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(A) The Reynolds number for a fluid flow is given by $R_e = \frac{2 v \rho r}{\eta}$,where $v$ is the velocity of flow,$\rho$ is the density of the fl

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$\frac{1}{e}$

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(A) The Reynolds number for a fluid flow is given by $R_e = \frac{2 v ho r}{\eta}$,where $v$ is the velocity of flow,$ho$ is the density of the fluid,$r$ is the radius of the tube,and $\eta$ is the viscosity of the fluid.From the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A = \pi r^2$. Thus,$\pi r_1^2 v_1 = \pi r_2^2 v_2$,which implies $\frac{v_1}{v_2} = \frac{r_2^2}{r_1^2}$.The ratio of Reynolds numbers is $\frac{R_1}{R_2} = \frac{v_1 r_1}{v_2 r_2} = \left(\frac{r_2^2}{r_1^2}ight) 	imes \left(\frac{r_1}{r_2}ight) = \frac{r_2}{r_1}$.Given $r = r_0 e^{-\alpha x}$,we have $r_1 = r_0 e^{-\alpha x_1}$ and $r_2 = r_0 e^{-\alpha (x_1 + 3)}$.Substituting these values,$\frac{R_1}{R_2} = \frac{r_0 e^{-\alpha (x_1 + 3)}}{r_0 e^{-\alpha x_1}} = e^{-\alpha (x_1 + 3) + \alpha x_1} = e^{-3 \alpha}$.Given $\alpha = \frac{1}{3} 	ext{ m}^{-1}$,we get $\frac{R_1}{R_2} = e^{-3(1/3)} = e^{-1} = \frac{1}{e}$.

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