For a container open to the atmosphere,derive the velocity of the liquid exiting a narrow hole in the container wall using Bernoulli's equation and obtain Torricelli's law.

(N/A) Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body.
Consider a tank containing a liquid of density $\rho$ with a small hole in its side at a height $y_{1}$ from the bottom. The air above the liquid,whose surface is at height $y_{2}$,is at pressure $P$.
The velocities $v_{1}$ and $v_{2}$ are at points $1$ and $2$ respectively. Using the equation of continuity for points $1$ and $2$:
$A_{1} v_{1} = A_{2} v_{2}$
$v_{2} = \frac{A_{1} v_{1}}{A_{2}}$
$A_{2}$ is the cross-sectional area of the tank and $A_{1}$ is the cross-sectional area of the hole.
Since $A_{2} \gg A_{1}$,it follows that $v_{2} \ll v_{1}$,so we can approximate $v_{2} \approx 0$.
Using Bernoulli's equation at points $1$ and $2$:
$P_{1} + \frac{1}{2} \rho v_{1}^{2} + \rho g y_{1} = P_{2} + \frac{1}{2} \rho v_{2}^{2} + \rho g y_{2}$
Here,$P_{1} = P_{a}$ (atmospheric pressure),$P_{2} = P$,and $v_{2} = 0$. Substituting these:
$P_{a} + \frac{1}{2} \rho v_{1}^{2} + \rho g y_{1} = P + \rho g y_{2}$
$\frac{1}{2} \rho v_{1}^{2} = (P - P_{a}) + \rho g (y_{2} - y_{1})$
Let $h = y_{2} - y_{1}$ be the height of the liquid above the hole.
$\frac{1}{2} \rho v_{1}^{2} = (P - P_{a}) + \rho g h$
$v_{1} = \sqrt{2g h + \frac{2(P - P_{a})}{\rho}}$
If the tank is open to the atmosphere,$P = P_{a}$,then:
$v_{1} = \sqrt{2gh}$
This is Torricelli's law.