Question

A motor of power ${{\text{P}}_{\text{o}}}$ is used to deliver water at a certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe n times, the power of the motor is increased to ${{\text{P}}_{1}}$. The ratio of ${{\text{P}}_{1}}$ to ${{\text{P}}_{\text{o}}}$ is
a) n:1
b) ${{\text{n}}^{\text{2}}}\text{:1}$
c) ${{\text{n}}^{3}}\text{:1}$
d) ${{\text{n}}^{4}}\text{:1}$

Answer 1

Hint: The rate of flow of water basically means how much mass of water comes out of the pipe in unit time. The power of the motor can be defined at the rate at which It works to pump water through the pipe which can be seen in the form of the kinetic energy of the water coming out of the pipe in a given interval of time. Hence we will see if the power of the motor is increased n times how the kinetic energy changes in the given unit time and compare the power with power for which n=1.

Formula used: $P= \dfrac{K.E}{t}$, $K.E =\dfrac{m{v^2}}{2}$

Complete step-by-step solution:
Let us consider the initial power of the motor to be ${{\text{P}}_{\text{o}}}=\dfrac{m{{v}^{2}}}{2t}$ where m is the mass of the water coming out in unit time t and v is the velocity of the water with which it comes out. Let us consider the density of water as $\rho $ and the area of cross-section of the pipe as A. Hence the mass of the water coming out in unit time is given by, $\rho Av...(1)$. Hence substituting this in the above equation of power of motor we get,
${{\text{P}}_{\text{o}}}=\dfrac{\rho Av{{v}^{2}}}{2t}=\dfrac{\rho A{{v}^{3}}}{2t}$.
Now let us calculate the power of the motor when the rate of flow is increased by n times,
The rate of flow is given by equation 1 i.e.$\rho Av$. Hence this, We can conclude that the rate of flow only increases if the velocity of the liquid is increased. Hence the power of the when the velocity of the liquid is increased by n times is given by,
${{\text{P}}_{1}}=\dfrac{\rho A{{(nv)}^{3}}}{2t}=\dfrac{{{n}^{3}}\rho A{{v}^{3}}}{2t}$ . now let us take the ratio of ${{\text{P}}_{1}}$ is to ${{\text{P}}_{\text{o}}}$.
$\dfrac{{{\text{P}}_{1}}}{{{\text{P}}_{\text{o}}}}=\dfrac{\dfrac{{{n}^{3}}\rho A{{v}^{3}}}{2t}}{\dfrac{\rho A{{v}^{3}}}{2t}}=\dfrac{{{n}^{3}}}{1}$
Hence the ratio of ${{\text{P}}_{1}}$:${{\text{P}}_{\text{o}}}$ is ${{n}^{3}}:1$. Hence the correct answer is option c.

Note: The electrical energy of the motor gets converted to the kinetic energy of the water. This is the reason why we consider the kinetic energy of the water to be proportional to the electrical output power delivered by the motor. If the efficiency of the motor is 100% then the electrical energy consumed by the motor will be numerically equal to the kinetic energy of the flowing water.