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JEE Main 2021 (25 Feb, Shift 1) · Simple Pendulum

Simple Pendulum question from JEE Main 2021 (25 Feb, Shift 1)

Q63JEE Main 2021 · 25 FEB · SHIFT 1Easy

Simple Pendulum

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

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Answer: (b)

  1. T=2πLg⇒g=4π2LT2T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow g = \frac{4\pi^2L}{T^2}
  2. g=4π2×222=2π2 m/s2g = \frac{4\pi^2 \times 2}{2^2} = 2\pi^2\text{ m/s}^2
  3. So the answer is (b) 2π2 m s−22\pi^2\text{ m s}^{-2}.
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