Question

Two capillary tubes of the same length but different radii $r_{1}and{r}_2$ are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the radius of a single tube that can replace the two tubes so that the rate of flow is same as before

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$V=V_1+V_2\Rightarrow \frac{\pi P{r}^4}{8nl}=\frac{\pi P{r}_1^4}{8nl}+\frac{\pi P{r}_2^4}{8nl}\Rightarrow r^4=r_1^4+r_2^4\therefore r=(r_1^4+r_2^4{)}^{\frac{1}{4}}$