Two capillary tubes of same length each of 50 cm but of different radii 4 mm and 2mm are connected in series. When water flows, the pressure, difference between the ends of the first tube is

Options

A. `(rho)/(2)`
B. `(rho)/(17)`
C. `(rho)/(4)`
D. `(rho)/(8)`

Answer

B

Explanation

According to the question,
According to poiseuille.s equation, water
resistance, `R = (8 etal)" ". . . (i)`
or `R prop (1)/(r^(4)) " ". . . (ii)`
Given , radius of first capillary tube , `r_(1) = 4` mm =` 4 xx 10^(3)` m and radius of second capillary tube, `r_(II) = 2m m = 2 xx 10^(-3) m `
`R_(I) prop (1)/((r_(I))^(4))prop (1)/(16 xx 16 xx 10^(-12)) " ". . . (ii)`
`R_(II) prop (1)/((r_(II))^(4)) prop (1)/(16 xx 10 ^(-12)) " ". . . (iv)`
From Eqs. (iii) and (iv), we get
`R_(II) = 16 R_(I)`
IF `R_(I) = R " then " R_(II) = 16 R`
Hence, the pressure difference between ends of the first tube ,`(p)/(R + 16 R) xx R = (p)/(17)`