Question

Equal volumes of two immiscible liquids of densities $\rho$ and $2\rho$ are filled in a vessel as shown figure. Two small holes are punches at depth $h/2$ and $3h/2$ from the surface of lighter liquid. If $v_1$ and $v_2$ are the velocities of a flux at these two holes then $v_1/v_2$ is :

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

Here, two small holes are punches at depth $\frac{h}{2}$ and $\frac{3h}{2}$ from the surface of lighter liquid. Hence, the height of first hole from the surface of lighter liquid is

$h_1=\frac{h}{2}$

and the height of second hole from the first hole is(i.e. from surface of remaining liquid)

$h_2=\frac{3h}{2}-\frac{h}{2}=h$

Now, we have the velocity of flux from first hole is

$v_1=\sqrt{2h_{1}g}$

$v_1=\sqrt{2\left(\frac{h}{2}\right)g}$

$v_1=\sqrt{hg}$

Applying Bernoulli's equation to point just inside and outside the second hole,

$P_a+\rho gh+2\rho g\frac{h}{2}=P_a+\frac{1}{2}2\rho v^2$

$v=\sqrt{2gh}$

So, the velocity of flux from second hole is

$v_2=\sqrt{2hg}$

Therefore,

$\frac{v_1}{v_2}=\frac{\sqrt{hg}}{\sqrt{2hg}}$

$\frac{v_1}{v_2}=\frac{1}{\sqrt{2}}$