AP EAMCET · PHYSICS · Mechanical Properties of Fluids
If water flows with a velocity of \(20 \mathrm{~cm} \mathrm{~s}^{-1}\) in a pipe of radius 2 cm, then the flow is
(The coefficient of viscosity of water is \(10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}\) and density of water is \(\left.10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right)\).
- A turbulant
- B steady flow
- C non-viscous
- D unsteady
Answer & Solution
Correct Answer
(A) turbulant
Step-by-step Solution
Detailed explanation
\( D = 2r = 2 \times 0.02 \mathrm{~m} = 0.04 \mathrm{~m} \) \( \mathrm{Re} = \frac{\rho v D}{\eta} \) \( \mathrm{Re} = \frac{(10^3 \mathrm{~kg} \mathrm{~m}^{-3})(0.20 \mathrm{~m} \mathrm{~s}^{-1})(0.04 \mathrm{~m})}{10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}} \)…
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