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 a pendulum clock with an iron pendulum behaves when the temperature rises, we will follow these steps:

### Step 1: Understand the relationship between length and temperature
The length of the pendulum changes with temperature due to linear expansion. The formula for linear expansion is given by:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
where:
- \( \Delta L \) is the change in length,
- \( L \) is the original length of the pendulum,
- \( \alpha \) is the coefficient of linear expansion,
- \( \Delta T \) is the change in temperature.

### Step 2: Calculate the change in length
Given:
- \( L = 1 \, \text{m} \)
- \( \alpha = 10^{-5} \, \text{°C}^{-1} \)
- \( \Delta T = 10 \, \text{°C} \)

Substituting the values into the linear expansion formula:
\[ \Delta L = 1 \cdot 10^{-5} \cdot 10 = 10^{-4} \, \text{m} \]

### Step 3: Determine the new length of the pendulum
The new length \( L' \) of the pendulum after the temperature increase is:
\[ L' = L + \Delta L = 1 + 10^{-4} = 1.0001 \, \text{m} \]

### Step 4: Calculate the time period of the pendulum
The time period \( T \) of a pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

The original time period \( T \) is:
\[ T = 2\pi \sqrt{\frac{1}{9.81}} \approx 2\pi \cdot 0.319 \approx 2.007 \, \text{s} \]

The new time period \( T' \) with the new length \( L' \) is:
\[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.0001}{9.81}} \approx 2\pi \cdot 0.319 \cdot \sqrt{1.0001} \]
Using the approximation \( \sqrt{1+x} \approx 1 + \frac{x}{2} \) for small \( x \):
\[ T' \approx 2\pi \cdot 0.319 \cdot \left(1 + \frac{0.0001}{2}\right) \approx 2.007 \cdot (1 + 0.00005) \approx 2.007 + 0.00010035 \approx 2.0071 \, \text{s} \]

### Step 5: Calculate the change in time period
The change in time period \( \Delta T \) is:
\[ \Delta T = T' - T \approx 2.0071 - 2.007 = 0.0001 \, \text{s} \]

### Step 6: Calculate the total change in time over a day
To find the total change in time over one day (86400 seconds):
\[ \Delta T_{\text{day}} = \Delta T \cdot \frac{86400 \, \text{s}}{T} \approx 0.0001 \cdot \frac{86400}{2.007} \approx 4.32 \, \text{s} \]

### Step 7: Conclusion
Since the time period has increased, the clock will lose time. Therefore, the clock loses approximately 4.32 seconds in one day.