1. Time period of a simple pendulum

For a pendulum of length $L$, the time period is $T = 2\pi \sqrt{\frac{L}{g}}.$

At $0^\circ\text{C}$, the time period is given as $T_0 = 2\ \text{s}.$

2. Effect of temperature on length

If the metallic wire has coefficient of linear expansion $\alpha$, then when temperature increases by $\theta$, its length becomes $L = L_0(1+\alpha \theta).$

3. New time period

So the new time period is $T = 2\pi \sqrt{\frac{L_0(1+\alpha \theta)}{g}} = T_0\sqrt{1+\alpha \theta}.$

Since $\alpha \theta$ is very small, use the approximation $\sqrt{1+x} \approx 1+\frac{x}{2}.$

Thus, $T \approx T_0\left(1+\frac{\alpha \theta}{2}\right).$

4. Change in time period

The change in time period is $\Delta T = T-T_0 = T_0\frac{\alpha \theta}{2}.$

Given $T_0=2\,$s, $\Delta T = 2\cdot \frac{\alpha \theta}{2} = \alpha \theta.$

5. Slope of graph

If $\Delta T$ is plotted against temperature $\theta$, then $\Delta T = \alpha \theta,$ which is a straight line with slope $S=\alpha.$

6. Option check

• A: $\alpha$ ✅
• B: $\dfrac{\alpha}{2}$ ❌
• C: $2\alpha$ ❌
• D: $\dfrac{1}{\alpha}$ ❌

Therefore, the correct answer is A.