An engine pumps water continously through a hose. Water leave the hose with a velocity `v` and `m` is the mass per unit length of the Water jet. What is the rate at Which kinetic energy is imparted to water?

Options

A. (a) `1/2kv^2`
B. (b) `1/2kv^3`
C. (c) `(v^2)/(2k)`
D. (d) `(v^3)/(2k)`

Answer

B

Explanation

`K=("Mass")/("Length")=(dm)/(dx)`, `v=`speed of water
`KE=1/2mv^2`
`implies(d)/(dt)(KE)=1/2((dm)/(dt))v^2`
`=1/2((dm)/(dx)*(dx)/(dt))v^2=1/2kv v^2=(kv^3)/(2)`
Alternative method:
Let in time t, L length of water come out.
Then `t=L/v`
Mass of water that comes out in time t, `m=kL`
KE imparted per unit time `=(1/2)(mv^2)/(t)`
`=(1/2)(kLv^2)/((L//v))=(1/2)kv^3`