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. (a)155.5 s
B. (b)15.55 s
C. (c)25.55 s
D. (d)18.55 s

Answer

To solve the problem, we need to determine how many seconds a clock with a pendulum rod made of brass will gain or lose per day when the temperature falls from \(20^\circ C\) to \(0^\circ C\). We will use the coefficient of linear expansion for brass, given as \(\alpha = 18 \times 10^{-6} \, ^\circ C^{-1}\).

### Step-by-Step Solution:

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

2. **Understand the Relationship Between Time 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. The time period changes with temperature due to the change in length of the pendulum.

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

4. **Calculate the Time Period at \(20^\circ C\):**
The time period \(T\) at \(20^\circ C\) is:
\[ T = 24 \text{ hours} = 24 \times 60 \times 60 \text{ seconds} = 86400 \text{ seconds} \]

5. **Substitute the Values into the Change in Time Period Formula:**
Now, substituting the values into the equation for \(\Delta T\):
\[ \Delta T = \frac{1}{2} \times (18 \times 10^{-6}) \times 20 \times 86400 \]

6. **Calculate \(\Delta T\):**
First, calculate the product:
\[ \Delta T = \frac{1}{2} \times 18 \times 10^{-6} \times 20 \times 86400 \]
\[ = 9 \times 10^{-6} \times 20 \times 86400 \]
\[ = 9 \times 20 \times 86400 \times 10^{-6} \]
\[ = 180 \times 86400 \times 10^{-6} \]
\[ = 15552 \times 10^{-6} \text{ seconds} \]
\[ = 0.015552 \text{ seconds} \]

7. **Convert to Seconds per Day:**
Since we want the change per day, we multiply by the number of seconds in a day:
\[ \Delta T = 15.552 \text{ seconds} \]

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