We have two (narrow) capillary tubes T1, and T2 . Their lengths are l1 and l2 and radii of cross-sections arc r1, and r2 respectively. The rate of flow of water under a pressure difference P through tube T1, is 8 cm3/s. If l1 =2 l2 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 (P) as before?
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
4cm3/s
b
(16/3)cm3/s
c
(8/17)cm3/m
d
none of the above
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Given that, V=πPr48ηl=8cm3/ s where l1=l For composite tube V1=πPr48η(l+l/2) ∵l1=l=2l2 or l2=l/2 ∴ V1=23×πPr48ηl =23×8cm3/s=163cm3/s