MHT CET · Physics · Rotational Motion
A disc and a ring both have same mass and radius. The ratio of moment of inertia of the disc about its diameter to that of a ring about a tangent in its plane is
- A \(1: 2\)
- B \(1: 4\)
- C \(1: 6\)
- D \(1: 8\)
Answer & Solution
Correct Answer
(C) \(1: 6\)
Step-by-step Solution
Detailed explanation
For disc,
\(\begin{aligned}
\mathrm{I}_{\mathrm{D}} & =\frac{\mathrm{I}_z}{2} \\
& =\frac{\frac{\mathrm{MR}^2}{2}}{2}=\frac{\mathrm{MR}^2}{4}
\end{aligned}\)
For ring,

by using perpendicular and parallel axis theorem,
\(\begin{array}{ll}
& \mathrm{I}_{\mathrm{R}}=\mathrm{I}_y+M \mathrm{x}^2=\frac{\mathrm{MR}^2}{2}+M R^2=\frac{3}{2} \mathrm{MR}^2 \\
\therefore \quad & \frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{R}}}=\frac{\mathrm{MR}^2}{4} \times \frac{2}{3 \mathrm{MR}^2} \\
\therefore \quad & \frac{I_D}{I_R}=\frac{1}{6}
\end{array}\)
\(\begin{aligned}
\mathrm{I}_{\mathrm{D}} & =\frac{\mathrm{I}_z}{2} \\
& =\frac{\frac{\mathrm{MR}^2}{2}}{2}=\frac{\mathrm{MR}^2}{4}
\end{aligned}\)
For ring,

by using perpendicular and parallel axis theorem,
\(\begin{array}{ll}
& \mathrm{I}_{\mathrm{R}}=\mathrm{I}_y+M \mathrm{x}^2=\frac{\mathrm{MR}^2}{2}+M R^2=\frac{3}{2} \mathrm{MR}^2 \\
\therefore \quad & \frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{R}}}=\frac{\mathrm{MR}^2}{4} \times \frac{2}{3 \mathrm{MR}^2} \\
\therefore \quad & \frac{I_D}{I_R}=\frac{1}{6}
\end{array}\)
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