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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Updated on: 10-Oct-2022

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