MHT CET · Physics · Rotational Motion
Let \(\mathrm{M}\) and \(\mathrm{L}\) be the mass and length of thin uniform rod respectively. In \(1^{\text {st }}\) case, axis of rotation is passing through centre and perpendicular to its length. In \(2^{\text {nd }}\) case, axis of rotation is passing through one end and perpendicular to its length. The ratio of radius of gyration in first case to second case is
- A \(3: 1\)
- B \(1: 2\)
- C \(2: 1\)
- D \(1: 3\)
Answer & Solution
Correct Answer
(B) \(1: 2\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}\mathrm{I}_{1}=\frac{\mathrm{ML}^{2}}{12}=\mathrm{MK}_{1}^{2} \\ \mathrm{I}_{2}=\frac{\mathrm{ML}^{2}}{3}=\mathrm{MK}_{2}^{2} \\ \mathrm{k}_{2}=\frac{\mathrm{L}}{\sqrt{3}} \\ \frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\frac{1}{2}\end{array}\)
\(\therefore \mathrm{K}_{1}=\frac{\mathrm{L}}{\sqrt{12}}=\frac{\mathrm{L}}{2 \sqrt{3}}\)
\(\therefore \mathrm{K}_{1}=\frac{\mathrm{L}}{\sqrt{12}}=\frac{\mathrm{L}}{2 \sqrt{3}}\)
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