A cylindrical tub of radius $ 12 \mathrm{~cm} $ contains water to a depth of $ 20 \mathrm{~cm} $. A spherical ball is dropped into the tub and the level of the water is raised by $ 6.75 \mathrm{~cm} $. Find the radius of the ball.

Given:

A cylindrical tub of radius \( 12 \mathrm{~cm} \) contains water to a depth of \( 20 \mathrm{~cm} \). A spherical ball is dropped into the tub and the level of the water is raised by \( 6.75 \mathrm{~cm} \).

To do:

We have to find the radius of the ball.

Solution:

Radius of the cylindrical tub $r=12 \mathrm{~cm}$

Depth of the water in the tub $h=20 \mathrm{~cm}$

Height of the water raised after dropping the ball $H=20+6.75$

$=26.75 \mathrm{~cm}$

Therefore,

Volume of the ball $= \pi r^{2}(H-h)$

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

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

Let the radius of the spherical ball be $R$.

Therefore,

$\frac{4}{3} \pi \mathrm{R}^{3}=144 \times 6.75 \pi$

$\mathrm{R}^{3}=\frac{144 \times 6.75 \pi \times 3}{4 \times \pi}$

$\mathrm{R}^{3}=\frac{36 \times 3 \times 675}{100}$

$\mathrm{R}^{3}=729$

$\mathrm{R}^{3}=(9)^{3}$

$\Rightarrow \mathrm{R}=9\ cm$

The radius of the ball is $9\ cm$.

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Updated on: 10-Oct-2022

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