Question

A pump motor is used to deliver water at a certain rate from a given pipe at the same level. To obtain 'n' times water from the same pipe in the same time the amount by which the power of the motor should be increased :

A) n$^2$
B) n$^3$
C) n$^4$
D) n$^{\dfrac{1}{2}}$

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Answer 1

For this we have to find the relation between the power of the motor and the amount of water delivered (i.e., volume of water).

The volume of the water delivered is given by,

$\longrightarrow\sf{V=Av\,\Delta t}$

where,

• $\sf{A=}$ cross sectional area of the pipe

• $\sf{v=}$ velocity of water flow

• $\sf{\Delta t=}$ time taken

In case of the same pipe and same time,

$\longrightarrow\sf{V\propto v\quad\quad\dots(1)}$

The power of the motor is,

$\longrightarrow\sf{P=Fv}$

where $\sf{F}$ is the force delivered by the motor. But since $\sf{F=ma,}$

$\longrightarrow\sf{P=mav}$

$\longrightarrow\sf{P=\rho Vav}$

where $\sf{\rho}$ is the density of water which is constant. Thus,

$\longrightarrow\sf{P\propto Vav}$

From (1), $\sf{v}$ should be replaced as,

$\longrightarrow\sf{P\propto V^2a\quad\quad\dots(2)}$

But the acceleration,

$\longrightarrow\sf{a=\dfrac{\Delta v}{\Delta t}}$

Velocity of water flow must be changed as a result of increasing the power of the motor. So for the same time,

$\longrightarrow\sf{a\propto v}$

From (1),

$\longrightarrow\sf{a\propto V}$

Hence (2) becomes,

$\longrightarrow\sf{\underline{\underline{P\propto V^3}}}$

I.e., to obtain $\sf{n}$ times water, the power should be increased $\bf{n^3}$ times.

Hence $\bf{(B)\ n^3}$ is the answer.

Answer 2

Explanation:

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