The period of oscillation of a simple pendulum is $$T = 2\pi\sqrt{\frac{L}{g}}$$. Measured value of $$L$$ is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of $$g$$ will be:

We are given the period of a simple pendulum as $$T = 2\pi\sqrt{\frac{L}{g}}$$. From this, we can express $$g$$ as $$g = \frac{4\pi^2 L}{T^2}$$.

Taking the logarithm and differentiating, the relative error in $$g$$ is given by $$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$$.

The measured length is $$L = 1.0$$ m and the least count of the meter scale is 1 mm, so $$\Delta L = 0.001$$ m. The measured time period is $$T = 1.95$$ s and the resolution of the stopwatch is 0.01 s, so $$\Delta T = 0.01$$ s.

Substituting these values, $$\frac{\Delta g}{g} = \frac{0.001}{1.0} + 2 \times \frac{0.01}{1.95} = 0.001 + 2 \times 0.005128 = 0.001 + 0.010256 = 0.011256$$.

Converting to percentage, the percentage error in $$g$$ is $$0.011256 \times 100 \approx 1.13\%$$.

The correct answer is 1.13%.