Question

A rectangular tank is placed on the 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. Find the distance $x$ from the bottom of the tank at which the water jet from the tank will hit the ground.
A) $2{\left[ {D\left( {H - D} \right)} \right]^{1/2}}$
B) $2{\left( {gD} \right)^{1/2}}$
C) $2{\left[ {D\left( {H + D} \right)} \right]^{1/2}}$
D) $\dfrac{1}{2}{\left( {DH} \right)^{1/2}}$

Answer 1

Torricelli described the water flowing from a hole in an open tank to be similar to a freely falling body. So the speed of the water jet will be the speed of a freely falling body and it will only depend on the acceleration due to gravity and height of the hole from the ground.

Formulas used:
-The velocity of efflux is given by, $v = \sqrt {2gh}$ where $g$ is the acceleration due to gravity and $h$ is the height from the surface at which the water flows out of the vessel.
-Newton’s first equation of motion gives the vertical displacement distance of a freely falling body as $s = \dfrac{1}{2}g{t^2}$ where $g$ is the acceleration due to gravity and $t$ is the time taken to cover the distance.
-The distance from the tank at which water hits can be expressed as $x = vt$ where $v$ is the velocity of the water jet and $t$ is the time taken.

Complete step by step answer.
Step 1: Sketch a figure of the tank describing the water flowing from the tank.

In the above figure, we see that the height of the tank is $H$ and at distance $D$ from the surface of the water in the tank a hole is made through which water flows out of the tank. So from the ground, the hole is at a distance $H - D$ as seen in the figure. We have to determine the distance $x$ at which water hits the ground.
In this setup, $H - D$ will be the vertical displacement of the water and $x$ will be the horizontal displacement of the water jet.
Step 2: Express the velocity of the water jet based on Toricelli’s law.
Torricelli gives the velocity of the efflux to be equal to the velocity of a freely falling body given by, $v = \sqrt {2gh}$ where $g$ is the acceleration due to gravity and $h$ is the height from the surface at which the water flows out of the vessel.
Here the hole is made at $D$ from the water surface. So by replacing $h$ in the above relation by $D$ we have the velocity of the water jet as $v = \sqrt {2gD}$ .
Step 3: Express the time taken for water to hit the ground using Newton’s first equation of motion.
Newton’s first equation of motion gives the vertical displacement of the water jet as $H - D = \dfrac{1}{2}g{t^2}$
$\Rightarrow {t^2} = \dfrac{{2\left( {H - D} \right)}}{g}$
Then the time taken can be expressed as $t = \sqrt {\dfrac{{2\left( {H - D} \right)}}{g}}$ .
Step 4: Express the relation for the horizontal distance $x$ covered by the jet to reach the ground.
The horizontal distance covered by the water jet to hit the ground can be expressed as
$x = vt$ ------- (1)
Substituting for $v = \sqrt {2gD}$ and $t = \sqrt {\dfrac{{2\left( {H - D} \right)}}{g}}$ in equation (1) we get, $x = \sqrt {2gD} \times \sqrt {\dfrac{{2\left( {H - D} \right)}}{g}}$
On simplifying the above expression becomes $x = 2\sqrt {D\left( {H - D} \right)}$
Thus the required distance will be $x = 2{\left[ {D\left( {H - D} \right)} \right]^{1/2}}$ .

So the correct option is A.

Note: Here we assumed that the tank was kept open to the atmosphere. As the water flowing from the tank is considered as a freely falling body, the initial velocity of the water jet will be zero. It is important to remember that the water jet is considered as a freely falling body only if the tank is kept open to the atmosphere.