Question

The height of a solid cylinder is $15cm$, and the diameter of its base is $7cm$. Two equal conical holes each of radius $3cm$, and height $4cm$ are cut off. Find the volume of the remaining solid.

Answer 1

Given,

Height of the cylinder $H=15cm$

Diameter of the cylinder $D=7cm$

Hence,Radius of the cylinder $R=\frac{D}{2}=3.5cm$

Two conical holes are cutoff from the cylinder.

Hight of the cone $h=4cm$

Radius of the cone $r=3cm$

Hence

Volume of the remaining solid, V = Volume of the cylinder - 2 ( Volume of the cone )

$\to V=\pi R^{2}H-2\times \frac{1}{3}\pi r^{2}h$

$\to V=3.14\times (3.5{)}^2\times 15-2\times \frac{1}{3}\times 3.14\times 3^2\times 4c{m}^3$

$\to V=3.14\times 12.25\times 15-2\times 3.14\times 3\times 4c{m}^3$

$\to V=576.975-75.36c{m}^3$

$\to V=501.61c{m}^3$

Hence, volume of the remaining solid is $501.61c{m}^3$