Physics-(general)
NEET Physics (general) MCQ Question
Type: MCQ-numerical-Medium-Class 11
What is the pressure difference between the two ends of a horizontal tube of length 1.5 m and radius 1.0 cm, if glycerine flows steadily through it with a mass flow rate of 4.0 × 10 –3 kg/s? (Density of glycerine = 1.3 × 10^3 kg/m^3, viscosity = 0.83 Pa·s).
A
13.3 Pa
B
26.6 Pa
C
53.2 Pa
D
106.4 Pa
Correct Answer
Option C
Detailed Explanation
The pressure difference can be calculated using Poiseuille's law for laminar flow: ΔP = (8ηLQ)/(πr⁴). Given values are: viscosity (η) = 0.83 Pa·s, length (L) = 1.5 m, flow rate (Q) = (mass flow rate/density) = (4.0 × 10^-3 kg/s)/(1.3 × 10^3 kg/m³), and radius (r) = 0.01 m. Substituting these values gives ΔP = 53.2 Pa.
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