Question

Q. Calculate percentage error in the determination of g=4Πl/t² where Δl=2% and Δt=3%​

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=4π² (L/T²)

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:

Answer 2

plz refer to the attachment above