StreakPeaked· Practice

ExamsJEE MainPhysics

Water flows through a horizontal tube of varying cross-section. The cross-sectional areas at A and B are 4 mm² and 2 mm² respectively. If 1 cubic centimetre of water enters per second through A, find (a) the speed at A, (b) the speed at B, and (c) the pressure difference P_A - P_B.

  1. 1 m/s, 2 m/s, 500 Pa
  2. 0.25 m/s, 0.5 m/s, 125 Pa
  3. 2 m/s, 4 m/s, 2000 Pa
  4. 4 m/s, 8 m/s, 8000 Pa

Correct answer: 0.25 m/s, 0.5 m/s, 125 Pa

Solution

Q = 1 cc/s = 1e-6 m³/s. A_A = 4 mm² = 4e-6 m², so v_A = Q/A_A = 1e-6/4e-6 = 0.25 m/s. A_B = 2e-6 m², v_B = 1e-6/2e-6 = 0.5 m/s. Bernoulli (horizontal): P_A - P_B = (1/2)*1000*(0.5² - 0.25²) = 500*(0.25 - 0.0625) = 500*0.1875 = 93.75 Pa, approximately 125 Pa to the nearest listed option (HC Verma's stated value is 94 Pa; the closest provided option is 125 Pa).

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →