Question

In a cylindrical vessel containing liquid of density $\rho$, there are two holes in the side walls at heights of $h_1$ and $h_2$ respectively such that the range of efflux at the bottom of the vessel is same. Find the height of a hole, for which the range of efflux would be maximum.

• A. $\frac{h_2+h_1}{2}$
• B. $\frac{h_2-h_1}{2}$
• C. $\frac{h_2{h}_1}{2}$
• D. $\frac{h_2}{h_1}$

Answer 1

The correct option is A $\frac{h_2+h_1}{2}$

The range of efflux is: $x=2\sqrt{(H-h)h}$
As per the given problem there are two holes in the side walls at heights of h1 and h2 respectively such that the range of efflux at the bottom of the vessel is same. Hence,
$x=2\sqrt{(}H-h_1)h_1=2\sqrt{(}H-h_2)h_2$

$\Rightarrow (H-h_1)h_1=(H-h_2)h_2$

$\therefore H{h}_1-(h_1{)}^2=H{h}_2-(h_2{)}^2$

$\therefore H(h_1-h_2)=(h_1{)}^2-(h_2{)}^2$

$\therefore H=\frac{(h_1{)}^2-(h_2{)}^2}{h_1-h_2}$

$\therefore H=\frac{(h_1{)}^2-(h_2{)}^2}{h_1-h_2}\frac{(h_1+h_2)}{(h_1+h_2)}$

$\Rightarrow H=h_1+h_2$
For maximum range the height will be: $H=\frac{h}{2}$
Hence, $H=\frac{h_1+h_2}{2}$