A tank is filled with water up to a height H. Water is allowed to come out of a hole P in one of the walls at a depth D below the surface of water. Express the horizontal distance x in terms of H and D

The correct option is: x = 2√{D(H-D)}.

To find the horizontal distance (range) $x$ traveled by the water coming out of the hole, we can break the motion down into two parts: finding the velocity of efflux and calculating the time of flight for the water droplets.

Step 1: Find the velocity of efflux ($v$)
According to Torricelli's Law, the velocity $v$ of a liquid flowing out of a hole at a depth $D$ below the free surface of the liquid is the same as the velocity acquired by a freely falling body from the same height. Therefore:
$v=\sqrt{2gD}$
where $g$ is the acceleration due to gravity.

Step 2: Find the time of flight ($t$)
The hole $P$ is located at a depth $D$ from the top of the water level in a tank of total height $H$. Therefore, the height $h$ of the hole above the bottom of the tank is:
$h=H-D$
Once the water leaves the hole horizontally, its initial vertical velocity is zero. Under the influence of gravity, the time $t$ it takes for the water to reach the ground is given by the equations of motion:
$h=\frac{1}{2}g{t}^2$
Rearranging the formula to solve for $t$:
$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(H-D)}{g}}$

Step 3: Calculate the horizontal range ($x$)
Since there is no horizontal acceleration, the horizontal distance $x$ covered by the water during the time of flight $t$ is:
$x=v\times t$
Substituting the expressions for $v$ and $t$ into the equation:
$x=\sqrt{2gD}\times \sqrt{\frac{2(H-D)}{g}}$
Simplifying the square roots:
$x=\sqrt{2gD\times \frac{2(H-D)}{g}}$
$x=\sqrt{4D(H-D)}$
$x=2\sqrt{D(H-D)}$