Solution

X_PHYSICS_3&5MARKS_PRACTICE QUESTIONS

Class 10 - Science
Section A
1. We know that pencil appears to be bent at the interface of air and water because of refraction of light. The degree of refection
depends on refractive index of a given liquid. Refraction indices of kerosene, water and other liquids would be different. Hence,
degree of bend would be different in case of different liquids.

2.

Yes, even when one-half of a convex lens is covered with a black paper, the lens will produce a complete image.
Take a live candle, keep it in front of a convex lens mounted on an optical bench.
Move the candle along the axis of bench and take its full image on a screen. Now cover the lower half of lens with a black paper
without changing the positions of candle, lens and screen.
You will observe that full image of candle is still seen on the screen, but the intensity of image is reduced. The reason is that a
large number of rays incident on the lens are blocked. In the case of covered lower half of lens with black paper, the rays that are
emerging from candle and incident on lens are refracted from upper part only and form the full image.
3. Let us assume that the window pane is between F2 and infinity from this lens and this is a convex lens. We know that when the
object is between infinity and F2, its inverted and real images is formed between 2F and 2F2.
Now, the distant building is at infinity from the lens. Its image would be formed at 2F. So, the screen needs to be moved towards
the lens in order to get a sharp image. Its approximate focal length is 10 cm (less than image distance in earlier case).
4. Atmospheric refraction:- The refraction of light caused by the earth's atmosphere (having their layers of varying optical
densities) is called atmospheric refraction.
Light from a star is refracted as it leaves space and enters the earth's atmosphere. Air higher up in the sky is rare but that near the
Earth's surface is denser. So, as the light from a star comes down the dense air bends the light more. Therefore, the apparent
position to the star is slightly different from its actual position.

5. Refractive indices of Air, Ice and Benzene are 1.003, 1.31 and 1.5 respectively. Velocity of light in a medium is inversely
proportional to refractive index of the medium. Light will travel fastest in air (having least refractive index i.e. 1.0003) and
slowest in Benzene (having maximum refractive index i.e. 1.5)

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 6. i. The position of object AB would have been beyond 2F1.

ii. Size of the object would have been bigger than the size of image.
7. The virtual image formed by a concave mirror is always magnified whereas the virtual image formed by a convex mirror is
diminished.

8. Convex Lens Concave Lens

1. It is a thicker at the center than at the edges. 1. It is thinner at the center than at the edges.
2. Is has real Focus. 2. It has virtual focus.
3. It converges a parallel beam of light on reflection through it. 3. It diverges a parallel beam of light on refraction through it.

4. 4.

9. Given focal length f of lens 20 cm

To obtain real and magnified image, the object should be placed between F1 and 2F1,So the range will be from 20 cm to 40 cm of
convex lens.
10. i. a. Object should be between pole and focus.
b. At the focus.
ii. a. Small head-convex mirror.
b. Legs of normal size-plane mirror.
11. A camera in many ways is similar to the human eye as both eye and camera has convex lens. But, there are some basic differences
in image formation between the two as follows:
i. In camera, the distance between the lens and the screen can be adjusted but not the focal length of the lens. However, in eye,
the ciliary muscles adjust the focal length keeping the distance between the lens and the retina constant.
ii. The image formed on the retina is temporary and its impression is recorded in the brain as memory. However, the image
formed on the film of camera is a permanent record.
12. This student is unable to see far off objects. This means that the student is suffering form myopia. Doctor will prescribe a concave
lens a suitable focal length.

Correction for myopia

13. There are mainly four common defect of vision that can be corrected by the use of suitable eyeglasses or spectacles. There are
i. Myopia or near-sightedness,
ii. Hypermetropia or far-sightedness,
iii. Presbyopia, and
iv. Astigmatism
14. When we are in bright sunlight the aperture of the pupil would be small to regulate the amount of light entering the eye preventing
glare, discomfort and damage to eyes. As we enter a dark room less amount of light would enter our eyes due to small size of
pupil, and we won't be able to see objects clearly. It takes some time to regulate the size of the pupil through iris. Hence, it
requires some time to see things.
15. 1. Functions of following parts of human eye are given below :
1. Cornea - It is a thin membrane which provides 67% of the eye's focussing power.
2. Iris - It controls amount of light entering the eye by controlling the size of pupil similar to the aperture of a camera which
has capacity to decrease or increase the amount of light entering eye.

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 3. Crystalline lens - It helps to focus light on retina for image formation.
4. Ciliary muscles - It contracts and relax in order to change the lens shape for focussing image at retina. when it contracts
the lens become thicker and when it relaxes the lens become flat.
2. 1. The objective of organising such compaigns is to guide, educate and help those people who are suffering from corneal
blindness that they can be cured by corneal replacement surgery.
2. 1. Come to participate in this campaign because, if someone get his vision through your eyes, it is an incredible help.
2. As eye is one of the most valuable sense organs through which an individual can achieve so many things in his/her life,
so try to realise the situation that these people are sufferinng from.
3. The persons who actively participate and contribute in such programme are strong hearted and very much helpful for the
people living in such situations.

