Two capillary of length L and 2L and of radius R and 2R are connected in series. The net rate of flow of fluid through them will be (given rate to the flow through single capillary,
`("X"=(pi"PR"^(4))/(8eta"L"))`

Options

A. `(8)/(9)X`
B. `(9)/(8) X`
C. `(5)/(7)X`
D. `(7)/(5)X`

Answer

A

Explanation

Fluid resistance is given by `R=(8 eta l)/(pi r^(4))`
When two capillary tubes of same size are joined in series, then equivalent fluid resistance is
`R_(e )=R_(1)+R_(2)=(8eta l)/(pi r^(4))+(8eta xx2l)/(pi (2R)^(4))=((8eta l)/(pi r^(4)))xx(9)/(8)`
Equivalent resistance becomes `9//8` times so rate of flow will be `(8//9)X`.