Question

Equal volumes of two immiscible liquids of densities $\delta$ and 3$\delta$ are filled in a vessel. Two small holes are punched at depth $h/3$ and $4h/3$ from upper surface of lighter liquid. If $V_1$ and $V_2$ are velocities of efflux at these two holes, then $V_1/V_2$ is :

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

Answer 1

Answer: (B)

$\frac{1}{\sqrt{2}}$

Applying Bernoulli's theorem to pints A( just at the surface of the liquid) and B( just outside the first hole in the vessel) taking the position of the hole as reference

$P_0+{\rho }_{1}g\frac{h}{3}+\frac{1}{2}{\rho }_1{v}^2=P_0+0+\frac{1}{2}{\rho }_1{v}_1^2$ .....(1)

$P_0$ - atmospheric pressure

$v$ - velocity of the liquid on the surface

$v_1$ - velocity of the liquid at the first hole

Applying Bernoulli's theorem to pints A( just at the surface of the liquid) and C( just outside the second hole in the vessel) taking the position of the hole as reference

$P_0+{\rho }_{1}gh+{\rho }_{2}g\frac{h}{3}+\frac{1}{2}{\rho }_1{v}^2=P_0+0+\frac{1}{2}{\rho }_2{v}_2^2$

Putting ${\rho }_2=3{\rho }_1$,

the above eqn becomes

$P_0+{\rho }_{1}gh+3{\rho }_{1}g\frac{h}{3}+\frac{1}{2}{\rho }_1{v}^2=P_0+\frac{1}{2}3{\rho }_1{v}_2^2$ ......(2)

$v_2$ - velocity of the liquid at the second hole

$v=0$ since the liquid is at rest on on the surface

Putting $v=0$ in eqns (1) we get

$v_1=\sqrt{\frac{2gh}{3}}$

Putting $v=0$ in eqns (2) we get

$v_2=\sqrt{\frac{4gh}{3}}$

$\therefore \frac{v_1}{v_2}=\sqrt{\frac{2}{4}}=\sqrt{\frac{1}{2}}$