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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