16.

i. Effective resistance is,

1 1 1 1
= + +
Reff R1 R2 R3

1 1 1 5+2+1 8
= + + = =
2 5 10 10 10
10
⇒ Ref f = = 1.25Ω
8

ii. Total current,

V
I =
Reff

10
=
1.25

= 8A

iii. Current through each resistor,

V 10
I1 = = = 5A,
R1 2

V 10
I2 = = = 2A
R2 5

and
I3 =
V

R3
=
10

10
= 1A .
17. i. I. RAB = R1 + R2 + R3
II. RA
1
=
1

R1
+
1

R2
+
1

R3
B

ii. According to ohm's law (V = IR). Slope of V-I graph gives resistance. So greater the slope of graph, greater will be its
resistance. Since we know that equivalent resistance in series is greater than the equivalent resistance in Parallel. So the first
(I) graph is correct representation of V-I graph in series and Parallel combination.
18. Given :-
Resistor, R₁ = 4 Ω
Resistor, R₂ = 8 Ω
Resistor, R₃ = 8 Ω
Potential Difference, V = 8 volts.
i. The potential difference across 4Ω resistors, V = IR = 1 × 4 = 4V
ii. Power dissipated in 4Ω resistors, P = I2R = (1A)2(4Ω) = 4W
19. i. ∴ Power of heater,
2
V 200×200
P = = = 500W
R 80

ii. ∴ Heat absorbed by water, H = mC θ R

∘ ∘ ∘ ∘
= 1 × 4200 × 40 [∵ θR = 60 − 20 = 40 C, C = 4200J/kg C]

= 168000J

= 168kJ

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 iii. H = P × t

168000 = 500 × t
168000
⇒ t = = 336s
500

20. Given, R1 = 2Ω , R2= 4Ω, t = 5s, V = 6V

∴ Net resistance, R = R1+ R2

=2Ω+4Ω
=6Ω
V 6
∴ Current, I = R
=
6

= 1A

In series, Current is same through both resistors.

∴ Heat dissipated, H = I2R1 × t
2
= (1) (4)(5) = 20J

21. Ohm's law: It states that the current flowing through a conductor between two points is directly proportional to the potential
difference across the two points.
The arrangement of components in a circuit is normally from the battery to the Ammeter then to the resistor. We have been given
the symbols for the Resistor, Ammeter and other components.
The correct order is therefore given as:
B, A, K, R, B
We, therefore conclude that student Y place them in the right order.
22. Resistors 3Ω and 2Ω are in series
The effective resistance of lower arm = 3 + 2 = 5 Ω
Resistance of upper arm is also 5Ω . The lower and upper arms are in parallel to each other.
Therefore, the total resistance is given by
1 1 1 2 5
= + = , Rt = = 2.5 Ω. . . . . . . . . . . (i)
R 5 5 5 2

The current through ammeter is equal to the total current.

V 40 8
∴ Total current =
4
= = = = 1.6 A
Rt 2.5 25 5

23. The two bulbs are connected in parallel and the complete circuit diagram is drawn below:

The reason for connecting the two bulbs in parallel is that (i) both the bulbs glow at the same voltage, and (ii) if one bulb stops
glowing, the other bulb remains unaffected.
24. Total resistance
Rs = R1 + R2 + R3
Rs = 5Ω + 8Ω + 12Ω = 25
V = 6 volts
V
∴ I = ( By Ohm's law )
R
6V
I =
25Ω
= 0.24 A
Since all resistances are connected in series therefore current through each resistance will be same. Hence, the reading of ammeter
is 0.24 ampere.
Now potential difference across 12Ω resistance
V = IR( By Ohm's law )
V = 0.24 × 12

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 V = 2.88volts

Since voltmeter connected across 12 ohm resistance, therefore the potential difference across 12 ohm resistance will be the
reading of the voltmeter. Hence, the reading of voltmeter is 2.88 volts.
25. Let the current through the circuit be I which is divided into I1 and I2 in the arms AB and CD respectively, then we have
I = I1 + I2
In the arm AB, the total resistance is
R1= 6Ω + 3Ω = 9Ω
and the total resistance in the arm CD is
R2 = 12Ω + 3Ω = 15Ω
i. Then current in the 6Q resistor i.e.,
I1 = V

R1

= 4

= 0.44 A
ii. Now the current through CD is
I2 = V

R2

= 4

15

= 0.27 A
The potential difference across 12Ω , resistor is
V1 = I 2 × 12Ω

V1 = 0.27 × 12
= 3.23 V
26. (i) Since R ∝ l (length of the conductor)
Length of wire C is more than A and B. Therefore wire C has higher resistance.
(ii) Resistivity of all wires is same as material of all the wires is same.Electrical Resistivity of a wire depends on the nature of the
material and not on the dimensions of a wire.

27. i.

If we connected two 9 ohm resistors in parallel and other 9 ohm resistor in series as shown in figure:
1 1 1 2
= + =
Rp 9 9 9

9
Rp = Ω
2
9
R = 9Ω + Ω = 13.5Ω
2

ii.

