# AIIMS Practice Paper Physics – 1

AIIMS Practice Paper Physics

1. Four metallic plates, each with a surface area of one side A are placed at a distance d apart from each other. The two inner plates are connected to point B and the other two plates to another point A as shown in figure. Then the capacitance of the system is

(a) ε0 A/d

(b) 2εA/d

(c) 3ε0 A/d

(d) 4εA/d

Solution

The arrangement can be shown as

So

2. Three masses 2, 3 and 4 kg are located at the corners of an equilateral triangle of side 2m. The centre of mass of the system is

(a) (0.77, 1.11)

(b) (1.11, 0.77)

(c) (1.77, 0.11)

(d) (0.11, 1.77)

The equilateral triangle lies in XY plane has a mass 2 kg at the origin. Let (x, y) be the co-ordinates  of centre of mass.

x₁ = 0,                 x₂ = 2 m,      x₃ = 1 m

m₁ = 2 kg,          m₂ = 3 kg     m₃ = 4 kg

y₁ = 0,              y₂ = 0,         y₃ = √3

m₁ = 2 kg,        m₂ = 3 kg    m₃ = 4 kg

3. Two cells of e.m.f.s 1.5 V and 2.0 V and internal resistances 2 ohm and 1 ohm respectively have their negative terminals joined by a wire of 6 ohm and positive terminals by a wire of 7 ohm resistance. A third resistance wire of 8 ohm connects middle points of these wires. Find the potential difference at the end of the third wire.

(a) 2.25 V

(b) 1.36 V

(c) 1.26 V

(d) 2.72 V

Solution

Using Kirchhoff’s law for loop I,

1.5 = 7 I1 + 8 (I1 + I2)

= 15 I1 + 8 I2     ………(1)

And for loop II,

2 = 6 I2 + 8(I1 + I2)

= 14 I2 + 8 I1

Or          1 = 7 I2 + 4 I1   ………(2)

Solving equations (1) and (2)

Voltage across 8Ω

= (I1 + I2) 8

4. Six charges are placed at the corner of a regular  hexagon as shown. If an electron is placed at its centre O, then the force on it will be :

(a) Zero

(b) Along OF

(c) Along OC

(d) None of these

Solution

Force due to (-2q & -2q) and (3q, 3q) pair cancel each other.

So F(-q)(-e) and F2q(-e)  are along OD

5. A particle of mass m and  charge q is thrown in a region where uniform gravitational  field and electric field are parallel. The path of particle

(a) may be a circle

(b) may be a parabola

(c) may be a hyperbola

(d) All of these

Solution

When the charge is projected perpendicular to the fields, the path is a parabola.

6. A body slides down a rough inclined plane. The co-efficient of friction between the body and plane is 0.5. The ratio of the net force required for the body to slide down and the normal reaction on the body is 1:2. Then the angle of the inclined plane is

(a) 15°

(b) 30°

(c) 45°

(d) 60°

Solution

7. Objects each of mass m kg are placed at 1m, 3m, 9m, 27m …… from point O. Find the ratio of gravitational potential to the field due to distribution of masses at O.

(a) 2:3

(b) 3:4

(c) 4:3

(d) 3:2

Solution

8. A liquid at 60℃ cools to 50℃ in 5 minutes and to 45℃ in 10 minutes. Find the temperature of the surrounding.

(a) 20℃

(b) 25℃

(c) 40℃

(d) 30℃

Solution

Case-1

Change in temperature in 1st case

dT = 60 – 50 = 10℃

time interval, dt = 5 min

Average temperature

Let To be the temperature of the surrounding

Case-II

Change in temperature dT = 50° − 45° = 5℃

dt = 10 – 5 = 5 min

Average temperature

Dividing (i) and (ii) and solving

T0 = 40℃

9. Three blocks of masses m, 3m and 5m are connected by massless strings and pulled by a force F on a frictionless surface as shown in the figure below. The tension P in the first string is 16 N

If the point of application of F is changed as given below,

the values of P’ and Q’ will be

(a) 16 N, 10N

(b) 10 N, 16 N

(c) 2 N, 8 N

(d) 10 N, 6 N

Solution

10. A car travels according to the acceleration time graph as given below.

Which of the following velocity – time graph best describes the acceleration – time graph?

