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Two capillary tubes of same radius r but...

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 to the bottom of a vessel. The pressure head is P. What should be the length of a single tube of same radius that can replace the two tubes so that the rate of flow is same as before?

A

`l_(1)+l_(2)`

B

`(l)/(l_(1))+(l)/(l_(2))`

C

`(l_(1)l_(2))/(l_(1)+l_(2))`

D

`(l)/(l_(1)+l_(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
For parallel combination `(1)/(R_(eff))=(1)/(R_(1))+(1)/(R_(2))implies(pir^(4))/(8etal)=(pir^(4))/(8etal_(1))+(pir^(4))/(8etal_(2))`
`implies (1)/(l)=(1)/(l_(1))+(1)/(l_(2)) " "therefore l = (l_(1)l_(2))/(l_(1)+l_(2))`
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