We have two (narrow) capillary tubes T1 and T2. Their lengths are l1 and l2 and radii of cross-section are r1 and r2 respectively. The rate of flow of water under a pressure difference P through tube T1 is 8cm3/sec. If l1 = 2l2 and r1 =r2, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

4 cm3/sec

(16/3) cm3/sec

(8/17) cm3/sec

None of these

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

V=πPr48ηl=8cm3secFor composite tube V1=Pπr48ηl+l2=23πPr48ηl=23×8=163cm3sec∵l1=l=2l2 or l2=l2

Watch 3-min video & get full concept clarity