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## QuestionPhysicsClass 12

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

C
4.6
4.6

## Solution

Given : height of tank $$H$$
The velocity of water stream coming out of $$P$$ is $$v = \sqrt{2gD}$$
The time taken by the water stream to fall through the height $$H-D$$ can be calculated as
$$s = ut + \frac{1}{2}at^2$$
i.e. $$H- D = \frac{1}{2}gt^2$$
$$t = \sqrt{\frac{2(H - D)}{g}}$$
So horizontal distance $$x =v\times t$$
$$x =\sqrt{2gD}\times \sqrt{\frac{2(H - D)}{g}}$$
$$x=2\sqrt {D(H-D)}$$