Search any question & find its solution
Question:
Answered & Verified by Expert
Consider a steady flow of oil in a pipeline. The cross-sectional radius of the pipeline decreases gradually as $r=r_0 e^{-\alpha x}$, where $\alpha=\frac{1}{3} \mathrm{~m}^{-1}$ and $x$ is the distance from the pipeline inlet. If $R_1$ is the Reynold's number
for a certain pipeline cross-section at a distance $x_1$ metre from the inlet and $R_2$ is for distance $\left(x_1+3\right)$ metre, then the ratio $\frac{R_1}{R_2}$ is
Options:
for a certain pipeline cross-section at a distance $x_1$ metre from the inlet and $R_2$ is for distance $\left(x_1+3\right)$ metre, then the ratio $\frac{R_1}{R_2}$ is
Solution:
1744 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{e}$
Reynold's number of flow is given by
$R_e=\frac{2 v \rho r}{\eta}$
where, $v=$ velocity of flow, $\rho=$ density of fluid, $r=$ radius of tube and $\eta=$ viscosity of flow Now, required ratio is
$\begin{aligned} & \frac{R_1}{R_2}=\left(\frac{2 v_1 \rho r_1}{\eta}\right) /\left(\frac{2 v_2 \rho r_2}{\eta}\right) \\ & \frac{R_1}{R_2}=\frac{v_1}{v_2} \times \frac{r_1}{r_2}\end{aligned}$
From Eqs. (i) and (ii), we get
$\begin{aligned} \frac{R_1}{R_2} & =\frac{r_2^2}{r_1^2} \times \frac{r_1}{r_2}=\frac{r_2}{r_1} \\ \Rightarrow \quad \frac{R_1}{R_2} & =\frac{r_0 e^{-\alpha\left(x_1+3\right)}}{r_0 e^{-\alpha x_1}}=\frac{e^{-\alpha x_1} \cdot e^{-3 \alpha}}{e^{-\alpha x_1}}=e^{-3 \alpha} \\ & =\frac{1}{e^{3 \alpha}}=\frac{1}{e} \quad\left(\because \alpha=\frac{1}{3} \mathrm{~m}^{-1}\right)\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.