Question

In a smooth hemispherical shell of radius $R,$ a rod of mass $\left(\sqrt{3}/2\right)kg$ is placed horizontally and to equilibrium. The length of rod is $R$ . Find the normal reaction at any end of the rod (in $N$ ). [ Take $g=10m/s^2$ ]

• A. $5$
• B. $\frac{5}{\sqrt{3}}$
• C. $5\sqrt{3}$
• D. $10$

Answer 1

The correct option is A $5$

The downward force acting at any one end of the rod is given as,

$F=\frac{mg}{2}$

$F=\frac{\frac{\sqrt{3}}{2}\times 10}{2}$

$F=\frac{10\sqrt{3}}{4}$

The normal reaction is given as,

$F=N\sin {60}^{\circ }$

$N=\frac{10\sqrt{3}}{4}\times \frac{2}{\sqrt{3}}$

$N=5N$