MHT CET · Physics · Mechanical Properties of Fluids
A glass tube of uniform cross-section is connected to a tap with a rubber tube. The tap is opened slowly. Initially the flow of water in the tube is streamline. The speed of flow of water to convert it into a turbulent flow is
[radius of tube \(=1 \mathrm{~cm}, \eta=1 \times 10^{-3} \frac{\mathrm{Ns}}{\mathrm{m}^2}, \mathrm{R}_{\mathrm{n}}=2500\) and density of water \(=10^3 \mathrm{~kg} / \mathrm{m}^3\) ]
- A 0.15 m/s
- B 0.125 m/s
- C 0.3 m/s
- D 0.2 m/s
Answer & Solution
Correct Answer
(B) 0.125 m/s
Step-by-step Solution
Detailed explanation
Reynold number is given by
\(
\begin{aligned}
& \mathrm{R}_{\mathrm{n}}=\frac{\mathrm{v}_{\mathrm{c}} \rho \mathrm{d}}{\eta} \\
& \therefore \mathrm{v}_{\mathrm{c}}=\frac{\mathrm{R}_{\mathrm{n}} \eta}{\rho \mathrm{d}} \\
& \mathrm{R}_{\mathrm{n}}=2500, \eta=10^{-3} \mathrm{Ns} / \mathrm{m}^2, \rho=10^3 \mathrm{~kg} / \mathrm{m}^3 \\
& \mathrm{~d}=2 \mathrm{r}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}
\end{aligned}
\)
Substituting these values and calculating we get critical velocity \(\mathrm{v}_{\mathrm{c}}=0.125 \mathrm{~m} / \mathrm{s}\)
\(
\begin{aligned}
& \mathrm{R}_{\mathrm{n}}=\frac{\mathrm{v}_{\mathrm{c}} \rho \mathrm{d}}{\eta} \\
& \therefore \mathrm{v}_{\mathrm{c}}=\frac{\mathrm{R}_{\mathrm{n}} \eta}{\rho \mathrm{d}} \\
& \mathrm{R}_{\mathrm{n}}=2500, \eta=10^{-3} \mathrm{Ns} / \mathrm{m}^2, \rho=10^3 \mathrm{~kg} / \mathrm{m}^3 \\
& \mathrm{~d}=2 \mathrm{r}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}
\end{aligned}
\)
Substituting these values and calculating we get critical velocity \(\mathrm{v}_{\mathrm{c}}=0.125 \mathrm{~m} / \mathrm{s}\)
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