Question

The rate of flow of liquid in a capillary tube under and constant pressure head is Q. If the diameter of the tube is reduced to half and its length is doubled, then the new rate of flow of liquid will be:

• A. $\frac{Q}{4}$
• B. $\frac{Q}{8}$
• C. $16Q$
• D. $\frac{Q}{32}$

Answer 1

Answer: (D)

$\frac{Q}{32}$

$Theflowthroughthepipeisgivenby,Q=\frac{\Delta P\pi R^4}{8\eta L}Now{R}^{‵}=\frac{R}{2},L^{‵}=2LKeepingallotherfactorsconstantandsubstitutingforRandL,Q^{‵}=\frac{\Delta P\pi }{8\eta }\left(\frac{\frac{R^4}{16}}{2L}\right)Q^{‵}=\frac{\Delta P\pi }{8\eta }\left(\frac{R^4}{32L}\right)Q^{‵}=\frac{\Delta P\pi R^4}{8\eta L}\times \frac{1}{32}\because Q^{‵}=\frac{Q}{32}$