JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 25)
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
8
Explanation
$${I_1} = {{M{L^2}} \over 3}$$ ..... (1)
For ring : $${I_2} = {{M{R^2}} \over 2}$$
and $$2\pi R = L$$
$$ \Rightarrow {I_2} = {M \over 2}\left( {{{{L^2}} \over {4{\pi ^2}}}} \right)$$ ...... (2)
$$ \Rightarrow {{{I_1}} \over {{I_2}}} = {{8{\pi ^2}} \over 3}$$
$$ \Rightarrow x = 8$$
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