A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical ball dropped into the tub and the level of the water is raised by 6.75 cm. Find the radius of the ball.
The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball into the tub, the height of the raised water is 6.75cm. Therefore, the volume of the raised water is
$V=\pi \times(12)^{2} \times 6.75 \mathrm{~cm}^{3}$
Let, the radius of the spherical ball isĀ r. Therefore, the volume of the spherical ball is
$V_{1}=\frac{4}{3} \pi \times r^{3} \mathrm{~cm}^{3}$
Since, the volume of the raised water is same as the volume of the spherical ball, we have
$V_{1}=V$
$\Rightarrow \frac{4}{3} \pi \times r^{3}=\pi \times(12)^{2} \times 6.75$
$\Rightarrow \quad r^{3}=\frac{(12)^{2} \times 6.75 \times 3}{4}$
$\Rightarrow \quad=12 \times 3 \times 6.75 \times 3$
$\Rightarrow \quad=9$
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