Question

A vessel is filled with water and kerosene oil. The vessel has a small hole in the bottom. Neglecting viscosity if the thickness of water layer is $h_1$ and kerosene layer is $h_2$ then the velocity $v$ of flow of water will be (Given : density of water is ${\rho }_1$ g/ cc and that of kerosene is ${\rho }_2$ g/cc, neglecting viscosity):

• A. $v=\sqrt{2g(h_1+h_2)}$
• B. $v=\sqrt{2g[h_1+h_2\left(\frac{{\rho }_2}{{\rho }_1}\right)]}$
• C. $v=\sqrt{2g(h_1{\rho }_1+h_2{\rho }_2)}$
• D. $v=\sqrt{2g[h_1\left(\frac{{\rho }_1}{{\rho }_2}\right)+h_2]}$

Answer 1

Answer: (B)

$v=\sqrt{2g[h_1+h_2\left(\frac{{\rho }_2}{{\rho }_1}\right)]}$

Berneoulli's Theorm : $\rho gh+\frac{1}{2}\rho v^2+P_o=constant$

Pressure at A and B, $P_A=P_B=P_o$ and velocity at B be $v$ and at A is zero.

Applying Berneoulli's Theorem at points A and B,

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

$⟹\frac{1}{2}{\rho }_1{v}^2={\rho }_{1}g{h}_1+{\rho }_{2}g{h}_2$

$v=\sqrt{[2g(h_1+h_2\frac{{\rho }_2}{{\rho }_1})]}$