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

Explanation

`V_(1)=(piPr^(4))/(8etal_(1)),V_(2)=(piPr^(4))/(8etal_(2)),V_(3)=(piPr^(4))/(8etal_(3))` and `V=(piPr^(4))/(8etal)`
Now `V=V_(1)+V_(2)+V_(3)implies(1)/(l)=(1)/(l_(1))+(1)/(l_(2))+(1)/(l_(3))`
Substituting the values we get
`l=(l_(1)l_(2)l_(3))/(l_(1)l_(2)+l_(1)l_(3)+l_(2)l_(3))=(1xx2xx3)/(1xx2+1xx3+2xx3)=(6)/(11)m`