A second's pendulum clock has a steel wire. The clock is calibrated at `20^@ C`. How much time does the clock lose or gain in one week when the temperature is increased to `30^@ C` ? `alpha_(steel = 1.2 xx 10^-5(^@ C)^-1`.

Explanation

To solve the problem of how much time a second's pendulum clock loses or gains in one week when the temperature increases from \(20^\circ C\) to \(30^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Pendulum Clock**: A second's pendulum clock has a time period of 2 seconds at \(20^\circ C\). 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. **Determine the Change in Temperature**: The change in temperature \(\Delta \theta\) is:
\[ \Delta \theta = 30^\circ C - 20^\circ C = 10^\circ C \]

3. **Calculate the Change in Length**: The change in length of the steel wire due to thermal expansion can be expressed as:
\[ \Delta L = \alpha L \Delta \theta \]
where \(\alpha\) is the coefficient of linear expansion for steel, given as \(1.2 \times 10^{-5} \, ^\circ C^{-1}\).

4. **Calculate the Change in Time Period**: The change in time period \(\Delta T\) due to the change in length can be approximated by:
\[ \Delta T = \frac{1}{2T} \alpha \Delta \theta \]
Substituting \(T = 2 \, \text{seconds}\), we get:
\[ \Delta T = \frac{1}{2 \times 2} \times (1.2 \times 10^{-5}) \times 10 = 1.2 \times 10^{-4} \, \text{seconds} \]

5. **Calculate the New Time Period**: The new time period \(T'\) is:
\[ T' = T + \Delta T = 2 + 1.2 \times 10^{-4} = 2.00012 \, \text{seconds} \]

6. **Calculate the Time Lost in One Week**: The time lost in one week can be calculated using the formula:
\[ \text{Loss in time} = \frac{\Delta T}{T'} \times \text{Total time in one week} \]
The total time in one week in seconds is:
\[ \text{Total time in one week} = 7 \times 24 \times 3600 = 604800 \, \text{seconds} \]
Thus, the loss in time is:
\[ \text{Loss in time} = \frac{1.2 \times 10^{-4}}{2.00012} \times 604800 \]

7. **Perform the Calculation**:
\[ \text{Loss in time} \approx \frac{1.2 \times 10^{-4}}{2} \times 604800 \approx 36.28 \, \text{seconds} \]

### Final Answer:
The clock will lose approximately **36.28 seconds** in one week when the temperature is increased to \(30^\circ C\).