Question

A tank with vertical walls is mounted so that its base is at a height $H$ above the horizontal ground. The tank is filled with water to a depth '$h$' . A hole is punched in the side wall of the tank at a depth ' $x$ ' below the water surface. To have maximum range of the emerging stream,the value of $x$ is

• A. $\frac{H+h}{4}$
• B. $\frac{H+h}{2}$
• C. $\frac{H+h}{3}$
• D. $\frac{3(H+h)}{4}$

Answer 1

Answer: (B)

$\frac{H+h}{2}$

Horizontal velocity of ejection, when a vessel is filled with

water to height h is $\sqrt{2gh}$

$\therefore$ velocity of ejection $=\sqrt{2g(h-x)}$

Time taken will be $\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(H+h-x)}{g}}$

$\therefore range=2\sqrt{(h-x)(H+h-x)}[\because R=ut]$

range is max , when $\frac{dR}{dx}=0$

$\Rightarrow x=\frac{H+h}{2}$