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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 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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