Gujarati
Hindi
9-1.Fluid Mechanics
normal

The height of water in a tank is $H$. The range of the liquid emerging out form a hole in the wall of the tank at a depth $\frac{{3H}}{4}$ from the upper surface of water, will be

A

$H$

B

$\frac{H}{2}$

C

$\frac{3H}{2}$

D

$\frac{{\sqrt 3 H}}{2}$

Solution

Range of a liquid $\mathrm{x}=2 \sqrt{\mathrm{h}(\mathrm{H}-\mathrm{h})}$

but $\mathrm{h}=\frac{3 \mathrm{H}}{4}$

$x=2 \sqrt{\frac{3 \mathrm{H}}{4}\left(\mathrm{H}-\frac{3 \mathrm{H}}{4}\right)}=2 \sqrt{\frac{3 \mathrm{H}}{4} \times \frac{\mathrm{H}}{4}}=\frac{\sqrt{3} \mathrm{H}}{2}$

Standard 11
Physics

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