JEE Main202124 Feb 2021Evening ShiftPhysicsMathematics in PhysicsActual
The period of oscillation of a simple pendulum is T = 2 π L g . Measured value of L is 1 . 0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1 . 95 s measured from stopwatch of 0 . 01 s resolution. The percentage error in the determination of ' g ' will be:
Options
- A1 . 03 %
- B1 . 33 %
- C1 . 30 %
- D1 . 13 %
Correct answer
D. 1 . 13 %
Step-by-step solution
T = 2 π ℓ g g = 4 π 2 ℓ T 2 Δ g g = Δ ℓ ℓ + 2 Δ T T Δ g g = 1 × 10 - 3 1 + 2 × 0 . 01 1 . 95 Δ g g = 0 . 0113 or 1 . 13 %