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 find the total upward force exerted by water on the cone is
Diffcult
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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
$=\rho g h \times \pi R^{2}$
Total upward force exerted by water on the cone $=\left(\rho g h \pi R^{2}\right)-\left(\frac{1}{3} \pi R^{2} h \rho g\right)$
$=\left(\frac{2}{3}\right) \pi R^{2} h \rho g$
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