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