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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