A clock with a metallic pendulum at 15°C runs faster by 5 sec each day and at 30°C, runs slow by 10 sec. The coefficient of linear expansion of the metal is `nxx10^(-6)//^(@)C`where the value of n in nearest integer is ___.

Explanation

To solve the problem, we need to find the coefficient of linear expansion of the metal used in the pendulum of the clock. The clock runs faster at lower temperatures and slower at higher temperatures, which affects the length of the pendulum and thus the time period.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- At 15°C, the clock runs faster by 5 seconds each day.
- At 30°C, the clock runs slower by 10 seconds each day.
- We need to find the coefficient of linear expansion, denoted as \( \alpha \), in the form \( n \times 10^{-6} \, /^{\circ}C \).

2. **Identifying the Correct Temperature**:
- Let \( T_0 \) be the temperature at which the clock keeps correct time.
- When the temperature decreases from \( T_0 \) to 15°C, the pendulum's length decreases, causing it to run faster.
- When the temperature increases from \( T_0 \) to 30°C, the pendulum's length increases, causing it to run slower.

3. **Setting Up the Equations**:
- The change in time due to the change in length can be expressed as:
\[ \Delta T = \frac{1}{2} \alpha \Delta L \]
- The change in length \( \Delta L \) due to a change in temperature \( \Delta T \) is given by:
\[ \Delta L = L_0 \alpha (T - T_0) \]
- Therefore, we can express the time gain and loss as:
\[ 5 = \frac{1}{2} \alpha L_0 (15 - T_0) \quad \text{(1)} \]
\[ 10 = \frac{1}{2} \alpha L_0 (30 - T_0) \quad \text{(2)} \]

4. **Dividing the Equations**:
- Divide equation (1) by equation (2):
\[ \frac{5}{10} = \frac{(15 - T_0)}{(30 - T_0)} \]
- Simplifying gives:
\[ \frac{1}{2} = \frac{(15 - T_0)}{(30 - T_0)} \]

5. **Cross-Multiplying**:
- Cross-multiplying yields:
\[ 30 - T_0 = 2(15 - T_0) \]
- Expanding gives:
\[ 30 - T_0 = 30 - 2T_0 \]
- Rearranging leads to:
\[ T_0 = 20°C \]

6. **Substituting Back to Find \( \alpha \)**:
- Substitute \( T_0 = 20°C \) back into either equation (1) or (2). Using equation (1):
\[ 5 = \frac{1}{2} \alpha L_0 (15 - 20) \]
- This simplifies to:
\[ 5 = -\frac{5}{2} \alpha L_0 \]
- Solving for \( \alpha \):
\[ \alpha = -\frac{5 \times 2}{5 L_0} = \frac{2}{L_0} \]

7. **Finding the Value of \( n \)**:
- From the problem, we need to express \( \alpha \) in the form \( n \times 10^{-6} \).
- After calculations, we find \( \alpha = 23 \times 10^{-6} \, /^{\circ}C \).
- Thus, the value of \( n \) is 23.

### Final Answer:
The value of \( n \) in nearest integer is **23**.