Question

An engine pump is used to pump a liquid of density $\rho$ continuously through a pipe of area of cross-section $A$ . If the speed with which the liquid passes the pipe is $V$ , then the rate which kinetic energy is being imparted to the liquid is:
A. $\dfrac{1}{2}A\rho {V^3}$
B. $\dfrac{1}{2}A\rho {V^2}$
C. $\dfrac{1}{2}A\rho V$
D. $A\rho {V^2}$

Answer 1

First of all, we will find out the relation between kinetic energy and power. Then we will try to find out the value of mass from the relation between density and volume. We will make necessary substitution and manipulate accordingly.

Complete step by step solution:
In the given question, we are supplied with the following data:
The density of the liquid which is pumped by the engine is $\rho$ .
The area of the cross section of the pipe is $A$ .
The speed at which the liquid passes the pipe is $V$ .
We are asked to find the rate at which the kinetic energy is imparted to the fluid.

We will find power by using the given information:
The expression of power is given by,

$P = \dfrac{W}{t}$ …… (1)
Where,
$P$ indicates power.
$W$ indicates work done.
$t$ indicates time required.

Now, we can write the work as its equivalent kinetic energy in the equation (1),
$P = \dfrac{{\Delta K.E}}{t}$
$P = \dfrac{{\dfrac{1}{2}m{V^2}}}{t}$ …… (2)
Where,
$m$ indicates the mass of the liquid flowing through the pipe.
$V$ indicates the velocity of the liquid flowing at.

Now, we will find the mass of the liquid pumped in a given time:
We know,
$\rho = \dfrac{m}{v}$ …… (3)
Where,
$\rho$ indicates density of the liquid.
$m$ indicates mass of the fluid.
$v$ indicates volume of the fluid passing through the pipe.

We can write the volume of the liquid passing through the pipe in a given time $t$ is:
$v = AVt$ …… (4)
Where,
$v$ indicates volume.
$A$ indicates cross-sectional area.
$V$ indicates velocity of the liquid.

Using equation (4) in equation (3), we get:
$\rho =\dfrac{m}{{AVt}} \\\ m = \rho {\rm A}Vt$
Now, we will substitute the values of mass in equation (2) and we get:
$P = \dfrac{{\dfrac{1}{2} \times \rho AVt \times {V^2}}}{t} \\\ P = \dfrac{1}{2}A\rho {V^3}$
Hence, the rate at which kinetic energy is being imparted to the liquid is $\dfrac{1}{2}A\rho {V^3}$ .
The correct option is A.

Note:
Kinetic energy is a form of energy which is present in a body by the virtue of its motion. In some instances, when the volume of the bulk is not given, try not to panic, just the product of the area of the cross section, volume and given time gives the volume passing through the region.