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 of the flow through single capillary, X=πPR4/8ηL)

Fluid resistance is given by R=8ηlπr4.When two capillary tubes of same size are joined in parallel, then equivalent fluid resistance is  Re=R1+R2=8ηLπr4+8η×2Lπ(2R)4=8ηLπr4×98Equivalent resistance becomes 98 times so rate of flow will be  89X