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TS EAMCET · Physics · Mechanical Properties of Fluids

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

  1. A \(\frac{1}{e}\)
  2. B \(e\)
  3. C \(\frac{1}{e^3}\)
  4. D \(\frac{1}{e^6}\)
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Answer & Solution

Correct Answer

(A) \(\frac{1}{e}\)

Step-by-step Solution

Detailed explanation

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…
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