JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 23)
The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is $n$ times higher than the moment of inertia of the given ring. Here, $\mathrm{n}=$ ________ Consider all the bodies have equal masses.
Answer
4
Explanation
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Given, $${I_1} = 2.5{I_2}$$ and $${I_3} = n{I_2}$$
we know, $${I_1} = {{MR_1^2} \over 4},{I_2} = {{MR_2^2} \over 2},{I_3} = {2 \over 5}MR_3^2$$
$$ \Rightarrow {{MR_1^2} \over 4} = 2.5{{MR_2^2} \over 2} \Rightarrow R_1^2 = 5R_2^2$$
Now, $${I_3} = n{I_2}$$
$$ \Rightarrow {2 \over 5}MR_1^2 = n{{MR_2^2} \over 2}$$ (As $${R_3} = {R_1}$$)
$$ \Rightarrow {2 \over 5} \times 5R_2^2 = {{nR_2^2} \over 2} \Rightarrow n = 4$$
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