Two capillary tubes of same radius $r$ but of lengths $l_1$ and $l_2$ are fitted in parallel to the bottom of a vessel. The pressure head is $P$. What should be the length 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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(c) For parallel combination $\frac{1}{{{R_{eff}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$
==> $\frac{{\pi {r^4}}}{{8\eta l}} = \frac{{\pi {r^4}}}{{8\eta {l_1}}} + \frac{{\pi {r^4}}}{{8\eta {l_2}}}$ ==> $\frac{1}{l} = \frac{1}{{{l_1}}} + \frac{1}{{{l_2}}}\;\;\;\;\therefore l = \frac{{{l_1}{l_2}}}{{{l_1} + {l_2}}}$
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