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Question

By what factor will the flow rate of a viscous fluid through a pipe change, if the pipe radius is doubled, the fluid viscosity is doubled, the length is doubled and the pressure drop is reduced to a quarter of its previous value?

A
4
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B
16
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C
14
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D
Doesn't change
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Solution

The correct option is D Doesn't change
We know,
Rate of flow of a viscous fluid,
V=πPr48ηl
where, P= Pressure drop across pipe
r= Internal radius of pipe
l= Length of pipe
η=Fluid viscosity
Given,
If P=P4,r=2r,η=2η,l=2l
Then, V=πP(r)48ηl=π(P4)(2r)48(2η)(2l)
VV=1
Volume flow rate remains same.

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