Question

In the figure shown, a rod of length $l(<2R)$ and mass $m$ is kept horizontally inside a smooth spherical shell as shown in the figure. The normal reaction when $l=R$ is $\frac{mg}{\sqrt{y}}.$ Find $y.(g=$ Acceleration due to gravity $)$

Answer 1

Answer: 3

$2N\sin \theta =mg$
$N=\frac{mg}{2}\text{cosec}\theta =\frac{mg}{2}\frac{R}{\sqrt{R^2-\frac{l^2}{4}}}$

When $l=R$,
$N=\frac{mg}{2}\frac{R}{\sqrt{\frac{3R^2}{4}}}=\frac{mg}{\sqrt{3}}$