500 persons have to dip in a rectangular tank which is 80 m long and 50 m broad.
Question:

500 persons have to dip in a rectangular tank which is $80 \mathrm{~m}$ long and $50 \mathrm{~m}$ broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is $0.04 \mathrm{~m}^{2}$ ?

Solution:

The average displacement of water by a person is $0.04$ cubic $m$. Hence, the total displacement of water in the rectangular tank by 500 persons is $V=500 \times 0.04=20$ Cubic $m$.

The length and width of the rectangular tank are 80m and 50m respectively. Upon dipping in the tank, let the height of the raised water is be m. Therefore, the volume of the raised water is

$V_{1}=80 \times 50 \times h$

$=4000 h$ cubic $m$

Since, the volume of the raised water is same as the volume of the water displaced by 500 persons, we have

$V_{1}=V$

$\Rightarrow 4000 h=20$

$\Rightarrow \quad h=\frac{20}{4000}$

$\Rightarrow \quad=0.005$

Therefore, the water will be raised by $0.005 \mathrm{~m}$ or $0.5 \mathrm{~cm}$

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