Question

A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of `3/2` cm and its depth is `8/9 `cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.

Answer 1

Given:
Radius of the cylinder, R = 3 cm
Height of the cylinder, H = 5 cm

∴ Volume of the cylinder \[= \pi R^2 H\]

\[= \pi \times \left( 3 \right)^2 \times \left( 5 \right)\]
\[ = 45\pi {cm}^3\]

Radius of the cone, r = \[\frac{3}{2}\] cm

Height of the cone, h = \[\frac{8}{9}\] cm

∴ Volume of the cone removed from the cylinder \[= \frac{1}{3}\pi r^2 h\]

\[= \frac{1}{3} \times \pi \times \left( \frac{3}{2} \right)^2 \times \left( \frac{8}{9} \right)\]
\[ = \frac{2\pi}{3} {cm}^3\]

According to the question,

Volume of the metal left in the cylinder = Volume of the cylinder − Volume of the cone

\[= 45\pi - \frac{2\pi}{3}\]

\[= \left( 45 - \frac{2}{3} \right)\pi\]

 

\[\therefore\frac{\text{Volume of metal left in the cylinder}}{\text{Volume of metal taken out in conical shape}} = \frac{\frac{133\pi}{3}}{\frac{2\pi}{3}} = \frac{133}{2}\]

Thus, the ratio of the volume of the metal left in the cylinder to the volume of the metal taken out in conical shape is 133 : 2.