Question

A cone of radius = height = $r$ is inside a liquid of density $d$. Its base is parallel to the free surface of the liquid at a depth $H$ from it as shown in the figure. What is the net force due to liquid on its curved surface ? (neglect atmospheric pressure)

• A. $\pi r^{2}dg(H-\frac{r}{3})$
• B. $2\pi r^{2}dg(H-\frac{2r}{3})$
• C. $\pi r^{2}dg(2H-\frac{r}{3})$
• D. $\pi r^{2}dg(2H-\frac{2r}{3})$

Answer 1

The correct option is A $\pi r^{2}dg(H-\frac{r}{3})$

Force due to liquid on curved surface is weight of the liquid above it.
$F=$ weight of liquid above the cone
$F=(\pi r^{2}H-\frac{1}{3}\pi r^3)dg=\pi r^{2}dg[H-\frac{r}{3}]$