At certain temperature the pendulum of clock keeps correct time. The coefficient of lienar expansion for the pendulum material is `alpha=1.85xx10^(-5)@c^(-1)`. How much will the clock gain or lose in 24 hours if the ambient temperature is `10^(@)C` higher?

Answer

To solve the problem of how much the clock will gain or lose in 24 hours due to a temperature increase of 10°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between temperature and the pendulum's time period**:
The time period \( T \) of a 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.

2. **Effect of temperature change on length**:
When the temperature increases, the length of the pendulum changes due to thermal expansion. The change in length \( \Delta L \) can be expressed as:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
where \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.

3. **Calculate the new length**:
The new length \( L' \) of the pendulum at the increased temperature is:
\[ L' = L + \Delta L = L + L \cdot \alpha \cdot \Delta T = L(1 + \alpha \cdot \Delta T) \]

4. **Calculate the new time period**:
The new time period \( T' \) at the new length \( L' \) is:
\[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{L(1 + \alpha \cdot \Delta T)}{g}} \]

5. **Find the change in time period**:
The change in time period \( \Delta T \) can be approximated using the relation:
\[ \frac{T'}{T} \approx 1 + \frac{1}{2} \alpha \Delta T \]
Therefore, the change in time period \( \Delta T \) is:
\[ \Delta T \approx \frac{1}{2} \alpha \Delta T \cdot T \]

6. **Substituting values**:
Given:
- \( \alpha = 1.85 \times 10^{-5} \, \text{°C}^{-1} \)
- \( \Delta T = 10 \, \text{°C} \)
- The time period \( T \) at the initial temperature (we can assume a standard pendulum period, e.g., 2 seconds for calculation).

The change in time period over 24 hours is:
\[ \Delta T = \frac{1}{2} \cdot (1.85 \times 10^{-5}) \cdot 10 \cdot T \]

7. **Calculate the total time gain/loss in 24 hours**:
Convert 24 hours to seconds:
\[ 24 \, \text{hours} = 24 \times 60 \times 60 = 86400 \, \text{seconds} \]
The total gain/loss in time over 24 hours is:
\[ \text{Total Gain/Loss} = \Delta T \cdot \frac{86400}{T} \]

8. **Final Calculation**:
Assuming \( T = 2 \, \text{seconds} \):
\[ \Delta T = \frac{1}{2} \cdot (1.85 \times 10^{-5}) \cdot 10 \cdot 2 = 1.85 \times 10^{-5} \, \text{seconds} \]
Therefore, the total gain/loss in 24 hours:
\[ \text{Total Gain/Loss} = 1.85 \times 10^{-5} \cdot \frac{86400}{2} \approx 7.992 \, \text{seconds} \]

### Conclusion:
The clock will gain approximately **7.992 seconds** in 24 hours due to the increase in temperature.