Q.
A liquid flows steadily through a series combination of three capillary tubes of radii r, 2r and 3r, all of the same length L. If the pressure difference across the combination is 28 cm of mercury, the pressure difference (in cm of Hg) across the tube of radius 2r is very nearly equal to____cm of Hg
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answer is 1.63.
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Detailed Solution
Q1=Q2=Q3. Therefore,πp1r48ηL=πp2(2r)48ηL=πp3(3r)48ηL⇒p1=16p2=81p3∴p1=16p2and p3=1681p2Given p1+p2+p3=28cm of Hg⇒16p2+p2+1681p2=28⇒ 16+1+1681p2=28⇒ p2=81×281393=1.628cm of Hg
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