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Water flows through a capillary tube of radius R and length L at a rate of $100 m^{3} s^{- 1}$ when pressure difference is 100 mmHg. If another tube of same length but twice radius is connected in parallel, then the total rate of flow of water through the two tube (for twice the pressure difference) will be:

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$1700 m^{3} s^{- 1}$

$(1700 / 16) m^{3} s^{- 1}$

$1600 m^{3} s^{- 1}$

$3400 m^{3} s^{- 1}$

answer is D.

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Detailed Solution

$Q = \frac{{πPr}^{4}}{8 η l}$

For parallel, pressure difference P is same and $Q_{net} = Q_{1} + Q_{2}$

here Q₂ is 16 Q₁ (P increases by 2 times and r increases by 2 times)

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