An engine pumps water continuously through a hole. Speed with which water passes through the hole nozzle is v, and k is the mass per unit length of the water jet as it leaves the nozzle. Find the rate at which kinetic energy is being imparted to the water.

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$\frac{v^{2}}{2 k}$

$\frac{1}{2} k v^{3}$

$\frac{1}{2} k v^{2}$

$\frac{v^{3}}{2 k}$

answer is B.

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Detailed Solution

Mass per unit length of water jet is

k = $\frac{Mass}{Length} = \frac{dm}{dx}$

Kinetic energy, KE = $\frac{1}{2} {mv}^{2}$

differentiate with respect to time,

$\frac{d}{dt} (K . E .) = \frac{1}{2} \frac{dm}{dt} v^{2}$

$\frac{d}{dt} (K . E .) = \frac{1}{2} \frac{dm}{dt} \times \frac{dx}{dx} v^{2}$

$\frac{d}{dt} (K . E .) = \frac{1}{2} \frac{dm}{dx} \times v \times v^{2} = \frac{1}{2} \frac{dm}{dx} v^{3}$

$\frac{d}{dt} (K . E .) = \frac{1}{2} {kv}^{3}$

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