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Two capillary tubes of the same length but different radii r1 and r2  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?

Option: 1

r_1+r_2


Option: 2

\frac{r_1r_2}{r_1+r_2}


Option: 3

\frac{r_1+r_2}{2}


Option: 4

None


Answers (1)

best_answer

\text { Use, } \quad V=V_1+V_2

\frac{\pi p r^4}{8 \eta l}=\frac{\rho r_1^4}{8 \eta l}+\frac{\pi p r_2{ }^4}{8}

r^4=r_1^4+r_2^4

r=\left(r_1 ^4+r_2^4\right)r^{4}

Posted by

Kuldeep Maurya

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