A pendulum clock loses 12 s a day if be 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 coefficient of linear expansion `alpha` of the pendulum shaft are, respectively

Options

A. `60^(@)C, alpha=1.85 xx 10^(-4)//""^(@)C`
B. `30^(@)C, alpha=1.85 xx 10^(-3)//""^(@)C`
C. `55^(@)C, alpha=1.85 xx 10^(-2)//""^(@)C`
D. `25^(@)C, alpha=1.85 xx 10^(-5)//""^(@)C`

Answer

To solve the problem, we need to determine the temperature at which the pendulum clock shows the correct time, as well as the coefficient of linear expansion (α) of the pendulum shaft.

### Step-by-Step Solution:

1. **Understand the Time Period of a Pendulum**:
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. The time period is directly proportional to the square root of the length of the pendulum.

2. **Relate Time Period Change to Length Change**:
The change in length of the pendulum due to temperature change can be expressed as:
\[ \Delta L = \alpha L \Delta T \]
where \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.

3. **Express Change in Time Period**:
The fractional change in time period can be related to the fractional change in length:
\[ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} \]
Substituting for \( \Delta L \):
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta T \]

4. **Set Up the Equations**:
- When the temperature is \( 40^\circ C \), the clock loses 12 seconds a day:
\[ \frac{-12}{86400} = \frac{1}{2} \alpha (40 - \theta_0) \]
- When the temperature is \( 20^\circ C \), the clock gains 4 seconds a day:
\[ \frac{4}{86400} = \frac{1}{2} \alpha (20 - \theta_0) \]

5. **Simplify the Equations**:
- From the first equation:
\[ -\frac{12}{86400} = \frac{1}{2} \alpha (40 - \theta_0) \]
This simplifies to:
\[ -\frac{1}{7200} = \frac{1}{2} \alpha (40 - \theta_0) \]
Multiplying both sides by -7200 gives:
\[ 1 = 3600 \alpha (40 - \theta_0) \quad \text{(Equation 1)} \]

- From the second equation:
\[ \frac{4}{86400} = \frac{1}{2} \alpha (20 - \theta_0) \]
This simplifies to:
\[ \frac{1}{21600} = \frac{1}{2} \alpha (20 - \theta_0) \]
Multiplying both sides by 21600 gives:
\[ 1 = 10800 \alpha (20 - \theta_0) \quad \text{(Equation 2)} \]

6. **Divide the Equations**:
Dividing Equation 1 by Equation 2:
\[ \frac{3600 \alpha (40 - \theta_0)}{10800 \alpha (20 - \theta_0)} = 1 \]
This simplifies to:
\[ \frac{40 - \theta_0}{20 - \theta_0} = \frac{1}{3} \]

7. **Cross-Multiply and Solve for \( \theta_0 \)**:
Cross-multiplying gives:
\[ 3(40 - \theta_0) = 20 - \theta_0 \]
Expanding and rearranging:
\[ 120 - 3\theta_0 = 20 - \theta_0 \]
\[ 2\theta_0 = 100 \implies \theta_0 = 50^\circ C \]

8. **Substitute \( \theta_0 \) Back to Find \( \alpha \)**:
Substituting \( \theta_0 = 50^\circ C \) back into either Equation 1 or Equation 2 to find \( \alpha \):
Using Equation 1:
\[ 1 = 3600 \alpha (40 - 50) \]
\[ 1 = -3600 \alpha \implies \alpha = -\frac{1}{3600} \]

### Final Answers:
- The temperature at which the clock shows the correct time: \( \theta_0 = 50^\circ C \)
- The coefficient of linear expansion \( \alpha = -\frac{1}{3600} \, \text{per degree Celsius} \)