Figure below shows a liquid being pushed out of the tube by a piston having area of cross section 2.0 cm². The area of cross section at the outlet is 10 mm². If the piston is pushed at a speed of 4 cm s⁻¹, the speed of outgoing fluid is __________ cm s⁻¹

Solution

Physics sir · 4 steps

Same volume per second through the wide part and the nozzle: put both areas in mm² first.

1. Continuity. The liquid cannot pile up inside the tube. So the volume pushed by the piston each second equals the volume leaving the outlet each second.

   A₁ v₁ = A₂ v₂

2. Match the units. 1 cm² = 100 mm², so the piston area is A₁ = 2.0 cm² = 200 mm². The outlet area is A₂ = 10 mm².
3. Put in numbers. v₂ = A₁/A₂ v₁ = 200/10 × 4 cm s⁻¹ = 20 × 4 cm s⁻¹.
4. So. The outlet is 20 times narrower, so the fluid comes out 20 times faster: v₂ = 80 cm s⁻¹.

The common slip: 0.8. Dividing 2.0 by 10 without converting cm² to mm² gives 0.8, not 80. A narrower outlet must make the fluid faster, never slower.

The idea to keep. In a pipe, speed goes inversely with area: v₂ = v₁ A₁ / A₂. Only the ratio of areas matters, so put both areas in the same unit before dividing.

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