A simple pendulum made of a bob of mass m and a metallic wire of a negligible mass has a time period of 2 s at $$T = 0°C$$. If the temperature of the wire is increased, and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope $$S$$. If the coefficient of linear expansion of metal is $$\alpha$$, then the value of $$S$$ is

Let the temperature above $$0^{\circ}\mathrm C$$ be denoted by $$\theta$$. At $$\theta = 0^{\circ}\mathrm C$$ the length of the wire is $$L_0$$ and the time period of the simple pendulum is given to be

$$T_0 = 2\ \text{s}$$

For any temperature $$\theta$$ the wire expands. The linear expansion formula is first stated:

$$L = L_0\left(1+\alpha\theta\right)$$

where $$\alpha$$ is the coefficient of linear expansion. The time-period formula of a simple pendulum is next stated:

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

Here $$g$$ is the acceleration due to gravity. Because the bob mass is unchanged, only the length matters, so the temperature dependence of the period comes solely through $$L$$.

We are asked for the slope $$S$$ of the graph of change in period versus temperature, i.e.

$$S = \frac{\Delta T}{\Delta\theta} \; \Longrightarrow \; S = \frac{\mathrm dT}{\mathrm d\theta}$$

The original length $$L_0$$ may be eliminated in favour of the known period $$T_0$$. Differentiate the period with respect to length first. Starting with

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

we write the differential:

$$\mathrm dT = 2\pi\;\dfrac{1}{2}\left(\dfrac{1}{\sqrt{Lg}}\right)\,\mathrm dL = \frac{\pi}{\sqrt{Lg}}\;\mathrm dL$$

But from the expansion formula, for a small change $$\mathrm d\theta$$ the length change is

$$\mathrm dL = \alpha L\;\mathrm d\theta$$

Substituting this $$\mathrm dL$$ in the expression for $$\mathrm dT$$ gives

$$\mathrm dT = \frac{\pi}{\sqrt{Lg}}\;\bigl(\alpha L\,\mathrm d\theta\bigr) = \alpha\frac{\pi L}{\sqrt{Lg}}\;\mathrm d\theta$$

Hence the required derivative (slope) is

$$\frac{\mathrm dT}{\mathrm d\theta} = \alpha\,\frac{\pi L}{\sqrt{Lg}}$$

To remove $$L$$ we relate it to the known period at $$0^{\circ}\mathrm C$$: from $$T_0 = 2\pi\sqrt{L_0/g}$$ we get

$$L_0 = \dfrac{g\,T_0^2}{4\pi^2}$$

Because at small temperatures $$L \approx L_0$$ and $$T \approx T_0$$ (the changes are small), we may put $$L = L_0$$ and $$T = T_0 = 2\ \text{s}$$ in the derivative. We first rewrite $$\dfrac{\pi L_0}{\sqrt{L_0 g}}$$ in terms of $$T_0$$.

From $$T_0 = 2\pi\sqrt{\dfrac{L_0}{g}}$$ we obtain $$\sqrt{L_0 g} = \dfrac{T_0 g}{2\pi}$$

Therefore

$$\frac{\pi L_0}{\sqrt{L_0 g}} = \pi L_0\,\frac{2\pi}{T_0 g} = \frac{2\pi^2 L_0}{T_0 g}$$

Now substitute $$L_0 = \dfrac{gT_0^2}{4\pi^2}$$:

$$\frac{2\pi^2}{T_0 g}\;\times\;\frac{gT_0^2}{4\pi^2} = \frac{2T_0^2}{4T_0} = \frac{T_0}{2}$$

Hence

$$\frac{\mathrm dT}{\mathrm d\theta} = \alpha\;\frac{T_0}{2}$$

Finally, putting $$T_0 = 2\ \text{s}$$, we get

$$\frac{\mathrm dT}{\mathrm d\theta} = \alpha\;\frac{2}{2} = \alpha$$

This derivative is exactly the slope $$S$$ of the straight line obtained in the experiment. Therefore

$$S = \alpha$$

Hence, the correct answer is Option C.