Question

2 capillary tubes of same length L but radius r1and r2 are fitted in parallel to the bottom of a vessel what should be the radius r of the single tube of same length that can replace the 2 tubes so that the rate of flow is same as before

Answer 1

Answer:

Explanation:

Assuming the pipes are at equal depths, applying bernoulli's principle

efflux velocity of fluid,V is same in both tubes = 1/2 ( row g h )

Rate of flow = cross section area× velocity

=(πr1^2 + πr2^2)V

Equivalent tube with same flow rate = πR^2V

(  πR^2)V=(πr1^2+πr2^2)V

R= (r1^2+r2^2)1/2

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