If we connected two 9 ohm resistors in series and other 9 ohm resistor in parallel as shown in figure:
Rs = 9Ω + 9Ω = 18Ω
1 1 1 3
= + =
R 18 9 18

∴ R = 6Ω

28. Given, power, P = 1500W, voltage, V = 230 V

i. ∴ Electric down drawn,
1500
= 6A
P
I = =
V 250

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 ii. ∴ Energy consumed, E = power ×
​ Time
= 1500 × 50

= 75000W h

= 75kW h [∴1 kW = 1000 W]

= 75 unit [∴ 1 unit = 1 kWh]

iii. ∴ Cost of energy consumed = 75​​​​​​×​​​​​6

= Rs 450.

29. i. Arm AB-Downward, Arm CD-Upward

ii. P and Q-Split ring/Commutator
iii. Arm AB upward, Arm CD downward/Direction of force will get reversed
iv. Fleming's left-hand rule
30. The resistance of an ammeter should be low. An ammeter has to be connected in series with the circuit to measure current. In case,
its resistance is not very low, its inclusion in the circuit will reduce the current to be measured. In fact, an ideal ammeter is one
which has zero resistance.
31. i. The magnitude of induced current increases.
ii. The induced current is zero.
32. i. The magnetic field and hence the magnetic line of force exist in all the planes all around the magnet.
ii. The relative strength of the magnetic field is shown by the degree of closeness of the field lines and the direction of the
magnetic field is obtained by tangent to the field lines at the point of intersect.
33. i. According to right-hand thumb rule,

a. The direction of magnetic field at a point above the power line. is from South to North.
b. The direction of magnetic field at a point below the power line is from North to South.
ii. a. If the current in the coil X is changed then some current will definitely be induced in the coil Y. When the current in coil X
is changed, the magnetic field associated with it also changes. As a result, the magnetic field around coil Y also changes.
This change in magnetic field lines around coil Y induces an electric current in it. This process is known electromagnetic
induction.
b. Fleming's Right-hand Rule: According to Fleming right-hand rule, Adjust your right-hand thumb, forefinger and middle
finger in such a way that all are perpendicular to one another. If forefinger shows the direction of magnetic field and
Thumb shows the direction of motion of the conductor then middle finger gives the direction of current.

34. A coil of insulated copper wire wound in the form of a cylinder is a solenoid. When current is passed through a solenoid, it
produces magnetic field lines like a bar magnet. The pattern of magnetic field lines of a solenoid is shown in the figure. Inside the
solenoid, field lines are parallel to each other This indicates that the magnetic field is uniform .

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35. i. The direction of force acting on the electron beam will be into the page.
ii. The angle between velocity and magnetic field is zero. Therefore, magnetic force on the particle is zero.
36. i. The direction of the magnetic field at a point can be found by placing a small magnetic compass at that point. The north end of
the needle of a compass indicates the direction of magnetic field at a point where it is placed.
ii. The direction of magnetic field at the centre of a current-carrying circular loop is perpendicular to the plane of the loop.
37. i.

ii. The direction of the magnetic field inside the solenoid always points from the induced South pole towards the induced North
pole.
38. According to Right Hand Thumb Rule, imagine the straight conductor in your right hand such that the thumb points in the
direction of the current, then the direction of curling of fingers of the right hand gives the direction of magnetic field lines. At P,
the direction of magnetic field is towards the paper and at point B, the direction of magnetic field is out of the paper. Since r1 > r2,
so the strength of magnetic field at P is less than that at Q.
39. i. Relative closeness of field lines indicates the strength of magnetic field. Since field lines are crowded around the ends of the
solenoid, hence these are the regions of strongest magnetism.
ii. The direction of the field will also be reversed.
40. Uniform magnetic field:The space or region where field is same everywhere is known as Uniform magnetic field.
Non-uniform magnetic field:The magnetic field which is unequal in magnitude and direction at every point in the space is called
non- uniform magnetic field.

Section B
41. When the object is placed between pole and principal focus of the concave mirror, an erect, enlarged, virtual image is formed
behind the mirror. Therefore range of distance is greater than zero and less than focal length i.e. between more than zero to less
than 15 cm.

42. i. The centre point of a lens is known as its optical centre. The optical centre is a point within the lens, directed to which incident
rays refract without any deviation in the path whether it is convex lens or concave lens as represented below:

ii. Given, there is a divergent lens( concave lens.)

Given, f =-20cm, h0 =4 cm , v = -10 cm
∵ By lens formula,

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 1 1 1
− =
v u f

−1 1 −1
⇒ − =
10 u 20
1 −1 1 1 −2 + 1
⇒ = + ⇒ =
u
10 20 u 20

⇒ u = - 20 cm
v hi
∴ Magnification, m = u
=
h0

hi −10
⇒ =
4
−20

⇒ hi = 2 cm
Size of the image, hi = 2 cm
iii.

Thus, the object is placed at 20 cm from the concave lens.

