A pump motor is used to deliver water at a certain rate from a given pipe. To obtain 'n' times water from the same pipe in the same time by what amount (a) the force and (b) power of the motor should be increased ?

Options

A. (a) `n^2` times
B. (b) `n^3` times
C. (c) `n` times
D. (d) `n^(3//2)` times

Answer

B

Explanation

If a liquid of density `rho` is flowing through a pipe of cross section A at speed V, the mass coming out per second will be
`((dm)/(dt))=rhoAV`
In order to get n times water in the same time, we get
`((dm)/(dt))^'=n((dm)/(dt))`
i.e., `A^'V^'rho^'=nAVrho`
But as pipe and liquid are same, `rho^'=rho`, `A^'=A`
`V^'=NV`,
So `F^'/F=(V^'((dm)/(dt)))/(V((dm)/(dt)))=(nV(ndm)/(dt))/(V((dm)/(dt)))=n^2`
or `F^'=n^2F`
`P^'/P=(F^'V^')/(FV)=((n^2F)(nV))/(FV)` or `P^'=n^3P`