Question

6. If the length and time period of an oscillating
pendulum have errors of 1% and 2% respectively,
what is the error in the estimate of g ?​

Answer 1

Answer:

Given :

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=4π^2(L/T^2)

Δg/g =ΔL/L +2Δ T/T

Δg/g=0.01 +2x0.02

=0.05

Δ(g/g)%=0.05x100=5%

∴The error in the estimation of g is 5%

Explanation:

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