43. i.

ii. U is -ve,V is -ve. By lens formula:

1 1 1
− =
v u
f
−1 1 1 −1 1 1
⇒ − (− )= ⇒ + =
v u f v u f

−u+v 1 uv
⇒ = ⇒ f =
uv f v−u

This is a required relation between u, v and f in the case when object is placed between optical centre and principal focus of
convex lens
iii. Given, m =-1
u = -20 cm
∴ m=
v v
⇒ −1 =
u −20

⇒ v = 20 cm
1 1
By lens formula, 1

v
−
u
=
f

1 −1 1 1 1 1
⇒ − ( ) = ⇒ + =
20
20 f 20 20 f

1 1
⇒ = ⇒ f = 10 cm
10 f
1 1
∴ P ower, p = = = 10D
f −2
10×10

⇒ P = 10 D
44. Images formed by concave mirror.
i. Object at Infinity. Two cases arise :
a. When mirror is in parallel plane to the object. In such a case, rays from infinity come parallel to principal axis. After
reflection they pass through principal focus F (Rule 1). Image is extremely small, it is real, inverted and at principal focus.

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 Object at infinity, real extremely diminished image is formed at principal focus.
b. When mirror is inclined so that the rays strike the mirror obliquely. The ray AB passing through F after reflection goes
parallel to principal axis towards BA' (Rule 2). Another ray DE through C striking the mirror at E is reflected back. The
two form an image at A' Image is real, inverted, extremely diminished and at F.

Object at infinity, image at F. It is real, inverted, very much diminished.

c. Object beyond C. A ray AD from A parallel to principal axis after reflection passes through F (Rule 1), Another ray from
A through C, ray AG is reflected back along the same path (Rule 3), forming real, diminished, inverted image of AB is
formed at A'B', between F and C.

Object beyond C, a real, inverted diminished image between F and C, inverted at C and is of same size as that of object.
d. Object at C i.e. at 2f. A ray AD from A parallel to principal axis after reflection from mirror passes through F (Rule 1).
Another ray AD' from A through F, goes parallel to principal axis i.e. towards D'A' (rule 2) forming real, inverted image of
AB at C i.e. at 2f. The image is of the same size as the object.

Object at C, Image is also at C. It is real.

e. Object between F and C (f and 2f) A ray AD from object going parallel to principal axis is reflected towards F (Rule 1).
Another ray AE through C is reflected back (Rule 3) forming image of A at A'. Similarly image of B is formed at B'.
Image is real, inverted, enlarged and beyond C (2f) i.e. as shown in fig.

f. Object at F. A ray AD parallel to principal axis passes through F. Another ray AE strikes the mirror normally at E is
reflected back as it passes through C (Rule 3). They form image of object AB at infinity. The image is very much enlarged
and is real and inverted.

Object at E. Real, inverted, extremely enlarged image is formed at infinity.

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 g. Object between F and P. A ray AD from A goes parallel to principal axis after reflection passes through F (rule 1). Another
ray AE striking the mirror normally through C is reflected back (rule 3). They form virtual image of the object behind the
mirror. The image is erect and enlarged.

Object between F and P. An erect, enlarged, virtual image is formed behind the mirror.
Images Formed by a Concave mirror

Position of Object Position of Image Size of the Image Nature of Image

At infinity At focus F Highly diminished Real and inverted

At C At C Same size Real and inverted

At F At infinity Highly Enlarged Real and inverted

Beyond C Between F and C Diminished Real and inverted

Between F and C Beyond C Enlarged Real and inverted

Between P and F Behind the mirror Enlarged Virtual and erect

45. i. a. Object is placed between F and 2F of thin converging lens .

b. Object is placed between optical centre and F.
ii. The ray diagrams for real magnified and virtual magnified images are as follows:
Part (a)

Part (b)

iii. a. There will be no change in focal length of converging lens.

b. Intensity will become one-fourth and brightness of lens will be less .
46. Laws of refraction are as follows:
i. Incident ray, refracted ray and normal at the point of incidence lie in the same plane.

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 ii. Ratio of sine of incidence and sine of refraction is constant for the given color and pair of media.

47. The ability of a lens to converge or diverge light rays is called power of the lens. It is defined as the reciprocal of focal length. It's
SI unit is dioptre (D). If focal length is expressed in metres, then power is expressed in dioptre. We can say, dioptre is the power of
a lens whose focal length is one metre. For concave lens P and f are negative. For convex lens P and f are positive. Lens A of
focal length + 10 cm is convex lens
100 100
and power, P = f (in cm)
=
10
= +10D

Lens B of focal length - 10 cm is concave lens

100 100
and power, P = = = −10D
f (in cm) −10

Lens A (i.e. convex lens) will form a virtual and magnified image of an object placed 8 cm from it, as shown.

48. i. Palmists use a convex lens because it shows an enlarged, virtual and erect image when the object is between F and O of a
convex lens.
ii. If the palmist wants a real and magnified image, he should put an object between F1 and F2 or on F. But in that case, he will
have to use a screen to see the image. So, for convenience, palmists.
iii. Given, f = 10 cm, u = -5 cm
1 1 1
− =
v u f

1 1 1
Or v
+
5
=
10

Or, 1

v
=
10
1
−
1

5
1−2
Or 1

v
=
10
= −
1

10

Or, v = -10 cm
The image is formed at 10 cm on the same side of the lens. It is erect and virtual.
Image size = = =2 v

u
10

Image is twice as big as object.

