We have two (narrow) capillary tubes $T_1$ and $T_2$. Their lengths are $l_1$ and $l_2$ and radii of cross-section are $r_1$ and $r_2$ respectively. The rate of flow of water under a pressure difference $ P$ through tube $T_1$ is $8cm ^3/sec$. If $l_1 = 2l_2$ and $ r_1 =r_2$, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before $ (= P)$
Diffcult
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(b)$V = \frac{{\pi {{\Pr }^4}}}{{8\eta l}} = \frac{{8c{m^3}}}{{\sec }}$
For composite tube
$ = \frac{2}{3} \times 8 = \frac{{16}}{3}\frac{{c{m^3}}}{{\sec }}$
For composite tube
$ = \frac{2}{3} \times 8 = \frac{{16}}{3}\frac{{c{m^3}}}{{\sec }}$
${V_1} = \frac{{P\pi {r^4}}}{{8\eta \left( {l + \frac{l}{2}} \right)}} = \frac{2}{3}\frac{{\pi {{\Pr }^4}}}{{8\eta l}}$ $ = \frac{2}{3} \times 8 = \frac{{16}}{3}\frac{{c{m^3}}}{{\sec }}$
$\left[ {\because \;\;{l_1} = l = 2{l_2}\;{\text{or}}\;{l_2} = \frac{l}{2}} \right]$
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