(a)

(b)

(c)

(d)

Solution

As magnitude of acceleration is higher than that of deceleration,from the velocity time graph, the gradient of the acceleration portion must be steeper than deceleration portion. In between two portions, velocity is constant.

11. A body of mass 10 kg is suspended by two strings making angles 30° and 60° with the horizontal. Find the tension T1 in the string.

(a)

(b)

(c)

(d)

Solution

Equating horizontal components to zero

T1 cos 30 − T2 cos 60 = 0

⇒ T1 cos 30 = T2 cos 60

⇒ T1 (√3/2) = T2 (1/2)

⇒ T1 = T2/√3

Equating vertical components to zero

T1 sin 30 + T2 sin 60 – mg = 0

12. A ball of mass 2 kg hangs in equilibrium from two strings OA and OB as shown in figure. What are the tensions in string OA and OB ? use g = 10 m/s2

(a) 10√3, 10

(b) 10, 10√3

(c) 20, 20√3

(d) 20√3, 20

Solution

13. The two slits at a distance of 1 mm are illuminated by the light of wavelength 6.5 × 10−7 m. The interference fringes are observed on a screen  placed at a distance of 1 m. The distance between third dark fringe and fifth bright fringe will be

(a) 0.65 cm

(b) 4.8 mm

(c) 1.63 mm

(d) 3.25 cm

Solution

The fringe width for nth bright fringe is

Where D → distance between screen and source

d → distance between coherent sources

The fringe width for nth dark fringe is

14. A particle positioned at x = 0 at time t = 0 starts moving along the positive x-directions with a velocity V which varies as V = β∛x. The displacement of the particles varies with time as

(a) t1/2

(b) t

(c) t3/4

(d) t3/2

Solution

15. A ball is thrown at an angle θ and another ball  with an angle 90 – θ with the horizontal direction from the same point with velocity of 19.6 m/s. Find the height of 2nd ball, if 2nd ball is at a height of 12 m higher than the1st ball.

(a) 17 m

(b) 14.7 m

(c) 15.8 m

(d) 16.2 m

Solution

Let θ be the angle of projection of 1st ball

and 90 – θ be the angle of projection of 2nd ball

Let h be the maximum height attained

For 2nd ball, the angle of projection is 90 – θ

Adding (1) and (2)

2h + 12 = 19.6

2h = 7.6

h = 3.8 m

h + 12 = 15.8 m

16. A particle is projected horizontally at a speed 10 m/s from the top of a plane inclined at an angle of 45° with the horizontal. How far from the  point of projection will the particle strike the plane?

(a) 20.2 m

(b) 14.14 m

(c) 28.28 m

(d) 32.45 m

Solution

The body is projected from O and reaches at P.

17. A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

(a) 1/2(mgl)

(b) (√3/2) mgl

(c) mgl

(d) 2mgl

Solution

vertical fallen height, h = l cos 30 = (√3/2)l

Loss of potential energy = mgh = gain in KE

= (√3/2) mgl

18. For particles each of mass m, 2m, 3m and 4m are located at the four corners of a parallelogram with each side equal to a and one of the angles between two adjacent sides is 30°. The parallelogram lies in the Y-Z plane with 2m at the origin and 3m on the Y-axis. The centre of mass of the arrangement will be located at :

(a) 1.122a, 0.25a

(b) 1.143a, 0.25a

(c) 1.133a, 0.15a

(d) 1.133a, 0.25a

Solution

m1 = 2m       m2 = m          m3 = 4m       m4 = 3m

Now position vector

Hence the centre of mass location is (1.133a, 0.25a)

19. Which of the following is not true regarding angular momentum of a particle moving on a straight line with a uniform velocity?

