Question

In a spherical container of radius $R$, liquid of density $\rho$ and surface tension $T$ is filled upto half volume of container. Consider ABC as imaginary half circular plane which divides the liquid into equal parts, then force applied by the left part liquid on the right part liquid will be:

• A. $(\rho g{R}^3+\frac{P_0\pi R^2}{2}-\pi RT)$
• B. $\mid \rho g{R}^3+P_0\pi R^2+TR\mid$
• C. $(\rho g{R}^3+\frac{P_0\pi R^2}{2}-2TR)$
• D. $\mid \frac{-2\rho }{3}g{R}^3+\frac{P_0\pi R^2}{2}-2TR\mid$

Answer 1

Answer: (D)

$\mid \frac{-2\rho }{3}g{R}^3+\frac{P_0\pi R^2}{2}-2TR\mid$

Consider a thin slice of width $dx$ at a depth $x$ from the top of the liquid surface.

Area of this slice is $dA\left(x\right)=2\sqrt{R^2-x^2}$

Pressure at depth x is $P\left(x\right)=P_0+\rho gx$

Force on this slice from by the liquid on the left is

$dF=P\left(x\right)\times dA\left(x\right)=(P_0+\rho gx)\times 2\sqrt{R^2-x^2}dx$

Integrating for $x$ between $0$ to $R$

$F={\int }_0^R(P_0+\rho gx)\times 2\sqrt{R^2-x^2}dx={\int }_0^{R}2P_0\sqrt{R^2-x^2}dx+{\int }_0^{R}2\rho gx\sqrt{R^2-x^2}dx$

$=2P_0({\frac{x\sqrt{R^2-x^2}}{2}+\frac{R^2}{2}{\sin }^{-1}\frac{x}{R}\left)\right|}_0^R-{\left(2\rho g\frac{{\sqrt{R^2-x^2}}^{\frac{3}{2}}}{\frac{3}{2}}\right)|}_0^R$

$\therefore F=\frac{P_0\pi R^2}{2}-\frac{2}{3}\rho g{R}^3$

The surface tension force is $2TR$ and acts opposite to the pressure force.

Hence, $F=\frac{P_0\pi R^2}{2}-\frac{2}{3}\rho g{R}^3-2TR$

$=|\frac{-2p}{3}g{R}^3+\frac{P_0\pi R^2}{2}-2TR|$