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`.

Explanation

To solve the problem step-by-step, we will calculate how much time the clock with an iron pendulum will lose or gain in a day when the temperature changes from \(20^\circ C\) to \(40^\circ C\).

### 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. When the temperature increases, the length of the pendulum will also increase due to thermal expansion.

### Step 2: Calculate the change in length due to thermal expansion
The change in length \(\Delta L\) due to thermal expansion can be calculated using the formula:
\[ \Delta L = L_0 \alpha \Delta T \]
where:
- \(L_0\) is the original length of the pendulum,
- \(\alpha\) is the coefficient of linear expansion (\(0.000012 \, \text{per}^\circ C\)),
- \(\Delta T\) is the change in temperature (\(40^\circ C - 20^\circ C = 20^\circ C\)).

### Step 3: Substitute values to find the new length
The new length \(L'\) of the pendulum at \(40^\circ C\) will be:
\[ L' = L_0 + \Delta L = L_0 + L_0 \alpha \Delta T = L_0(1 + \alpha \Delta T) \]

### Step 4: Calculate the new time period
The new time period \(T'\) at \(40^\circ C\) will be:
\[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{L_0(1 + \alpha \Delta T)}{g}} \]

### Step 5: Find the change in time period
The change in time period \(\Delta T = T' - T\) can be expressed as:
\[ T' - T = 2\pi \sqrt{\frac{L_0}{g}} \left( \sqrt{1 + \alpha \Delta T} - 1 \right) \]
Using the binomial approximation \(\sqrt{1 + x} \approx 1 + \frac{x}{2}\) for small \(x\):
\[ \sqrt{1 + \alpha \Delta T} \approx 1 + \frac{\alpha \Delta T}{2} \]
Thus, the change in time period becomes:
\[ \Delta T \approx 2\pi \sqrt{\frac{L_0}{g}} \left( \frac{\alpha \Delta T}{2} \right) \]

### Step 6: Calculate the total time lost in one day
The total time lost in one day (in seconds) is:
\[ \text{Time lost in one day} = \Delta T \times \text{Number of periods in a day} \]
The number of seconds in a day is \(24 \times 60 \times 60 = 86400\) seconds. Therefore:
\[ \text{Time lost} = 86400 \times \Delta T \]

### Step 7: Substitute values and calculate
Substituting \(\alpha = 0.000012\) and \(\Delta T = 20\):
\[ \Delta T \approx 86400 \times \left( 2\pi \sqrt{\frac{L_0}{g}} \cdot \frac{0.000012 \times 20}{2} \right) \]
This simplifies to:
\[ \Delta T \approx 86400 \times \left( 2\pi \sqrt{\frac{L_0}{g}} \cdot 0.00012 \right) \]
Assuming \(L_0\) and \(g\) are constants, we can calculate the final time lost.

### Final Calculation
After performing the calculations, we find that the clock will lose approximately \(10.36\) seconds in one day.