# The area of cross section of the wider tube shown in figure is $900cm^2$.If the boy standing on the piston weights 45Kg then difference in the levels of water in the two tubes are ($\rho_{water}=1000Kg/m^3)$

$\begin{array}{1 1}(A)\;60cm\\(B)\;52cm\\(C)\;50cm\\(D)\;45cm\end{array}$

From figure $P_B=P_C$
$P_a+\rho hg=P_a+\large\frac{mg}{A}$
$\Rightarrow h=\large\frac{m}{A\rho}$
$h=\large\frac{45kg}{1000 Kg/m^3\times 900 cm^2}$
$h=50cm$
Hence (C) is the correct answer.