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, `(Deltag //g) = (Delta L//L) + 2(Delta T//T)`
`= (0.1)/( 20.0) +2 ((1)/(90) ) = 0.027`
Thus, the percentage error in g is
`100 (Delta g//g) = 100(Delta L//L) + 2 xx 100 (Delta T //T) = 3`