Question

If the radius of solid sphere is $\frac{1}{2}\sqrt{\frac{5}{7}}$ m, Thin its radius of gyration when the axis is along the tangent is -

• A. $\frac{5}{14}$ m
• B. $\frac{1}{2}$ m
• C. $\frac{5}{7}$ m
• D. 1 m

Answer 1

Answer: (B)

$\frac{1}{2}$ m

$\begin{array}{c}\text{Given, radius of solid sphere}=\frac{1}{2}\sqrt{\frac{5}{7}}\text{m.}\left(r\right)\text{.[}\mathrm{mass}=m]\\ \text{we know, moment of inertia about diameter}=\frac{2}{5}m{r}^2\text{.}\end{array}$

$\begin{array}{c}\text{By parallel axis theorem:}\\ \text{moment of inertia about tangent}\left(I\right)=\frac{2}{5}m{r}^2+m{r}^2=\frac{7}{5}m{r}^2\end{array}$

$\begin{array}{c}\text{Let, radius of gyration is}k\text{.}\\ \text{Then, we can write,}I=m{k}^2\Rightarrow k=\sqrt{\frac{I}{m}}\\ \text{So,}k=\sqrt{\frac{\frac{7}{5}m{r}^2}{m}}=\sqrt{\frac{7}{5}}r=\frac{1}{2}m\end{array}$