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 the figure. 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 the figure. The density of the water is $\rho$. In equilibrium, the height $H$ of the water column in the cylinder satisfies

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

Answer 1

Answer: (C)

Applying Boyle's law, we have
$P_1{V}_1=P_2{V}_2$
That is, $P_0\left(\pi R^2{L}_0\right)=P\left[\pi R^2\left(L_0-H\right)\right](1)$
where $P=P_0+\left(L_0-H\right)\rho g(2)$
From Eq. (1), we get $P=\frac{P_0{L}_0}{L_0-H}$
Therefore,
$\rho =\frac{P_0{L}_0}{L_0-H}=P_0+\left(L_0-H\right)\rho g$
$\Rightarrow \rho g{\left(L_0-H\right)}^2+P_0\left(L_0-H\right)-P_0{L}_0=0$