Question

Two iron solid discs of negligible thickness have radii \(R_{1}\) 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 \(1 / \mathrm{x}\), where \(\mathrm{x}=\) _____

Answer 1

Answer is : 16

 

Given \( \mathrm{R}_{2}=2 \mathrm{R}_{1}\) 

\(M_{1}=\sigma \times \pi R_{1}^{2}=M_{0}\) 

\(\mathrm{M}_{2}=\sigma \times \pi \mathrm{R}_{2}^{2}=\mathrm{M}_{\mathrm{0}}\) 

\(\mathrm{M}_{2}=\sigma \times \pi \mathrm{R}_{2}^{2}=\sigma \times \pi\left[2 \mathrm{R}_{1}\right]^{2}=\sigma \times 4 \pi \mathrm{R}_{1}^{2}=4 \mathrm{M}_{\mathrm{0}}\) 

\( \frac{I_{1}}{I_{2}}=\frac{\frac{\mathrm{M}_{1} \mathrm{R}_{1}^{2}}{2}}{\frac{\mathrm{M}_{2} \mathrm{R}_{2}^{2}}{2}}=\frac{\mathrm{M}_{1} \mathrm{R}_{1}^{2}}{\mathrm{M}_{2} \mathrm{R}_{2}^{2}}=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}\)