Question

the acceleration due to gravity is measured on earths surface by using a simple pendulum . if time period of the pendulum T= [4.00+- 0.01] s and length of pendulum l=[4.00+-0.01]m were measured then value of g upto correct significant figures is

Answer 1

Answer:

9.87±0.03 m/s²

Explanation:

T = 2π√L/g

T = 2π√L/g

g = 4π²L/T² = π² = 9.87 m/s²

error calculation Δg/g = ΔL/L - 2ΔT/T = 0.01/4 - .02/4 = 0.01/4

Δg = 0.01/4 x 9.87 = 0.1/4 = 0.025 ≈ 0.03

g = 9.87±0.03 m/s²