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Physics MCQs for NEET — Practice Questions with Answers

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The moment of inertia of a rod about an axis through its centre and perpendicular to it is $\frac{1}{12} M L^{2}$ (where M is the mass and L is the length of the rod). The rod is bent in the middle so that the two halves make an angle of 60 $°$ . The moment of inertia of the bent rod about the same axis would be

$\frac{1}{48} M L^{2}$

$\frac{1}{12} M L^{2}$

$\frac{1}{24} M L^{2}$

$\frac{M L^{2}}{8 \sqrt{3}}$

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A man is sitting on a rotating table with his arms stretched outwards. When he suddenly folds his arms inside, then

his angular velocity will decrease

his angular velocity remains constant

his moment of inertia decreases

angular momentum increases

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Explanation

Decrease of distance of masses from the axis results in decrease in M.I. as $I = \sum_{} m r^{2}$

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A body of mass M and radius R is rolling horizontally without slipping with speed v. It then rolls up a hill to a maximum height h. If $h = \frac{5 v^{2}}{6 g}$ , what is the M.I of the body?

$\frac{M R^{2}}{2}$

$\frac{2}{3} M R^{2}$

$\frac{3}{4} M R^{2}$

$\frac{2}{5} M R^{2}$

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Explanation

$M g h = \frac{1}{2} M v^{2} + \frac{1}{2} l ω^{2} = \frac{1}{2} M v^{2} + \frac{1}{2} \times I ω^{2} = \frac{1}{2} M v^{2} + \frac{1}{2} \frac{I v^{2}}{R^{2}}; a s h = \frac{5 v^{2}}{6 g} m g \times \frac{5 v^{2}}{6 g} = \frac{1}{2} M v^{2} + \frac{1}{2} \frac{I v^{2}}{R^{2}} (\frac{5 M v^{2}}{6} - \frac{M v^{2}}{2}) \frac{2 R^{2}}{v^{2}} = I M (\frac{5}{6} - \frac{3}{6}) \times 2 R^{2} = I, ∴ \frac{2}{3} M R^{2} = I w h i c h g i v e s I = \frac{2}{3} M R^{2} (c h o i c e 2)$

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One projectile moving with velocity v in space, gets burst into 2 parts of masses in the ratio 1:3. The smaller part becomes stationary. What is the velocity of the other part?

$\frac{4 v}{3}$

$\frac{3 v}{4}$

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Explanation

$m v = (\frac{3}{4} m) v' ∴ v' = \frac{4 v}{3}$

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One circular ring and one circular disc both having the same mass and radius. The ratio of their moments of inertia about the axes passing through their centres and perpendicular to planes will be

1:1

2:1

1:2

4:1

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Explanation

$I_{r i n g} = m R^{2}, I_{d i s c} = \frac{1}{2} m R^{2}$

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A wheel of radius R rolls on the ground with a uniform velocity v. The velocity of topmost point relative to the bottommost point is

v/2

zero

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Explanation

In rolling motion, the topmost point has an instantaneous velocity equal to the velocity of the center of the wheel plus the peripheral velocity. Since the center's velocity is v and the peripheral velocity is also v, the velocity of the topmost point relative to the bottommost point is 2v.

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The moment of inertia of a cube of mass m and side a about one of its edges is equal to

$\frac{2}{3} m a^{2}$

$\frac{4}{3} m a^{2}$

$3 m a^{2}$

$\frac{8}{3} m a^{2}$

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If the net external forces acting on the system of particles is zero, then which of the following may vary ?

Momentum of the system

Velocity of centre of mass

Position of centre of mass

None of the above

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A constant torque of 1000 N-m turns a wheel of moment of inertial 200 $kg - m^{2}$ about an axis through its centre. Its angular velocity after 3 s is

1 rad/s

5 rad/s

10 rad/s

15 rad/s

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Explanation

Angular acceleration

$α = \frac{τ}{I} = \frac{1000}{200} = 5 rad / s^{2} ω = αt = (5) (3) = 15 rad / s$

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A wheel is rotating about an axis through its centre at 720 r.p.m. It is acted on by a constant torque opposing its motion for 8 seconds to bring it to rest finally. The value of torque is Nm is (given l = $\frac{24}{π}$ kg $m^{2}$ )

120

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Explanation

$n = 720 rpm = \frac{720}{60} = 12 rps ∴ ω_{0} = 2 π x 12 = 24 π rad / \sec and ω = 0 applying ω = ω_{0} + α t 0 = 24 π + α x 8 α = - 3 π rad / \sec^{2} τ = I α = \frac{24}{π} x 3 π = 72 Nm$

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