Question

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 $\frac{H}{2}$ as shown in figure. The lower density liquid is open to the atmosphere having pressure $P_0$. A homogeneous solid cylinder of length $L(L<\frac{H}{2})$ cross-sectional area $\frac{A}{5}$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with the length $\frac{L}{4}$ in the denser liquid. The total pressure at the bottom of the container is given as $p=p_o+\frac{xH+L}{4}$. Find $x$.

Answer 1

Applying Bernoulli's theorem
$P_0+[\frac{H}{2}\times d\times g+(\frac{H}{2}-h)2d\times g]=P_0+\frac{1}{2}\left(2d\right)V^2$
$\Rightarrow V=\sqrt{(3H-4h)/4g}$

Horizontal distance x
$u_x=v$ time$=t$, $x=vt⟶\left(i\right)$
For vertical motion of liquid falling from hole
$u_y=0,S_y=x,a_y=g,t_y=t$
$S=ut+\frac{1}{2}a{t}^2$
$\Rightarrow h=\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{2h/g}⟶\left(ii\right)$

From $\left(i\right)\&\left(ii\right)$
$x=u_y\times \sqrt{2h/g}=\sqrt{(3H-4h)/4g}\times \sqrt{2h/g}$
$x=\sqrt{(3H-4h)h}⟶\left(iii\right)$

Total pressure at the bottom of the cylinder$=$Atmospheric pressure $+$ Pressure due to liquid of density d $+$ Pressure due to liquid of density $2d$ $+$ Pressure of cylinder
$P=P_0+\frac{H}{2}dg+\frac{H}{2}\times 2d\times g+\frac{A}{5}\times L\times D\times \frac{g}{A}$
$=P_0+(\frac{3H}{2}+\frac{L}{4})dg$
$=P_0+\left(\frac{6H+L}{4}\right)dg$
So, the value of $x=6$