An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is :

Solution

Physics sir · 4 steps

Velocity goes as one over area, and area goes as diameter squared, so square the diameter ratio.

1. Continuity. For an ideal fluid the volume flowing per second is the same at every cross-section, so A₁ v₁ = A₂ v₂. A narrower section forces the fluid to move faster.
2. Match the extremes. Minimum velocity occurs where the pipe is widest (d = 6.4 cm). Maximum velocity occurs where it is narrowest (d = 4.8 cm).

   vₘᵢₙ/vₘₐₓ = A_(narrow)/A_(wide) = (d_(narrow)/d_(wide))² = (4.8/6.4)²

3. Put in numbers. 4.8/6.4 = 3/4, and squaring gives 9/16.
4. So. The ratio of minimum to maximum velocity is 9/16, which is option (b).

The idea to keep. In continuity problems, v ∝ 1/A ∝ 1/d². Always check whether you are given diameters or areas, and square the diameter ratio.

Answer from the NTA final answer key. Solution by PrepWiser.