JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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The moment of inertia of a circular ring of radius

15 \mathrm{~cm}

and mass

150 \mathrm{~g}

is _____

View answer and explanation

Correct option: D

0.003 \mathrm{~kg} \cdot \mathrm{m}^2

Option D (

0.003 \mathrm{~kg} \cdot \mathrm{m}^2

) is correct because it is calculated using the formula for the moment of inertia of a circular ring,

I = mr^2

More Questions from SYSTEM OF PARTICLES

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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)