Question

A solid in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is $2.1cm$ and the height of the cone is $4cm$.The solid is placed in a cylindrical vessel ,full of water ,in such a way that the whole solid is submerged in water.If the radius of the cylindrical vessel is $5cm$ and its height is $9.8cm$ ,find the volume of water left in the tub to the nearest $c{m}^3$.

Answer 1

Original volume of water in the cylindrical tub
= Volume of Cylinder
$=\pi r^{2}h$
$=\frac{22}{7}\times 5^2\times 9.8$
$=22\times 25\times 1.4$
$=770c{m}^3$
Given that Radius of hemisphere, $R=2.1cm$
and height of cone, $h=4cm$
Volume of solid = Volume of cone+ Volume of hemisphere
$=\frac{1}{3}\pi R^{2}h+\frac{2}{3}\pi R^3$
$=\frac{1}{3}\pi R^2(h+2R)$
$=\frac{1}{3}\times \frac{22}{7}\times 2.1\times 2.1(4+2\times 2.1)$
$=\frac{1}{3}\times 22\times 0.3\times 2.1\times 8.2$
$=37.884c{m}^3$
$\therefore$ Volume of water displaced ( removed )$=37.884c{m}^3$
Hence, the required volume of the water left in the cylindrical tub $=770-37.884$
$=732.116c{m}^3$