NEET Physics Bernoulli's Principle MCQ Question
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.
2000 Pa
4000 Pa
6000 Pa
8000 Pa
Correct Answer
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.
Found an issue with this question?
Related Questions
More from
A metal block of area 0.10 m² is connected to a 0.01 kg mass via a string that passes over a massless and frictionless pulley as shown in figure. A li...
Which of the following statements about pressure in fluids is correct?
Match Column-I with Column-II. Column-I Column-II (a) Streamline flow (i) Path of a fluid particle (b) Capillary action (ii) Rise of l...