# Find the value of $g$ (in $m\;s^{-2}$) up to approximate significant figures stating the uncertainty in the value of $g$, given the following information:

The time period of a simple pendulum is given by $T = 2\pi \sqrt {\large\frac{L}{g}}$. The measured value of $L$ is $20.0 \;cm$ using a scale of least count $1\;mm$. The time $t$ for $100$ oscillations is found to be $90\;s$ using a watch of least count $1\;s$.

Given $T = 2\pi \sqrt {\large\frac{L}{g}}$ where $T = \large\frac{t}{n}$ where $t$ is the time taken for $n$, the number of oscillations,
$\large\frac{t}{n}$$= 2\pi \sqrt {\large\frac{L}{g}}$$ \rightarrow$ (Squaring and solving for $g$) $g = \large\frac{4\pi^2Ln^2}{t^2}$
Substituting, we get $g = \large\frac{4 \times \pi^2 \times 0.2 \times 100^2}{90^2}$$= 9.74\;m\;s^{-2} Given g = \large\frac{4\pi^2Ln^2}{t^2}, the relative error in g is: \large\frac{\Delta g}{g}$$ = \large\frac{\Delta L}{L}$$+ \large\frac{2\Delta t}{t} \Rightarrow \large\frac{\Delta g}{g}$$ = \large\frac{0.1}{20} $$+ \large\frac{2\times 1}{90}$$ = 0.005+0.022 = 0.027$
$\Rightarrow \Delta g = 0.027 \times 9.74 = 0.26\;m\;s^{-2}$
Rounding off the first significant digit, we get $\Delta g = 0.3 \;m\;s^{-2}$
Therefore, $g$ must also be rounded off as $9.7 \;m\;s^{-2}$.
Therefore, $g = (9.7 \pm 0.3) \;m\;s^{-2}$