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.
There fore `(Deltag//g)=(DeltaL//L)+2(DeltaT//T)`
`=(0.1)/(20.0)+2((1)/(90))=0.027`
Thus the percentage error in g is
`100(Deltag//g)=100 (DeltaL//L)+2xx100(DeltaT//T)=3`