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A vessel in the shape of a cuboid contains some water. If three indentical spheres are immersed in the water, the level of water is increased by $ 2 \mathrm{~cm} $. If the area of the base of the cuboid is $ 160 \mathrm{~cm}^{2} $ and its height $ 12 \mathrm{~cm} $, determine the radius of any of the spheres.
Given:
A vessel in the shape of a cuboid contains some water.
Three identical spheres are immersed in the water, the level of water is increased by \( 2 \mathrm{~cm} \).
The area of the base of the cuboid is \( 160 \mathrm{~cm}^{2} \) and its height \( 12 \mathrm{~cm} \).
To do:
We have to find the radius of any of the spheres.
Solution:
Area of the base of the cuboid $=160 \mathrm{~cm}^{2}$
Height of the cuboid $h=12 \mathrm{~cm}$
Level of water raised when 3 spheres are immersed $=2 \mathrm{~cm}$
This implies,
Volume of water raised $=$ Area of the base $\times$ Height of the water raised
$=160 \times 2$
$=320 \mathrm{~cm}^{3}$
Therefore,
Volume of 3 spheres $=$ Volume of water raised
$=320 \mathrm{~cm}^{3}$
Volume of each sphere $=\frac{320}{3} \mathrm{~cm}^{3}$
Let $r$ be the radius of each sphere.
This implies,
$\frac{4}{3} \pi r^{3}=\frac{320}{3}$
$\Rightarrow \frac{4}{3} \times \frac{22}{7} \times r^{3}=\frac{320}{3}$
$\Rightarrow r^{3}=\frac{320 \times 3 \times 7}{3 \times 4 \times 22}$
$\Rightarrow r^{3}=\frac{280}{11}=25.45$
$\Rightarrow r=\sqrt[3]{25.45}$
$\Rightarrow r=2.94 \mathrm{~cm}$
The radius of each sphere is $2.94 \mathrm{~cm}$.
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