The time period of a simple pendulum is given by $$T = 2\pi\sqrt{\frac{l}{g}}$$. The measured value of the length of the pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of $$g$$ using this pendulum is $$x$$. The value of $$x$$ to the nearest integer is:-

The time period is given by $$T = 2\pi\sqrt{\frac{l}{g}}$$. Squaring both sides: $$T^2 = 4\pi^2 \frac{l}{g}$$, so $$g = 4\pi^2 \frac{l}{T^2}$$.

The percentage error in $$g$$ is given by $$\frac{\Delta g}{g} \times 100 = \frac{\Delta l}{l} \times 100 + 2 \times \frac{\Delta T}{T} \times 100$$.

The length $$l = 10$$ cm is known to 1 mm accuracy, so $$\Delta l = 1$$ mm $$= 0.1$$ cm. Therefore $$\frac{\Delta l}{l} = \frac{0.1}{10} = 0.01 = 1\%$$.

The time for 200 oscillations is 100 s with a resolution of 1 s. So the period of one oscillation is $$T = \frac{100}{200} = 0.5$$ s, and the error in the total time is $$\Delta t = 1$$ s. The error in the period is $$\Delta T = \frac{1}{200}$$ s. Therefore $$\frac{\Delta T}{T} = \frac{1/200}{100/200} = \frac{1}{100} = 1\%$$.

The percentage error in $$g$$ is $$1\% + 2 \times 1\% = 3\%$$.

So $$x = 3$$, which matches Option B: 3%.