49. a. Complete ray diagram-

b. Ratio of speed of light in air/vaccum to speed of light in the given medium is known as absolute refractive index.
Mathematical expression,
n= c/v
c. Mirror-1 is Concave
Uses : By dentists/Shaving Mirror/Torch search light/Vehicle headlight /solar furnace.

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 Mirror- 2 is convex mirror.
Uses : Rear view mirror in vehicles/ for security purpose
50. When two plane mirrors are at 90o then incident ray and reflected ray will always be parallel as explained below:

ED and DC are two plane mirrors placed at 900:

∠AN M = ∠M N Q (angle i=angle r)

∠N QP = ∠P QB (same as above)
∠M N Q + ∠QN D = 90 (MN is normal on CD) ...(1)
∘

∠QDN = 90 (mirrors are at right angle)

∘

Hence, in ΔQDN ;
∘
∠QN D + ∠N QD = 90

(acute angles of a right triangle are complementary) …(2)

From equations (1) and (2);
∠M N Q = ∠N QD

Hence, ∠QN D = ∠N QP
(because ∠N QP and ∠N QD are complementary)
Now, ∠P QB and ∠P QD are supplementary (PQ is normal on DE)
So, ∠BQE = N QD = ∠M N Q = ∠AN M
Or, ∠BQE = ∠AN M
Since corresponding angles are equal
Hence, BQ∥AN proved
This means that incident ray and reflected ray will always be parallel; irrespective of value of angle of incidence.
51. a. The defect of vision the person is suffering from is hypermetropia or farsightedness
Causes:
i. Shortening of eyeball
ii. focal length of eye lens increases

b.

c. Convex lens
1/f = 1/v - 1/u
= 1/(-50cm) - 1/(-25cm)
= 1/50cm
Hence, f = 50 cm = 0.5 m
There fore power = (1/0.5)D = 2D
d. Correction of Hypermetropia

52. When a person is unable to clearly see distant objects, he is considered a myopic person. Such a person is suffering from myopia
or short-sightedness or nearsightedness. This happens when image is formed in front of the retina.
When a person is unable to clearly see a nearby object, he is considered a hypermetropic person. Such a person is suffering from
hypermetropia or longsightedness. This happen when image is formed behind the retina but not on it.
Correction of Myopia: A person suffering from myopia needs to use a concave lens of suitable focal length. The concave lens

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 diverges the rays coming from infinity. After refraction from the concave lens, the rays appear to be coming from the far point of
this person’s eye.

Correction of Hypermetropia: A hypermetropic person needs to use a convex lens of suitable focal length. The convex lens
converges the light rays coming from a nearby object. As a result, these light rays appear to be coming from the near point of this
person’s eyes. Due to this, a clear image of nearby object is made on the retina of that person.

53. (i) Power, P = -5.5 D, Focal length, f = ?

p= 1

f (inm)

or -5.5 = 1

or f = −5.5
1

= -0.18 m for correcting the ditant vision.

The negative (-) sign indicates that the lens is concave.
(ii) Power, P = + 1.5 D
Focal length, f = ?
1
p= f (inm)

1
or f = +1.5

or f = +0.67 m for correcting the near vision.

The positive sign indicates that the lens is convex.
54. Let PE is the incident light on the prism, EF is the refracted ray, and FS is the emergent ray. When a ray of light is entered through
surface AB from air to glass, then the light ray suffers refraction and bend towards the normal ray. At the second surface AC, the
rays enter from glass to air which results in bending away from the normal ray. The angle of incidence and the angle of refraction
are equal. The emergent ray bend at an angle towards the direction of the incident ray. This angle is known as the Angle of
deviation.
It can be shown with the following ray diagram:

55. Myopia: The defect of an eye in which it cannot see the distant objects clearly is called myopia. A person with myopia can see
nearby objects clearly.
Hypermetropia: Hypermetropia is also known as long-sightedness. In this defect, a person can see the distant objects clearly but
cannot see the nearby objects clearly.
Causes of Hypermetropia:
i. The focal length of the eye lens is too long.
ii. The eyeball has become too small.
In hypermetropia, the image of a distant object is formed behind the retina and not on the retina. The defect is corrected by using
the Convex lens of suitable power so that the lens will bring the image back on to the retina. The ray diagram of hypermetropia

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 eye are as follows:

56. Series Combination Parallel Combination

The main current from the source divides itself in different arms.
The current has a single path for its flow, hence, same
The current in each resistors is inversely proportional to its
current flows through each resistor.
resistance.

The potential difference across the entire circuit is equal to The potential difference across each resistor is same and it is equal
the sum of the potential difference across the individual to the potential difference across the terminals of the battery (or
resistor. source).

The equivalent resistance in series combination is greater The equivalent resistance in parallel combination is less than the
than the highest resistance in the series combination. least resistance in the parallel combination.

