Question

A rectangular tank is placed on a horizontal ground and is filled with water to a height $H$ above the base. A small hole is made on one vertical side at a depth $D$ below the level of water in the tank. The distance $x$ from the bottom of the tank at which the water jet from the tank will hit the ground is

• A. $2\sqrt{D(H-D)}$
• B. $2\sqrt{DH}$
• C. $2\sqrt{D(H+D)}$
• D. $\frac{1}{2}\sqrt{DH}$

Answer 1

The correct option is A $2\sqrt{D(H-D)}$

This setup can be visualised as a horizontal projectile from height h .

here $h=$ height of the hole above the bottom of the tank.

so $h=H-D$

let Horizontal velocity of water be $u_x$.

so velocity of liquid falling at depth d is given by $\sqrt{2gd}$

here $d=D$ so $u_x=\sqrt{2gD}$

Time taken to fall at surface can be found using

$s=ut+\frac{1}{2}a{t}^2$

$h=u_{y}t+\frac{1}{2}g{t}^2$

here $u_y=0$, no component of velocity in vertical direction.

and $h=H-D$

$H-D=\frac{1}{2}g{t}^2$

$t=\sqrt{\frac{2(H-D)}{g}}$

since there is no acceleration in horizontal direction

Range in Horizontal direction= $u_{x}t$

putting values from above

$r=\sqrt{2gD}\times$$\sqrt{\frac{2(H-D)}{g}}$

So, range of the water stream $x=2\sqrt{D(H-D)}$