Question

A cone of radius $R$ and height $H$, is hanging inside a liquid of density $\rho$ by means of a string as shown in the figure. The force, due to the liquid acting on the slant surface of the cone is (Neglect atmosphere pressure)

• A. $\rho \pi g{HR}^2$
• B. $\pi \rho {HR}^2$
• C. $\frac{4}{3}\pi \rho g{HR}^2$
• D. $\frac{2}{3}\pi \rho g{HR}^2$

Answer 1

The correct option is D $\frac{2}{3}\pi \rho g{HR}^2$

Buoyancy force = - Weight of fluid displaced by body
= Resultant of forces acting on the body due to pressure differences
Weight of fluid displaced acting downwards = $\left(\frac{1}{3}\pi R^{2}H\right)\times \rho g$
Force acting on the bottom of cone acting upwards =$\left(\rho gH\right)\times \pi R^2$ (Assuming tip of cone is just touching the surface)
Net force acting sidewards on the cone is zero by symmetry.
Weight of fluid displaced by body(downwards) = Force acting on the bottom(upwards) - Force acting on the slant surface of cone
The force acting on the slant surface of the cone = ($\rho gH\pi R^2$) - ($\frac{1}{3}\pi R^{2}H\rho g$)
= $\frac{2}{3}\pi \rho gH{R}^2$