Water flows in a horizontal tube (see figure).
The pressure of water changes by 700 Nm –2
between A and B where the area of cross section
are 40 cm² and 20 cm², respectively. Find the
rate of flow of water through the tube.
(density of water = 1000 kgm –3 )

Solution

Physics sir · 4 steps

The narrow part at B is half the area, so water runs twice as fast there. That speed-up costs 700 Pa.

1. Continuity first. The same volume passes every section each second, so A_A v_A = A_B v_B. With 40 cm² at A and 20 cm² at B, this gives v_B = 2v_A.
2. Bernoulli, horizontal tube. There is no height change, so the pressure drop from A to B pays only for the gain in kinetic energy per unit volume.

   P_A - P_B = 1/2ρ(v_B² - v_A²) = 1/2ρ(4v_A² - v_A²) = 3/2ρ v_A²

3. Find the speed. 700 = 3/2 × 1000 × v_A², so v_A² = 7/15 = 0.467 m² s⁻². That gives v_A = 0.683 m s⁻¹ = 68.3 cm s⁻¹.
4. So. Rate of flow Q = A_A v_A = 40 cm² × 68.3 cm s⁻¹ ≈ 2730 cm³ s⁻¹. The closest option is (d) 2720 cm³/s.

The idea to keep. Pressure difference across a constriction: use continuity to write both speeds in terms of one, then use Bernoulli. Remember that both speeds appear in v_B² - v_A², not only the faster one.

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