**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 *h *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 $0.5 \mathrm{~cm}$