Question

A cylinder of height $H$ is filled with water to a height $h_0(<H)$, and is placed on a horizontal floor. Two small holes are punched at time $t=0$ on the vertical line along the length of the cylinder, one at a height $h_1$ from the bottom & the other a depth $h_2$, below the level of water in the cylinder. Find the relation between $h_0,h_1$ & $h_2$ such that the instantaneous water jets emerging fall at the same point on the floor.

• A. $h_0=2h_1-h_2$
• B. $h_0=2h_2-h_1$
• C. $h_0=h_1+h_2$
• D. $h_0=2h_1+h_2$

Answer 1

The correct option is C $h_0=h_1+h_2$


Let $v_1$ be the speed of efflux from a hole which is at height $h_1$ from ground and $v_2$ be the speed of efflux from a hole which is at depth $h_2$ from level of water.

Applying Torricelli's law

$v_1=\sqrt{2g(h_0-h_1)}$

$v_2=\sqrt{2g{h}_2}$

Given that the range of the efflux is same,

$v_1\sqrt{\frac{2h_1}{g}}=v_2\sqrt{\frac{2(h_0-h_2)}{g}}$

$v_1^2{h}_1=v_2^2(h_0-h_2)$

$2g(h_0-h_1)h_1=2g{h}_2(h_0-h_2)$

$h_0(h_1-h_2)=h_1^2-h_2^2=(h_1-h_2)(h_1+h_2)$

$\therefore h_0=h_1+h_2$

Hence, option $\left(\text{C}\right)$ is correct.