A rectangular tank \( 15 \mathrm{~m} \) long and \( 11 \mathrm{~m} \) broad is required to receive entire liquid contents from a full cylindrical tank of internal diameter \( 21 \mathrm{~m} \) and length \( 5 \mathrm{~m} \). Find the least height of the tank that will serve the purpose.
Given:
A rectangular tank \( 15 \mathrm{~m} \) long and \( 11 \mathrm{~m} \) broad is required to receive entire liquid contents from a full cylindrical tank of internal diameter \( 21 \mathrm{~m} \) and length \( 5 \mathrm{~m} \).
To do:
We have to find the least height of the tank that will serve the purpose.
Solution:
Internal diameter of the cylindrical tank $=21 \mathrm{~m}$
This implies,
Radius of the cylindrical tank $r=\frac{21}{2} \mathrm{~m}$
Length of the cylindrical tank $\mathrm{H})=5 \mathrm{~m}$
Therefore,
Volume of water filled in the tank $=\pi r^{2} h$
$=\frac{22}{7} \times(\frac{21}{2})^{2} \times 5$
$=\frac{22}{7} \times \frac{441}{4} \times 5$
$=\frac{3465}{2} \mathrm{~m}^{3}$
Volume of water in the rectangular tank $=$ Volume of water in the cylindrical tank
$=\frac{3465}{2} \mathrm{~m}^{3}$
Length of the rectangular tank $l=15 \mathrm{~m}$
Breadth of the rectangular tank $b=11 \mathrm{~m}$
Let the height of the rectangular tank be $h$.
Therefore,
$15 \times 11 \times h=\frac{3465}{2}$
$\Rightarrow h=\frac{3465}{2} \times \frac{1}{15 \times 11}$
$\Rightarrow h=\frac{21}{2}$
$\Rightarrow h=10.5 \mathrm{~m}$
The least height of the tank that will serve the purpose is $10.5 \mathrm{~m}$.
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