1. **State the problem:** We are given the formula for the period of a pendulum: $$T = 2\pi \sqrt{\frac{l}{g}}$$ We want to plot length $L$ against $T^2$, find the slope of the graph, calculate gravitational acceleration $g$ from the slope, and then find the percentage error compared to the standard value of $g = 9.8$ m/s². 2. **Rewrite the formula:** Square both sides: $$T^2 = 4\pi^2 \frac{l}{g}$$ Rearranged as: $$l = \frac{g}{4\pi^2} T^2$$ This shows $L$ is proportional to $T^2$ with slope $m = \frac{g}{4\pi^2}$. 3. **Plot $L$ against $T^2$ using the table data:** | Length $L$ (cm) | $T^2$ (s²) | |-----------------|------------| | 90.00 | 3.80 | | 80.00 | 3.57 | | 70.00 | 3.06 | | 60.00 | 2.56 | | 50.00 | 2.19 | | 40.00 | 1.79 | 4. **Calculate the slope $m$ of the line $L$ vs $T^2$:** Using two points, for example $(T^2_1, L_1) = (3.80, 90.00)$ and $(T^2_2, L_2) = (1.79, 40.00)$: $$m = \frac{L_1 - L_2}{T^2_1 - T^2_2} = \frac{90.00 - 40.00}{3.80 - 1.79} = \frac{50.00}{2.01} \approx 24.88$$ 5. **Calculate $g$ from the slope:** Recall: $$m = \frac{g}{4\pi^2} \Rightarrow g = m \times 4\pi^2$$ Calculate: $$g = 24.88 \times 4 \times \pi^2 = 24.88 \times 4 \times 9.8696 \approx 24.88 \times 39.4784 = 981.9$$ Since length $L$ is in cm, convert to meters by dividing by 100: $$g = \frac{981.9}{100} = 9.819 \text{ m/s}^2$$ 6. **Calculate percentage error:** Given standard value $Sv = 9.8$ m/s² and experimental value $Ev = 9.819$ m/s²: $$\text{Percentage error} = \frac{Sv - Ev}{Sv} \times 100 = \frac{9.8 - 9.819}{9.8} \times 100 = \frac{-0.019}{9.8} \times 100 \approx -0.19\%$$ 7. **Conclusion:** The acceleration due to gravity is approximately $9.82$ m/s².