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 $H/2$ as shown in the figure. The lower density liquid is open to the atmosphere having pressure $P_0.A$ 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.
The cylinder is then removed and the original arrangement is restored. A tiny hole of area $s(s<<A)$ is punched on the vertical side of the container at a height $h(h<H/2)$. As a result of this, liquid starts flowing out of the hole with a range $x$ on the horizontal surface.

The horizontal distance traveled by the liquid, initially, is :

• A. $\sqrt{(3H+4h)h}$
• B. $\sqrt{(3h+4H)h}$
• C. $\sqrt{(3H-4h)h}$
• D. $\sqrt{(3H-3h)h}$

Answer 1

The correct option is C $\sqrt{(3H-4h)h}$

Applying Bernoulli's equation just inside and just outside the hole,$P_0+\left(\frac{H}{2}\right)\left(d\right)g+(\frac{H}{2}-h)\left(2d\right)\left(g\right)=\frac{1}{2}\left(2d\right)v^2+P_0$

$\therefore$$v=\sqrt{\frac{g}{2}(3H-4h)}$

$t=\sqrt{\frac{2h}{g}}$

$\therefore$ $x=vt=\sqrt{(3H-4h)h}$