1.

Three capillary tubes of some radius 1 cm but of lengths 1m, 2m and 3m are fitted horizontally to the bottom of a long vessel containing a liquid at constant pressure land flowing through these tubes. What is the length of a single tube which can replace the three capillaries.

Answer»

Solution :`V_(1)=(piPr^(4))/(8 eta1_(1)),V_(2)=(piPr^(4))/(8 eta 1_(2)),V_(3)=(piPr^(4))/(8 eta I_(3))` and `V=(piPr^(4))/(8etal)`
Now `V=V_(1)+V_(2)+V_(3)implies1/l+1/(l_(1))+1/(l_(2))+1/(l_(3))`
Susbtituting the VALUES we get
`l=(l_(1)l_(2)_(3))/(l_(1)l_(2)+l_(1)l_(3)+l_(2)l_(3))=(1xx2xx3)/(1xx2+1xx3+2xx3)=6/11` m


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