A clock with a metallic pendulum gains `6` seconds each day when the temperature is `20^(@)C` and loses `6` second when the temperature is `40^(@)C`. Find the coefficient of linear expansions of the metal.

Options

A. `1.4xx10^(-5) "^(@)C^(-1)`
B. `1.4xx10^(-6) "^(@)C^(-1)`
C. `1.4xx10^(-4) "^(@)C^(-1)`
D. `1.4xx10^(-6) "^(@)C^(-1)`

Explanation

To find the coefficient of linear expansion of the metal in the pendulum, we can follow these steps:

### Step 1: Understand the problem
The clock gains 6 seconds each day at 20°C and loses 6 seconds each day at 40°C. We need to find the coefficient of linear expansion (α) of the metal used in the pendulum.

### Step 2: Relate time period to length
The time period (T) of a pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
Where:
- \( T \) = time period
- \( L \) = length of the pendulum
- \( g \) = acceleration due to gravity

### Step 3: Change in time period due to change in length
The change in time period can be related to the change in length (ΔL) as follows:
\[ \frac{\Delta T}{T} = \frac{\Delta L}{2L} \]
Where:
- \( \Delta T \) = change in time period
- \( \Delta L \) = change in length

### Step 4: Relate change in length to temperature
The change in length due to temperature change can be expressed as:
\[ \Delta L = \alpha L \Delta T \]
Where:
- \( \alpha \) = coefficient of linear expansion
- \( \Delta T \) = change in temperature

### Step 5: Combine the equations
Substituting the expression for ΔL into the equation for ΔT, we have:
\[ \frac{\Delta T}{T} = \frac{\alpha L \Delta T}{2L} \]
This simplifies to:
\[ \frac{\Delta T}{T} = \frac{\alpha \Delta T}{2} \]
Thus, we can express ΔT as:
\[ \Delta T = \frac{2T}{\alpha} \]

### Step 6: Calculate the time gain/loss
Given that the clock gains 6 seconds at 20°C and loses 6 seconds at 40°C, we can set up the equations for both temperatures:
1. At 20°C:
\[ 6 = \frac{1}{2} \alpha (20 - T) \times 86400 \]
2. At 40°C:
\[ -6 = \frac{1}{2} \alpha (40 - T) \times 86400 \]

### Step 7: Solve the equations
From the first equation:
\[ 6 = \frac{1}{2} \alpha (20 - T) \times 86400 \]
Rearranging gives:
\[ \alpha (20 - T) = \frac{12}{86400} \]
\[ \alpha (20 - T) = \frac{1}{7200} \]

From the second equation:
\[ -6 = \frac{1}{2} \alpha (40 - T) \times 86400 \]
Rearranging gives:
\[ \alpha (40 - T) = -\frac{12}{86400} \]
\[ \alpha (40 - T) = -\frac{1}{7200} \]

### Step 8: Set up the ratio
Now we can set up the ratio of the two equations:
\[ \frac{20 - T}{40 - T} = \frac{1}{-1} \]
This leads to:
\[ 20 - T = -(40 - T) \]
Solving gives:
\[ 20 - T = -40 + T \]
\[ 2T = 60 \]
\[ T = 30°C \]

### Step 9: Substitute back to find α
Substituting T = 30°C back into either equation:
Using the first equation:
\[ 6 = \frac{1}{2} \alpha (20 - 30) \times 86400 \]
\[ 6 = -\frac{1}{2} \alpha \times 86400 \]
\[ \alpha = -\frac{12}{86400} \]
\[ \alpha = \frac{12}{86400} = \frac{1}{7200} \]

### Step 10: Final calculation
Converting to the coefficient of linear expansion:
\[ \alpha = \frac{1}{7200} \text{ °C}^{-1} \approx 1.38 \times 10^{-5} \text{ °C}^{-1} \]

### Conclusion
The coefficient of linear expansion of the metal is approximately:
\[ \alpha \approx 1.38 \times 10^{-5} \text{ °C}^{-1} \]