CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Water is flowing through a capillary tube at the rate of 20×106m3/s. Another tube of same radius and double the length is connected in series to the first tube. Now the rate of flow of water in m3s1 is:

A
10×106
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3.33×106
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
6.67×106
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
20×106
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 6.67×106
Qflow=πr4ΔP8ηL=20×106m3/s ----------(1)
As now 2 pipes are connected in series, the q through both the pipes will be same and let the pressure be P1 at the function,
πr4(PP1)8ηL=πr4(P1)8η2L
2P2P1=P12P3=P1-------------(2)
So, From when both pipe are connected in series
=πr4P18η2L=13×20×106m3/s........[from(1)]
=6.67×106m3/s

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Variation of Pressure in Fluids
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon