Answer:
Let
a = area of cross section of pipe, v = velocity of flow of water, p = density
of water.
\[\therefore \] Mass of water flowing out/sec = \[\frac{dm}{dt}=\text{av}\rho
\]. To get n times water in the same time,
\[(\text{dm}/\text{dt})'=\text{n}\left(
\text{dm}/\text{dt} \right)\] \[\therefore \] \[\text{a}'\text{v}'\rho
'\text{ }=\text{n }(\text{av}\rho )\]
But \[\rho '=\rho \] and a' = a (pipe is same) \[\therefore
\] \[~\upsilon '=\text{n}\upsilon \] ...(i)
As \[\text{F}=\upsilon \frac{dm}{dt}\] \[\therefore \] \[\frac{F'}{F}=\frac{\upsilon
'(dm/dt)'}{\upsilon (dm/dt)}=\frac{n\upsilon \left( n\frac{dm}{dt}
\right)}{\upsilon \,dm/dt}={{n}^{2}}\] \[\therefore \]\[F'={{n}^{2}}F\]
Also, \[\frac{P'}{P}=\frac{F'\times \upsilon '}{F\times \upsilon
}=\frac{{{n}^{2}}F\times n\upsilon }{F\times \upsilon }={{n}^{3}}\]
Therefore, \[\text{P}'={{\text{n}}^{\text{3}}}\text{P}\]
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