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NEET 2026 · Simple Pendulum

Simple Pendulum question from NEET 2026

Q5NEET 2026Easy

Simple Pendulum

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately:

(Consider mass of the bob =20 g= 20\text{ g})

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Answer: (b)

  1. At the equilibrium position the whole energy is kinetic: 12mv2=E\frac{1}{2}mv^2 = E.
  2. 12(0.02)v2=0.02⇒v2=2\frac{1}{2}(0.02)v^2 = 0.02 \Rightarrow v^2 = 2
  3. v=2≈1.41 m/sv = \sqrt{2} \approx 1.41\text{ m/s}
  4. So the answer is (b) 1.41 m/s1.41\text{ m/s}.
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