If a small orifice is made at a height of $0.25\, m$ from the ground, the horizontal range of water stream will be, (in $cm$)
AIIMS 2019, Diffcult
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The initial velocity of the water stream is given as,

$v=\sqrt{2 g(H-h)}$

The time taken by stream to hit the surface is,

$h=\frac{1}{2} g t^{2}$

$t=\sqrt{\frac{2 h}{g}}$

The horizontal range can be calculated as,

Horizontal range $=v t$

$=\sqrt{2 g(H-h)} \times \sqrt{\frac{2 h}{g}}$

$=2 \sqrt{(H-h) h}$

$=2 \sqrt{(1-0.25)(0.25)}$

$=0.866 m$

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