Question

A cylinder of height h filled with water and is kept on lock 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 height 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 on the plane PQ is the hole number.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

We know from Torricelli's theorem, that the range of the liquid falling from a certain height is given by:

$R=2\times \sqrt{h(H-h)}$

where $H$ is the total height of the container and

$h$ is the height where the hole is.

For $R=R_{max}$;

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

$\frac{dR}{dh}=2\times (\frac{1}{2\sqrt{h}}\sqrt{H-h}+\sqrt{h}\frac{-1}{2\sqrt{H-h}})$

$\frac{dR}{dh}=(\frac{\sqrt{H-h}}{\sqrt{h}}+\frac{-\sqrt{h}}{\sqrt{H-h}})$

$\frac{dR}{dh}=\frac{H-h-h}{\sqrt{h(H-h)}}=0$
$\Rightarrow H-h-h=0$

$\Rightarrow H=2h$

For $R=R_{max}$

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

Taking PQ as the reference,

$H=h+\frac{h}{2}=\frac{3h}{2}$

So hole must be at height

$\frac{H}{2}=\frac{(\frac{3h}{2})}{2}=\frac{3h}{4}$

For hole 1,

$h_1=\frac{h}{2}+0=\frac{h}{2}$

For hole 2,

$h_2=\frac{h}{2}+\frac{h}{4}=\frac{3h}{4}$

For hole 3,

$h_3=\frac{h}{2}+\frac{h}{2}=h$

For hole 4,

$h_4=\frac{h}{2}+\frac{3h}{4}=\frac{5h}{4}$

Since, hole 2 is at the $\frac{H}{2}$ height required for longest range.

$\therefore$ Hole 2 is from which water will reach farthest distance on the plane PQ.

Hence, the correct answer is OPTION B.