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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2025
  • Easy
  • Numerical

Two iron solid discs of negligible thickness have radius \(R_1\) and \(R_2\) and moment of intertia \(I_1\) and \(I_2\), respectively. For \(R_2=2 R_1\), the ratio of \(I_1\) and \(I_2\) would be \(1 / x\), where \(\mathrm{x}=\) ________ -

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Answer & Solution

Correct Answer

(B) 16

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Detailed explanation




Given \(R_2=2 R_1\)
\(\begin{aligned}
& M_1=\sigma \times \pi R_1^2=M_o \\
& M_2=\sigma \times \pi R_2^2=M_o \\
& M_2=\sigma \times \pi R_2^2=\sigma \times \pi\left[2 R_1\right]^2=\sigma \times 4 \pi R_1^2=4 M_o
\end{aligned}\)
\(\frac{I_1}{I_2}=\frac{\frac{M_1 R_1^2}{2}}{\frac{M_2 R_2^2}{2}}=\frac{M_1 R_1^2}{M_2 R_2^2}=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}\)
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