NTA Abhyas JEE Main2020PhysicsUnits and DimensionsPractice
The time period of oscillation of a simple pendulum is given by T = 2 π l g . The length of the pendulum is measured as l = 10 ± 0.01   c m   and the time period as T = 0.5 ± 0.02   s . The percentage error in the value of g is
Options
- A5 %
- B8%
- C7 %
- DNone of these
Correct answer
B. 8%
Step-by-step solution
Δ g g × 100 = Δ l l × 100 + 2 Δ T T × 100 Δ l = 0.01 c m Δ T = 0.02 s l = 10 c m T = 0.5 s Δ l l × 100 = 0.01 10 × 100 = 0.1 Δ T T × 100 = 0.02 0.5 × 100 = 4 Δg g × 100 = 0.1 + 2 × 4 ≈ 8 %