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Indian Navy 10+2 B.Tech Cadet Surface Tension — practice questions

Explore 4 free Surface Tension questions in Indian Navy 10+2 B.Tech Cadet 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 1–4 of 4.

A pendulum of mass m swings to amplitude angle θ. At the lowest point of swing, the TENSION in the string equals:Read the answer and worked solution →A car of mass 1000 kg moving at 36 km/h is stopped by brakes in 5 seconds. Average force exerted by brakes:Read the answer and worked solution →A body in equilibrium under several forces hasRead the answer and worked solution →A 0.5 kg ball hits a wall at 10 m/s and rebounds at 10 m/s. The impulse on the ball isRead the answer and worked solution →

Worked practice questions

A pendulum of mass m swings to amplitude angle θ. At the lowest point of swing, the TENSION in the string equals:

  1. mg
  2. mg(2 − cos θ)
  3. mg(3 − 2cos θ)
  4. 2mg sin θ
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Correct answer: C. mg(3 − 2cos θ)

**At the lowest point:** T − mg = mv²/L (centripetal). v² at bottom = 2gL(1 − cos θ) (energy conservation from amplitude). We have So T = mg + 2mg(1 − cos θ). So So T = mg(3 − 2cos θ). For θ = 60°: T = mg(3 − 1) = 2mg.

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A car of mass 1000 kg moving at 36 km/h is stopped by brakes in 5 seconds. Average force exerted by brakes:

  1. 1000 N
  2. 2000 N
  3. 5000 N
  4. 10000 N
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Correct answer: B. 2000 N

**Convert:** 36 km/h = 10 m/s. Deceleration a = (0 - 10)/5 = -2 m/s². Braking force F = ma = 1000 × 2 = 2000 N.

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A body in equilibrium under several forces has

  1. zero mass
  2. increasing momentum
  3. maximum acceleration
  4. zero net force
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Correct answer: D. zero net force

Equilibrium means the vector sum of forces is zero, so acceleration is zero.

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