The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $$N$$%. The value of $$N$$ is:

We need to find the accuracy (percentage error) in the measurement of $$g$$ using a simple pendulum.

Length: $$l = 20$$ cm, $$\Delta l = 2$$ mm = 0.2 cm

Time for 50 oscillations: $$t = 40$$ s, $$\Delta t = 1$$ s

From $$T = 2\pi\sqrt{l/g}$$, we get: $$g = \frac{4\pi^2 l}{T^2}$$

Since $$T = t/50$$: $$g = \frac{4\pi^2 l \times 2500}{t^2} = \frac{10000\pi^2 l}{t^2}$$

Taking logarithmic differentiation:

$$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} = \frac{\Delta l}{l} + 2\frac{\Delta t}{t}$$

(Since $$T = t/50$$, $$\Delta T/T = \Delta t/t$$.)

$$\frac{\Delta g}{g} = \frac{0.2}{20} + 2 \times \frac{1}{40} = 0.01 + 0.05 = 0.06 = 6\%$$

Therefore $$N = 6$$.

The correct answer is Option 3: 6.