A clock is calibrated at a temperature of `20^@C` Assume that the penduum is a thin brass rod of negligible mass with a heavy bob attached to the end `(alpha_("brass")=19xx10^(-6)//K)`

Options

A. On a hot day at `30^@C` the clock gains 8.2 s
B. On a hot day at `30^@C` the clock loses 8.2 s
C. On a cold day at `10^@C` the clock gains 8.2 s
D. On a cold day at `10^@C` the clock loses 8.2 s

Explanation

To solve the problem regarding the clock calibrated at a temperature of \(20^\circ C\) and the effects of temperature changes on the pendulum made of brass, we will follow these steps:

### Step 1: Understand the relationship between temperature and time period
The time period \(T\) of a pendulum is affected by the length of the pendulum, which changes with temperature due to thermal expansion. The change in time period can be expressed as:
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta T \]
where:
- \(\Delta T\) is the change in time period,
- \(T\) is the original time period at the calibrated temperature,
- \(\alpha\) is the coefficient of linear expansion of brass,
- \(\Delta T\) is the change in temperature.

### Step 2: Determine the change in temperature
We need to analyze two scenarios: when the temperature increases to \(30^\circ C\) and when it decreases to \(10^\circ C\).

**For \(30^\circ C\):**
\[ \Delta T = 30 - 20 = 10^\circ C \]

**For \(10^\circ C\):**
\[ \Delta T = 10 - 20 = -10^\circ C \]

### Step 3: Calculate the change in time period for both scenarios
Using the formula for the change in time period, we can calculate the change for both temperatures.

1. **At \(30^\circ C\)**:
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha (10) \]
Given \(\alpha = 19.8 \times 10^{-6} \, K^{-1}\):
\[ \frac{\Delta T}{T} = \frac{1}{2} \times 19.8 \times 10^{-6} \times 10 = 9.9 \times 10^{-6} \]

2. **At \(10^\circ C\)**:
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha (-10) \]
\[ \frac{\Delta T}{T} = \frac{1}{2} \times 19.8 \times 10^{-6} \times (-10) = -9.9 \times 10^{-6} \]

### Step 4: Convert the change in time period to seconds lost or gained in a day
The total time in a day is \(86400\) seconds. We can now calculate the total time lost or gained in a day for both scenarios.

1. **At \(30^\circ C\)**:
\[ \text{Time lost in a day} = 86400 \times 9.9 \times 10^{-6} \approx 0.8544 \, \text{seconds} \]

2. **At \(10^\circ C\)**:
\[ \text{Time gained in a day} = 86400 \times (-9.9 \times 10^{-6}) \approx -0.8544 \, \text{seconds} \]

### Conclusion
- On a hot day at \(30^\circ C\), the clock loses approximately \(0.8544\) seconds.
- On a cold day at \(10^\circ C\), the clock gains approximately \(0.8544\) seconds.