Water is flowing through a tube of nonuniform cross-section, ratio of the radius at entry and exit end of the pipe is 3: 2. Then the ratio of velocities at entry and exit of liquid is
Correct answer: A. 4:9
- A. 4:9
- B. 9:4
- C. 8:27
- D. 1:1
Explanation
To solve this problem, we apply the principle of continuity, which states that the product of the cross-sectional area and velocity of fluid flow is constant throughout the tube. This can be expressed as A₁V₁ = A₂V₂, where A₁ and A₂ are the cross-sectional areas and V₁ and V₂ are the velocities at entry and exit, respectively. The cross-sectional area is proportional to the square of the radius, so the ratio of cross-sectional areas is (r₁/r₂)² = (3/2)² = 9/4. Therefore, the ratio of velocities is the inverse, which is 4:9.
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About Fluid Dynamics
Fluid flow is described through speed, pressure and density, including steady flow and the equation of continuity for conserving mass. Bernoulli's equation relates pressure, speed and height, while fluid drag and terminal velocity explain why a falling object eventually moves at constant speed when drag balances its weight.
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