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In an experiment to determine the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period squared is plotted against the string length of the pendulum in the figure. What is the value of g at the place?
$$T = 2\pi\sqrt{\frac{L}{g}}$$
$$T^2 = \left(\frac{4\pi^2}{g}\right)L$$
This equation represents a straight line passing through the origin in a $$T^2$$ vs $$L$$ graph, where the slope is $$m = \frac{4\pi^2}{g}$$
$$m = \frac{\Delta T^2}{\Delta L} = \frac{8.0 - 4.0}{2.0 - 1.0} = 4\text{ s}^2/\text{m}$$
$$\frac{4\pi^2}{g} = 4$$
$$g = \pi^2$$
$$g \approx 9.87\text{ m s}^{-2}$$
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