A clock which keeps correct time at `20^@C` has a pendulum rod made of brass . How many seconds will it gain or lose per day when temperature falls to `0^@C` ? `[alpha=18xx10^(-6)//^@C]`

Options

A. 155.5 s
B. 15.55 s
C. 25.55 s
D. 18.55 s

Answer

To solve the problem of how many seconds a clock will gain or lose per day when the temperature falls from \(20^\circ C\) to \(0^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Coefficient of linear expansion of brass, \(\alpha = 18 \times 10^{-6} \, ^\circ C^{-1}\)
- Initial temperature, \(T_1 = 20^\circ C\)
- Final temperature, \(T_2 = 0^\circ C\)
- Change in temperature, \(\Delta \theta = T_2 - T_1 = 0 - 20 = -20^\circ C\)

2. **Understand the Relation of Time Period and Temperature:**
The time period \(T\) of a pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity. When the temperature changes, the length \(L\) changes due to thermal expansion.

3. **Calculate the Change in Time Period:**
The change in time period \(\Delta T\) due to the change in temperature can be expressed as:
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta \]
Rearranging gives:
\[ \Delta T = T \cdot \frac{1}{2} \alpha \Delta \theta \]

4. **Calculate the Original Time Period at \(20^\circ C\):**
Since we need to find the change in time period, we can assume \(T\) is approximately constant for small changes. For simplicity, we can denote the original time period as \(T\).

5. **Substituting Values:**
Substitute \(\alpha\) and \(\Delta \theta\) into the equation:
\[ \Delta T = T \cdot \frac{1}{2} \cdot (18 \times 10^{-6}) \cdot (-20) \]
\[ \Delta T = T \cdot (-18 \times 10^{-5}) = -0.00018 T \]

6. **Convert Time Period Change to Seconds per Day:**
We need to find out how this change affects the clock over one day. There are \(86400\) seconds in a day (24 hours):
\[ \text{Total change in time over one day} = \Delta T \cdot \frac{86400}{T} \]
Since \(T\) is not specified, we can assume it is a constant value, and we will find the relative change.

7. **Calculate the Total Gain or Loss:**
The clock will lose time due to the negative sign:
\[ \text{Total change in seconds} = -0.00018 \cdot 86400 \]
\[ \text{Total change in seconds} = -15.552 \]
Rounding this, we find that the clock will lose approximately \(15.55\) seconds per day.

### Final Answer:
The clock will lose approximately **15.55 seconds per day** when the temperature falls to \(0^\circ C\).