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Physics NEET Questions — Free MCQ Practice

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A uniform force of (3i + j) N acts on a particle of mass 2 kg. Hence the particle is displaced from position (2i+k) m to position (4i+3j-k) m. The work done by the force on the particle is-

A.

9J

B.

6J

C.

13J

D.

15J

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Explanation

(a) Given, force F=3i+j

∴ r₁=(2i+k)m and r₂=(4i+3j-k)m

∴ s=r₂-r₁=(4i+3j-k)-(2i+k)=(2i+3j-2k)m

W=F $.$ s=(3i+j) $.$ (2i+3j-2k)

=3x2+3+0

=6+3=9J

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A body of mass m is taken from the earth’s surface to the height equal to twice the radius (R) of the earth. The change is potential energy of body will be

A.

mg2R

B.

2/3mgR

C.

3mgR

D.

1/3mgR

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Explanation

(b) Change in potential energy:

ΔU = $= - \frac{G M m}{R + 2 r} - (- \frac{G M m}{R}) = \frac{2 G M m}{3 R} = \frac{2}{3} m g R$

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The potential energy of a particle in a force field is $U = \frac{A}{r^{2}} - \frac{B}{r},$ where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

A.

B/2A

B.

2A/B

C.

A/B

D.

B/A

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Explanation

Given, the potential energy of a particle in a force field $U = \frac{A}{r^{2}} - \frac{B}{r^{1}}$

For stable equilibrium, F= $- \frac{d U}{d r} = 0$

0=- $\frac{2 A}{r^{3}} + \frac{B}{r^{2}}$

or $\frac{2 A}{r} = B$

The distance of particle from the centre of the field

$r = \frac{2 A}{B}$

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The potential energy of a system increases if work is done

A.

by the system against a conservative force

B.

by the system against a nonconservative force

C.

upon the system by a conservative force

D.

upon the system by a nonconservative force

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Explanation

The potential energy of a system increases if work is done by the system against a conservative force.

$∆ U = - W_{conservative force} = W_{system}$

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A ball moving with velocity $2 m s^{- 1}$ collides head on with another stationery ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in $m s^{- 1}$ ) after collision will be

A.

0,1

B.

1,1

C.

1,0.5

D.

0,2

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Explanation

If two bodies collide head on with coefficient of restitution

$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$ ...(i)

From the law of conservation of linear momentum

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{1} v_{2}$

$\Rightarrow v_{1} = [\frac{m_{1} - e m_{2}}{m_{1} + m_{2}}] u_{1} + [\frac{(1 + e) m_{2}}{m_{1} + m_{2}}] u_{2}$

Substituting $u_{1} = 2 m s^{- 1}, u_{2} = 0, m_{1} = m a n d m_{2} = 2 m, e = 0.5$

we get $v_{1} = [\frac{m - m}{m + 2 m}] \times 2$

$\Rightarrow v_{1} = 0$

Similarly,

$v_{2} = [\frac{(1 + e) m_{1}}{m_{1} + m_{2}}] u_{1} + [\frac{m_{2} - e m_{1}}{m_{1} + m_{2}}] u_{2}$

$= [\frac{1.5 \times m}{3 m}] \times 2$

= $1 m s^{^{- 1}}$

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An engine pumps water through a hosepipe. Water passes through the pipe and leaves it with a velocity of 2 $m s^{- 1} .$ The mass per unit length of water in the pipe is 100 $k g m^{- 1} .$ What is the power of the engine?

A.

400W

B.

200W

C.

100W

D.

800W

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Explanation

$F = \frac{m (v - u)}{t} = \frac{mv}{t} if u = 0 . P = F . v = \frac{{mv}^{2}}{t} As \frac{m}{t} = linear density \times velocity = μv P = {μv}^{3} = 800 W$

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A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is v, the power delivered to the particle is

A.

$\frac{{Mv}^{2}}{T}$

B.

$\frac{1}{2} \frac{{Mv}^{2}}{T^{2}}$

C.

$\frac{{Mv}^{2}}{T^{2}}$

D.

$\frac{1}{2} \frac{{Mv}^{2}}{T}$

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Explanation

$The kinetic energy of particle = \frac{1}{2} {Mv}^{2} Power = \frac{Energy}{Time} P = \frac{1}{2} \frac{{Mv}^{2}}{T}$

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A body of mass 1 kg is thrown upwards with a velocity $20 {ms}^{- 1} .$ It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? $(g = 10 {ms}^{- 2})$

A.

20 J

B.

30 J

C.

40 J

D.

10 J

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Explanation

Key Idea The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.

Initially, body posses only kinetic energy and after attaining a height the kinetic energy is zero.

Therefore, loss of energy=KE-PE

$= \frac{1}{2} {mv}^{2} - mgh = \frac{1}{2} \times 1 \times 400 - 1 \times 18 \times 10 = 200 - 180 = 20 J$

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A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

A.

Mg/k

B.

2Mg/k

C.

4 Mg/k

D.

Mg/2k

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Explanation

Use the law of conservation of energy. Let x be the extension in the spring.

Applying conservation of energy-

$m g x - \frac{1}{2} k x^{2} = 0 - 0$

$\Rightarrow x = \frac{2 m g}{k}$

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K is the force constant of a spring. The work done in increasing its extension from $l_{1}$ to $l_{2}$ will be

A.

$K (l_{2} - l_{1})$

B.

$\frac{K}{2} (l_{2} + l_{1})$

C.

$K (l_{2}^{2} - l_{1}^{2})$

D.

$\frac{K}{2} (l_{2}^{2} - l_{1}^{2})$

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Explanation

(d)

$A t e x t e n s i o n l_{1}, t h e s t o r e d e n e r g y = \frac{1}{2} K l_{1}^{2} A t e x t e n s i o n l_{2}, t h e s t o r e d e n e r g y = \frac{1}{2} K l_{2}^{2} W o r k d o n e i n i n c r e a \sin g i t s e x t e n s i o n f r o m l_{1} t o l_{2} = \frac{1}{2} K (l_{2}^{2} - l_{1}^{2})$

Physics MCQs for NEET — Practice Questions with Answers

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