(a) always zero

(b) is zero at a point on the straight line

(c) is not zero about a point away from the straight line

(d) about any given point remains constant

Solution

Angular momentum becomes zero if the point (reference) is on the line of motion of the particle, and the line of action of momentum passes through the point of reference. If the reference point does not lie on the line, both momentum & lever arm of momentum are non zero and thus angular momentum is non zero.

20. An artificial satellite is moving in a circular orbit around the earth with a speed equal to one third the magnitude of escape velocity of earth. The height of the satellite above the surface of earth is

(a) 6400 km

(b) 22400 km

(c) 12800 km

(d) 3200  km

Solution

9 gR2 = 2gR2 + 2gRh

7gR2 = 2gRh

h = (7/2) R

= 7/2 × 6400 = 22400 km

21. Infinite number of masses each of 2 kg are  placed along the X-axis at x = ±1m, ±2m, ±4m ……… .The magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin (x = 0) is

(a) G2

(b) 2G

(c) 4G

(d) 8G

Solution

22. Two bodies of masses 2kg and 3 kg are connected by a steel wire of cross section 3 cm2 moves over a smooth pulley. What will be the longitudinal strain in the wire if Y = 2 × 1011 N/m2?

(a) 4× 10−7

(b) 4 × 10−3

(c) 4× 10−5

(d) 6 × 10−6

Solution

From the free body diagram

30 – T = 3a     ——-(1)

and T – 20 = 2a ——-(2)

Solving these two we have T = 24

23. Which of the following graph represents the variation of surface tension with temperature over small temperature ranges for water?

(a)

(b)

(c)

(d)

Solution

Surface tension decreases with increase in temperature.

24. Which of the following substances has the highest elasticity?

(a) Steel

(b) Copper

(c) rubber

Solution

Steel requires a very high stress for the same strain compared to other given materials. Hence steel has the highest modulus of elasticity & thus  highest elasticity.

25. Water flows in a river. If the velocity of a layer at a distance 15 cm from the bottom is 25 cm/s, the velocity of layer at a height of 45 cm from the bottom is

(a) 8.33 cm/s

(b) 25 cm/s

(c) 50 cm/s

(d) 75 cm/s

Solution

We  have

Velocity of layer of bottom is zero. Hence

As F is constant so x ∝ v

26. A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio of  of the mixture is

(a) 1.4

(b) 1.54

(c) 1.59

(d) 1.62

Solution

27. An ideal gas undergoes a cycle of processes as shown in the P-V diagram below. Which statement correctly describe the situation?

(a) The internal energy of the gas increases over one complete cycle

(b) The gas gives out more heat than it absorbs over the whole cycle

(c) Over the entire cycle, the gas absorbs heat and does not work on its environment

(d) The two curved portions of the graph represent adiabatic process

Solution

Out of the 4 processes there is no work done where the arrows are vertical. The arrow to the right represents expansion which results in workdone by the gas where arrow on the left shows workdone on the gas, which encloses a larger area and hence net workdone is positive.

∆U = Q + W

As ∆U = 0,  Q = −W

Q, −ve ⇒ gas gives out heat.

28. Find the change in entropy from the following process. 80 gm of ice at 0℃ melts when dropped in a container of water at 40℃.

(a) 3.008 cal/k

(b) 4.04 cal/k

(c) 2.792 cal/k

(d) 4.132 cal/k

Solution

29. For an ideal gas if ∆V is the increased in volume and ∆T is the increase in temp. at constant pressure P, if , then Z – T graph is

(a)

(b)

(c)

(d)

Solution

For an ideal gas PV = RT

At constant P, P ∆ V = R ∆ T

which is obvious from graph c.

