Two capillary tubes of same radius r but of lengths $l_{1}$ and $l_{2}$ are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the length of a single tube that can replace the two tubes so that the rate of flow is same as before

l_{1} + l_{2}

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\frac{1}{l_{1}} + \frac{1}{l_{2}}

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\frac{l_{1} l_{2}}{l_{1} + l_{2}}

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

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Solution

The correct option is C $\frac{l_{1} l_{2}}{l_{1} + l_{2}}$
For parallel combination $\frac{1}{R_{e f f}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\Rightarrow \frac{π r^{4}}{8 η l} = \frac{π r^{4}}{8 η l_{1}} + \frac{π r^{4}}{8 η l_{2}} \Rightarrow \frac{1}{l} = \frac{1}{l_{1}} + \frac{1}{l_{2}}$ $∴ l = \frac{l_{1} l_{2}}{l_{1} + l_{2}}$

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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 of a single tube that can replace the two tubes so that the rate of flow is same as before

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