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Rotational Motion MCQs for NEET — Physics Questions with Answers

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

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If a body of mass m collides head on, elastically with velocity u with another identical body at rest. After collision velocity of the second body will be

zero

data insufficient

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Explanation

Between elastic collision of two identical masses, velocities are interchanged.

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If a person standing on a rotating disc stretches out his hands, the angular speed will

increase

decrease

remain same

None of these

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Explanation

$Iω = constant ω \propto \frac{1}{I} On stretching hands, I will increase, therefore ω will decrease .$

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If the earth were to suddenly contract to $\frac{1}{n}$ th of its present size without any change in its mass, the duration of the new day will be nearly

24/n hours

24 n hours

24/ $n^{2}$ hours

24 $n^{2}$ hours

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Explanation

Using the principle of conservation of angular momentum, $I_{1} ω_{1} = I_{2} ω_{2}$

$\frac{2}{5} {MR}^{2} \times \frac{2 π}{24} = \frac{2}{5} M {(\frac{R}{n})}^{2} \frac{2 π}{T}; T = \frac{24}{n^{2}} hours$

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A constant torque of 31.4 N-m is exerted on a pivoted wheel. If angular acceleration of wheel is $4 π rad / \sec^{2}$ , then the moment of inertia of the wheel is

2.5 $k g - m^{2}$

$3.5 k g - m^{2}$

$4.5 k g - m^{2}$

$5.5 k g - m^{2}$

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Explanation

$I = \frac{τ}{α} = \frac{31.4}{4 π} = 2.5 k g m^{2}$

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The moment of inertia of a flywheel having kinetic energy 360 J and angular speed of 20 rad/s is

18 kg $m^{2}$

1.8 kg $m^{2}$

2.5 kg $m^{2}$

9 kg $m^{2}$

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Explanation

$(b) Given : kinetic energy K = 360 J Angular speed ω = 20 rad / s ∴ K = \frac{1}{2} {Iω}^{2} Where I = moment of inertia \Rightarrow I = \frac{2 K}{ω^{2}} = \frac{2 \times 360}{20 \times 20} = 1.8 kg m^{- 2} .$

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A particle performs uniform circular motion with an angular momentum L. If the frequency of the particle motion is double, the angular momentum becomes

$\frac{L}{2}$

$\frac{L}{4}$

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Explanation

$L = I ω = I (2 π f)$

Frequency f is doubled. Hence angular momentum will become 2L.

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A 20 kg solid disc rolls on a horizontal surface at the rate of 4 $m s^{- 1}$ . Its total kinetic energy is

60 J

120 J

240 J

80 J

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Explanation

$\begin{aligned} (KE) & = (KE)_{vrensialonal} + (KE)_{retational} \\ = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} \\ = \frac{1}{2} m v^{2} + \frac{1}{2} \times \frac{1}{2} m r^{2} ω^{2} \\ = \frac{1}{2} m v^{2} + \frac{1}{4} m v^{2} = \frac{3}{4} m v^{2} \\ = \frac{3}{4} \times 20 \times 4^{2} = 240 J \end{aligned}$

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A sphere rolls down on an inclined plane of inclination $θ$ . What is the acceleration as the sphere reaches the bottom?

$\frac{5}{7} g \sin θ$

$\frac{3}{5} g \sin θ$

$\frac{2}{7} g \sin θ$

$\frac{2}{5} g \sin θ$

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Explanation

Acceleration of a rolling body down an inclined plane is $a = \frac{gsin θ}{1 + k^{2} / r^{2}} = \frac{gsin θ}{1 + \frac{2}{5}} (1 = \frac{2}{5} {mr}^{2} = {mk}^{2}$

$\frac{k^{2}}{r^{2}} = \frac{2}{5} = \frac{5}{7} g \sin θ$ Choice A is correct.

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Two bodies having masses $m_{1}$ and $m_{2}$ and velocities $u_{1}$ and $u_{2}$ collide form a composite system. If $m_{1}$ $u_{1}$ + $m_{2}$ $u_{2}$ = ( $m_{1}$ $\neq$ $m_{2}$ ), the velocity of composite system will be

$u_{1}$ - $u_{2}$

$u_{1}$ + $u_{2}$

$\frac{u_{1} + u_{2}}{2}$

Zero

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Explanation

$\begin{aligned} u_{CM} = \frac{m_{1} u_{1} + m_{2} u_{2}}{m_{1} + m_{2}} = 0 \\ as m_{1} u_{1} + m_{2} u_{2} = 0 is given. \end{aligned}$

Hence velocity of the composite system will be zero.

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Centre of the mass of 3 particles 10 kg, 20kg, and 30kg is at (0.0,0). Where should a particle of mass 40 kg be placed so that the combined center of mass will be at (3, 3, 3)?

(0, 0, 0)

(7, 5, 7, 5, 7, 5)

(1, 2, 3)

(4, 4, 4)

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Explanation

$\vec{r_{c m}} = \frac{m_{1} {\vec{r}}_{1} + m_{4} {\vec{r}}_{2}}{m_{1} + m_{2}} = \frac{60 \times (0 i + 0 j + 0 k) + 40 {\vec{r}}_{2}}{100} {\vec{r}}_{2} = \frac{5}{2} (3 i + 3 j + 3 k) = 7.5 i + 7.5 j + 7.5 k$

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