Question

Two capillary tubes of same length and radii $r_1,r_2$ are fitted horizontally side by side(in parallel) to the bottom of a vessel containing water. The radius of a single tube that can replace the two tubes such that the rate of steady flow through this tube equals the combined rate of flow through the tube is

• A. $(r_1^2+r_2^2{)}^{\frac{1}{2}}$
• B. $(r_1^3+r_2^3{)}^{\frac{1}{3}}$
• C. $(r_1+r_2{)}^{\frac{1}{2}}$
• D. $(r_1^4+r_2^4{)}^{\frac{1}{4}}$

Answer 1

Answer: (D)

$(r_1^4+r_2^4{)}^{\frac{1}{4}}$

flow through two capillary $V=\frac{\pi P{r}_1^4}{8\eta l}+\frac{\pi P{r}_2^4}{8\eta l}$ .....(1)

let the radius of final pipe $r=r_3$

V will be same for each case

$V=\frac{\pi P{r}_3^4}{8\eta l}$ .....(2)

from 1 and 2 equation

$r_3=(r_1^4+r_2^4{)}^{1/4}$