Question

A uniform rod of length $2m$, specific gravity $0.5$ and mass $2kg$ is hinged at one end to the bottom of a tank of water (specific gravity $=1.0$) filled upto a height of $1m$ as shown in the figure. Taking the case $\theta \ne 0^o$, the force exerted by the hinge on the rod is :

• A. $10.2N$ upwards
• B. $4.2N$ downwards
• C. $8.3N$ downwards
• D. $6.2N$ upwards

Answer 1

The correct option is C $8.3N$ downwards

Length of rod inside the water $=1\sec \theta =\sec \theta$

Upthrust, $F=\left(\frac{2}{2}\right)\left(\sec \theta \right)\left(\frac{1}{500}\right)\left(1000\right)\left(10\right)$

$\Rightarrow F=20\sec \theta$

Weight of rod, $w=20\times 10=20N$

For rotational equilibrium of rod, net torque about $O$ should be zero.

$\therefore F\left(\frac{\sec \theta }{2}\right)\left(\sin \theta \right)=w=\left(1\sin \theta \right)$

$\Rightarrow \frac{20}{2}{\sec }^2\theta =20$

$\Rightarrow \theta ={45}^o$

$\Rightarrow F=20\sec {45}^o$

$F=20\sqrt{2}N$

For vertical equilibrium of rod, force exerted by the hinge on the rod will be $(20\sqrt{2}-20)N$ downwards i.e. $8.3N$ downwards.