A pendulum clock has an iron pendulum 1 m long `(alpha_("iron")=10^(-5) //^(@)C)` . If the temperature rises by `10^(@)C`, the clock

Options

A. will lose 8 seconds per day
B. will lose 4.32 seconds per day
C. will gain 8 seconds per day
D. will gain 4.32 second per day

Explanation

To solve the problem of how much time a pendulum clock will lose due to a temperature rise, we can follow these steps:

### Step 1: Understand the pendulum clock's time period formula
The time period \( T \) of a simple 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.

### Step 2: Identify the change in length due to thermal expansion
When the temperature increases, the length of the pendulum will change due to thermal expansion. The change in length \( \Delta L \) can be calculated using the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
where \( \alpha \) is the coefficient of linear expansion and \( \Delta T \) is the change in temperature.

### Step 3: Calculate the new length of the pendulum
Given:
- \( L = 1 \, \text{m} \)
- \( \alpha = 10^{-5} \, \text{°C}^{-1} \)
- \( \Delta T = 10 \, \text{°C} \)

We can substitute these values into the equation for \( \Delta L \):
\[ \Delta L = 1 \cdot 10^{-5} \cdot 10 = 10^{-4} \, \text{m} \]
Thus, the new length \( L' \) will be:
\[ L' = L + \Delta L = 1 + 10^{-4} = 1.0001 \, \text{m} \]

### Step 4: Calculate the new time period
Now we can calculate the new time period \( T' \) using the new length \( L' \):
\[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.0001}{g}} \]

### Step 5: Compare the new time period with the original time period
The original time period \( T \) was:
\[ T = 2\pi \sqrt{\frac{1}{g}} \]
The change in time period \( \Delta T \) can be expressed as:
\[ \Delta T = T' - T \]

### Step 6: Use the binomial approximation
For small changes, we can use the binomial approximation:
\[ T' \approx T \left(1 + \frac{1}{2} \frac{\Delta L}{L}\right) \]
Substituting \( \Delta L = 10^{-4} \, \text{m} \) and \( L = 1 \, \text{m} \):
\[ T' \approx T \left(1 + \frac{1}{2} \cdot 10^{-4}\right) \]

### Step 7: Calculate the change in time period
The change in time period can be approximated as:
\[ \Delta T \approx T \cdot \frac{1}{2} \cdot 10^{-5} \cdot 10 = 5 \times 10^{-5} T \]

### Step 8: Calculate the time lost per day
To find out how much time the clock loses in a day, we need to convert this change per second into a daily loss:
\[ \text{Time lost per day} = \Delta T \cdot \text{seconds in a day} = 5 \times 10^{-5} T \cdot 86400 \]
Calculating this gives:
\[ \text{Time lost per day} = 5 \times 10^{-5} \cdot 86400 = 4.32 \, \text{seconds} \]

### Conclusion
The pendulum clock will lose approximately **4.32 seconds per day** due to the temperature rise.

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