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Q9.13: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF FLUIDS 9.13 Solution: Glycerine flows steadily through a horizontal tube of length…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the radius exponent in Poiseuille's law, using $r^2$ (from the area formula) instead of the correct $r^4$, which leads to a massive underestimation of the pressure drop. Always verify that the radius is converted to meters and raised to the fourth power in the denominator.

Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0×10−34.0 \times 10^{-3} kg s__, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3×1031.3 \times 10^3 kg m__ and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Step-by-Step Verified Solution

**Answer:** Pressure difference = 9.75×1029.75 \times 10^2 Pa. Laminar flow assumption is valid (Re � 0.5). **Step-by-Step Explanation:** Volume flow rate: Q=mρ=4.0×10−31.3×103=3.08×10−6Q = \frac{m}{\rho} = \frac{4.0 \times 10^{-3}}{1.3 \times 10^3} = 3.08 \times 10^{-6} m_/s. Using Poiseuille's equation:
ΔP=8ηLQπr4=8×0.83×1.5×3.08×10−6π×(0.01)4=9.75×102 Pa\Delta P = \frac{8 \eta L Q}{\pi r^4} = \frac{8 \times 0.83 \times 1.5 \times 3.08 \times 10^{-6}}{\pi \times (0.01)^4} = 9.75 \times 10^2 \text{ Pa}
Reynolds number: v=Q/(πr2)=0.0098v = Q/(\pi r^2) = 0.0098 m/s. Re=ρvd/η=1.3×103×0.0098×0.02/0.83=0.31<2000Re = \rho v d/\eta = 1.3 \times 10^3 \times 0.0098 \times 0.02 / 0.83 = 0.31 < 2000. Laminar flow is valid.

All Questions in this Chapter

Q9.1 Q9.2 Q9.3 Q9.4 Q9.5 Q9.6 Q9.7 Q9.8 Q9.9 Q9.10 Q9.11 Q9.12 Q9.13 Q9.14 Q9.15 Q9.16 Q9.17 Q9.18 Q9.19 Q9.20

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