Question

A tank is filled with water up to 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 :

• A. $x=\sqrt{D(H-D)}$
• B. $x=\sqrt{\frac{D(H-D)}{2}}$
• C. $x=\sqrt[2]{2D(H-D)}$
• D. $x=\sqrt[4]{4D(H-D)}$

Answer 1

The correct option is C $x=\sqrt[2]{2D(H-D)}$

Answer is C.
The velocity of water stream coming out of P = $\sqrt{2gD}$.
The time taken by the water stream to fall through the height H-D = $\sqrt{\frac{2(H-D)}{g}}$.
Therefore, $x=\sqrt{2gD}\times \sqrt{\frac{2(H-D)}{g}}=2\sqrt{D(H-D)}$.
Hence, the horizontal distance x in terms of H and D is $2\sqrt{D(H-D)}$.