The period of oscillation of a simple pendulum is $T=2 \pi \sqrt{\frac{\bar{L}}{g}}$. Measured value of ' $L^{\prime}$ is $1.0 \mathrm{~m}$ from meter scale
having a minimum division of $1 \mathrm{~mm}$ and time of one complete oscillation is $1.95 \mathrm{~s}$ measured from stopwatch of $0.01 \mathrm{~s}$ resolution. The percentage error in the determination of ' $g$ ' will be :
Correct Option: , 3
$(3)$
$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$
$\mathrm{T}^{2}=4 \pi^{2}\left[\frac{\ell}{\mathrm{g}}\right]$
$\mathrm{g}=4 \pi^{2}\left[\frac{\ell}{\mathrm{T}^{2}}\right]$
$\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{T}}{\mathrm{T}}$
$=\left[\frac{1 \mathrm{~mm}}{1 \mathrm{~m}}+\frac{2\left(10 \times 10^{-3}\right)}{1.95}\right] \times 100$
$=1.13 \%$