Question

A fixed thermally conducting cylinder has a radius $R$ and height $L_0$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass $M$ is held at a distance $L$ from the top surface as shown in Fig. The atmospheric pressure is $p_0$.
The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in Fig. The density of the water is $\rho$. In equilibrium, the height $H$ of the water column in the cylinder satisfies.

• A. $\rho g(L_0-H{)}^2+P_0(L_0-H)+L_0{P}_0=0$
• B. $\rho g(L_0-H{)}^2-P_0(L_0-H)-L_0{P}_0=0$
• C. $\rho g(L_0-H{)}^2+P_0(L_0-H)-L_0{P}_0=0$
• D. $\rho g(L_0-H{)}^2-P_0(L_0-H)+L_0{P}_0=0$

Answer 1

The correct option is C $\rho g(L_0-H{)}^2+P_0(L_0-H)-L_0{P}_0=0$

$p_1=p_2$
$p_0+\rho g(L_0-H)=p...\left(i\right)$
Now, applying $p_1{V}_1=p_2{V}_2$ for the air inside the cylinder, we have
$p_0\left(L_0\right)=p(L_0-H)$
$p=\frac{p_0{L}_0}{L_0-H}$
Substituting in Eq. (i), we have
$p_0+\rho g(L_0-H)=\frac{p_0{L}_0}{L_0-H}$
$\Rightarrow \rho g(L_0-H{)}^2+P_0(L_0-H)-L_0{P}_0=0$
Therefore, option (c) is correct.