A clock which keeps correct time at `20^(@)C` is subjected to `40^(@)C.` If coefficient of linear expansion of the pendulum is `12xx10^(-6)//"^(@)C.` How much will it gain or loss in time ?

Options

A. `10.3" s day"^(-1)`
B. `20.6" s day"^(-1)`
C. `5" s day"^(-1)`
D. `20" min day"^(-1)`

Answer

To solve the problem of how much time a clock will gain or lose when subjected to a temperature change, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Temperatures:**
- The initial temperature \( T_1 = 20^\circ C \)
- The final temperature \( T_2 = 40^\circ C \)

2. **Calculate the Change in Temperature:**
\[ \Delta T = T_2 - T_1 = 40^\circ C - 20^\circ C = 20^\circ C \]

3. **Use the Coefficient of Linear Expansion:**
- The coefficient of linear expansion \( \alpha = 12 \times 10^{-6} / ^\circ C \)

4. **Relate the Change in Time to the Change in Temperature:**
The formula relating the change in time \( \Delta t \) to the change in temperature is given by:
\[ \frac{\Delta t}{T} = \frac{\alpha \Delta T}{2} \]
where \( T \) is the original time period of the clock.

5. **Calculate the Original Time Period:**
- The original time period of the clock is 24 hours, which can be converted into seconds:
\[ T = 24 \times 60 \times 60 = 86400 \text{ seconds} \]

6. **Substituting the Values into the Formula:**
\[ \Delta t = \frac{\alpha \Delta T}{2} \times T \]
Substitute \( \alpha = 12 \times 10^{-6} \), \( \Delta T = 20 \), and \( T = 86400 \):
\[ \Delta t = \frac{12 \times 10^{-6} \times 20}{2} \times 86400 \]

7. **Perform the Calculation:**
\[ \Delta t = \frac{12 \times 20 \times 86400}{2 \times 10^6} \]
\[ \Delta t = \frac{240 \times 86400}{2 \times 10^6} = \frac{20736000}{2000000} = 10.368 \text{ seconds} \]

8. **Conclusion:**
The clock will gain approximately **10.36 seconds per day**.