A cylinder of radius $12\ cm$ contains water to a depth of $20\ cm$. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by $6.75\ cm$. Find the radius of the ball. (Use $\pi = \frac{22}{7}$).

 Given:

A cylinder of radius $12\ cm$ contains water to a depth of $20\ cm$. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by $6.75\ cm$. 

To do:

We have to find the radius of the ball.

Solution:

Radius of the cylinder $(r) = 12\ cm$

Depth of the water in the cylinder $(h) = 20\ cm$

The water level rises by $6.75\ cm$.

Therefore,

The volume of the water raised $=\pi r^{2} h$

$=\pi(12)^{2} \times 6.75$

$=144 \times 6.75 \pi$

$=972 \pi \mathrm{cm}^{3}$

This implies,

Volume of the ball $=972 \pi \mathrm{cm}^{3}$

The radius of the ball $=\sqrt[3]{\frac{\text { Volume }}{\frac{4}{3} \pi}}$

$=\sqrt[3]{\frac{972 \pi \times 3}{4 \pi}}$

$=\sqrt[3]{243 \times 3}$

$=\sqrt[3]{729}$

$=\sqrt[3]{(9)^{3}}$

$=9 \mathrm{~cm}$

Hence, the radius of the ball is $9 \mathrm{~cm}$.

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Updated on: 10-Oct-2022

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