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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2025 · 28 January · Morning Shift · Physics · Rotational Motion

The moment of inertia of a solid disc rotating along its diameter...

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.The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a

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The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a
I1=MR124,\mathrm{I}_1=\frac{\mathrm{MR}_1^2}{4},\mathrm{I}2=MR222,_2=\frac{\mathrm{MR}_2^2}{2},\mathrm{I}3=2MR125_3=\frac{2 \mathrm{MR}_1^2}{5} According to problem \frac{\mathrm{I}_1}{\mathrm{I}_2}=2.5 \Rightarrow \frac{\frac{\mathrm{MR}_1^2}{4}}{\frac{\mathrm{MR}_2^2}{2}}=\frac{5}{2} \Rightarrow \frac{\mathrm{R}_1^2}{\mathrm{R}_2^2}=5 \ldots Now we are provided with information that I3I2=n⇒2MR125MR222=n⇒4R125R22=n\begin{aligned} & \frac{\mathrm{I}_3}{\mathrm{I}_2}=\mathrm{n} \\ & \Rightarrow \frac{\frac{2 \mathrm{MR}_1^2}{5}}{\frac{\mathrm{MR}_2^2}{2}}=\mathrm{n} \Rightarrow \frac{4 \mathrm{R}_1^2}{5 \mathrm{R}_2^2}=\mathrm{n} \end{aligned} From Eq', (1) and (2) ⇒n=4\Rightarrow n=4
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