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
(8/9) X
b
(9/8) X
c
(5/7) X
d
(7/5) X
answer is A.
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Detailed Solution
Fluid resistance R=8ηlπr4When two capillary tubes connected in series thenR=R1+R2=8ηlπR4=8η(2L)π(2R)4=8ηLπR4×98Equivalent resistance becomes 98times, so rate of flow will be 89X.