Question

A number of holes are drilled along a diameter of a disc of radius $R$. To get minimum time period of oscillations, the disc should be suspended from a horizontal axis passing through a hole whose distance from the centre should be

• A. $\frac{R}{2}$
• B. $\frac{R}{\sqrt{2}}$
• C. $\frac{R}{2 \sqrt{2}}$
• D. Zero

Answer 1

Answer: (B)

$T=2 \pi \sqrt{\frac{L}{g}} $
where $L=\frac{l^{2}+k^{2}}{l}$
Here $k^{2}=\frac{R^{2}}{2}$
$\therefore L=\frac{l^{2}+\frac{R^{2}}{2}}{l}=l+\frac{R^{2}}{2 l}$
For minimum time period, $L$ should be minimum.
Hence $\frac{d L}{d l}=0 $
$\Rightarrow \frac{d}{d l}\left(l+\frac{R^{2}}{2 l}\right)=0$
$\Rightarrow 1+\frac{R^{2}}{2}\left(\frac{-1}{l^{2}}\right)=1-\frac{R^{2}}{2 l^{2}}=0 $
$\Rightarrow l=\frac{R}{\sqrt{2}}$