A boy of $45 k g$ weight is standing on the piston of cross sectional area $900 c m^{2}$ . Find the difference in the levels of water in the two tubes.

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Solution

Given area, $A = 900 {c m}^{2} = \frac{900}{10^{4}} m^{2}$
Density of water, $ρ = \frac{1000 k g}{m^{3}}$

Let the difference in the height of two columns be $H$ .

Using Pascal's Law, $P_{A} = P_{B}$
$\frac{M g}{a} + P_{a t m} = P_{a t m} + ρ g H$
$\frac{45}{\frac{900}{10^{4}}} = 1000 \times H$
$H = 0.50 m = 50 c m$

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