Let $\mathrm{M}$ and $\mathrm{L}$ be the mass and length of thin uniform rod respectively. In $1^{ ext {st }}$ case, axis of rotation is passing through centre and perpendicular to its length. In $2^{ ext {nd }}$ case, axis of rotation is passing through one end and perpendicular to its length.…

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Let $\mathrm{M}$ and $\mathrm{L}$ be the mass and length of thin uniform rod respectively. In $1^{	ext {st }}$ case, axis of rotation is passing through centre and perpendicular to its length. In $2^{	ext {nd }}$ 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

Aicharya · Accepted Answer

$\begin{array}{l}\mathrm{I}_{1}=\frac{\mathrm{ML}^{2}}{12}=\mathrm{MK}_{1}^{2} \ \mathrm{I}_{2}=\frac{\mathrm{ML}^{2}}{3}=\mathrm{MK}_{2}^{2} \ \mathrm{k}_{2}=\frac{\mathrm{L}}{\sqrt{3}} \ \frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\frac{1}{2}\end{array} \quad 	herefore \mathrm{K}_{1}=\frac{\mathrm{L}}{\sqrt{12}}=\frac{\mathrm{L}}{2 \sqrt{3}}$

Let $\mathrm{M}$ and $\mathrm{L}$ be the mass and length of thin uniform rod respectively. In $1^{\text {st…

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