In an experiment to determine the gravitational acceleration g of a...

$T = 2\pi\sqrt{\dfrac{L}{g}}$

$⇒ g = \dfrac{4{π}^{2}L}{T2}$

$\dfrac{T2}{L} = \dfrac{4{π}^{2}}{g} = tan \theta =$slope of graph

$⇒ g = \dfrac{4{π}^{2}}{tanθ}$

$∵ tan \theta = 8 - 2$$\begin{array}{ll}2 - 0.5\end{array}$$= \dfrac{6}{1.5} = 4$

$⇒ g = \dfrac{4{π}^{2}}{4} = {( 3.14 )}^{2}= {π}^{2} = \left(3 . 14\right)2 = 9 . 87ms^{- 2}$