A container of large uniform cross-sectional area A: Bernoulli's Equation

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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 p0.  (IIT JEE 1995)

A container of large uniform cross-sectional area A
  1. 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,
    1. the density of the solid.
    2. the total pressure at the bottom of the container.
  2. 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,
    1. the initial speed of efflux of the liquid at the hole.
    2. the horizontal distance $x$ travelled by the liquid initially.
    3. 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 $h$.

A container of large uniform cross-sectional area A solution

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 $x_m={3H}/{4}$.

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