A pendulum clock is 5 sec. Slow at a temperature `30^(@)C` and `10` sec. fast at a temperature of `15^(@)C`, At what temperature does it give the correct time-

Options

A. (a)`18^@C`
B. (b)`22^@C`
C. (c)`20^@C`
D. (d)`25^@C`

Answer

To solve the problem of determining the temperature at which the pendulum clock gives the correct time, we can follow these steps:

### Step 1: Define the Variables
Let \( \theta \) be the temperature at which the clock gives the correct time. We know that:
- At \( 30^\circ C \), the clock is 5 seconds slow.
- At \( 15^\circ C \), the clock is 10 seconds fast.

### Step 2: Establish the Relationships
The change in time (\( \Delta T \)) can be expressed in terms of the coefficient of linear expansion (\( \alpha \)) and the change in temperature (\( \Delta \theta \)):
\[ \Delta T = \alpha \cdot \Delta \theta \cdot T \]
Where \( T \) is the particular daytime.

### Step 3: Set Up the Equations
1. For the situation when the clock is slow at \( 30^\circ C \):
- The change in temperature is \( (30 - \theta) \).
- The clock is 5 seconds slow, so:
\[ 5 = \alpha \cdot (30 - \theta) \cdot T \]
Rearranging gives us:
\[ \alpha = \frac{5}{(30 - \theta) \cdot T} \quad \text{(Equation 1)} \]

2. For the situation when the clock is fast at \( 15^\circ C \):
- The change in temperature is \( (\theta - 15) \).
- The clock is 10 seconds fast, so:
\[ 10 = \alpha \cdot (\theta - 15) \cdot T \]
Rearranging gives us:
\[ \alpha = \frac{10}{(\theta - 15) \cdot T} \quad \text{(Equation 2)} \]

### Step 4: Equate the Two Expressions for \( \alpha \)
Since both expressions for \( \alpha \) must be equal, we can set them equal to each other:
\[ \frac{5}{(30 - \theta) \cdot T} = \frac{10}{(\theta - 15) \cdot T} \]

### Step 5: Simplify the Equation
Cancel \( T \) from both sides (assuming \( T \neq 0 \)):
\[ \frac{5}{30 - \theta} = \frac{10}{\theta - 15} \]
Cross-multiplying gives:
\[ 5(\theta - 15) = 10(30 - \theta) \]

### Step 6: Expand and Rearrange
Expanding both sides:
\[ 5\theta - 75 = 300 - 10\theta \]
Rearranging gives:
\[ 5\theta + 10\theta = 300 + 75 \]
\[ 15\theta = 375 \]

### Step 7: Solve for \( \theta \)
Dividing both sides by 15:
\[ \theta = \frac{375}{15} = 25^\circ C \]

### Conclusion
The temperature at which the pendulum clock gives the correct time is \( 25^\circ C \).