A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemi-sphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub (Take π = 22/7)
To find the volume of the water left in the tube, we have to subtract the volume of the hemisphere and cone from volume of the cylinder.
For right circular cylinder, we have
$r=5 \mathrm{~cm}$
$h=9.8 \mathrm{~cm}$
The volume of the cylinder is
$V_{1}=\pi r^{2} h$
$=\frac{22}{7} \times 5^{2} \times 9.8$
$=770 \mathrm{~cm}^{3}$
For hemisphere and cone, we have
$r=3.5 \mathrm{~cm}$
$h=5 \mathrm{~cm}$
Therefore the total volume of the cone and hemisphere is
$V_{2}=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}$
$=\frac{1}{3} \times \frac{22}{7} \times 3.5^{2} \times 5+\frac{2}{3} \times \frac{22}{7} \times 3.5^{3}$
$=154 \mathrm{~cm}^{3}$
The volume of the water left in the tube is
$V=V_{1}-V_{2}$
$=770-154$
$=616 \mathrm{~cm}^{3}$
Hence, the volume of the water left in the tube is $V=616 \mathrm{~cm}^{3}$