A clock with an iron pendulum keeps correct time at `15^(@)C`. If the room temperature rises to `20^(@)C`, the error in sconds per day will be (coefficient of linear expansion for iron is `0.000012//^(@)C)`

Options

A. `2.5sec`
B. `2.6sec`
C. `2.4sec`
D. `2.2sec`

Explanation

To solve the problem, we need to determine the error in timekeeping of a clock with an iron pendulum when the temperature changes from 15°C to 20°C. The coefficient of linear expansion for iron is given as \( \alpha = 0.000012 \, \text{°C}^{-1} \).

### Step-by-Step Solution:

1. **Identify the change in temperature (\( \Delta T \))**:
\[ \Delta T = T_{final} - T_{initial} = 20°C - 15°C = 5°C \]

2. **Use the formula for time error due to thermal expansion**:
The formula for the error in timekeeping due to temperature change is given by:
\[ \text{Error} = \frac{1}{2} \alpha \Delta T \times T \]
where \( T \) is the total time period in seconds for one day.

3. **Calculate the total time in seconds for one day**:
\[ T = 24 \, \text{hours} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 86400 \, \text{seconds} \]

4. **Substitute the values into the error formula**:
\[ \text{Error} = \frac{1}{2} \times (12 \times 10^{-6} \, \text{°C}^{-1}) \times (5 \, \text{°C}) \times (86400 \, \text{seconds}) \]

5. **Calculate the error**:
\[ \text{Error} = \frac{1}{2} \times 12 \times 10^{-6} \times 5 \times 86400 \]
\[ = \frac{1}{2} \times 12 \times 5 \times 86400 \times 10^{-6} \]
\[ = \frac{1}{2} \times 60 \times 86400 \times 10^{-6} \]
\[ = 30 \times 86400 \times 10^{-6} \]
\[ = 2592 \times 10^{-6} \, \text{seconds} \]
\[ = 0.002592 \, \text{seconds} \approx 2.592 \, \text{seconds} \]

6. **Final answer**:
The error in seconds per day is approximately \( 2.6 \, \text{seconds} \).