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QJEEM25Phd8fc: JEE Main 2025 Physics - Class 11 JEEM25Phd8fc Solution: The moment of inertia of a solid disc rotating along its…

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 nn times higher than the moment of inertia of the given ring. Here, n=\mathrm{n}= ________ Consider all the bodies have equal masses.

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JEE Main 2025 (Online) 28th January Morning Shift Physics - Rotational Motion Question 2 English Explanation

Given, I1=2.5I2{I_1} = 2.5{I_2} and I3=nI2{I_3} = n{I_2}

we know, I1=MR124,I2=MR222,I3=25MR32{I_1} = {{MR_1^2} \over 4},{I_2} = {{MR_2^2} \over 2},{I_3} = {2 \over 5}MR_3^2

⇒MR124=2.5MR222⇒R12=5R22\Rightarrow {{MR_1^2} \over 4} = 2.5{{MR_2^2} \over 2} \Rightarrow R_1^2 = 5R_2^2

Now, I3=nI2{I_3} = n{I_2}

⇒25MR12=nMR222\Rightarrow {2 \over 5}MR_1^2 = n{{MR_2^2} \over 2} (As R3=R1{R_3} = {R_1} )

⇒25×5R22=nR222⇒n=4\Rightarrow {2 \over 5} \times 5R_2^2 = {{nR_2^2} \over 2} \Rightarrow n = 4

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