A cylindrical tub of radius $12\ cm$ contains water to a depth of $20\ cm$. A spherical form ball is dropped into the tub and thus the level of water is raised by $6.75\ cm$. What is the radius of the ball?

 Given:

A cylindrical tub of radius $12\ cm$ contains water to a depth of $20\ cm$. A spherical form ball is dropped into the tub 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 cylindrical tub $(r) = 12\ cm$

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

The level of water raised $=6.75\ cm$

Therefore,

Volume of the ball $=\pi r^{2} h$

$=\pi \times 12 \times 12 \times 6.75$

$=6.75 \times 144 \pi \mathrm{cm}^{3}$

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

$=\sqrt[3]{\frac{3 \times 6.75 \times 144 \pi}{4 \pi}}$

$=\sqrt[3]{36 \times 3 \times 6.75}$

$=\sqrt[3]{729}$

$=9 \mathrm{~cm}$

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Updated on: 10-Oct-2022

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