In an experiment to find acceleration due to gravity $$g$$ using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy. The accuracy in the determination of $$g$$ is found to be $$x\%$$. The value of $$x$$ is

We are given a simple pendulum experiment where the time period is $$T = 0.5 \text{ s}$$ (measured from 100 oscillations using a watch with 1 s resolution), and the length is $$L = 10 \text{ cm}$$ known to 1 mm accuracy. We need to find the percentage accuracy in the determination of $$g$$.

The formula relating $$g$$ to the pendulum parameters is $$g = 4\pi^2 \frac{L}{T^2}$$. Taking the relative error, we get $$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$$.

Now, the error in length is $$\Delta L = 1 \text{ mm} = 0.1 \text{ cm}$$, so $$\frac{\Delta L}{L} = \frac{0.1}{10} = 0.01$$.

For the time period, the total time for 100 oscillations is $$100 \times 0.5 = 50 \text{ s}$$. Since the watch has a resolution of 1 s, the error in measuring this total time is $$\Delta t_{\text{total}} = 1 \text{ s}$$. The time period is obtained by dividing the total time by 100, so the error in the time period itself is $$\Delta T = \frac{1}{100} = 0.01 \text{ s}$$. Therefore, $$\frac{\Delta T}{T} = \frac{0.01}{0.5} = 0.02$$.

Substituting these into the error formula, we get $$\frac{\Delta g}{g} = 0.01 + 2(0.02) = 0.01 + 0.04 = 0.05$$.

Converting to percentage, the accuracy in the determination of $$g$$ is $$0.05 \times 100 = 5\%$$.

Hence, the value of $$x$$ is $$\textbf{5}$$.