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Laws of Motion MCQs for NEET — Physics Questions with Answers

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For a particle in a non-uniform accelerated circular motion

Velocity is radial and acceleration is transverse only

Velocity is transverse and acceleration is radial only

Velocity is radial and acceleration has both radial and transverse components

Velocity is transverse and acceleration has both radial and transverse components

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Explanation

In a non-uniform circular motion, the particle possesses both centripetal as well as tangential acceleration.

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The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t + θ_{2} t^{2}$ . Then the angular acceleration of the body is

θ₁

θ₂

2θ₁

2θ₂

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Explanation

_{$ω = \frac{d θ}{d t}$}

Angular acceleration $= \frac{d ω}{d t}$ = 2θ₂

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A particle moves in the x-y plane according to rule $x = a \sin ωt$ and $y = a \cos ωt$ . The particle follows

an elliptical path

a circular path

a parabolic path

a straight line path inclined equally to x and y-axis

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Explanation

$x = a \sin ωt \Rightarrow \frac{x}{a} = \sin ωt x = a \cos ωt \Rightarrow \frac{y}{a} = \cos ωt \to x^{2} + y^{2} = a^{2}$

This is an equation of a circle, so the particle follows a circular path.

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The coordinates of a moving particle at a time t, are given by, x=5 sin 10t, y=5 cos 10t. The speed of the particle is:

$50 \sqrt{2}$

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Explanation

(B)

$x = 5 \sin 10 t y = 5 \cos 10 t v_{x} = \frac{dx}{dt} v_{y} = \frac{dy}{dt} = 5 \times 10 \cos l 0 t = - 50 \sin l 0 t = 50 \cos l 0 t Velocity \vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = 50 \cos l 0 t \hat{i} - 50 \sin l 0 t \hat{j} Speed = Magnitude of velocity = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{{(50 \cos l 0 t)}^{2} + {(- 50 \sin l 0 t)}^{2}} = 50 \sqrt{\cos^{2} 10 t + \sin^{2} 10 t} units = 50 units$

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For a rocket moving in free space, the fraction of mass to be disposed of Off to attain a speed equal to two times the exhaust speed is given by (given $e^{2}$ = 7.4)

0.40

0.37

0.50

0.86

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Explanation

$\begin{aligned} v & = v_{r} \ln (\frac{m_{0}}{m}) \end{aligned} o r \begin{aligned} 2 v_{r} & = v_{r} \ln (\frac{m_{0}}{m}) \end{aligned} ∴ \begin{aligned} 2 & = \ln (\frac{m_{0}}{m}) \end{aligned} ∴ \begin{aligned} \frac{m_{0}}{m} & = e^{2} = 7.4 \end{aligned} o r \begin{aligned} \frac{m}{m_{0}} & = 0.14 \end{aligned}$

This is a fraction of mass remaining. Hence fraction of mass disposed of = 0.86.

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A simple pendulum hangs from the roof of a train moving on horizontal rails. If the string is inclined towards the front of the train, then train is

Moving with constant velocity

In accelerated motion

In retarded motion

At rest

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Explanation

Using concept of inertia

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A lighter body and heavier body both are moving with same momentum and applying same retarding force. Their stoping distance are $s_{1} a n d s_{2}$ respectively. Then the correct relation between $s_{1} a n d s_{2}$ .

$s_{1} > s_{2}$

$s_{1} < s_{2}$

$s_{1} = s_{2}$

None of these

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Explanation

$s = \frac{m u^{2}}{2 F} = \frac{P^{2}}{2 F m} = \frac{K E}{F} s α \frac{1}{m} {P a n d F i s g i v e n} s_{1} > s_{2}$

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A plastic box of mass 5 kg is found to accelerate up at the rate of g / 6, when placed deep inside the water. How much sand should be put inside the box so that it may accelerate down at the rate of g / 6?

2kg

3kg

4kg

5kg

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Explanation

$B - Mg = Ma . . . (1) (M + m) g - B = (M + m) a . . . (2) mg = (2 M + m) a [a = g / 6 \dots \dots given] mg = (10 + m) \frac{g}{6} 6 m - m = 10 m = 2 kg$

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Let $θ$ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of bob is m. The tension in the string is mg cos $θ$

always

never

at extreme

at mean position

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A metal ring of mass m and radius R is placed on smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with a speed v. Tension in the ring is:

$\frac{m v^{2}}{R}$

$\frac{m v^{2}}{2 R}$

$\frac{m v^{2}}{2 π R}$

None of these

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Explanation

For a ring of mass m and radius R rotating with a uniform speed v, the tension in the ring is given by (mv^2)/(2Ï€R). This is because the centripetal force required to keep each point of the ring in circular motion is provided by the tension in the ring.

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