Q.

Two capillaries of lengths L and 2L and of radii R and 2R are connected in series. The net rate of flow of fluid through them will be: (given rate of the flow through single capillary X=πPR4/8ηL))

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a

 (9/8) X

b

(5/7) X 

c

(7/5) X

d

 (8/9) X 

answer is A.

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Detailed Solution

Fluid resistance R=8ηlπr4
When two capillary tubes connected in series then
R=R1+R2=8ηlπR4=8η(2L)π(2R)4=8ηLπR4×98
Equivalent resistance becomes 98times, so rate of flow will be 89X.

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