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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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A uniform rod of length 2L is placed with one end in contact with horizontal and is then inclined at an angle $ \alpha $ to the horizontal and allowed to fall without slipping at contact point. When it becomes horizontal, its angular velocity will be…..

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Explanation

By Conservation of Energy P.E. of rod = Rotational K.E. $ M.g {1 \over 2} sin d = { 1 \over 2} I \omega^2 = { 1 \over 2} { mL^2 \over 3} \omega^2 $ $ \therefore \omega = \sqrt { 3 g sin \alpha \over L } $ As here l = 2 L $ \omega = \sqrt { 3g sin \alpha \over 2L } $

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A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity w. Two objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity

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Explanation

Initial angular momentum of ring $ = I \omega = MR^2 \omega $ Final angular momentum of ring and particles $ = ( MR^2 + 2 mR^2 ) \omega' $ As No external forque so According to Law of conservation of angular momentum. $ MR^2 \omega = ( MR^2 + 2 mR^2 ) \omega' $ $ \therefore \omega' = { wM \over (M + 2m ) } $

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A smooth sphere A is moving on a frictionless horizontal plane with angular speed $ \omega $ and centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are $ \omega_A $ and $ \omega_B $ respectively, Then

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Explanation

As it is head-on elastic collision between two idential balls there fore they will exchange their linear vecocity is A comes to rest and B starts moving with linear velocity V. As there is no friction any where, forque on both the spheres about their centre of mass is zero and their angular velocities remains unchanged $ \therefore \omega_A = \omega and \omega = 0_B $

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Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m
and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of …..

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Explanation

$ I = 0.3 x^2 + 0.7 ( 1.4 - x)^2 $ For minimum work moment of inertia of the system should be minimum is $ { dI \over dx} = 0 = 03 \times 2x -0.7 \times 2 ( 1.4 - x) = 0 $

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A child is standing with folded hands at the centre of a platform rotating about its central axis the kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is

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Explanation

$ E = { L^2 \over 2I } = K given K \alpha { 1 \over I } $ If L is constant when child stretches his arms the moment of inertia of system get doubled so kinetic energy will becomes half i.e $ { K \over 2 } $

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If the earth is treated as a sphere of radius R and mass M. Its angular momentum about the axis of rotation with period T is…..

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Explanation

Moment of inertia of sphere $ I = { 2 \over 5 } MR^2 $ about its axis $ L = I \omega $

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If the angular momentum of any rotating body increases by 200%, then the increase in its kinetic energy will be…..

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Explanation

$ E = { L^2 \over 2 I } $ $ E \alpha L^2 $

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The M.I. of a body about the given axis is 1.2 kgm2 initially the body is at rest. In order to produce a rotational kinetic energy of 1500 J. an angular acceleration of $ 25 rad / sec ^2 $ must be applied about that axis for duration of ….

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Explanation

$ Rotational K.E. = { 1 \over 2} I \omega^2 = 1500 $ $ According to w = wo + \alpha t $

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An automobile engine develops 100kw when rotating at a speed of 1800 r.p.m. what torque does it deliver ?

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Explanation

$ \omega = 1800 rpm = { 2 \pi \times 1800 \over 60 } = 60 \pi rad /sec $

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The moment of inertia of two rotating bodies A and B are $ I_A $ and $I_B $ . $ ( I_A \gt I_B ) $ and their angular momentum are equal. If their K.E. be $ K_A and K_B $ respectively

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Explanation

$ As K = { I^2 \over 2I } , K_A = { J^2 \over 2I_A}, and K_B = { J^2 \over 2 I_B} $

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