JEE PHYSICS MCQ: SYSTEM OF PARTICLES

1 Medium

1 of 1

Application +4 | -1

A body of radius

R

and mass

M

is rolling horizontally without slipping with speed

v

, it then rolls up a hill to a maximum height

h

. If

h=3 v^2 / 4 \mathrm{~g}

, what is the moment of inertia of the body?

View answer and explanation

Correct option: A

\frac{\mathrm{MR}^2}{2}

Option A (

\frac{\mathrm{MR}^2}{2}

) is correct because it satisfies the energy conservation equation when the rolling body climbs the hill to the given height.

More Questions from SYSTEM OF PARTICLES

Practice All Chapter MCQs →

\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)