We have two narrow capillary tubes T_1 and T_ | vedclass

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(B) According to Poiseuille's equation,the rate of flow $V$ is given by $V = \frac{\pi P r^4}{8 \eta l}$.For tube $T_1$,$V_1 = \frac{\pi P r_1^4}{8 \e

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$(16/3) \ cm^3/sec$

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(B) According to Poiseuille's equation,the rate of flow $V$ is given by $V = \frac{\pi P r^4}{8 \eta l}$.For tube $T_1$,$V_1 = \frac{\pi P r_1^4}{8 \eta l_1} = 8 \ cm^3/sec$.Given $l_1 = 2l_2$ and $r_1 = r_2 = r$,we have $l_2 = l_1/2$.When connected in series,the equivalent resistance $R_{eq} = R_1 + R_2$,where $R = \frac{8 \eta l}{\pi r^4}$.$R_1 = \frac{8 \eta l_1}{\pi r^4}$ and $R_2 = \frac{8 \eta l_2}{\pi r^4} = \frac{8 \eta (l_1/2)}{\pi r^4} = \frac{1}{2} R_1$.$R_{eq} = R_1 + \frac{1}{2} R_1 = \frac{3}{2} R_1$.The new rate of flow $V' = \frac{P}{R_{eq}} = \frac{P}{(3/2) R_1} = \frac{2}{3} \left( \frac{P}{R_1} ight) = \frac{2}{3} V_1$.$V' = \frac{2}{3} 	imes 8 = \frac{16}{3} \ cm^3/sec$.

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