A cylindrical tub of radius $ 12 \mathrm{~cm} $ contains water to a depth of $ 20 \mathrm{~cm} $. A spherical form ball of radius $ 9 \mathrm{~cm} $ is dropped into the tub and thus the level of water is raised by $ h \mathrm{~cm} $. What is the value of $ h $?

Given:

A cylindrical tub of radius \( 12 \mathrm{~cm} \) contains water to a depth of \( 20 \mathrm{~cm} \).

A spherical form ball of radius \( 9 \mathrm{~cm} \) is dropped into the tub and thus the level of water is raised by \( h \mathrm{~cm} \).

To do:

We have to find the value of \( h \).

Solution:

Radius of the spherical ball $r=9 \mathrm{~cm}$

This implies,

Volume of the spherical ball $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi \times(9)^{3}$

$=4 \times 243 \pi$

$=972 \pi \mathrm{cm}^{3}$

Volume of the water raised in the tub $=$ Volume of the spherical ball

$=972 \pi \mathrm{cm}^{3}$
Radius of the tub $\mathrm{R})=12 \mathrm{~cm}$

Rise in the level of water $=h \mathrm{~cm}$

Therefore,

Volume of the water raised in the tub $=\pi \mathrm{R}^{2} h$

$\pi R^{2} h=972 \pi$

$\Rightarrow \pi(12)^{2} h=972 \pi$

$\Rightarrow 144 \pi h=972 \pi$

$\Rightarrow h=\frac{972 \pi}{144 \pi}$

$\Rightarrow h=\frac{27}{4} \mathrm{~cm}$

$\Rightarrow h=6.75 \mathrm{~cm}$

The value of \( h \) is $6.75\ cm$.

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Updated on: 10-Oct-2022

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