The period of oscillation of a simple pendulum is $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$. Measured value of 'L' 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: 1
$T=2 \pi \sqrt{\frac{\ell}{g}}$
$\mathrm{g}=\frac{4 \pi^{2} \ell}{\mathrm{T}^{2}}$
$\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{T}}{\mathrm{T}}$
$\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{1 \times 10^{-3}}{1}+2 \times \frac{0.01}{1.95}$
$\frac{\Delta \mathrm{g}}{\mathrm{g}}=0.0113$ or $1.13 \%$
option (1) is correct
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