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Physics MCQs for NEET — Practice Questions with Answers

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A car is moving with a constant speed the wheels of the car make 120 rotations per minute the breaks are applied and the car comes to rest in 8 sec how many rotation are completed by the wheels before the car is brought to rest.

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Explanation

$ \omega_0 = { 2 \pi \times 120 \over 60} = 4 \pi rad /sec $ Now $ \omega = \omega_0 + \alpha t $ Total angle descrited in 8 second is $ \theta = w_0 t + { 1 \over 2} \alpha t^2 $

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The angular momentum of a wheel changes from 2L to 5L in 3 seconds what is the magnitudes of torque acting on it?

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Explanation

$ \tau = { dL \over dt} = { 5L - 2L \over 3 } ={ 3L \over 3 } + L $

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A uniform disc of mass 500kg and radius 2 m is rotating at the rate of 600 r.p.m. what is the torque required to rotate the disc in the opposite direction with the same angular speed in a time of 100 sec ?

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Explanation

$ For disc I = { MR^2 \over 2} = { 500 \times 4 \over 2 } $ $ angular speed = \omega = { 2 pi \times 600 \over 60 } $ so angular momentum $ L = I \omega$ final angular momentum in opposite direction $ = -1000 \times 20 \pi kgm^2 / sec $ So change in angular momentum = $ \triangle = 2 \times 1000 \times 20 \pi kg m^2 / sec $ $ \tau = { dL \over dt } = { 2 times 1000 \times 20 \pi \over 100 } = 400 \pi N.m$

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The moment of inertia of a meter scale of mass 0.6kg about an axis perpendicular to the scale and passing through 30 cm position on the scale is given by (Breath of scale is negligible).

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Explanation

By Parallel axis therom solved problem

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How much constant force should be applied tangential to equator of the earth to stop its rotation in one day ?

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Explanation

$ \omega_1 = 2 \pi rad/day and \omega_2 = 0 , t =1 day $ $ \alpha = { w_2 - w_1 \over t } $ Torque required to stop the earth = T = $ I \alpha $ = F.R $ F = { I.\alpha \over R } $

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A constant torque of 1500 Nm turns a wheel of moment of inertia 300 kg m2 about an axis passing through its centre the angular velocity of the wheel after 3 sec will be….. rad/sec

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Explanation

$ \tau = I \alpha = I { d \omega \over dt} $

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A mass m is moving with a constant velocity along the line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin

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Explanation

$ L = Momentum \times perpenlicular distance between point of rotation and line of action $ = m.V.y all remain constant L = remaing constant

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A body is rolling down an incline plane. If the rotational K.E. of the body is 40% of its translational K.E. then the body is ….

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Explanation

R.K.E = 40 /100 T.K.E $ {1 \over 2} I \omega^2 = { 4 \over 10 } \times { 1 \over 2} mv^2 = { 1 \over 5} mv^2 $ $ \therefore { 1 \over 2} mk^2 \times {V^2 \over r^2 } = { 2 \over 5} mv^2 $

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A spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is

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Explanation

$ Total energy E = { 1 \over 2} I \omega^2 + {1 \over 2} mv^2 = { 1 \over 2} \times { 2 \over 5 } mr^2 \omega^2 + { 1 \over 2} mr^2 \omega^2 $

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A binary star consist of two stars A (2.2 Ms) and B(mass 11Ms) where Ms is the mass of sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B. about the centre of mass is

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Explanation

Here C.M wrt A $ = \left( { 11M_5 \over 11M_5 + 2.2 M_5 } \right) d = {5d \over 6 } $ $ L_A = 2.2 M_5 w \left( { 5d \over 6} \right) ^2 = 55M_5 w { d^2 \over 36 } $ $L_B = 11 M_5 \omega \left( { d \over 6} \right)^2 = 11 M_5 \omega { d^2 \over 36 } $

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