NEET 2026 (Phase 1) · PhysicsPrevious Year Question
The sum of the kinetic energy and potential energy of a simple pendulum bob is 0.02 J. The speed of the bob at its equilibrium position is approximately (mass of the bob = 20 g)
- A.0.2 m/s
- B.1.41 m/s✓
- C.14.1 m/s
- D.2.0 m/s
Correct Answer
(B) 1.41 m/s
Solution & Explanation
Correct Answer (B): 1.41 m/s Step-by-step solution: 1. Total mechanical energy E = K.E. + P.E. = 0.02 J (constant). 2. At the equilibrium (mean) position, P.E. = 0, so all the energy is kinetic: ½mv² = E. 3. v = √(2E/m) = √(2 × 0.02 / 0.020). 4. v = √(0.04/0.020) = √2 ≈ 1.41 m/s.
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