if the length and period of an oscillation pendulum have errors of 1% and 2% respectively,what is the percentage in the estimate of g
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ΔL/L =1% =1/100=0.01
ΔT/T=2%=2/100=0.02
T=2π(√L/g)
g=4π square (LT) square
Δg/g =ΔL/L +2Δ T/T
Δg/g=0.01+2 into(*) 0.02 = 0.05
Δ(g/g)%=0.05 into (*) 100 = 5% answer
ΔT/T=2%=2/100=0.02
T=2π(√L/g)
g=4π square (LT) square
Δg/g =ΔL/L +2Δ T/T
Δg/g=0.01+2 into(*) 0.02 = 0.05
Δ(g/g)%=0.05 into (*) 100 = 5% answer
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