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Q.

Two capillary tubes of 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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a

$\sqrt{l_{1} l_{2}}$

b

$l_{1} + l_{2}$

c

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

d

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

answer is C.

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Detailed Solution

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

Rate of flow of liquid, $V = P / R; r_{1} = r_{2}$
$\frac{1}{R} = \frac{π}{8 η} (\frac{r_{1}^{4}}{ℓ_{1}} + \frac{r_{2}^{4}}{ℓ_{2}}); \Rightarrow \frac{{πr}_{}^{4}}{8 ηℓ} = \frac{π}{8 η} (\frac{r_{}^{4}}{ℓ_{1}} + \frac{r_{}^{4}}{ℓ_{2}}); \Rightarrow \frac{ℓ_{1} ℓ_{2}}{ℓ_{1} + ℓ_{2}} = ℓ$

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