JEE Main 2021 · 25 February · morning · Single correct · +4 / −1
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 :
- (a) 16 m/s²
- (b) 2π² ms⁻²
- (c) π² ms⁻²
- (d) 9.8 ms⁻²
Answer and solution
Answer (b) 2π² ms⁻²
Solution
Physics sir · 3 steps
Turn the period formula around: square both sides, and g falls out in one line.
- Start from the formula. For a simple pendulum, T = 2π√(L/g). Square both sides to remove the root: T² = 4π² L/g.
g = (4π² L)/T²
- Put in numbers. With L = 2 m and T = 2 s, we get g = (4π² × 2)/2² = 8π²/4 = 2π² m s⁻².
- So. g = 2π² m s⁻², about 19.7 m s⁻², which is option (b). The pendulum is at a place where gravity is stronger than on Earth's surface.
The idea to keep. Whenever a pendulum's length and period are given, rearrange to g = 4π² L/T² and use only the given data. Do not assume g is 9.8.
Answer from the NTA final answer key. Solution by PrepWiser.