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WBJEE Surface Tension — practice questions — page 3

Explore 66 free Surface Tension questions in WBJEE Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Surface Tension questions and exercises with answers

Showing questions 61–66 of 66.

For a 3 kg block connected over a frictionless pulley to a 20 kg trolley (μk = 0.04, g = 10 m s⁻²), the tension in the connecting string is:Read the answer and worked solution →A net force F = 6t newton (t in seconds) acts on a 2 kg body initially at rest. The speed of the body at t = 4 s is:Read the answer and worked solution →A block slides down a rough incline of 30° with coefficient of kinetic friction 0.2 (g = 10 m s⁻²). Its acceleration down the incline is:Read the answer and worked solution →A lift descends with a downward acceleration equal to g. The apparent weight of a passenger standing inside is:Read the answer and worked solution →A 0.2 kg ball strikes the floor vertically at 10 m s⁻¹ and rebounds at 8 m s⁻¹. The magnitude of the impulse imparted by the floor on the ball is:Read the answer and worked solution →The optimum (design) speed for a car on a frictionless road banked at 45° along a curve of radius 100 m is (g = 10 m s⁻²):Read the answer and worked solution →

Worked practice questions

For a 3 kg block connected over a frictionless pulley to a 20 kg trolley (μk = 0.04, g = 10 m s⁻²), the tension in the connecting string is:

  1. 8.0 N
  2. 22.0 N
  3. 27.1 N
  4. 30.0 N
Show answer and solution

Correct answer: C. 27.1 N

Using a = 0.96 m s⁻² from the system equations, T = 3(g − a) = 3(10 − 0.96) = 27.1 N (equivalently T = 8 + 20a).

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A net force F = 6t newton (t in seconds) acts on a 2 kg body initially at rest. The speed of the body at t = 4 s is:

  1. 48 m s⁻¹
  2. 6 m s⁻¹
  3. 24 m s⁻¹
  4. 12 m s⁻¹
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Correct answer: C. 24 m s⁻¹

Impulse = ∫₀⁴ 6t dt = 3t²|₀⁴ = 48 N s = Δp hence v = Δp/m = 48/2 = 24 m s⁻¹. Distractors drop the time-integration or the mass.

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A block slides down a rough incline of 30° with coefficient of kinetic friction 0.2 (g = 10 m s⁻²). Its acceleration down the incline is:

  1. 3.3 m s⁻²
  2. 1.7 m s⁻²
  3. 5.0 m s⁻²
  4. 2.7 m s⁻²
Show answer and solution

Correct answer: A. 3.3 m s⁻²

Down the incline a = g(sinθ − μk cosθ) = 10(0.5 − 0.2×0.866) = 10(0.5 − 0.173) = 3.3 m s⁻². The value 5.0 wrongly ignores friction.

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