The rate of steady volume flow of water through of capillary tube of length 1 and radius r, under a pressure difference of p is V. This tube is connected with another tube of the same length but half the radius, in series. Then, the rate of steady volume flow through them is (The pressure difference across the combination is p)

Options

A. `V/16`
B. `V/17`
C. `(16V)/17`
D. `(17V)/16`

Answer

B

Explanation

Rate of flow,`V=(piPr^(4))/(8nl)`
pressure difference,` P=(V(8nl))/(pir^(4))`
`P=P_(1) + P_(2))` [`because` Series connection]
`implies (V8nl)/(pir^(4)) = (V^(')(8l))/(pir^(4)) + (V^(')(8nl))/(pi(r//2)^(4))`
`implies V/r^(4) = V^(')/r^(4) + (V^(')xx16)/r^(4)`
[ `because` Flow rate is same in series combination ]
`therefore V^(') =V/17`