A clock pendulum made of invar has a period of `0.5sec` at `20^(@)C`. If the clock is used in a climate where the temperature average to `30^(@)C`, how much time does the clock loose in each oscilliation. For innar `alpha = 9 xx 10^(-7) ^(@)C^(-1)`

Options

A. `2.25 xx 10^(-6) sec`
B. `2.5 xx 10^(-7) sec`
C. `5 xx 10^(-7) sec`
D. `1.125 xx 10^(-6) sec`

Explanation

To solve the problem, we need to determine how much time the clock loses in each oscillation when the temperature changes from \(20^\circ C\) to \(30^\circ C\). We will use the formula for the time loss or gain in a pendulum due to temperature change.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial period of the pendulum, \(T_0 = 0.5 \, \text{seconds}\)
- Coefficient of linear expansion for invar, \(\alpha = 9 \times 10^{-7} \, ^\circ C^{-1}\)
- Initial temperature, \(T_i = 20^\circ C\)
- Final temperature, \(T_f = 30^\circ C\)

2. **Calculate the Change in Temperature:**
\[ \Delta T = T_f - T_i = 30^\circ C - 20^\circ C = 10^\circ C \]

3. **Use the Formula for Time Loss:**
The formula for the time loss or gain in a pendulum due to temperature change is given by:
\[ \Delta t = \frac{1}{2} T_0 \cdot \alpha \cdot \Delta T \]

4. **Substitute the Values into the Formula:**
\[ \Delta t = \frac{1}{2} \times 0.5 \, \text{s} \times 9 \times 10^{-7} \, ^\circ C^{-1} \times 10^\circ C \]

5. **Perform the Calculation:**
\[ \Delta t = \frac{1}{2} \times 0.5 \times 9 \times 10^{-7} \times 10 \]
\[ = \frac{1}{2} \times 0.5 \times 9 \times 10^{-6} \]
\[ = 0.25 \times 9 \times 10^{-6} \]
\[ = 2.25 \times 10^{-6} \, \text{seconds} \]

6. **Conclusion:**
The clock loses \(2.25 \times 10^{-6}\) seconds in each oscillation when the temperature increases from \(20^\circ C\) to \(30^\circ C\).

### Final Answer:
The clock loses \(2.25 \times 10^{-6}\) seconds in each oscillation.