A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion ($$\alpha$$) of the metal of the pendulum shaft are respectively:

For a simple pendulum the time-period is given by the well-known relation

$$T \;=\;2\pi\sqrt{\dfrac{l}{g}}.$$

Here $$l$$ is the length of the pendulum. The clock counts one second every time the bob completes one oscillation, so the reading of the clock depends directly on this period.

The metal rod of the pendulum expands or contracts with temperature, so its length changes according to the linear expansion formula

$$l \;=\;l_0\bigl[1+\alpha\,(t-\theta)\bigr],$$

where $$l_0$$ is the length at some reference temperature $$\theta$$, $$t$$ is the new temperature and $$\alpha$$ is the coefficient of linear expansion.

Because $$T\propto\sqrt{l}$$, the fractional change in the period is one half of the fractional change in the length:

$$\frac{\Delta T}{T} \;=\;\dfrac12\,\frac{\Delta l}{l}\;=\;\dfrac12\,\alpha\,(t-\theta).$$

Let $$\epsilon(t)$$ denote this fractional change at temperature $$t$$, viz.

$$\epsilon(t)\;=\;\dfrac{\Delta T}{T}\;=\;\dfrac12\,\alpha\,(t-\theta). \quad -(1)$$

Next we must connect the fractional change of period to the time gained or lost in one day (86400 s).

If the period becomes $$T(1+\epsilon)$$ instead of $$T,$$ then in one real day (86400 s) the clock will make

$$N \;=\;\dfrac{86400}{T(1+\epsilon)} \;\approx\;\dfrac{86400}{T}\,(1-\epsilon)$$

oscillations and will therefore indicate

$$86400(1-\epsilon)\;\text{s}$$

instead of the correct 86400 s. Hence the daily error equals

$$\text{error} \;=\; -\,86400\,\epsilon.$$

Thus

$$\epsilon \;=\; -\,\frac{\text{error per day}}{86400}. \quad -(2)$$

At 40 °C the clock loses 12 s per day, so “error” = -12 s. Substituting in (2),

$$\epsilon_{40} \;=\; -\frac{-12}{86400} \;=\;\frac{12}{86400} \;=\;\frac1{7200}.$$

At 20 °C the clock gains 4 s per day, so “error” = +4 s. Hence

$$\epsilon_{20} \;=\; -\frac{4}{86400} \;=\; -\,\frac1{21600}.$$

Now we feed these two values into equation (1).

For 40 °C:

$$\dfrac12\,\alpha\,(40-\theta) \;=\;\frac1{7200}.$$

Multiplying both sides by 2,

$$\alpha\,(40-\theta) \;=\;\frac1{3600}. \quad -(3)$$

For 20 °C:

$$\dfrac12\,\alpha\,(20-\theta) \;=\;-\,\frac1{21600}.$$

Again multiplying by 2,

$$\alpha\,(20-\theta) \;=\;-\,\frac1{10800}. \quad -(4)$$

We have two simultaneous equations, (3) and (4), in the two unknowns $$\alpha$$ and $$\theta$$. To eliminate $$\alpha,$$ we divide (3) by (4):

$$\frac{40-\theta}{20-\theta} \;=\; \frac{1/3600}{-1/10800} \;=\; -\,\frac{10800}{3600} \;=\; -3.$$

Cross-multiplying,

$$40-\theta \;=\; -3\,(20-\theta) \;=\; -60 + 3\theta.$$

Adding $$\theta$$ to both sides and adding 60 to both sides,

$$40+60 \;=\;3\theta+\theta$$

$$100 \;=\;4\theta$$

$$\theta \;=\;25^{\circ}\mathrm{C}.$$

This is the temperature at which the clock will keep correct time.

To find $$\alpha$$ we substitute $$\theta = 25^{\circ}\mathrm{C}$$ into equation (3):

$$\alpha\,(40-25) \;=\;\frac1{3600}$$

$$\alpha \times 15 \;=\;\frac1{3600}$$

$$\alpha \;=\;\frac1{3600 \times 15} \;=\;\frac1{54000}$$

$$\alpha \;=\;1.85 \times 10^{-5}\;\text{per }^{\circ}\mathrm{C}.$$

Both the required temperature and the coefficient of linear expansion match Option C.

Hence, the correct answer is Option C.