The rate of steady volume flow of water through a capillary tube of length ' l ' 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

C

Explanation

`V= (pi P r^(4))/(8 eta l)` , when tubes are connected in series, then the rate of flow of water through each tube `(V_(1))` is same. Let `(P - P_(1))` be the diffrence in pressure across the first tube and `P_(1)` be the difference in pressure across the second tube. Then
`V_(1) =(pi(P - P_(1)r^(4)))/(8eta l) =(pi P_(1)(r//2)^(4))/(8eta l)` ....(i)
or `P - P_(1)=P_(1)//16)` or `P_(1)=16P//17`
From (i),
`V_(1) =(pi(16 P//17)(r^(4)//16))/(8eta l) =(1)/(17)xx(pi P r^(4))/(8eta l) =(V)/(17)`
Rate of decrease of volume `=V - (V)/(17) = (16V)/(17)`