A clock pendulum made of invar has a period of `2` s at `20^(@)C`. If the clock is used in a climate where average temperature is `40^(@)C`, what correction (in seconds ) may be necessary at the end of `10` days to the time given by clock? `(alpha_("invar") = 7 xx 10^(-7) "^(@)C, 1` "day" =`8.64xx10^(4)s`)

Explanation

To solve the problem, we need to find the correction in seconds necessary for a clock pendulum made of invar when the temperature changes from 20°C to 40°C over a period of 10 days. The steps to solve this problem are as follows:

### Step 1: Identify the given data
- Period of the pendulum at 20°C, \( T = 2 \) s
- Average temperature change, \( \Delta \theta = 40°C - 20°C = 20°C \)
- Coefficient of linear expansion for invar, \( \alpha = 7 \times 10^{-7} \, °C^{-1} \)
- Number of seconds in one day, \( 1 \text{ day} = 8.64 \times 10^4 \, s \)
- Total time in seconds for 10 days, \( T_{10 \text{ days}} = 10 \times 8.64 \times 10^4 \, s \)

### Step 2: Calculate the total time in seconds for 10 days
\[ T_{10 \text{ days}} = 10 \times 8.64 \times 10^4 = 864000 \, s \]

### Step 3: Determine the change in time period due to temperature change
The change in time period \( \Delta T \) can be calculated using the formula:
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta \]
Substituting the values:
\[ \Delta T = \frac{T}{2} \cdot \alpha \cdot \Delta \theta \]

### Step 4: Substitute the known values into the equation
\[ \Delta T = \frac{2}{2} \cdot (7 \times 10^{-7}) \cdot (20) \]
\[ \Delta T = 1 \cdot (7 \times 10^{-7}) \cdot (20) \]
\[ \Delta T = 1.4 \times 10^{-6} \, s \]

### Step 5: Calculate the total correction over 10 days
Now, we need to find out how many periods fit into the total time of 10 days:
\[ \text{Number of periods in 10 days} = \frac{T_{10 \text{ days}}}{T} = \frac{864000}{2} = 432000 \]

### Step 6: Calculate the total correction
The total correction over 10 days will be:
\[ \text{Total correction} = \Delta T \times \text{Number of periods} \]
\[ \text{Total correction} = (1.4 \times 10^{-6}) \times 432000 \]
\[ \text{Total correction} = 0.6048 \, s \approx 0.605 \, s \]

### Conclusion
The necessary correction at the end of 10 days is approximately **0.605 seconds**.

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