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Step-by-Step Detailed Solution
Step 1: Understand the Problem
A pendulum clock loses time or gains time based on the change in temperature. We must determine the temperature at which it shows the correct time, and then find the coefficient of linear expansion ($\alpha$) of the material of the pendulum shaft.
Step 2: Formulate the Effect of Temperature on Pendulum Length
The time period of a pendulum ($T$) depends on its length ($L$):
$$ T = 2\pi \sqrt{\frac{L}{g}}. $$
When the temperature changes, the length of the pendulum changes due to thermal expansion:
$$ L = L_0 \left(1 + \alpha \,\Delta \theta\right), $$
where $\alpha$ is the coefficient of linear expansion and $\Delta \theta$ is the change in temperature.
A small change in length leads to a small change in the time period of the pendulum. Over a full day (86400 seconds), this change accumulates, causing the clock to lose or gain a certain number of seconds.
Step 3: Express Time Lost/Gained in Terms of Temperature Deviation
We are told the clock loses 12 seconds a day at $40^\circ \mathrm{C}$, and gains 4 seconds a day at $20^\circ \mathrm{C}$. Let $\theta$ be the temperature at which the clock keeps correct time. The time lost or gained per day is proportional to the difference from that correct temperature ($\theta$), i.e.:
$$ \text{Time lost/gained per day} \;=\; \frac{1}{2}\,\alpha\,\bigl( \Delta \theta \bigr)\,86400. $$
This factor $\tfrac{1}{2}$ comes from the approximate relation for small fractional changes in length affecting the pendulum period.
Step 4: Write Equations for Losing and Gaining Time
1) At $40^\circ \mathrm{C}$, the clock loses 12 seconds/day:
$$ 12 \;=\; \frac{1}{2}\,\alpha\,\bigl(40 - \theta\bigr)\,86400. \quad (i) $$
2) At $20^\circ \mathrm{C}$, the clock gains 4 seconds/day:
$$ 4 \;=\; \frac{1}{2}\,\alpha\,\bigl(\theta - 20\bigr)\,86400. \quad (ii) $$
Step 5: Find the Temperature $\theta$
Divide equation (i) by equation (ii):
$$
\frac{12}{4} \;=\; \frac{\frac{1}{2}\,\alpha\,\bigl(40 - \theta\bigr)\,86400}{\frac{1}{2}\,\alpha\,\bigl(\theta - 20\bigr)\,86400}
\;\;\Rightarrow\;\;
3 \;=\; \frac{40 - \theta}{\theta - 20}.
$$
Cross-multiply and solve for $\theta$:
$$
3(\theta - 20) \;=\; 40 - \theta,
$$
$$
3\theta - 60 \;=\; 40 - \theta,
$$
$$
4\theta \;=\; 100,
$$
$$
\theta \;=\; 25^\circ \mathrm{C}.
$$
Thus, the pendulum keeps correct time at $25^\circ \mathrm{C}$.
Step 6: Find the Coefficient of Linear Expansion $\alpha$
Substitute $\theta = 25^\circ \mathrm{C}$ back into either (i) or (ii). Using equation (i):
$$
12
= \frac{1}{2}\,\alpha\,\bigl(40 - 25\bigr)\,86400
= \frac{1}{2}\,\alpha \,(15)\,86400.
$$
Solve for $\alpha$:
$$
12 = \frac{1}{2} \cdot 15 \cdot 86400 \cdot \alpha,
$$
$$
12 = \frac{15 \times 86400}{2} \;\alpha,
$$
$$
\alpha = \frac{12 \times 2}{15 \times 86400}.
$$
When you calculate this value, it simplifies to:
$$
\alpha = 1.85 \times 10^{-5}\,/^\circ \mathrm{C}.
$$
Step 7: Final Answer
• The temperature at which the pendulum clock keeps correct time is $25^\circ \mathrm{C}.$
• The coefficient of linear expansion $\alpha$ of the metal is $1.85 \times 10^{-5}/^\circ \mathrm{C}.$
