A uniform rod of length 2.0m 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 1.0m as shown in fig. Taking the case $\mathrm{\theta }\ne 0º$ find the force exerted by the hinge on the rod. (g=10m/s²)

Rod length in the water = $1\mathrm{sec\theta }=\mathrm{sec\theta }$
Upthrust,

$\mathrm{F}=\left(\frac{2}{2}\right)\left(\mathrm{sec\theta }\right)\left(\frac{1}{500}\right)\left(1000\right)\left(10\right)=20\mathrm{sec\theta }$

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

Net torque about O should be zero for the rod to be in rotational equilibrium.

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

$\frac{20}{2}{\sec }^2\mathrm{\theta }=20 \Rightarrow \mathrm{\theta }={45}^{\mathrm{o}}$

$\Rightarrow \mathrm{F}=20\sec {45}^{\mathrm{o}} =20\sqrt{2}\mathrm{N}$

The force applied to the rod by the hinge to maintain vertical balance will be $\left(20\sqrt{2}-20\right)$N downwards, or 8.3N downwards.

Hence the correct answer is 8.3N.