical centre is called the focal length (f).

Solution 3 (MCQ).

Solution 4.

The term magnification means a comparison between the size of the image formed by a lens with respect to the size of the object.
For a lens: Magnification ‘m’ is the ratio of the height of the image to the height of the object.

Solution 4 MCQ.

Power of a lens is +1.0 D. The positive sign indicates that the focal length of the lens is positive which indicates the lens is a convex lens.

Solution 5.

(i) Positive sign of magnification indicates that the image is virtual while negative sign indicates that the image is real.
(ii) Positive sign of magnification indicates that the image is erect while negative sign indicates that the image is inverted.

Solution 6.

The power of a lens is a measure of deviation produced by it in the path of rays refracted through it.
Its unit is Dioptre (D).

Solution 7.

Solution 8.

If focal length of a lens doubled then its power gets halved.

Solution 9.

The sign of power depends on the direction in which a light ray is deviated by the lens. The power could be positive or negative. If a lens deviates a ray towards its centre (converges), the power is positive and if it deviates the ray away from its centre (diverges), the power is negative.

Solution 10.

It is a concave.

Solution 9 (MCQ).

Solution 10 (MCQ).

Solution 1 (Num).

Solution 2 (Num).

Solution 3 (Num).

Solution 4 (Num).

Solution 5 (Num).

Solution 6 (Num).

Solution 7 (Num).

Solution 8 (Num).

Solution 9 (Num).

Solution 10 (Num).

Solution 11 (Num).

Solution 11 (Num).

Solution 13 (Num).

Exercise 5(D)

Solution 1.

Magnifying glass is a convex lens of short focal length. It is mounted in a lens holder for practical use.
It is used to see and read the small letters and figures. It is used by watch makers to see the small parts and screws of the watch.

Solution 2.

Let the object (AB) is situated between focal length and optical centre of a convex lens then its image (A’B’) will form on the same side of lens.

The image formed will be virtual, magnified and erect.

Solution 3.

The object is placed between the lens and principal focus.
The image is obtained between the lens and principal focus.

Solution 4.

The magnifying power of the microscope is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object (assumed to be placed at the least distance of distinct vision D = 25 cm) at the eye, i.e.,

where F is the focal length of the lens.
The magnifying power of a microscope can be increased by using the lens of short focal length. But it cannot be increased indefinitely.

Solution 5.

The two applications of a convex lens are:-

1. It is used as an objective lens in a telescope, camera, slide projector, etc.
2. With its short focal length it is also used as a magnifying glass.

The two applications of a concave lens are:-

1. A person suffering from short sightedness or myopia wears spectacles having concave lens.
2. A concave lens is used as eye lens in a Galilean telescope to obtain an erect final image of the object.

Solution 6.

The approximate focal length of a convex lens can be determined by using the principle that a beam of parallel rays incident from a distant object is converged in the focal plane of the lens.
In an open space, against a white wall, a metre scale is placed horizontally with its 0 cm end touching the wall.

By moving the convex lens to and fro along the scale, focus a distant object on wall. The image which forms on the wall is very near to the focus of the lens and the distance of the lens from the image is read directly by the metre scale. This gives the approximate focal length of the lens.

Solution 7.

Solution 8.

To determine focal length by using plane mirror we need a vertical stand, a plane mirror, a lens and a pin.
Place the lens L on a plane mirror MM’ horizontally. Arrange a pin P on the clamp of a vertical stand such that the tip of pin is vertically above the centre O of the lens.

Adjust the height of the pin until it has no parallax (i.e., when the pin and its image shift together) with its inverted image as seen from vertically above the pin.
Now measure the distance x of the pin from the lens and the distance y of the pin from the mirror, using a metre scale and a plumb line. Calculate the average of the two distances. This gives the focal length of the lens, i.e.,

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