The period of oscillation of a simple pendulum is T =`2pisqrt(L/g)`. Measuted value of L is 20.0 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ?

Explanation

`g=4pi^(2)L//T^(2), " Here, "T=t/n and DeltaT=(Deltat)/n`.
Therefore, `(DeltaT)/T=(Deltat)/t`.
The errors in both L and t are the least count errors. Therefore, `(Deltag/g)=(DeltaL//L)+22(DeltaT//T)`
`" "=(0.1)/(20.0)+2(1/90)=0.027`
Thus, the percentage error in g is 100 `(Deltag//g)`
`=100(DeltaL//L)+2 times 100(DeltaT//T)=3`.