JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

If a body of moment of inertia

2\, {~kg} \,{~m}^2

revolves about its own axis making 2 rotations per second, then its angular momentum (in

{J} {s}

) is

View answer and explanation

Correct option: D

8 \pi

Option D (

8 \pi

) is correct because angular momentum is the product of moment of inertia and angular velocity.

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\mathrm{AB}

\mathrm{L}

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

x

\mathrm{A}

\mathrm{CM}

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

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

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

t = 1\,sec

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