Explanation

Time period of a simple pendulum $T = 2\pi\sqrt{\frac{L}{8}}\$ $Or\ \ g = \frac{4\pi^{2}L}{T^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$ Differentiating Eq. (i), we have $\frac{\mathrm{\Delta}g}{g} = \frac{\mathrm{\Delta}L}{L} + \frac{2\mathrm{\Delta}T}{T}\ \ \ \ \ \ \ \ \ldots\ldots\ldots(ii)$ Given, L=100 cm, T=2s, $\mathrm{\Delta}T = \frac{0.1}{100} = 0.001s,\$ $\mathrm{\Delta}L = 1mm = 0.1\ cm$ Substituting the in Eq. (ii), we have $\therefore\left| \frac{\mathrm{\Delta}g}{g} \right|_{\max\ } = \frac{\mathrm{\Delta}L}{L} + \frac{2\mathrm{\Delta}T}{T}$ $= \frac{0.1}{100} + 2 \times \frac{0.001}{2}$ Thus, maximum percentage eror $\left| \frac{\mathrm{\Delta}g}{g} \right|_{\max\ } \times 100 = \left( \frac{0.1}{100} \times 100 \right) + \left( \frac{2 \times 0.001}{2} \times 100 \right)$ $= 0.1\% + 0.1\% = 0.2\%$