30. The displacement x of a particle is given by x(x – 6) = 1 – 10 cos ωt

(a) The motion of the particle is oscillatory but not SHM

(b) The particle executes SHM

(c) The motion of the particle is neither oscillatory nor harmonic

(d) The particle is not acted upon by a fore when it is at x = 6

Solution

x(x – 6) = 1 – 10 cosωt

⇒ x2 – 6x + 9 = 1 – 10 cosωt

⇒(x – 3)2 = 10(1 – cosωt)

31. In a series resonant L-C-R circuit, the voltage across R is 120 V and R is 960 Ω. The capacitance of capacitor is 2.4 × 10−6 F and angular frequency of AC is 180 rad s−1. The potential difference across the inductance coil is

(a) 275 V

(b) 289 V

(c) 310 V

(d) 302 V

Solution

32. Photons of energy 8 eV can emit photoelectrons from the surface of metal with maximum kinetic energy of 6 eV. The stopping potential is

(a) 8 V

(b) 6 V

(c) 4 V

(d) 10 V

Solution

Stopping potential in volts = kinetic energy of the emitted photoelectrons in eV.

Hence stopping potential is = 6V.

33. What is the maximum height attained by a body projected with velocity equal to three fourth of the escape velocity from the surface of the earth?

(a)

(b)

(c)

(d)

Solution

34. Pressure of CHin container is given  by The mass of the gas in the container is given by

(a) 8 g

(b) 16 g

(c) 32 g

(d) 48 g

Solution

Vander Waal’s gas equation of n mole of real gas is

Mass of gas = n × M

= 2 × 16

= 32 g

35. Angle of dip at a place is 60°. If BH is the horizontal component of earth’s magnetic field, the total field intensity is

(a)

(b) 2 BH

(c)

(d)

Solution

BH = B cos δ

Where B is total field intensity.

BH is horizontal component’s earth’s magnetic field

δ = 60°

36. An object executing S.H.M. attached to a spring has displacement meter. Find the time at which the maximum speed 1st occurs.

(a) 0.5 s

(b) 1.5 s

(c) 2s

(d) 2.5 s

Solution

37. 7 mole of an ideal tri atomic gas (γ = 1.33) are heated at constant pressure. If 266 J of heat energy is supplied to the gas find the work-done by the gas.

(a) 66 J

(b) 34 J

(c) 432 J

(d) 102 J

Solution

From 1st law of thermodynamics

∆W  = ∆Q – ∆U

= 266 – 200

= 66 J

38. By the introduction of glass plate (μ = 1.5), the fringes of young’s double slit experiment get displaced by distance x. When this plate is replaced by another plate of same thickness, the shifting of fringes becomes . The refractive index of the 2nd plate is

(a) 1.58

(b) 1.45

(c) 1.75

(d) 1.25

Solution

39. The ratio of intensities between two coherent sound sources is 9 : 4. The difference of loudness in decibel between maximum and minimum intensity when they interfere in space is

(a) 15 log 5

(b) 20 log 5

(c) 10 log 5

(d) 5 log 5

Solution

40. An engine of power 160 KW draws a train of mass 18 × 104 kg up in an inclination of 1 in 40. The frictional resistance is 5 kg wt/1000 kg. Find the maximum speed.

(a) 2.962 m/s

(b) 1.48 m/s

(c) 3.14 m/s

(d) 4.2 m/s

Solution

Power, P = 160 KW

= 160 × 103 W

mass, m = 18 × 104 kg

sin θ = 1/40

As both force & velocity are in same direction, θ = 0, cos θ = 1

Power, P = F’V

41. A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from the plates and then reinserted. The net work done by the system in this process is:

(a) Zero

(b)

(c)

(d)

Solution

Potential energy of a charged capacitor

With insertion of dielectric, the energy = q2/2KC, once is taken out, the energy increases to old value.

Hence the work done = zero.

42. A 25 W and 100W bulb are joined in series and connected to the mains. Which bulb will glow brighter?

(a) 25 W bulb

(b) 100 W bulb

(c) Both bulbs will glow brighter

(d) None will glow brighter

Solution

R = V2/P

At const. V, R ∝ 1/P

Resistance of 25 W bulb is greater than that of 100 W. In series, current flows the same. Heat produced H = I2Rt

⇒ H ∝ R

So heat of 25 W bulb is greater than 100 W bulb.

Hence 25 W will glow brighter.

