A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place.

(3) Given, Length of simple pendulum, $l=25.0 \mathrm{~cm}$

Time of 40 oscillation, $T=50 \mathrm{~s}$

Time period of pendulum

$T=2 \pi \sqrt{\frac{\ell}{g}}$

$\Rightarrow T^{2}=\frac{4 \pi^{2} \ell}{g} \Rightarrow g=\frac{4 \pi^{2} \ell}{T^{2}}$

$\Rightarrow$ Fractional error in $g=\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2 \Delta T}{T}$

$\Rightarrow \frac{\Delta g}{g}=\left(\frac{0.1}{25.0}\right)+2\left(\frac{1}{50}\right)=0.044$

$\therefore$ Percentage error in $g=\frac{\Delta g}{g} \times 100=4.4 \%$