Physics-Bernoulli's Principle
NEET Physics Bernoulli's Principle MCQ Question
Type: MCQ-numerical-Hard-Class 11
In a horizontal pipe of varying cross-section, water flows steadily. At a point where the cross-sectional area is reduced to half, what is the pressure difference between this point and a point in the wider section, if the velocity of water in the wider section is 2 m/s? Assume water density is 1000 kg/m³.
A
2000 Pa
B
4000 Pa
C
6000 Pa
D
8000 Pa
Correct Answer
Option A
Detailed Explanation
According to Bernoulli's principle, the pressure difference can be computed using the equation: P1 + 0.5 * ρ * v1² = P2 + 0.5 * ρ * v2². With the velocity v2 in the narrower section as 4 m/s (due to continuity, A1v1 = A2v2), the pressure difference is calculated as 2000 Pa.
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