A clock with an iron pendulum keeps correct time at `20^(@)C`. How much time will it lose or gain in a day if the temperature changes to `40^(@)C`. Thermal coefficient of liner expansion `alpha = 0.000012 per^(@)C`.

Options

A. 10.3 second/day
B. 20.6 second/day
C. 5 second/day
D. 20 minute/day

Explanation

To solve the problem of how much time a clock with an iron pendulum will gain or lose in a day when the temperature changes from 20°C to 40°C, we can follow these steps:

### Step 1: Understand the relationship between temperature and time period
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 effect of temperature on the length of the pendulum
The length of the pendulum changes with temperature due to thermal expansion. The change in length \( \Delta L \) can be expressed as:
\[ \Delta L = L \alpha \Delta \theta \]
where \( \alpha \) is the coefficient of linear expansion and \( \Delta \theta \) is the change in temperature.

### Step 3: Calculate the change in temperature
Given:
- Initial temperature \( \theta_1 = 20°C \)
- Final temperature \( \theta_2 = 40°C \)

The change in temperature \( \Delta \theta \) is:
\[ \Delta \theta = \theta_2 - \theta_1 = 40°C - 20°C = 20°C \]

### Step 4: Relate the change in time period to the change in length
Using the concept of error analysis, we can express the relative change in time period \( \frac{\Delta T}{T} \) in terms of the relative change in length \( \frac{\Delta L}{L} \):
\[ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} \]

### Step 5: Substitute the expression for \( \Delta L \)
Substituting for \( \Delta L \):
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta \]

### Step 6: Calculate the total time period in a day
The total time in a day is:
\[ T_{\text{day}} = 24 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 86400 \text{ seconds} \]

### Step 7: Substitute values into the equation
Now, substituting the values:
- \( \alpha = 0.000012 \, \text{per °C} \)
- \( \Delta \theta = 20°C \)

We get:
\[ \frac{\Delta T}{T} = \frac{1}{2} \times 0.000012 \times 20 \]
\[ \frac{\Delta T}{T} = 0.00012 \]

### Step 8: Calculate the change in time \( \Delta T \)
Now, we can calculate \( \Delta T \):
\[ \Delta T = T_{\text{day}} \times \frac{\Delta T}{T} = 86400 \times 0.00012 \]
\[ \Delta T = 10.368 \text{ seconds} \]

### Conclusion
Thus, the clock will gain approximately **10.368 seconds** in a day when the temperature increases from 20°C to 40°C.

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