A pendulum clock ( fitted with a small heavy bob that is connected with a metal rod ) is 3 seconds fast each day at a temperature of `15^(@)C` and 2 seconds slow at a temperature of `30^(@)C`. Find the temperature ( in `""^(@)C` ) at which will show the correct time.

Answer

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

### Step 1: Understand the problem
The pendulum clock is 3 seconds fast at 15°C and 2 seconds slow at 30°C. We need to find a temperature \( T \) between 15°C and 30°C where the clock shows the correct time.

### Step 2: Set up the equations
We can denote the fractional loss of time as:
- At 15°C: The clock is fast by 3 seconds per day.
- At 30°C: The clock is slow by 2 seconds per day.

The total time in a day is 86400 seconds (24 hours).

We can express the time error in terms of the temperature:
1. For 15°C:
\[ \frac{3 \text{ seconds}}{86400 \text{ seconds}} = \frac{1}{2} \alpha (T - 15) \]
2. For 30°C:
\[ \frac{-2 \text{ seconds}}{86400 \text{ seconds}} = \frac{1}{2} \alpha (30 - T) \]

### Step 3: Simplify the equations
From the first equation:
\[ \frac{3}{86400} = \frac{1}{2} \alpha (T - 15) \]
Multiplying both sides by \( 86400 \):
\[ 3 = \frac{1}{2} \alpha (T - 15) \cdot 86400 \]
This simplifies to:
\[ \alpha (T - 15) = \frac{6}{86400} = \frac{1}{14400} \]
(Equation 1)

From the second equation:
\[ \frac{-2}{86400} = \frac{1}{2} \alpha (30 - T) \]
Multiplying both sides by \( 86400 \):
\[ -2 = \frac{1}{2} \alpha (30 - T) \cdot 86400 \]
This simplifies to:
\[ \alpha (30 - T) = -\frac{4}{86400} = -\frac{1}{21600} \]
(Equation 2)

### Step 4: Solve the equations
Now we have two equations:
1. \( \alpha (T - 15) = \frac{1}{14400} \)
2. \( \alpha (30 - T) = -\frac{1}{21600} \)

Dividing Equation 1 by Equation 2:
\[ \frac{T - 15}{30 - T} = \frac{\frac{1}{14400}}{-\frac{1}{21600}} = -\frac{21600}{14400} = -\frac{3}{2} \]

Cross-multiplying gives:
\[ 2(T - 15) = -3(30 - T) \]
Expanding:
\[ 2T - 30 = -90 + 3T \]
Rearranging:
\[ T = 60 \]
\[ T - 3T = -90 + 30 \]
\[ -T = -60 \implies T = 24 \]

### Step 5: Conclusion
The temperature at which the pendulum clock shows the correct time is \( T = 24°C \).