In a cement factory, a viscous liquid flows through a pipe with an area of cross-section 6 m² at a velocity of 8 m/s.When this fluid moves forward, the area of cross-section of the pipe decre velocity of the liquid will now be 4 m² The
Correct answer: B. 12 m/s.
- A. 3 m/s.
- B. 12 m/s.
- C. 18 m/s.
- D. 24 m/s.
Explanation
The principle of conservation of mass for a fluid in a closed system is given by the equation: A1 * V1 = A2 * V2, where A1 and V1 are the initial area and velocity, and A2 and V2 are the final area and velocity. Initially, the fluid flows with an area of 6 m² and a velocity of 8 m/s, so A1 * V1 = 6 * 8 = 48. When the area decreases to 4 m², applying the equation gives 48 = 4 * V2, solving for V2 gives V2 = 12 m/s. Therefore, the correct answer is Option B: 12 m/s. The other options do not satisfy this equation and therefore are incorrect.
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About Equation of Continuity
The equation of continuity expresses conservation of mass in fluid flow. For steady incompressible flow, Av remains constant, so fluid speed increases when the cross-sectional area decreases; for compressible flow, ρAv is constant. This relation describes flow rate and area-speed changes without including pressure effects from Bernoulli’s equation.
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