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

Explanation

`g = 4 pi^2 L//T^2`
Here , `T = t/n and Delta T = (Delta t)/n`
Therefore, `(Delta T)/T = (Delta t)/t`
The erros 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`
Thus, the percentage error in g is
`100(Delta g//g) = 100(Delta L//L) + 2 xx 100 (Delta T//T) = 3`.