PYQ NEET- Rotational Motion L-1

Question: The ratio of radius of gyration of a solid sphere of mass $M$ and radius $R$ about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is :-#

A) $5: 3$

B) $2: 5$

C) $\sqrt{5}: \sqrt{3}$

D) $\sqrt{3}: \sqrt{5}$

Answer: $\sqrt{3}: \sqrt{5}$#

Solution:#

To solve this problem, we need to find the ratio of the radius of gyration for a solid sphere $\left(\mathrm{K}_1\right)$ to the radius of gyration for a thin hollow sphere $\left(\mathrm{K}_2\right)$ of the same mass $M$ and radius $R$.

The moment of inertia (I) of a solid sphere about its own axis is given by : $$ I_{\text {solid }}=\frac{2}{5} M R^2 $$

The radius of gyration $(\mathrm{K})$ is related to the moment of inertia $(\mathrm{I})$ and mass $(\mathrm{M})$ by the formula : $$ I=M K^2 $$

So for the solid sphere, we can find $\mathrm{K}1$ using : $$ \begin{aligned} & K_1^2=\frac{I{\text {solid }}}{M}=\frac{2}{5} R^2 \ & K_1=R \sqrt{\frac{2}{5}} \end{aligned} $$

Now, for a thin hollow sphere, the moment of inertia about its axis is given by : $$ I_{\text {hollow }}=\frac{2}{3} M R^2 $$

We can find $\mathrm{K}2$ using : $$ K_2^2=\frac{I{\text {hollow }}}{M}=\frac{2}{3} R^2 $$ $$ K_2=R \sqrt{\frac{2}{3}} $$

Now, we need to find the ratio $K_1: K_2$ : $$ \frac{K_1}{K_2}=\frac{R \sqrt{\frac{2}{5}}}{R \sqrt{\frac{2}{3}}} $$

The R terms cancel out, and we are left with : $$ \frac{K_1}{K_2}=\frac{\sqrt{\frac{2}{5}}}{\sqrt{\frac{2}{3}}} $$

Simplify by taking the square root of the fraction : $$ \frac{K_1}{K_2}=\frac{\sqrt{2} \sqrt{3}}{\sqrt{5} \sqrt{2}} $$

Now, the square root of 2 terms cancel out, giving us : $$ \frac{K_1}{K_2}=\frac{\sqrt{3}}{\sqrt{5}} $$