A pendulum clock, made of a material having coefficient of linear expansion `alpha=9xx10^(-7)//.^(@)C` has a period of 0.500 sec at `20^(@)C`. If the clock is used in a climate where temperature averages `30^(@)C`, what correction is necessary at the end of 30 days to the time given by clock?

Options

A. `11.66s`
B. `3.88s`
C. `0.100s`
D. `2.0s`

Answer

To solve the problem, we need to determine the correction necessary for the pendulum clock after 30 days due to the change in temperature. Here’s a step-by-step solution:

### Step 1: Understand the relationship between the period of a pendulum and temperature
The 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. The length \( L \) changes with temperature due to linear expansion.

### Step 2: Determine the change in length \( \Delta L \)
The change in length \( \Delta L \) due to a change in temperature \( \Delta t \) is given by:
\[ \Delta L = \alpha L \Delta t \]
where \( \alpha \) is the coefficient of linear expansion.

### Step 3: Calculate the change in temperature \( \Delta t \)
The change in temperature is:
\[ \Delta t = 30^\circ C - 20^\circ C = 10^\circ C \]

### Step 4: Substitute values into the equation for \( \Delta L \)
Given \( \alpha = 9 \times 10^{-7} \, ^\circ C^{-1} \) and \( \Delta t = 10^\circ C \), we can express \( \Delta L \) as:
\[ \Delta L = \alpha L \Delta t \]

### Step 5: Relate the change in period \( \Delta T \) to the change in length
The change in period \( \Delta T \) can be approximated by:
\[ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} \]
This implies:
\[ \Delta T = \frac{1}{2} \frac{\Delta L}{L} T \]

### Step 6: Substitute \( \Delta L \) into the equation for \( \Delta T \)
Substituting \( \Delta L = \alpha L \Delta t \) into the equation for \( \Delta T \):
\[ \Delta T = \frac{1}{2} \frac{\alpha L \Delta t}{L} T = \frac{1}{2} \alpha \Delta t T \]

### Step 7: Calculate \( \Delta T \)
Now, we need to calculate \( \Delta T \) using \( T = 0.5 \, \text{sec} \):
\[ \Delta T = \frac{1}{2} \times (9 \times 10^{-7}) \times (10) \times (0.5) \]
\[ \Delta T = \frac{1}{2} \times 9 \times 10^{-7} \times 10 \times 0.5 = 2.25 \times 10^{-7} \, \text{sec} \]

### Step 8: Convert the time correction to 30 days
Now, we need to find the total correction over 30 days. There are \( 30 \times 24 \times 60 \times 60 = 2,592,000 \) seconds in 30 days:
\[ \text{Total correction} = \Delta T \times \text{Total seconds} \]
\[ \text{Total correction} = 2.25 \times 10^{-7} \times 2,592,000 \]
\[ \text{Total correction} \approx 0.583 \, \text{seconds} \]

### Final Answer
The necessary correction at the end of 30 days to the time given by the clock is approximately **0.583 seconds**.

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