To find the value of `g` using simple pendulum. `T=2.00 sec`, `1=1.00m` was measured. Estimate maximum permissible error in `g`. (use `pi^(2)=10`)

Options

A. `0.1m//s^(2)`
B. `0.2m//s^(2)`
C. `0.3m//s^(2)`
D. `0.4m//s^(2)`

Answer

B

Explanation

`T=2pisqrt((l)/(g)) implies g=(4pi^(2)l)/(T^(2))`
`((dg)/(g))_(max)=(Deltal)/(l)+2(DeltaT)/(T)=((0.01)/(1.00)+2(0.01)/(2.00))xx100%=2%`
value of `g=(4pi^(2)l)/(T^(2))=(4xx10xx1.00)/((2.00)^(2))=10.0m//s^(2)`
`((dg)/(g))_(max)=2//100` so `(dg_(max))/(10.0)`
`=(2)/(100) so `(dg)_(max)=0.2=`max "error" in `g`
so `g=(10.0+-0.2)m//s^(2)`