Question

A heavy hollow cone of radius $R$ and height $h$ is placed on a horizontal table surface, with its flat base on the table. The whole volume inside the cone is filled with water of density $\rho$. The circular rim of the cone's base has a watertight seal with the table's surface and the top apex of the cone has a small hole. Neglecting atmospheric pressure, the total upward force exerted by water on the cone is

• A. $\left(2/3\right)\pi R^{2}h\rho g$
• B. $\left(1/3\right)\pi R^{2}h\rho g$
• C. $\pi R^{2}h\rho g$
• D. None

Answer 1

Answer: (A)

$\left(2/3\right)\pi R^{2}h\rho g$

Consider the equilibrium of water,
Weight of water(downwards) + Force due to table(upwards) + Force due to side walls(downwards)= 0
Weight of water = $\left(\frac{1}{3}\pi R^{2}h\right)\rho g$
Force due to table on water = Pressure at the bottom$\times$ Area
= $\rho gh\times \pi R^2$
Total upward force exerted by water on the cone = $\left(\rho gh\pi R^2\right)$- $\left(\frac{1}{3}\pi R^{2}h\rho g\right)$
= $\left(\frac{2}{3}\right)\pi R^{2}h\rho g$