Question

A solid sphere of radius $r$ is floating at the interfare of two immiscible liquids of densities $p_1$ and $p_2(p_2>p_1)$, half of its volume lying in each. The height of the upper liquid column from the interfare of the two liquids is $h$. The force exerted on the upper liquid is (atmosphere pressure $=p_o$ and acceleration due to gravity is $g$):

• A. $p_0\pi r^2+(h-\frac{2}{3r})\pi r^2{p}_{1}g$
• B. $(h-\frac{2}{3r})\pi r^2{p}_{1}g$
• C. $\frac{2}{3}\pi r^3{p}_{1}g$
• D. $P_0\times \pi r^2$

Answer 1

The correct option is A $p_0\pi r^2+(h-\frac{2}{3r})\pi r^2{p}_{1}g$

$\therefore$ Force at upper surface of hemisphere (base area)=Weight of liquid above it $+$Force due to ATM pressure.

$\begin{array}{c}mg+p_0\left(\pi r^2\right)\leftarrow projectedarea\\ =\rho .V.g\left\{V=vol.ofwater\right\}\\ =\rho \left\{\pi r^{2}h-\frac{2}{3}\pi r^3\right\}.g\\ =\left(h-\frac{2}{3}r\right)\pi r^2\rho g\end{array}$

Force at upper surface of hemisphere by water is of same magnitude as force exerted by upper surface on water by Action.

Reaction pair

$F=p_0\left(\pi r^2\right)+\left(h-\frac{2}{3}r\right)\pi r^2\rho .g$