Medium INTEGER +4 / -1 PYQ · JEE Mains 2022

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I2. If ${{{I_1}} \over {{I_2}}}$ is ${{x{\pi ^2}} \over 3}$, then the value of x will be ____________.

Answer (integer) 8

Solution

<p>${I_1} = {{M{L^2}} \over 3}$ ..... (1)</p> <p>For ring : ${I_2} = {{M{R^2}} \over 2}$</p> <p>and $2\pi R = L$</p> <p>$\Rightarrow {I_2} = {M \over 2}\left( {{{{L^2}} \over {4{\pi ^2}}}} \right)$ ...... (2)</p> <p>$\Rightarrow {{{I_1}} \over {{I_2}}} = {{8{\pi ^2}} \over 3}$</p> <p>$\Rightarrow x = 8$</p>

About this question

Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass

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