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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, $X = {πPR}^{4} / 8 ηL)$

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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

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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)

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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, $X = {πPR}^{4} / 8 ηL)$