A cylindrical tub of radius $16\ cm$ contains water to a depth of $30\ cm$. A spherical iron ball is dropped into the tub and thus level of water is raised by $9\ cm$. What is the radius of the ball?

 Given:

A cylindrical tub of radius $16\ cm$ contains water to a depth of $30\ cm$. A spherical iron ball is dropped into the tub and thus level of water is raised by $9\ cm$. 

To do:

We have to find the radius of the ball.

Solution:

Radius of the cylindrical tub $(r) = 16\ cm$

Height of water in the tub $(h) = 30\ cm$

Therefore,

Volume of the water in the tub $=\pi r^{2} h$

$=\pi(16)^{2} \times 30$

$=\pi \times 256 \times 30$

$=7680 \pi \mathrm{cm}^{3}$

Water level raised $=9 \mathrm{~cm}$

The volume of the raised water $=\pi \times(16)^{2} \times 9$

$=256 \times 9 \times \pi$

$=2304 \pi \mathrm{cm}^{3}$

Volume of the ball $=2304 \pi \mathrm{cm}^{3}$

This implies,

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

$=\sqrt[3]{\frac{2304 \pi \times 3}{4 \times \pi}}$

$=\sqrt[3]{576 \times 3}$

$=\sqrt[3]{1728}$

$=\sqrt[3]{(12)^{3}}$

$=12 \mathrm{~cm}$

Hence, the radius of the ball is $12 \mathrm{~cm}$.

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Updated on: 10-Oct-2022

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