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(B) According to Poiseuille's equation,the rate of flow $V$ is given by:$V = \frac{P \pi r^4}{8 \eta l}$Since $P, \pi, \eta,$ and $l$ are constant,we

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$46$

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(B) According to Poiseuille's equation,the rate of flow $V$ is given by:$V = \frac{P \pi r^4}{8 \eta l}$Since $P, \pi, \eta,$ and $l$ are constant,we have $V \propto r^4$.Given that the radius $r$ increases by $10\%$,the new radius $r_2 = r_1 + 0.1r_1 = 1.1r_1$.The ratio of the new flow rate $V_2$ to the initial flow rate $V_1$ is:$\frac{V_2}{V_1} = \left( \frac{r_2}{r_1} ight)^4 = (1.1)^4 = 1.4641$.Therefore,$V_2 = 1.4641 V_1$.The percentage change in the rate of flow is:$\frac{\Delta V}{V_1} 	imes 100 = \frac{V_2 - V_1}{V_1} 	imes 100 = (1.4641 - 1) 	imes 100 = 46.41\% \approx 46\%$.

Column - $I$	Column - $II$
$(a)$ $R_e > 2000$	$(i)$ Streamline
$(b)$ $1000 < R_e < 2000$	$(ii)$ Turbulent
	$(iii)$ Unsteady

$A$ capillary tube is connected to the bottom of a container. If its radius is increased by $10\%$,what is the percentage change in the rate of flow of the liquid (in $\%$)?

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