1.

Two capillary tubes of equal length and inner radii 2r and 4r respectively are added in series and a liquid flows through it. If the pressure difference between the ends of the whole system is 8.5 cm of mercury, find the pressure difference between the ends of the first capillary tube.

Answer» Let `P , P_(1), P_(2)` be the pressure at the beginning of a tube, at the joint of two tubes and at the end of second tube. Since the two tubes are in series, so
`(pi(P-P_(1))(2r)^(4))/(8eta l) = (pi(P_(1)-P_(2))(4r)^(4))/(8eta l)`
or `P-P_(1) = (P_(1)-P_(2)) 16`
or `P=17 P_(1) = 16 P_(2)` ...(i)
Given, `P-P_(2) = 8.5` ...(ii)
Fron (i) and (ii) , we have
`P_(2) +8.5 = 17 P_(1) - 16P_(2)`
or `P_(1)-P_(2) = 8.5//17 = 0.5 cm" of "Hg`.
`:.` pressure difference between the ends of the first tube
`=P-P_(1)=(P-P_(2)) - (P_(1)-P_(2))`
`=8.5 - 0.5 = 8.0" cm of "Hg`.


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