Question

From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid.

Answer 1

Height of the cylinder, h = 14 cm
Base diameter = 7 cm
So, base radius, r = 7/2 = 3.5 cm
Volume of the cylinder $=\pi r^{2}h=\frac{22}{7}\times {3.5}^2\times 14=539c{m}^2$
Radius of the conical holes, r' = 2.1 cm
Height of the conical hole, h' = 4 cm
Volume of each conical hole $=\frac{1}{3}\pi r^{\prime 2}{h}^{\prime }=\frac{1}{3}\times \frac{22}{7}\times {2.1}^2\times 4=18.48c{m}^2$
Total volume of the two conical holes $2\times 18.48=36.96c{m}^3$
Hence, volume of the remaining part = Volume of the cylinder - Volume of two conical hole
$=539-36.96=502.04c{m}^3$