Gujarati
Hindi
9-1.Fluid Mechanics
normal

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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