91Ó°ÊÓ

The face lengths of the rectangular block is $\text{a}=35\text{cm}=35×{10}^{−2}\text{m}$ and $\text{b}=45\text{cm}=45×{10}^{−2}\text{m}$

Use the parallel axis theorem and expression of the period for physical pendulum.

Formula:

$\begin{array}{c}\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{com}}\mathbf{+}{\mathbf{mh}}^{\mathbf{2}}\\ \mathbf{T}\mathbf{=}\mathbf{2}\mathbf{Ï€}\sqrt{\frac{\mathbf{I}}{\mathbf{mgh}}}\end{array}$

(a)

According to the parallel axis theorem,

$\text{I}={\text{I}}_{\mathrm{com}}+{\text{mh}}^2……………………………………………………………………………………….\left(1\right)$

Here, the distance of the hole from the center of the rectangular block is $h=r$

The axis of rotation passing through the center of the rectangular block and perpendicular to its plane is

${\text{I}}_{\mathrm{com}}=\frac{\text{m}({\text{a}}^2+{\text{b}}^2)}{12}$

The equation (1) becomes as,

$\text{I}=\frac{\text{m}({\text{a}}^2+{\text{b}}^2)}{12}+{\text{mr}}^2$

The period of oscillation of the physical pendulum is

$\begin{array}{c}\text{T}=2\mathrm{π}\sqrt{\frac{\text{I}}{\text{mgh}}}\\ ⇒\text{T}=2\mathrm{π}\sqrt{\frac{\frac{\text{m}({\text{a}}^2+{\text{b}}^2)}{12}+{\text{mr}}^2}{\text{mgr}}}\\ ⇒\text{T}=2\mathrm{π}\sqrt{\frac{\frac{({\text{a}}^2+{\text{b}}^2)}{12}+{\text{r}}^2}{\text{gr}}}\\ ⇒\text{T}=\frac{2\mathrm{π}}{\sqrt{\text{g}}}\sqrt{\frac{({\text{a}}^2+{\text{b}}^2)}{12\text{r}}+\text{r}}\end{array}$

We plot the graph $T$ verses$\mathrm{r}$ for the given $a$ and $b$ as

(b)

For the minimum value of $\mathrm{T}$, its first derivative is zero.

$\begin{array}{c}\frac{\text{dT}}{\text{dr}}=\frac{2\mathrm{π}}{\sqrt{\text{g}}}\sqrt{\frac{({\text{a}}^2+{\text{b}}^2)}{12}−{\text{r}}^2}\\ ⇒\text{0}=\frac{2\mathrm{π}}{\sqrt{\text{g}}}\sqrt{\frac{({\text{a}}^2+{\text{b}}^2)}{12}−{\text{r}}^2}\\ ⇒0=\sqrt{\frac{({\mathrm{a}}^2+{\mathrm{b}}^2)}{12}}−\mathrm{r}\\ ⇒\text{r}=\sqrt{\frac{[{(35×{10}^{−2}\mathrm{m})}^2+{(45×{10}^{−2}\mathrm{m})}^2]}{12}}\end{array}$

$⇒\text{r}=0.16\text{m}$

(c)

The direction from the center is not important. Hence, the locus of the point is a circle around the center of radius
$\sqrt{\frac{({\text{a}}^2+{\text{b}}^2)}{12}}$