Question:
The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm and height 4 cm are cut off. Find the volume of the remaining solid.
Solution:
The height of cylinder h = 15 cm
Radius of cylinder $r=\frac{7}{2}$
The volume of cylinder
$=\pi r^{2} h$
$=\pi \times\left(\frac{7}{2}\right)^{2} \times 15 \mathrm{~cm}^{2}$
$=183.75 \pi$
The radius of conical holes = 3 cm
Height of conical holes = 4 cm.
The volume of conical holes
$=\frac{1}{3} \pi r^{2} h$
$=12 \pi \mathrm{cm}^{3}$
Clearly,
The volume of remaining solid
= vol. of cylinder − 2 × vol. of cone
$=183.75 \pi-24 \pi$
$=501.6 \mathrm{~cm}^{3}$