Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 at which kinetic energy is imparted to water?

$\frac{1}{2}mv^3$

$mv^3$

$\frac{1}{2}mv^2$

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

Step-by-Step Solution

Identify the Goal: We need to find the rate at which kinetic energy is imparted, which is the Power ( $P$ ).
Formula for Power: Power is the rate of change of kinetic energy: $P = \frac{dK}{dt}$ .
Kinetic Energy Formula: The kinetic energy of a mass $M$ moving with velocity $v$ is $K = \frac{1}{2} M v^2$ .
Differentiate: Since $v$ is constant, the rate of change is: $\frac{dK}{dt} = \frac{1}{2} v^2 \frac{dM}{dt}$ .
Determine Mass Flow Rate: We are given $m$ as the mass per unit length (linear density). The mass $dM$ flowing in time $dt$ corresponds to a length $dx = v dt$ . Thus, $dM = m dx = m (v dt)$ . The mass flow rate is $\frac{dM}{dt} = mv$ .
Substitute and Solve: $P = \frac{1}{2} v^2 (mv) = \frac{1}{2} m v^3$

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