An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate which kinetic energy is imparted to water ?

Correct answer: B. (1)/(2) mv^3 mv= dm du × du dt = dm dt = Rate of flowing mass aligned & F_ av = dm dt × v 2 = (mv) v 2 = mv^2 2 & p= dK dt = mv^2 2 × v= mv^3 3 aligned

NEET 2009

Exam: NEET
Year: 2009
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

NEET 2009 · Physics · Center of Mass, Momentum and Collision

An engine pumps water continuously through a hose. Water leaves the...

An engine pumps water continuously through a hose. Water leaves the hose with a velocity

\mathrm{v}

and

\mathrm{m}

is the mass per unit length of the water jet. What is the rate which kinetic energy is imparted to water ?

Options

A $\frac{1}{2} m^2 v^2$
B $\frac{1}{2} \mathrm{mv}^3$
C $\mathrm{mv}^3$
D $\frac{1}{2} \mathrm{mv}^2$

Correct answer

B. $\frac{1}{2} \mathrm{mv}^3$

Step-by-step solution

\mathrm{mv}=\frac{\mathrm{dm}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}}=

Rate of flowing mass

\begin{aligned}& \mathrm{F}_{\mathrm{av}}=\frac{\mathrm{dm}}{\mathrm{dt}} \times \frac{\mathrm{v}}{2}=\frac{(\mathrm{mv}) \mathrm{v}}{2}=\frac{\mathrm{mv}^2}{2}\\ & \mathrm{p}=\frac{\mathrm{dK}}{\mathrm{dt}}=\frac{\mathrm{mv}^2}{2} \times \mathrm{v}=\frac{\mathrm{mv}^3}{3} \end{aligned}

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