A second's pendulum gives correct time at `25^(@)C` . The pendulum shaft is thin and is made of steel. How many second will it lose per day at g`35^(@)C` ? `(alpha_(steel)=11xx10^(-5)//^(@)C)`

Options

A. `1.75s`
B. `2.5s`
C. `3.5s`
D. `4.75s`

Explanation

To solve the problem of how many seconds a second's pendulum will lose per day when the temperature changes from 25°C to 35°C, we can follow these steps:

### Step 1: Understand the Pendulum's 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.

### Step 2: Determine the Change in Length Due to Temperature
When the temperature changes, the length of the pendulum changes according to the formula for linear thermal expansion:
\[ L_1 = L_0 (1 + \alpha \Delta \theta) \]
where:
- \( L_0 \) is the original length of the pendulum,
- \( \alpha \) is the coefficient of linear expansion (for steel, \( \alpha = 11 \times 10^{-5} \, ^\circ C^{-1} \)),
- \( \Delta \theta = T_{\text{final}} - T_{\text{initial}} = 35^\circ C - 25^\circ C = 10^\circ C \).

### Step 3: Calculate the Change in Length
Substituting the values into the thermal expansion formula:
\[ L_1 = L_0 (1 + 11 \times 10^{-5} \times 10) \]
\[ L_1 = L_0 (1 + 1.1 \times 10^{-4}) = L_0 (1.00011) \]

### Step 4: Calculate the New Time Period
Now, we can calculate the new time period \( T_1 \):
\[ T_1 = 2\pi \sqrt{\frac{L_1}{g}} = 2\pi \sqrt{\frac{L_0 (1.00011)}{g}} = T_0 \sqrt{1.00011} \]
Using the binomial approximation for small \( x \):
\[ \sqrt{1+x} \approx 1 + \frac{x}{2} \]
we can approximate:
\[ T_1 \approx T_0 \left(1 + \frac{1.1 \times 10^{-4}}{2}\right) = T_0 (1 + 5.5 \times 10^{-5}) \]

### Step 5: Calculate the Time Lost
The change in time period can be expressed as:
\[ \Delta T = T_1 - T_0 \approx T_0 \cdot 5.5 \times 10^{-5} \]
Since \( T_0 = 2 \) seconds (for a second's pendulum), we find:
\[ \Delta T \approx 2 \cdot 5.5 \times 10^{-5} = 1.1 \times 10^{-4} \text{ seconds} \]

### Step 6: Calculate the Loss in Time Over One Day
To find the total loss in time over one day (24 hours):
\[ \text{Loss per day} = \Delta T \times \text{Number of oscillations in one day} \]
The number of oscillations in one day is:
\[ \text{Number of oscillations} = \frac{86400 \text{ seconds}}{T_0} = \frac{86400}{2} = 43200 \]
Thus, the total loss in time is:
\[ \text{Total loss} = 1.1 \times 10^{-4} \times 43200 \approx 4.75 \text{ seconds} \]

### Final Answer
The pendulum will lose approximately **4.75 seconds** per day at 35°C.

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