Solution

In a simple pendulum,

Time period is dependent on the length directly.

Time period is directly proportional to the square root of its effective length.
i.e., T $$ \alpha \sqrt{1}$$

The acceleration due to gravity (g) can be calculated from the above mentioned graph:

To find the slope of the straight line, two points P and Q can be taken on the straight line. Value of T2 can be noted at a and b. To note the value at 'I', consider the points c and d.
Slope= $$ \dfrac{AC}{BC} $$

$$= \dfrac{T_{1}^{2} - T_{2}^{2}}{1_{1} - 1_{2}}$$

The slope is observed to be constant at a point which is equal to
$$ \dfrac{4x^{2}}{g}$$,

g = acceleration due to gravity at that place.

Hence 'g' can be determined at a place with the help of these measurements with the help of this relation:

$$g = \frac{4 \pi ^{2}}{Slope of T^{^{2}} vs I graph}$$