If one component breaks down, the whole circuit will burn Other components will function even if one component breaks
out. down, each has its own independent circuit
57. Combination of resistors: Resistors of all values of resistances are not available. Hence resistors are connected in a number of
ways to increase or decrease the combined resistance. There are two distinct ways in which resistors can be connected. They are
resistors in series and resistors in parallel.
Resistors connected in series
Resistors are said to be in series if they are joined end to end so that the same current flows through each one of them in
succession. Since there is a single path for the moving charge, the same current must flow through each resistor. Let the conductor
AB, BC and CD having resistances equal to R1, R2 and R3 respectively, be joined in series and let the current passing through
them be I.

Let V1, V2 and V3 be the potential difference between the ends of the first, second and third conductor respectively.
By Ohm's law,
V1 = IR1; V2 = IR2 and V3 = IR3
If V is the total potential difference between the ends A and D and Rs is the effective resistance of the combination of all the
resistors, then
V = IRs But V = V1 + V2 + V3

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 OR IRs = IR1 + IR2 + IR3
OR Rs = R1 + R2 + R3
The above result holds good for any number of resistors joined in series.

Thus when some resistors are joined in series, the total resistance is the sum of individual resistances.
58. i. Given:
Length of first wire x = 30 cm
Length of second wire y = 10 cm
Radius of first wire = 2 cm
Radius of second wire = 1 cm
To find :
Ratio of resistance of both wires
Solution :
Radius of first wire = 2 cm
So, Area of cross section of first wire = πr = π × 2 = 4πcm
2 2 2

Radius of second wire = 1 cm

So, Area of cross section of second wire = πr = π × 1 = πcm
2 2 2

l
General formula of resistance R = p a

Resistivity of both wires is same because both wires are made of same copper material.
30
So, Resistance of first wire = r 1= p(
4π
)Ω

10
and Resistance of second wire = r 2= p( π )Ω
30
p
r1
So, the ratio of both resistance = r2
=
4π

10
=
30

40
=
3

4
p
π

So the ratio of first and second wire = 3 : 4

ii. The resistance of the electric bulb is 2 ohms
Explanation:
Given that,
Current flowing in the circuit, I = 500 mA = 500 × 10-3 A = 0.5 A
The resistance of the conductor, Rc = 10 π
The equivalent resistance of the series combination of an electric lamp and a conductor is given by:
R = Re + Rl
Voltage of the circuit, V = 6 V
Let R is the effective resistance. It is given by:
V
R=
I
6
R=
0.5

R = 12 V
In a series combination, the current flowing through each circuit is the same. The resistance of the lamp is given by:
Rl = R - Re
Rl = 12 - 10
Rl = 2 Ω
So, the resistance of the electric bulb is 2 ohms. Hence, this is the required solution.
59. a. Power is defined as rate of doing work/ rate at which energy is consumed/ rate at which electric energy is dissipated in an
electric circuit. In simpler terms, it measures how quickly energy is converted from one form to another or how quickly work
is done.
S.I unit of Power is watt
i. P = VI
= 5 volt × 500 mA

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 = 5 volt × 500

1000
A
= 2.5 watt
2

ii. P = V

5 volt ×5 volt
or R =
2.5 watt
R= 250

25
= 10Ω
iii. Energy Consumed = Power × Time
= 2.5 W × 2.5 h
= 6.25 Wh
60. a. In this case : Length L becomes l/2 then Area of cross section A become 2A
When the resistance wire is doubled on itself, it essentially creates two equal segments of wire connected in parallel. In this
configuration, the total resistance of the wire decreases.
L
R = ρ = 6Ω
A
L/2
′
R = ρ
2A
′ 1 L
R = ρ
4 A
′ 1
R = × 6
4
′ 3
R = Ω
2

b. No of resistors = 3
Each resistor has resistance = 2 Ω In this arrangement where total resistance is 3 Ω
Resistor A of 2 Ω is connected in series with a parallel combination of the resistors B (2 Ω ) and C (2 Ω ).
Arrangement :

Total resistance R = 2Ω + 2Ω×2Ω

2Ω+2Ω

R = 2Ω + 4Ω

4Ω

R = 2Ω + 1Ω
R = 3Ω
61. Modified circuit is as shown. Since 5Ω , 8Ω and 12Ω are in series, therefore the total resistance in series.
Rs = R1 + R2 + R3 = 5 + 8 + 12 = 25Ω

V 6
Current through circuit I = R
=
25
= 0.24A

∴ Reading of ammeter = 0.24A

= I.R. = 0.24 × 12
= 2.88Ω
62. A. R = R1 + R2
R = 1Ω + 2Ω
R = 3Ω
V = IR
I= V

6 V
I= 3Ω
= 2 Ampere or 2 A

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 P = I2R
=2×2×2
=8W
V2
B. P = R

P= 4×4

P=8W
Hence, the power used in 2 Ω resistor in each of the following circuits is same.
63. i. V = IR, where V is potential difference, I is current and R is constant for a given resistor as its resistance. Hence, current I is
directly proportional to V.
Labelled circuit diagram to verify this relationship is as follows;

ii. An ammeter has a low resistance, which makes it easy for all the current to flow through the circuit without energy loss from
heat losses. This means that it does not alter the flow of electric current.
iii. We know that V = IR or = 1

