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Physics MCQs for NEET — Practice Questions with Answers

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A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizotal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be

$\sqrt{\frac{2 g h}{I + m r}}$

$\sqrt{\frac{2 m g h}{I + m r^{2}}}$

${[\frac{2 m g h}{I + 2 m}]}^{\frac{1}{2}}$

$\sqrt{2 g h}$

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Explanation

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} m g h = \frac{1}{2} m r^{2} ω^{2} + \frac{1}{2} I ω^{2} 2 m g h = ω^{2} (m r^{2} + I) ω = \sqrt{\frac{2 m g h}{I + m r^{2}}}$

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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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A particle of mass m at rest is acted upon by a constant force F for a time t. Its kinetic energy after an interval t is

$\frac{F^{2} t^{2}}{m}$

$\frac{F^{2} t^{2}}{2 m}$

$\frac{F^{2} t^{2}}{3 m}$

$\frac{Ft}{2 m}$

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Explanation

$\begin{aligned} Kinetic energy E & = \frac{p^{2}}{2 m} \\ = \frac{(Ft)^{2}}{2 m} = \frac{F^{2} t^{2}}{2 m} \end{aligned} (As p = Ft)$

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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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