JEE PHYSICS MCQ: SYSTEM OF PARTICLES

1 Easy

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Application +4 | -1

Moment of inertia of a ring of mass

m = 4\,gm

and radius

r = 1\,cm\,

about an axis passing through its edge and parallel to its natural axis is

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Correct option: A

8\,g - c{m^2}

Option A (

8\,g - cm^2

) is correct because it is calculated using the parallel axis theorem for the moment of inertia of a ring.

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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)