A pendulum clock runs fast by `5` seconds per day at `20^(0)c` and goes slow by `10` seconds per day at `35^(0)C`. It shows correct time at a temperature of

Options

A. `27.5^(0)C`
B. `25.^(0)C`
C. `30.^(0)C`
D. `33.^(0)C`

Explanation

To determine the temperature at which the pendulum clock shows the correct time, we can analyze the problem step by step.

### Step 1: Understand the problem
The pendulum clock runs fast by 5 seconds per day at 20°C and slow by 10 seconds per day at 35°C. We need to find the temperature at which the clock runs accurately.

### Step 2: Set up the equations
Let \( t \) be the temperature at which the clock shows the correct time. The clock runs fast at 20°C, meaning it gains time, and runs slow at 35°C, meaning it loses time.

1. At 20°C:
- The clock gains 5 seconds per day.
- The time gained can be expressed as:
\[ \text{Gain} = \frac{1}{2} \alpha (t - 20) \cdot T_0 \]
where \( T_0 \) is the time period in seconds for one day (86400 seconds).

2. At 35°C:
- The clock loses 10 seconds per day.
- The time lost can be expressed as:
\[ \text{Loss} = \frac{1}{2} \alpha (35 - t) \cdot T_0 \]

### Step 3: Relate the gain and loss
Since the clock shows the correct time at temperature \( t \), we can set the gain equal to the loss:
\[ \frac{5}{10} = \frac{t - 20}{35 - t} \]
This simplifies to:
\[ \frac{1}{2} = \frac{t - 20}{35 - t} \]

### Step 4: Cross-multiply and solve for \( t \)
Cross-multiplying gives:
\[ 1 \cdot (35 - t) = 2(t - 20) \]
Expanding both sides:
\[ 35 - t = 2t - 40 \]
Rearranging the equation:
\[ 35 + 40 = 2t + t \]
\[ 75 = 3t \]
Dividing both sides by 3:
\[ t = 25 \]

### Step 5: Conclusion
The temperature at which the pendulum clock shows the correct time is \( 25°C \).