Prove that the velocity of efflux through an orifice is equal to the velocity attained by a freely falling body from the surface of the liquid to the orifice. Also find the horizontal range in terms of height. ( 5marks)

Dear Student,
H = Height of container
h = Height from the surface of the container to the orifice made
Let atmospheric pressure be P , density of liquid in tank be $\rho$and velocity of liquid at efflux be v
Applying Bernoulli's equation at the surface of container and at the orifice we get
$$\begin{gathered} P+\rho gH=P+\frac{1}{2}\rho v^2+\rho g\left(H-h\right) \\ \Rightarrow 0=\frac{1}{2}\rho v^2-\rho gh \\ \Rightarrow \frac{1}{2}\rho v^2=\rho gh \\ \Rightarrow v=\sqrt{2gh} \end{gathered}$$
Thus the velocity of efflux is $v=\sqrt{2gh}$
$$\begin{gathered} And velocity of a body falling freely has same velocity \\ as by third equation of motion \\ {v}^2=0+2gh \\ v=\sqrt{2gh} \\ and for range \\ time to reach ground \\ t=\sqrt{\frac{2(H-h)}{g}} \\ and range R=vt=\sqrt{2gh}\times \sqrt{\frac{2(H-h)}{g}}=2\sqrt{h(H-h)} \end{gathered}$$
Regards