Question

The area of cross section of the wider tube shown in fig. is 900 cm${}^2$. If the body standing on the position weighs 45 kg, find the difference in the levels of water in the two tubes.

Answer 1

Given that,

m= 45 kg = 45000g

A = 900 cm²

Ρ = 1000 kg/m³ = 1 g/cm³

Let, the difference in the height of the water level be $hunits$.

Now,

According to the figure

$P_a+h\rho g=P_a+\frac{mg}{A}$

$h=\frac{m}{A\rho }$

Put the values of all elements

$h=\frac{45\times 1000}{900\times 1}$

$h=50cm$

Hence, the difference in the levels of water in the two tubes is $50$cm.