Correct Answer - `(pi P)/(8 eta(l_1//r_1^4 + l_2//r_2^4))`
Let `P_(1)` and `P_(2)` be the pressure difference across the first and second capillary tube respectively. Let `R_(1), R_(2)` be the second capiliary tube respectively. Then
`R_(1)=(8 etal_(1))/(pi r_(1)^(4))` and `R_(2)=(8 etal_(2))/(pir_(2)^(4))`
As the two capillary tubes are connected in series total liquid resistance,
`R=R_(1)+R_(2)=(8 eta l_(2))/(pi r_(1)^(4))=(8 eta)/(pi)[(l_(1))/(r_(1)^(4))+(l_(2))/(r_(2)^(4))]`
Rate of flow loquid through the tubes is
`V=(P)/(R)=(P)/(8 eta)/(pi)[(l_(1))/(r_(1)^(4))+(l_(2))/(r_(2)^(4))]=(pi P)/(8 eta[(l_(1))/(r_(1)^(4))+(i_(2))/(r_(2)^(4))]`