Question

Water level is maintained in a cylindrical vessel up to a fixed height H. The vessel is kept on a horizontal plane. At what height above the bottom should a hole be made in the vessel so that the water stream coming out of the hole strikes the horizontal plane at the greatest distance from the vessel?

• A. H/4
• B. H/√2
• C. H/2
• D. H/3

Answer 1

The correct option is C

H/2

Let the hole is made at a height 'h' above the bottom.

Let the water spurts out through the hole with velocity 'v'.

By Torricelli's equation

$v=\sqrt{2g(H-h)}$
Time taken by this get to reach the bottom,
$h=\frac{1}{2}g{t}^2$
$\Rightarrow t=\sqrt{\frac{2h}{g}}$
Now horizontal distance covered by stream of water in this time interval,
$x=\left(\sqrt{2g(H-h)}\right)x\sqrt{\frac{2h}{g}}$
$x=\sqrt{2h(H-h)}$
Maximizing x is same as maximizing $x^2$ as it is the distance.
$\frac{d\left(x^2\right)}{dh}=2H-4h=0$
H=$\frac{H}{2}$