Question

A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is $3.5cm$ and the height of the cone is $4cm$. The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water. If the radius of the cylinder is $5cm$ and its height is $10.5cm$, find the volume of water left in the cylindrical tub $(Use\pi =\frac{22}{7})$

Answer 1

We have, $VO=4cm,OA=OB=O{O}^{\prime }=3.5cm$

Volume of solid = volume of its conical part + volume of its semi-spherical part

Volume of the solid $\{\frac{1}{3}\times \frac{22}{7}\times {\left(3.5\right)}^2\times 4+\frac{2}{3}\times \frac{22}{7}\times {\left(3.5\right)}^3\}{cm}^3$

$=\frac{1}{3}\times \frac{22}{7}\times {\left(3.5\right)}^2(4+2\times 3.5){cm}^3=\{\frac{1}{3}\times \frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\times 11\}{cm}^3$

Clearly when the solid is submerged in the cylindrical tub the volume of water that flows out of the cylinder is equal to the volume of the solid

hence,
Volume of water left in the cylinder = volume of cylinder - volume of the solid

$=\{\frac{22}{7}\times {\left(5\right)}^2\times 10.5-\frac{1}{3}\times \frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\times 11\}{cm}^3$

$=683.83{cm}^3$