Two pendulum clocks, one having an iron pendulum and the other having a brass pendulu, are keeping correct time at `5^(@)C`. How much per day will they differ at `25^(@)C`?
`alpha_("iron")=12xx10^(-5).^(@)C^(-1), alpha_("brass")=18.7xx10^(-6).^(@)C^(-1)`

Answer

To solve the problem of how much the two pendulum clocks will differ per day at 25°C, we will follow these steps:

### Step 1: Understand the relationship between temperature change and time period
The time period \( T \) of a 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. When the temperature changes, the length \( L \) changes due to thermal expansion, which affects the time period.

### Step 2: Calculate the change in time period for the iron pendulum
The change in time period due to a change in temperature can be expressed as:
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta \]
where:
- \( \Delta T \) is the change in time period,
- \( T \) is the original time period,
- \( \alpha \) is the coefficient of linear expansion,
- \( \Delta \theta \) is the change in temperature.

For the iron pendulum:
- \( \alpha_{\text{iron}} = 12 \times 10^{-5} \, ^\circ C^{-1} \)
- \( \Delta \theta = 25 - 5 = 20 \, ^\circ C \)

Substituting these values:
\[ \frac{\Delta T_{\text{iron}}}{T} = \frac{1}{2} \times (12 \times 10^{-5}) \times 20 \]
Calculating \( \Delta T_{\text{iron}} \):
\[ \Delta T_{\text{iron}} = \frac{1}{2} \times 12 \times 10^{-5} \times 20 \times T \]

### Step 3: Calculate the change in time period for the brass pendulum
For the brass pendulum:
- \( \alpha_{\text{brass}} = 18.7 \times 10^{-6} \, ^\circ C^{-1} \)

Using the same formula:
\[ \frac{\Delta T_{\text{brass}}}{T} = \frac{1}{2} \times (18.7 \times 10^{-6}) \times 20 \]
Calculating \( \Delta T_{\text{brass}} \):
\[ \Delta T_{\text{brass}} = \frac{1}{2} \times 18.7 \times 10^{-6} \times 20 \times T \]

### Step 4: Convert the time period change to seconds per day
To find the total change in time over one day (86400 seconds):
\[ \Delta T_{\text{iron}} = \left(\frac{1}{2} \times 12 \times 10^{-5} \times 20\right) \times 86400 \]
Calculating:
\[ \Delta T_{\text{iron}} = 10.368 \, \text{seconds} \]

For brass:
\[ \Delta T_{\text{brass}} = \left(\frac{1}{2} \times 18.7 \times 10^{-6} \times 20\right) \times 86400 \]
Calculating:
\[ \Delta T_{\text{brass}} = 16.1568 \, \text{seconds} \]

### Step 5: Calculate the difference in time between the two pendulums
Now, we find the difference in time:
\[ \Delta T_{\text{difference}} = \Delta T_{\text{brass}} - \Delta T_{\text{iron}} \]
Calculating:
\[ \Delta T_{\text{difference}} = 16.1568 - 10.368 = 5.7888 \, \text{seconds} \]

### Final Answer
The two pendulum clocks will differ by approximately **5.79 seconds per day**.

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