Question

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

Pressure inside the piston must be equal to the pressure at the bottom of the piston exerted by water to stay in equilibrium=$P_0+\rho g(L_0-H)$

Applying Boyle's Law, $PV=constant$

$P_0\left(A{L}_0\right)=(P_0+(L_0-H\left)\rho g\right)\left(A\right(L_0-H\left)\right)$

$⟹\rho g(L_0-H{)}^2+P_0(L_0-H)-L_0{P}_0=0$