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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.

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