A P watt motor is used to pump the water in a vertical height of h meters out of a flooded basement through a cylindrical pipe. The water is ejected through the end of the pipe at a speed of v m/s. Ignoring friction and assuming that all of the energy of the motor goes to the water, then the radius of the pipe is (Take g acceleration due to gravity density of water $ρ$ )

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$r = \sqrt{\frac{P}{ρ π v (g h - \frac{v^{2}}{2})}}$

$r = \sqrt{\frac{2 P}{ρ π v (g h - \frac{v^{2}}{2})}}$

$r = \sqrt{\frac{P}{ρ π v (g h + \frac{v^{2}}{2})}}$

$r = \sqrt{\frac{2 P}{ρ π v (g h + \frac{v^{2}}{2})}}$

answer is C.

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Detailed Solution

Mass of water pumped in time dt is $d m = ρ π r^{2} v d t$
So, $P d t = d m (g h + \frac{v^{2}}{2}) = ρ π r^{2} v d t (g h + \frac{v^{2}}{2})$
So, $P = ρ π r^{2} v (g h + \frac{v^{2}}{2})$
So, $r = \sqrt{\frac{P}{ρ π v (g h + \frac{v^{2}}{2})}}$