A clock with an Iron Pendulum keeps correct time at `20^(@)C` How much will it lose or gain if temperature changes to `40^(@)C`? [Given cubical expansion of iron `=36xx10^(-6)""^(@)C^(-1)`]

Options

A. 10.368 sec gain
B. 10.368 sec loss
C. 5.184 sec gain
D. 5.184 sec loss

Explanation

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

### Step 1: Understand the relationship between temperature and the length of the pendulum
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: Determine the change in temperature
The change in temperature \(\Delta T\) is:
\[ \Delta T = 40^\circ C - 20^\circ C = 20^\circ C \]

### Step 3: Calculate the new length of the pendulum due to thermal expansion
The length of the pendulum will change due to thermal expansion. The new length \(L'\) can be calculated using the formula for linear expansion:
\[ L' = L(1 + \alpha \Delta T) \]
where \(\alpha\) is the coefficient of linear expansion. Given that the cubical expansion of iron is \(36 \times 10^{-6} \, ^\circ C^{-1}\), we can find \(\alpha\) as:
\[ \alpha = \frac{\gamma}{3} = \frac{36 \times 10^{-6}}{3} = 12 \times 10^{-6} \, ^\circ C^{-1} \]

### Step 4: Substitute the values into the length equation
Substituting \(\alpha\) and \(\Delta T\) into the length equation:
\[ L' = L(1 + 12 \times 10^{-6} \times 20) = L(1 + 240 \times 10^{-6}) = L(1 + 0.00024) \]

### Step 5: Calculate the new time period \(T'\)
Now, substituting \(L'\) into the time period formula:
\[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{L(1 + 0.00024)}{g}} \]
This can be approximated using the binomial expansion:
\[ T' \approx 2\pi \sqrt{\frac{L}{g}} \sqrt{1 + 0.00024} \approx T \left(1 + \frac{0.00024}{2}\right) = T(1 + 0.00012) \]

### Step 6: Calculate the gain in time
The gain in time per day can be calculated as:
\[ T' - T = T \cdot 0.00012 \]
Given that \(T\) (the time period for one day) is \(24 \times 3600 \, \text{seconds}\):
\[ T = 86400 \, \text{seconds} \]
Thus, the gain in time is:
\[ T' - T = 86400 \cdot 0.00012 = 10.368 \, \text{seconds} \]

### Conclusion
The clock will gain approximately \(10.368\) seconds per day when the temperature changes from \(20^\circ C\) to \(40^\circ C\).