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Two capillary tubes of same diameter but of lengths 20 cm and 30 cm are fitted horizontally side by side at the bottom of a vessel containing water. The length of a single tube of same diameter that can replace the two tubes such that the volume flow through it is equal to the total volume flow through the two tubes in one second is

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25 cm

50 cm

15 cm

12 cm

answer is D.

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Detailed Solution

$here both the tubes are at the same level so they have same pressure$

$and V_{1} + V_{2} = V where V_{1} and V_{2} are the volumes of water flowing through the two pipes in the 1 st case$

$while V is the volume of water flowing in the second case$

$(\frac{1}{l_{1}} + \frac{1}{l_{2}}) \frac{π P r^{4}}{8 η} = \frac{π P r^{4}}{8 η} \frac{1}{l}$

$l = \frac{l_{1} l_{2}}{l_{1} + l_{2}} = \frac{30 \times 20}{50} = 12 c m$

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