First slide

Fluid dynamics

Question

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

Moderate

A

$x = \sqrt{D (H - D)}$
B

$x = \sqrt{\frac{D (H - D)}{2}}$
C

$x = 2 \sqrt{D (H - D)}$
D

$x = 4 \sqrt{D (H - D)}$

Solution

Time taken by water to reach the bottom
$t = \sqrt{\frac{2 (H - D)}{g}}$
and velocity of water coming out of hole, $v = \sqrt{2 g D}$
$∴ Horizontal distance covered x = v \times t$
$= \sqrt{2 g D} \times \sqrt{\frac{2 (H - D)}{g}} = 2 \sqrt{D (H - D)}$

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