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
Diffcult
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(d)$V = {V_1} + {V_2}$
==>$\frac{{\pi {{\Pr }^4}}}{{8\eta l}} = \frac{{\pi \Pr _1^4}}{{8\eta l}} + \frac{{\pi pr_2^4}}{{8\eta l}}$ ==> ${r^4} = r_1^4 + r_2^4$
$r = {(r_1^4 + r_2^4)^{1/4}}$
==>$\frac{{\pi {{\Pr }^4}}}{{8\eta l}} = \frac{{\pi \Pr _1^4}}{{8\eta l}} + \frac{{\pi pr_2^4}}{{8\eta l}}$ ==> ${r^4} = r_1^4 + r_2^4$
$r = {(r_1^4 + r_2^4)^{1/4}}$
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