R
= slope of graph given.
1

The resistance increases as the slope decreases. Graph B clearly has a greater slope than graph A. As a result, B has less
resistance than A. As far as we are aware, combined series resistance is consistently greater than combined parallel resistance.
As a result, A and B are series and parallel combinations, respectively.
64. a. All electrical appliances in common domestic electric circuits are connected in parallel because:
i. All appliances will be operated by different switches.
ii. If one appliance stop working it will not affect another.
iii. All appliances will get the same voltage hence it will be easy to run all the appliances simultaneously.
b. The two different circuits in a house lighting circuit with 5 A fuse and a power circuit with 15 A fuse. The lighting circuit is
for running low power rating devices such as electric bulbs, fans, radio, etc. The power circuit is for running high power rating
devices like Electric iron.
c. Electric short circuit - if the insulation of the live wire and the neutral wire gets stone then the two wires touch each other.
Live wire and neutral wire come in contact with each other. the current flowing through the wires become very large and hits
the wire to a high temperature and fire may start.
65. a. If the potential difference across the two ends of a conductor is 1 volt and the current through it is 1 ampere, then the resitance
1 volt
of the conductor is 1 Ω / 1Ω =
1 ampere
b. Electric Power is defined as the rate at which electric energy is consumed /dissipated.
2
v
P =
R

c. To determine how many 132 ohm resistors in parallel are required to carry 5 A on a 220 V line, we can use Ohm's Law to find
the total resistance needed for the circuit:
V = I×R
V
R= I

220 volt
=
5 ampere
R = 44 Ω
132Ω
Net Resistance = n
, where n is the number of bulbs in parallel.
R = 44 = 132

n= 132

44
= 3 resistors
66. i. According to this rule, stretch the thumb, forefinger and middle finger of left hand such that they are mutually perpendicular.
If the middle finger points in the direction of the magnetic field, the fore finger points in the direction of the flow of current,
then the thumb points in the direction of motion.
ii. Three characteristics of electric current use in our home are as follows:
a. the current supplied in our homes is alternating current.

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 b. the current supplied in our homes is at 220 V.
c. the neutral wire and the live wire carry the current in our homes.
iii. Fuse is a safety device used in a circuit to prevent damage due to overloading/short-circuiting. It protects the circuit by
stopping the flow of any unduly high electric current. If current larger than the specified value flows through the circuit, due to
Joule’s heating effect the fuse wire melts and breaks the circuit.
iv. When live wire touches the metallic appliance then electric current flows through the casing to the earth instead of the human
body and thus we prevent ourselves from getting shocked. It is necessary to earth metallic casing of the appliance because it
saves electrical appliance from burning and electric shock.
67. A solenoid is a long circular coil containing a large number of close turns of insulated copper wire. When a electric current is
passed through the solenoid, it produces magnetic field around it as shown in fig. Magnetic field produced by a current carrying
solenoid is similar to the magnetic field produced by a bar magnet. As is clear from the figure, the lines enter from the left side
and leave out from the right side. If we look from left side, the current appears to be passing in the coil in clockwise direction and
hence it acts as a south pole according to clock rule. If the coil is viewed from right side, the current appears to be in anticlockwise
direction. Hence, left-hand side face behaves as if this were a north pole. If the coil is left free, it will point South and North.
Since the current in the turns of the solenoid flows in the same direction, the magnetic field produced by each turn of the solenoid
adds up, giving a very strong resultant field inside the solenoid. Hence, a solenoid may be used in making electromagnets.

Strength of the magnetic field produced depends upon the following three factors :
Number of turns: Large the number of turns, stronger will be the magnetic field produced.
Strength of the current in the solenoid: Larger the current, stronger will be the magnetic field produced.
Nature of core of solenoid: The strength of the field depends upon the core on which the coil is wound. For air core, field is very
mild whereas for iron-core, the field is very strong.
68. Safety Fuse or fuse : Usually the wire chosen for electric circuit are such that these allow a certain maximum current to pass
through them without excessive heating of the circuits. However, incidentally there is a short-circuiting or over-loading, the
current exceeds this maximum permissible value. The wires may get over-heated and catch a fire. Sparking at the points of short-
circuit may also cause fire. Many precautions and safety measures are taken to protect the circuits against damage due to over-
heating. All wirers used in electric circuits are coated with layer of insulating materials. In addition these are coated with rubber or
plastic layer.
The most important safety device used these days is safety fuse or fuse. Fuse is a piece of wire of a material with a low melting
point. Good fuse wire is always made of pure tin but cheaper variety is made of alloy of tin and copper or tin and lead (63% tin
and 37% lead)
Fuse is always connected to the live wire. When current of value more than maximum permissible is passed through the circuit,
the fuse wire melts due to excessive heating. This way the circuit is broken to ensure safety of the circuit. It is due to this fact that
the fuse is usually called safety fuse. The thickness, length and material of the fuse wire depends upon the maximum current
permitted through the circuit. For proper protection, a fuse of proper value is must. Fuse of improper rating is a curse instead of
being a safety device.
69. a. Two magnetic field lines can never cross each other because it would mean that at the point of intersection the compass needle
would point towards two directions simultaneously which is not possible.
b. The magnetic field lines inside a current carrying solenoid are in the form of parallel straight lines. This indicates that the
magnetic field is the same (uniform) at all points inside the solenoid.
c. Fleming’s left hand rule: Stretch the thumb, forefinger and middle finger of your left hand such that they are mutually
perpendicular. If the first finger points in the direction of the magnetic field, the second finger in the direction of current, then
the thumb will point in the direction of motion.
d. Any two factors:
i. Strength of electromagnet
ii. Large number of coil/turns of the conducting wire
iii. A soft iron core on which the coil is wound.

