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Physics MCQs for NEET — Practice Questions with Answers

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Two bodies of masses m and 4 m are moving with equal K.E. The ratio of their linear momentums is

4 : 1

1 : 1

1 : 2

1 : 4

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Explanation

$p = \sqrt{2 m E}$

∴ $P \propto \sqrt{m}$

⇒ $\frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}} = \sqrt{\frac{m}{4 m}} = \frac{1}{2}$

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A particle of mass m₁ is moving with a velocity v₁ and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum but their different kinetic energies are E₁ and E₂ respectively. If m₁ > m₂ then

E₁ < E₂

$\frac{E_{1}}{E_{2}} = \frac{m_{1}}{m_{2}}$

E₁ > E₂

E₁ = E₂

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Explanation

$E = \frac{P^{2}}{2 m}$ if P = constant then $E \propto \frac{1}{m}$

According to problem $m_{1} > m_{2}$ ∴ $E_{1} < E_{2}$

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Four particles are given, having same momentum. Which one has maximum kinetic energy-

Proton

Electron

Deutron

α-particles

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Explanation

$E = \frac{P^{2}}{2 m}$

∴ $E \propto \frac{1}{m}$ (If P = constant)

i.e. the lightest particle will possess maximum kinetic energy and in the given option mass of electron is minimum.

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If a man increase his speed by 2 m/s , his K.E. is doubled, the original speed of the man is

$(1 + 2 \sqrt{2}) m / s$

4 m/s

$(2 + 2 \sqrt{2}) m / s$

$(2 + \sqrt{2}) m / s$

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Explanation

Initial kinetic energy $E = \frac{1}{2} m v^{2}$ …(i)

Final kinetic energy $2 E = \frac{1}{2} m {(v + 2)}^{2}$ …(ii)

by solving equation (i) and (ii) we get

$v = (2 + 2 \sqrt{2}) m / s$

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A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It slides down a smooth surface to the ground, then climbs up another hill of height 30 m and finally slides down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is

10 m/s

$10 \sqrt{30}$ m/s

40 m/s

20 m/s

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A body of mass m accelerates uniformly from rest to v₁ in time t₁. As a function of time t, the instantaneous power delivered to the body is

$\frac{m v_{1} t}{t_{1}}$

$\frac{m v_{1}^{2} t}{t_{1}}$

$\frac{m v_{1} t^{2}}{t_{1}}$

$\frac{m v_{1}^{2} t}{t_{1}^{2}}$

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Explanation

$P = \vec{F} . \vec{v} = m a \times a t = m a^{2} t$ [as u = 0]

$= m {(\frac{v_{1}}{t_{1}})}^{2} t = \frac{m v_{1}^{2} t}{t_{1}^{2}}$ $[As a = v_{1} / t_{1}]$

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A weight lifter lifts 300 kg from the ground to a height of 2 meter in 3 second. The average power generated by him is

5880 watt

4410 watt

2205 watt

1960 watt

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Explanation

P $= \frac{Workdone}{Time}$ = $\frac{m g h}{t} = \frac{300 \times 9 .8 \times 2}{3} = 1960 W$

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The average power required to lift a 100 kg mass through a height of 50 metres in approximately 50 seconds would be

50 J/s

5000 J/s

100 J/s

980 J/s

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Explanation

$P = \frac{m g h}{t}$ = $\frac{100 \times 9 .8 \times 50}{50} = 980 J / s$

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The power of a pump, which can pump 200kg of water to a height of 200m in 10sec is (g = 10 m/s²)

40 kW

80 kW

400 kW

960 kW

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Explanation

$p = \frac{m g h}{t} = \frac{200 \times 10 \times 200}{10} = 40 k W$

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A 60 kg man runs up a staircase in 12 seconds while a 50 kg man runs up the same staircase in 11, seconds, the ratio of the rate of doing their work is

6 : 5

12 : 11

11 : 10

10 : 11

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Explanation

$P = \frac{m g h}{t}$

⇒ $\frac{P_{1}}{P_{2}} = \frac{m_{1}}{m_{2}} \times \frac{t_{2}}{t_{1}}$ (As h = constant)

∴ $\frac{P_{1}}{P_{2}} = \frac{60}{50} \times \frac{11}{12} = \frac{11}{10}$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.