JEE PHYSICS MCQ: SYSTEM OF PARTICLES

1 Easy

1 of 1

Application +4 | -1

Two rigid bodies A and B rotate with rotational kinetic energies

{E_A}\,and\,{E_B}

respectively. The moments of inertia of A and B about the axis of rotation are

{I_A}\,and\,{I_B}

respectively. If

{I_A} = {I_B}/4\,and\,{E_A} = 100\;{E_B}

the ratio of angular momentum

\left( {{L_A}} \right)

of A to the angular momentum

\left( {{L_B}} \right)

of B is

View answer and explanation

Correct option: C

Option C (5) is correct because the ratio of angular momentum is calculated using the relationship between rotational kinetic energy, moment of inertia, and angular momentum.

More Questions from SYSTEM OF PARTICLES

Practice All Chapter MCQs →

It is easier for a swimmer jumping into water from a height to describe a loop in the air by A uniform thin rod

\mathrm{AB}

of length

\mathrm{L}

has linear mass density

\mu(x)=a+\frac{b x}{L}

, where

x

is measured from

\mathrm{A}

. If the

\mathrm{CM}

of the rod lies at a distance of

\left(\frac{7}{12}\right) \mathrm{L}

from

\mathrm{A}

, then a and

\mathrm{b}

1.0 \mathrm{~g}

is rotating in a circle of diameter

8 \mathrm{~cm}

. Its moment of inertia will be The angular velocity of a particle is given by

\omega = {t^4} + 2{t^3}

then its angular acceleration of the particle at

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)