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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A fire hose has a diameter of 2.5 cm and is required to direct jet of water to height of at least 40 m. The minimum power of pump needed is

\[(a)\;21.5\;KW \quad (b)\;40\;KW \quad (c)\;36.5\;KW \quad (d)\;48\;KW\]
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1 Answer

+1 vote
The speed of water leaving hose $=\sqrt {2gh}$
where h is height to be reached by water.
Let diameter of pipe='d', then
$A \times V=$Volume of water ejected per second
$=\large\frac{1}{4}$$ \pi d^2 \times \sqrt {2gh} \large\frac{m^3}{s}$
Mass of water ejected = Volume $\times$ density
$'m'=\large\frac{1}{4}$$ \pi d^2 \times \sqrt {2gh} \times density$
KE per unit time =Power of pump
$=\large\frac{1}{2}$$ mv^2=\large\frac{1}{8}$$\pi d^2 \times (2gh)^{3/2} \times density$
substituting values for h,g,density,diameter of pipe , we get
$Power =21.5\; KW$
Hence a is the correct answer. 


answered Jul 19, 2013 by meena.p
edited Feb 10, 2014 by meena.p
But the answer is coming.. 21.5/4 after doing the above mentioned calculation:(

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