Question

If two equal conical holes of diameter 6 cm and height 4 cm are drilled from a cylinder of diameter 8 cm and height 40 cm, find the volume of the remaining solids.

Answer 1

Answer:

Given:

Cylinder:

• $Diameter = 8 \ cm$
• $Height = 40 \ cm$

Cones :

• $Diameter = 6 \ cm$
• $Height = 4 \ cm$

Solution

Cone:

$\boxed{\boxed{Volume = \dfrac{1}{3} \pi r^2 h}}$

$\therefore Volume \ of \ cones = 2 \times \dfrac{1}{3}\pi r^2 h$

$V_1 = \dfrac{2}{3} \pi 3^2(4)$

Cylinder:

$\boxed{\boxed{Volume = \pi r^2 h}}$

$\therefore Volume = \pi r^2 h$

$V_2 = \pi 4^2(40)$

Remaining volume:

$\implies V_2 - V_1$

$\implies \pi 4^2(40) - \dfrac{2}{3} \pi 3^2(4)$

$\implies \pi \left(16(40) - \dfrac{2}{3} 9(4)\right)$

$\implies \pi (640 - 24)$

$\implies \dfrac{22}{7} \times 616$

$\boxed{\implies{\boxed{1936 \ cm^3}}}$