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Q.

Two capillary tubes of same radius r but of lengths l₁ and l₂ are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the length l of the single tube that can replace the two tubes, so that the rate of flow is same as before :

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a

$l = l_{1} + l_{2}$

b

$l = \frac{l_{1} l_{2}}{l_{1} + l_{2}}$

c

$l = \frac{1}{l_{1}} + \frac{1}{l_{2}}$

d

$l = \frac{1}{l_{1} + l_{2}}$

answer is B.

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Detailed Solution

In parallel combination $P_{1} = P_{2} = P$ and rate of flow $Q = Q_{1} + Q_{2}$
$\frac{{πPr}^{4}}{8 ηl} = \frac{{πP}_{1} r^{4}}{8 {ηl}_{1}} + \frac{{πP}_{2} r^{4}}{8 {ηl}_{2}} \Rightarrow \frac{1}{l} = \frac{1}{l_{1}} + \frac{1}{l_{2}} \Rightarrow l = \frac{l_{1} l_{2}}{l_{2} + l_{1}}$

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