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Laws of Motion MCQs for NEET — Physics Questions with Answers

Practice free Laws of Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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A body, whose momentum is constant, must have constant

Force

Velocity

Acceleration

All of these

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Explanation

For a given mass $P \propto v$ . If the momentum is constant then it's velocity must have constant.

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Rocket engines lift a rocket from the earth surface because hot gas with high velocity

Push against the earth

Push against the air

React against the rocket and push it up

Heat up the air which lifts the rocket

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Explanation

It works on the principle of conservation of momentum.

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A force of 50 dynes is acted on a body of mass 5 g which is at rest for an interval of 3 seconds, then impulse is

$0 .15 \times 10^{- 3} N s$

$0 .98 \times 10^{- 3} N s$

$1 .5 \times 10^{- 3} N s$

$2 .5 \times 10^{- 3} N s$

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Explanation

Impulse, $I = F \times Δ t = 50 \times 10^{- 5} \times 3$ = $1 .5 \times 10^{- 3} N - s$

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A bullet of mass 0.1 kg is fired with a speed of 100 m/sec, the mass of gun is 50 kg. The velocity of recoil is

0.2 m/sec

0.1 m/sec

0.5 m/sec

0.05 m/sec

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Explanation

According to principle of conservation of linear momentum, $m_{G} v_{G} = m_{B} v_{B}$

$\Rightarrow v_{G} = \frac{m_{B} v_{B}}{m_{G}} = \frac{0 .1 \times 10^{2}}{50} = 0 .2 m / s$

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Two forces of magnitude F have a resultant of the same magnitude F. The angle between the two forces is

45°

120°

150°

60°

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Explanation

$F = \sqrt{{(F)}^{2} + {(F)}^{2} + 2 F . F \cos θ}$ ⇒ $θ = 120 °$

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Two forces with equal magnitudes F act on a body and the magnitude of the resultant force is F/3. The angle between the two forces is

$\cos^{- 1} (- \frac{17}{18})$

$\cos^{- 1} (- \frac{1}{3})$

$\cos^{- 1} (\frac{2}{3})$

$\cos^{- 1} (\frac{8}{9})$

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Explanation

$F_{_{n e t}}^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ$

⇒ ${(\frac{F}{3})}^{2} = F^{2} + F^{2} + 2 F^{2} \cos θ$

⇒ $\cos θ = (- \frac{17}{18})$

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If two forces of 5 N each are acting along X and Y axes, then the magnitude and direction of resultant is

$5 \sqrt{2}, π / 3$

$5 \sqrt{2}, π / 4$

$- 5 \sqrt{2}, π / 3$

$- 5 \sqrt{2}, π / 4$

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Explanation

$| \vec{F} | = \sqrt{5^{2} + 5^{2}} = 5 \sqrt{2} N .$

and $\tan θ = \frac{5}{5}$ =1

$\Rightarrow θ = π / 4 .$

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A block is kept on a frictionless inclined surface with angle of inclination 'α'. The incline is given an acceleration 'a' to keep the block stationary. Then a is equal to

g tan α

g / tan α

g cosec α

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Explanation

For a block to remain stationary on a frictionless inclined surface with angle Î±, the acceleration a given to the incline must be equal to g tan Î±, where g is the acceleration due to gravity. This acceleration provides just enough force component along the incline to balance the weight component.

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Two masses 2 kg and 3 kg are attached to the end of the string passed over a pulley fixed at the top. The tension and acceleration are

$\frac{7 g}{8}; \frac{g}{8}$

$\frac{21 g}{8}; \frac{g}{8}$

$\frac{21 g}{8}; \frac{g}{5}$

$\frac{12 g}{5}; \frac{g}{5}$

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Explanation

$T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} g = \frac{2 \times 2 \times 3}{2 + 3} g = \frac{12}{5} g$

$a = (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) g = (\frac{3 - 2}{3 + 2}) g = \frac{g}{5}$

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Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is (g = 9.8 m/s²)

4.9 m/s²

2.45 m/s²

1.4 m/s²

9.5 m/s²

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Explanation

Acceleration $= \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$

$= \frac{4 - 3}{4 + 3} \times 9 .8 = \frac{9 .8}{7} = 1 .4 m / \sec^{2}$

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