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

At equilibrium ,$p=P$

$P=P_0(L_0-H)Pg$...........(1)

Also

$P_{0}x\left(\pi R^2{L}_0\right)=P\left[\pi R^2\right(L_0-H\left)\right)$

$P=\frac{L_0{P}_0}{L_0-H}$............(ii)

form eqs (i) and (ii) we get

$\frac{L_0{P}_0}{L_0-H}=P_o+(L_0-H)pg$

$L_0{P}_0=P_0(l_0-H)+(L_0-H{)}^{2}pg$

$Pg(L_0-H{)}^2+P_0(L_0-H)-L_0{P}_0=0$