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9-1.Fluid Mechanics
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The height of water in a tank is $H$. The range of the liquid emerging out from a hole in the wall of the tank at a depth $\frac {3H}{4}$ form the upper surface of water, will be
A
$H$
B
$\frac {H}{2}$
C
$\frac {3H}{2}$
D
$\frac {\sqrt 3H}{2}$
Solution
$t=\sqrt{\frac{2\left(H-\frac{3 H}{4}\right)}{g}}=\sqrt{\frac{H}{2 g}}$
Range $=v t=\sqrt{2 g\left(\frac{3 H}{4}\right)} \sqrt{\frac{H}{g}}=\frac{\sqrt{3}}{2} H$
Standard 11
Physics
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