A cylindrical tank is filled with two liquids of density ${\rho }_1=900kg{m}^{-3} \& {\rho }_2=600kg{m}^{-3},$ to a height h = 60cm each as shown in figure. A small hole having area $a=4.9c{m}^2$ is made in right vertical wall at height y = 20 cm from the bottom. $(g=10m/s^2)$

The velocity of efflux is (in m/s)

Since area a of hole is very small in comparison to base area A of the cylinder, therefore, velocity of liquid inside the cylinder is negligible. Let velocity of efflux be v and atmospheric pressure $P_0.$ Consider two points A (inside the cylinder) and B (just outside the hole) in the same horizontal line as shown in figure

Pressure at A, $P_A=P_0+h{\rho }_{2}g+(h–y){\rho }_{1}g$

Pressure at B, $P_B = P_0.$

According to Bernoulli's theorem, pressure energy at A= pressure energy at B + kinetic energy at B

$\therefore P_A=P_B+\frac{1}{2}{\rho }_1{v}^2,$

$\therefore v=4{\text{ms}}^{\text{-1}}$