500 men took a dip in a tank which is 80m long and 50m broad. What is the rise in the water level if the average displacement of water by a man is $4m^{3}$.

A) 25 cm

B) 20 cm

C) 45 cm

D) 50 cm


Option D


Volume of water displayed by 1 man = $4m^{3}$

Volume of water displayed by 500 men =$4\times 500$


Let xm be the rise in water level in the tank.

Therefore, Volume of water in the tank = $80\times 50\times xm^{3}$

$80\times 50\times x=2000$

x=$\frac{2000}{80\times 50}$


= 50 cm