1. Time period of a pendulum and temperature dependence

For a pendulum, $T = 2\pi \sqrt{\frac{l}{g}}$

If temperature changes, the length changes as $l = l_0(1+\alpha \theta)$

Hence, for small changes, $\frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta l}{l} = \frac{1}{2}\alpha \Delta \theta$

So the fractional change in period is $\frac{T'-T}{T} = \frac{1}{2}\alpha (\theta-\theta_0)$

A larger period means the clock runs slow (loses time), and a smaller period means it runs fast (gains time).

2. Relate daily gain/loss to fractional change in period

Let the clock keep correct time at temperature $\theta$.

Then at $40^\circ C$, it loses $12\text{ s}$ per day. So, $\frac{\Delta T}{T} = \frac{12}{86400}$

Thus, $\frac{1}{2}\alpha (40-\theta)=\frac{12}{86400} \qquad (1)$

At $20^\circ C$, it gains $4\text{ s}$ per day, so its period is smaller than correct. Thus, $\frac{1}{2}\alpha (20-\theta)= -\frac{4}{86400} \qquad (2)$

3. Solve for correct temperature

From (1): $\frac{1}{2}\alpha (40-\theta)=\frac{12}{86400}$

From (2): $\frac{1}{2}\alpha (20-\theta)=-\frac{4}{86400}$

Divide (1) by (2): $\frac{40-\theta}{20-\theta}=\frac{12}{-4}=-3$

So, $40-\theta = -3(20-\theta)$ $40-\theta = -60 + 3\theta$ $100 = 4\theta$ $\theta = 25^\circ C$

4. Find coefficient of linear expansion

Using equation (1): $\frac{1}{2}\alpha (40-25)=\frac{12}{86400}$ $\frac{1}{2}\alpha (15)=\frac{12}{86400}$ $7.5\alpha = \frac{12}{86400}$ $\alpha = \frac{12}{86400\times 7.5}$

Now, $86400\times 7.5 = 648000$

Hence, $\alpha = \frac{12}{648000} = 1.85\times 10^{-5}/^\circ C$

5. Check options

• A: $30^\circ C$, wrong temperature and wrong $\alpha$
• B: $55^\circ C$, wrong
• C: $25^\circ C$, $\alpha = 1.85\times 10^{-5}/^\circ C$ ✅
• D: wrong

Therefore, the correct option is: $\boxed{\text{C}}$