**Problem:**
A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immiscible, non-viscous and incompressible liquids of densities d and 2d, each of height H/2 as shown in the figure. The lower density liquid is open to the atmosphere having pressure p_{0}.
(IIT JEE 1995)

- A homogeneous solid cylinder of length L, (L < H/2), cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid. Determine,
- the density of the solid.
- the total pressure at the bottom of the container.

- The cylinder is removed and the original arrangement is restored. A tiny hole of area $s\,(s\ll A)$ is punched on the vertical side of the container at a height $h\,(h < H/2)$. Determine,
- the initial speed of efflux of the liquid at the hole.
- the horizontal distance $x$ travelled by the liquid initially.
- the height $h_m$ at which the hole should be punched so that the liquid travels the maximum distance $x_m$ initially. Also calculate $x_m$.

**Solution:**
Let point A be at the top of the cylinder, point B be on the interface and point C be at the bottom of the cylinder. Let the depth of A be $h$.

The pressures at A, B and C are \begin{align} p_\text{A}&=p_0+hdg, \nonumber\\ p_\text{B}&=p_\text{A}+(\tfrac{3L}{4})dg, \nonumber\\ p_\text{C}&=p_\text{B}+(\tfrac{L}{4})(2d)g \nonumber\\ &=p_\text{A}+\tfrac{5}{4}Ldg. \nonumber \end{align} The forces on the cylinder are its weight $W=D (LA/5) g$ (downwards), hydrostatic force at the top surface $F_\text{A}=p_\text{A} (A/5)$ (downwards), and hydrostatic force at the bottom surface $F_\text{C}=p_\text{C} (A/5)$ (upwards). In equilibrium, $F_\text{A}+W=F_\text{C}$ i.e., \begin{align} p_\text{A} A/5+DLgA/5=(p_\text{A}+\tfrac{5}{4}Ldg)A/5, \nonumber \end{align} which gives \begin{align} D={5d}/{4}. \end{align} Let $p$ be the pressure at the bottom of the container. The forces on the liquid-cylinder system are $pA$ at the bottom surface (upwards), $p_0A$ at the top surface (downwards), weight of the upper liquid $\frac{1}{2}HAdg$ (downwards), weight of the lower liquid $HAdg$ (downwards), and weight of the cylinder $L(\frac{A}{5})(\frac{5d}{4})g=\frac{1}{4}LAdg$ (downwards). In equilibrium, \begin{align} pA=p_0A+\tfrac{1}{2}HAdg+HAdg+\tfrac{1}{4}LAdg,\nonumber \end{align} which gives \begin{align} p=p_0+\frac{1}{4}{(L+6H)dg}. \end{align}

Apply Bernoulli's equation between the interface surface and the hole to get \begin{align} \left(p_0+\tfrac{H}{2}dg\right)+\left(\tfrac{H}{2}-h\right)(2d)g+0\nonumber\\ =p_0+\tfrac{1}{2}(2d)v^2, \nonumber \end{align} which gives the velocity of efflux, \begin{align} v=\sqrt{({3H}/{4}-h)2g}. \nonumber \end{align} The time taken by the water to fall by a vertical distance $h$ is \begin{align} t=\sqrt{2h/g}. \nonumber \end{align} The horizontal distance covered in time $t$ is given by \begin{align} x=vt=\sqrt{h(3H-4h)}. \nonumber \end{align} The distance $x$ is maximum when \begin{align} \frac{\mathrm{d}x}{\mathrm{d}h}=\frac{1}{2\sqrt{h(3H-4h)}}(3H-8h)=0, \nonumber \end{align} which gives $h_m={3H}/{8}$ and maximum value of $x$ as $x_m={3H}/{4}$.