Question

find the relative error in "g" where T=2 pi sqrt 1/g​

Answer 1

Explanation:

Time period of Simple pendulum =2π(√L/g) where L is the length of pendulum. g =acceleration due to gravity. Δ L/L=1%=1/100=0.01. ΔT/T=2%=2/100 =0.02.

T=2π(√L/g) g=4π² (L/T²) Δg/g =ΔL/L +2Δ T/T. Δg/g=0.01 +2x0.02. =0.05.

Δ(g/g)%=0.05x100=5%....

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Answer 2

Hope this answer helps.