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 Delta T= (Delta t)/(n)`
Therefore, `(Delta T)/(T) = (Delta t)/(t)`
The errors in both L and t are the least count errors.
Therefore, `(Delta g//g) = (Delta L//L) + 2 (Delta T//T)`
`=(0.1)/(20.0) +2 ((1)/(90)) = 0.027`
Theus, the percentage error in g is
`100 (Delta g//g)= 100 (DeltaL//L) + 2 xx 100 (Delta T//T)= 3`