Question

The area of cross section of the wider tube shown in figure is 900 cm². If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes.

Answer 1

Given:
Area of the wider tube, A = 900 cm²
Weight of the boy, m = 45 kg
Density of water, $\rho$ = 10³ kgm⁻³
Let h be the difference in the levels of water in the tubes and p_a be the atmospheric pressure.


As per the figure, we have:
$p_a\mathit{+}h\rho g\mathit{=}{p}_a\mathit{+}\frac{mg}{A}$
$$\begin{gathered} \Rightarrow h\rho g\mathit{=}\frac{mg}{A} \\ \Rightarrow h\mathit{=}\frac{m}{\rho A} \\ \Rightarrow h=\frac{m}{1000\times \mathrm{A}} \\ =\frac{45}{1000\times 900\times {10}^{-4}}=\frac{1}{2}\mathrm{m}=50\mathrm{cm} \end{gathered}$$