The period of oscillation of a simple pendulum is $T = 2\pi \sqrt {{L \over g}}$. 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 :

Given, $T = 2\pi \sqrt {{L \over g}}$ .... (i)<br/><br/>where, time period, T = 1.95 s<br/><br/>Length of string, l = 1 m<br/><br/>Acceleration due to gravity = g<br/><br/>Error in time period, $\Delta$T = 0.01 s = 10^$-$2 s<br/><br/>Error in length, $\Delta$L = 1 mm = 1 $\times$ 10^$-$3 m<br/><br/>Squaring Eq. (i) on both sides, we get<br/><br/>${T^2} = 4{\pi ^2}{L \over g}$<br/><br/>$\Rightarrow g = 4{\pi ^2}{L \over {{T^2}}}$<br/><br/>$$ \Rightarrow {{\Delta g} \over g} = {{\Delta L} \over L} + {{2\Delta T} \over T} = {{{{10}^{ - 3}}} \over 1} + {{2 \times {{10}^{ - 2}}} \over {1.95}}$$<br/><br/>$= {10^{ - 3}} + 1.025 \times {10^{ - 2}}$<br/><br/>$= {10^{ - 3}} + 10.25 \times {10^{ - 3}}$<br/><br/>$= 11.25 \times {10^{ - 3}}$<br/><br/>$\because$ $\Delta g/g \times 100 = 11.25 \times {10^{ - 3}} \times {10^2}$<br/><br/>$= 1.125\% \simeq 1.13\%$