43. What is the de-Broglie wavelength of an electron in the 2nd orbit of a hydrogen atom?

(a)

(b) πr

(c) 2πr

(d) πr2

Solution

From Bohr’s theory

For n = 2,      2πr = 2λ

⇒ λ = πr

44. A coil has a self inductance of 10 mH. What is the maximum magnitude of induced emf in the inductor when a current I = 0.1 sin 200 t ampere is sent through it?

(a) 0.2 V

(b) 0.3 V

(c) 0.5 V

(d) 0.4 V

Solution

L = 10 mH = 102 M

I = 0.1 sin 200 t

45. A 100 PF capacitor is connected to a 230 V, 50 Hz a.c. source. What is the r.m.s. value of the conduction current?

(a) 4.8 μA

(b) 7.2 μA

(c) 8.4 μA

(d) 6.7 μA

Solution

46. The electrostatic potential inside a charged spherical ball is given by Φ = ar2 + b where r is distance from the centre a, b are constants. Then the charge density inside the ball is

(a) −6aε0r

(b) −24πaε0

(c) −6ε0a

(d) −24πaε0r

Solution

47. Pressure of a gas decreases by 30% at a constant temperature. Find the percentage change in volume.

(a)

(b)

(c)

(d)

Solution

From Boyle’s law, P1V1 = P2V2

As pressure gets decreased by 30%

48. A force of 45 N is just able to move a block of wood weighing 9 kg on a rough horizontal surface. Find its co-efficient of friction. (use g = 10 m/s2)

(a) 1

(b) 0.5

(c) 0.4

(d) 0.3

Solution

49. A small sphere of volume V falls in a viscous fluid and attains a terminal velocity vt.What will be the terminal velocity of a sphere of volume 27 V, of same material falling in the same fluid?

(a) (3/2)vt

(b) 3vt

(c) 9vt

(d) 6vt

Solution

50. The extension of a wire by application of a load is 3 mm. The extension in a wire of same material and length but one third of the radius by applying same load is

(a) 36 mm

(b) 27 mm

(c) 9 mm

(d) 18 mm

Solution

⇒ ∆ℓ2 = 9 ∆ℓ1

= 9 × 3

= 27 mm

51. Statement I: Propagation of light through an optical fibre is due to total internal reflection taking place at the core – clad interface.

Statement II: Refractive index of the material of the core of the optical fibre is greater than that of air.

(a) Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I.

(b) Statement I is true; Statement II is false

(c) Statement I is false; Statement II is true

(d) Statement I is true; Statement II is true ; Statement II is the correct explanation for statement I.

Solution

self explanatory

52. Statement I: Two heater coils made of same material are connected in parallel across the mains. The length and diameter of the first wire is double the second wire. The first wire will produce less heat.

Statement II: Length is immaterial for electric fuse of given current strength say I amp.

(a) Statement I is true; Statement II true; Statement II is not a correct explanation for Statement I.

(b) Statement I is true; Statement II is false.

(c) Statement I is false; Statement II is true.

(d) Statement I is true; Statement II is true ; Statement II is the correct explanation for statement I.

Solution

Resistance of first wire,

Resistance of second wire,

Heat produced

As the second coil has resistance equal to half of the first coil, so heat produced in first coil is half of that second coil.

Statement-II

Heat produced in a fuse wire in time t is

If m is mass, S = specific heat, d= density and ∆θ = rise in temperature, then heat produced

Q = mS∆θ

= (Vol × density) S ∆θ

= AℓdS∆θ …………..(2)

Equating (1) and (2) we get

Which is independent of length .Hence length is immaterial for electric fuse.

53. Assertion: When a bottle of cold carbonated drink is opened, a slight fog forms around the opening.

Reason: Adiabatic expansion of gas causes lowering of temperature and condensation of water vapour.

(a) If both assertion and reason are true and reason is a correct explanation of the assertion.

(b) If both assertion and reason are true and reason is not a correct explanation of the assertion.

(c) If assertion is true but reason is false.

(d) If assertion is false but reason is true.

