A pendulum clock with a pendulum made of Invar `(a = 0.7 xx10^(-6)//""^@C)` has a period of 0.5 and is accurate at `25^@C` . If the clock is used in a country where the temperature average `35^@C` , What correction is necessary at the end of a month (30 days) to the time given by the clock?

Answer

To solve the problem step by step, we will follow the concepts of thermal expansion and the physics of a pendulum clock.

### Step 1: Understand the Problem
We have a pendulum clock made of Invar with a coefficient of linear expansion \( \alpha = 0.7 \times 10^{-6} \, \text{°C}^{-1} \). The clock is accurate at \( 25 \, \text{°C} \) and has a period of \( 0.5 \, \text{s} \). We need to find the correction necessary when the clock is used in an environment with an average temperature of \( 35 \, \text{°C} \).

### Step 2: Calculate the Change in Temperature
The change in temperature \( \Delta T \) is given by:
\[ \Delta T = 35 \, \text{°C} - 25 \, \text{°C} = 10 \, \text{°C} \]

### Step 3: Relate Change in Length to Change in Temperature
The change in length \( \Delta L \) of the pendulum due to thermal expansion can be expressed as:
\[ \Delta L = \alpha L \Delta T \]
where \( L \) is the original length of the pendulum.

### Step 4: Find the Change in Time Period
The formula for the time period \( T \) of a pendulum is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
The change in time period \( \Delta T \) due to the change in length can be approximated as:
\[ \Delta T = \frac{1}{2} \frac{\Delta L}{L} T \]
Substituting \( \Delta L \):
\[ \Delta T = \frac{1}{2} \frac{\alpha L \Delta T}{L} T = \frac{1}{2} \alpha T \Delta T \]

### Step 5: Substitute Known Values
Now, substituting the known values:
- \( \alpha = 0.7 \times 10^{-6} \, \text{°C}^{-1} \)
- \( T = 0.5 \, \text{s} \)
- \( \Delta T = 10 \, \text{°C} \)

We get:
\[ \Delta T = \frac{1}{2} \times (0.7 \times 10^{-6}) \times (0.5) \times (10) \]
Calculating this:
\[ \Delta T = \frac{1}{2} \times 0.7 \times 10^{-6} \times 0.5 \times 10 = 0.5 \times 0.7 \times 10^{-5} = 3.5 \times 10^{-6} \, \text{s} \]

### Step 6: Calculate the Total Correction Over 30 Days
Now, we need to find the total correction over 30 days. There are \( 30 \times 86400 \) seconds in 30 days:
\[ \text{Total seconds in 30 days} = 30 \times 86400 = 2592000 \, \text{s} \]
The total correction in time over 30 days will be:
\[ \text{Total Correction} = \Delta T \times \text{Total seconds} \]
Substituting in the values:
\[ \text{Total Correction} = (3.5 \times 10^{-6}) \times 2592000 \]

Calculating this gives:
\[ \text{Total Correction} \approx 9.072 \, \text{s} \]

### Final Answer
Thus, the correction necessary at the end of a month (30 days) to the time given by the clock is approximately **9.072 seconds**.

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