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70. Very strong electromagnets can be produced by binding an insulated copper wire on a soft iron core. Electromagnets work on the
magnetic effects of electric current. We have studied in previous article that when a current is passed through a long solenoid, a
magnetic field is produced. Now if a soft iron core is placed inside the solenoid, the strength of the magnetic field becomes very
large. The reason for a large increase in magnetic field is due to the fact that iron gets magnetized by induction. The combination
of soft iron core and a current carrying insulated copper wire wound over it is called an electromagnet.

A simple electromagnet is as shown in fig. To make an electromagnet, a soft iron core is taken and insulated copper wire is wound
over it. The two ends of it are connected to a battery and a key. The electromagnet gets demagnetized when key is removed.

Strength of electromagnet depends upon

i. Number of turns. Strength of electromagnets is directly proportional to the number of turns.
ii. Current flowing. Larger the current flowing through the wire, stronger is the electromagnet. Strong electromagnets are used in
cranes to lift the cars not obeying traffic rules.
iii. Length of the air gap. Lesser the length of air gap between poles, stronger is the electromagnet.
Air gap between the poles of a U-shaped electromagnet is small, hence it is very strong.
71. a. The magnetic field lines produced around a current-carrying straight conductor passing through cardboard is shown below.

A right-hand thumb rule is applied to find the direction of these field lines. Imagine that you are holding a current-carrying
straight conductor in your right hand such that the thumb points towards the direction of the current. Then your fingers will
wrap around the conductor in the direction of the field lines of the magnetic field.
b. When we move away from the straight wire, the deflection of the needle decreases which implies the strength of the magnetic
field decreases. The reason is that the concentric circles representing the magnetic field around a current-carrying straight wire
become larger and longer as the distance increases.
72. The activity to demonstrate that a current-carrying conductor experiences a force perpendicular to its length and the external
magnetic field can be explained as follows:
Activity: To show the effect of magnetic field on a current-carrying conductor
Materials Required: For this, we need to take a small aluminum rod, a horseshoe magnet, battery, plug key, wires, and a stand.
i. Suspend an aluminum rod horizontally from the stand and two wires at the ends of it are tied. The wires are connected to a
Rheostat, battery and a key so that a circuit is completed,
ii. Place a horseshoe magnet in such a manner that the aluminum rod is between the poles of a magnet.
Assume that the above the aluminum rod is South pole of the magnet and below, the north pole of the magnet. Insert the plug key
and current is supplied to the rod.
Observation: the aluminum rod is deflected towards the left direction
On changing the direction of the current, the rod is deflection in the right direction.
Hence, it demonstrates that a current-carrying conductor experiences a force perpendicular to its length and the external magnetic
field

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 The direction of the magnetic field can find out with the help of Fleming's left-hand rule. Let current is moving in an
anticlockwise direction, then the direction of the magnetic field will be in clockwise direction i.e. at the top of the loop whereas
vice-versa in case of the clockwise direction of the current.
73. a. If the current is increased in the conductor the deflection of the compass needle increases. This is because the strength of
magnetic field varies directly as the magnitude of the electric current, or the current passing through the wire. It indicates the
presence of magnetic field.
b. Right hand thumb rule states that, “If you imagine holding a current carrying wire in your right-hand with your thumb
pointing towards the direction of electric current flow then the direction in which your fingers curl, gives the direction of lines
of force of the magnetic field”.
74. a. Fleming's Left Hand Rule states that if we arrange our thumb, forefinger and middle finger of the left-hand perpendicular to
each other, then the thumb points towards the direction of the force experienced by the conductor, the forefinger points
towards the direction of the magnetic field and the middle finger points.
b. Three characteristic features of the electric current used in our homes are:
Appliances to be connected in parallel.
Each of the appliances has a separate switch to ON/Off the flow of current through it.
The potential difference between a live wire and the neutral wire is about
c. A fuse is an electrical safety device that protects circuits from excessive current by melting and breaking the circuit when too
much current flows through it.
d. It's necessary to earth metallic electric appliances to prevent electric shock and other hazards, such as fire and damage to
equipment.
75. Solenoid is a coil of a number of turns of insulated copper wire closely wrapped in shape of a cylinder. Magnetic field around a
current carrying solenoid is shown in fig.

These appear to be similar to that of a bar magnet shown in below fig.

One end [right end] of solenoid behaves like north pole and the other and [left end] behaves like south pole. Magnetic field lines
inside the solenoid are in the form of parallel straight lines. This means that the field is the same at all points inside the solenoid.
When a soft iron rod is placed inside the solenoid, it behaves like a electromagnet.

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