The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at A is 1.5 cm², and B is 25 mm², if the speed of liquid at B is 60 cm / s then (P_A-P_B) is :
(Given P_A and P_B are liquid pressures at A and B% points.
density ρ=1000 kg m⁻³
A and B are on the axis of tube

Solution

Physics sir · 4 steps

Continuity gives the speed at A, then Bernoulli turns the speed difference into pressure.

1. Same units first. Convert both areas to the same unit: 1.5 cm² = 150 mm², so A_A = 150 mm² and A_B = 25 mm². Also v_B = 60 cm s⁻¹ = 0.6 m s⁻¹.
2. Continuity. The same volume flows past A and B each second, so A_A v_A = A_B v_B. This gives v_A = 25/150 × 0.6 = 0.1 m s⁻¹.

   P_A - P_B = 1/2ρ(v_B² - v_A²)

3. Put in numbers. The tube is horizontal and A and B are on its axis, so the height terms cancel. P_A - P_B = 1/2 × 1000 × (0.36 - 0.01) = 500 × 0.35.
4. So. P_A - P_B = 175 Pa, option (b). The pressure is higher at the wide end A, where the liquid moves more slowly.

The idea to keep. Pipe questions with two areas always need two equations: continuity gives the missing speed, and Bernoulli gives the pressure. Put every quantity in SI units before you square any speed.

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