Question

A cylinder of height h is filled with water and is kept on a block of height h/2. The level of water in the cylinder is kept constant. Four holes numbered 1,2,3 and 4 are at the side of the cylinder and at heights 0, h/4, h/2 and 3h/4 respectively.When all four holes are opened together, the hole from which water will reach farthest distance is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

Range of water for height of hole $h$ and height of tank $H,x=2\sqrt{h(H-h)}$

Differentiate w.r.t $h$ to find maximum of $x$

$=>\frac{dx}{dh}=\frac{d}{dh}\left(2\sqrt{(H-h)\times h}\right)=\frac{d}{dh}\left(2\sqrt{Hh-h^2}\right)=\frac{2(H-2h)}{2\sqrt{Hh-h^2}}=\frac{H-2h}{\sqrt{Hh-h^2}}$

For $x$ to be maximum,

$\frac{dx}{dh}=0$

$=>\frac{H-2h}{\sqrt{Hh-h^2}}=0$

$=>H-2h=0=>h=\frac{H}{2}$

$h=\frac{H}{2}$ corresponds to hole $3$