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Simple Harmonic Motion question

2021 · 18 Mar · Shift 1 · Q71
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  5. /2021 · 18 Mar · Shift 1 · Q71

Simple Harmonic Motion question

2021 · 18 Mar · Shift 1 · Q71

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
The time period of a simple pendulum is given by T=2πlgT = 2\pi \sqrt {{l \over g}}T=2πgl​​. The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-
  1. A
    2%
  2. B
    3%
  3. C
    5%
  4. D
    4%
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