Solution

Cold drinks are produced by injecting carbon dioxide into the drink at several atmospheres. Carbon dioxide dissolves readily even at normal atmospheric pressure. When the pressure is released (that is bottle is opened) carbon dioxide comes out of solution forming numerous bubbles and releasing the carbon dioxide back into the atmosphere. Also, in this process no heat is lost or gained, hence it is an adiabatic expansion.

54. Assertion: In a movie, ordinarily 24 frames are projected per second from one end to the other of the complete film.

Reason: The image formed on retina of eye is sustained upto 1/10 s after the removal of stimulus.

(a) If both the assertion and reason are true and reason is the correct explanation of the assertion.

(b) If both the assertion and reason are true but the reason is not correct explanation of the assertion.

(c) If the assertion is true but the reason is false.

(d) If the assertion is false but the reason is true.

Solution

When the movie is screened, the frames are displayed on after the other at the rate of 24 frames per second. The image of an object is formed on the retina. From when it is sent to the brain. If the object moves, the brain continues to show the old image for about one sixteenth of a second.

55. Assertion : A tube light emits white light.

Reason : Emission of light in a tube takes place at a very high temperature.

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.

(c) If assertion is true but the reason is false.

(d) If both assertion and reason are false.

Solution

Tube light is basically a fluorescent lamp that uses electricity to excite mercury vapour in argon or neon gas, resulting is plasma that produces short-wave ultraviolet light. This light then causes a phosphor to fluorescence producing visible light. The emission takes place at low pressure and not at height temperature, since thermionic emission takes place.

56. Assertion : A red object appears dark in the yellow light.

Reason : The red colour is scattered less.

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.

(c) If assertion is true but the reason is false.

(d) If both assertion and reason are false.

Solution

A red object looks red because it reflects only red color and absorbs all other colors present in the white light. Hence, when red object is seen through yellow light then it absorbs yellow color falling on its and appears dark. This assertion has got nothing to do with scattering. Also from Rayleigh’s criteria of scattering.

57. Assertion : By roughening the surface of a glass sheet its transparency can be reduced.

Reason : Glass sheet with rough surface absorbs more light.

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.

(c) If assertion is true but the reason is false.

(d) If both assertion and reason are false.

Solution

When glass surface is made rough then the light falling on it is scattered in different directions reducing its transparency. Also since rough surfaces scatter light more, it is absorbed less.

58. STATEMENT-1 : Total energy of revolving electron in any stationary orbit is negative.

STATEMENT-2 : Energy is a scalar quantity. It can have positive or negative values.

(a) Statement -1 is True , Statement – 2 is True ; Statement – 2 is a CORRECT explanation for Statement – 1

(b) Statement-1 is True, Statement-2 is True ; Statement-2 is a NOT CORRECT explanation for Statement-1

(c) Statement – 1 is True, Statement – 2 is False

(d) Statement-1 is False, Statement-2 is True

Solution

Self Explanatory

59. STATEMENT-1:TheBohr’smodel cannot differentiate between the spectra of hydrogen and deuterium.

STATEMENT-2 : The Bohr’s model considers the nucleus as infinitely massive in comparison to the orbiting electrons.

(a) Statement-1 isTrue, Statement-2 is True; Statement-2 is a CORRECT explanation for Statement-1

(b) Statement-1 isTrue,Statement-2 is True; Statement-2 is a NOT CORRECT explanation for Statement-1

(c) Statement – 1 is True, Statement – 2 is False

(d) Statement-1 is False, Statement-2 is True

Solution

Self Explanatory

60. STATEMENT.1 : A ball is projected at an angle θ with horizontal from a platform with constant velocity relative to platform. The platform is moving in horizontal direction as shown. The time of flight of ball is independent of the horizontal velocity of platform. (Neglect the air friction)

STATEMENT.2 : The time of flight of projectile depends only on vertical component of initial velocity and acceleration in vertical direction.

(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a CORRECT explanation for Statement-1

(b) Statement-1 is True,Statement-2 is True ;Statement-2 is a NOT CORRECT explanation for Statement-1

(c) Statement – 1 is True, Statement – 2 is False

(d) Statement-1 is False, Statement-2 is True