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 $\frac{3}{2}$ cm and its depth is $\frac{8}{9}$ cm. Calculate the ratio in the volume of metal left in the cylinder to the volume of the metal taken out in conical shape.

Given:

Radius of the cylinder $R=3\ cm$

Height of the cylinder $H=5\ cm$
Radius of the cone $r=\frac{3}{2}$ cm

Height of the cone $h=\frac{8}{9}$ cm

To do:

We have to calculate the ratio in the volume of metal left in the cylinder to the volume of the metal taken out in conical shape.

Solution:

We know that,

Volume of a cylinder of height $H$ and radius $R=\pi R^2H$

Volume of a cone of height $h$ and radius $r=\frac{1}{3}\pi r^2h$

Therefore,

Volume of the metallic cylinder$=\frac{22}{7}\times(3)^2\times5\ cm^3$

$=\frac{22}{7}\times45\ cm^3$

Volume of the cone removed$=\frac{1}{3}\times\frac{22}{7}\times(\frac{3}{2})^2\times\frac{8}{9}\ cm^3$

$=\frac{22}{7}\times(\frac{9\times8}{3\times4\times9})\ cm^3$

$=\frac{22}{7}\times\frac{2}{3}\ cm^3$

Volume of the metal left in the metallic cylinder$=$ Volume of the metallic cylinder $-$ Volume of the cone removed

$=(\frac{22}{7}\times45-\frac{22}{7}\times\frac{2}{3})\ cm^3$

$=\frac{22}{7}(45-\frac{2}{3})\ cm^3$

$=\frac{22}{7}(\frac{45\times3-2}{3})\ cm^3$

$=\frac{22}{7}\times\frac{133}{3}\ cm^3$

The ratio of the volume of metal left in the cylinder to the volume of the metal taken out in conical shape

$=\frac{\frac{22}{7}\times\frac{133}{3}}{\frac{22}{7}\times\frac{2}{3}}$

$=\frac{133}{3}\times\frac{3}{2}$

$=\frac{133}{2}$

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

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Updated on: 10-Oct-2022

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