A clock while keeps correct time at `30^(@)C` has a pendulum rod made of brass. The number of seconds it gains (or) looses per second when the temperature falls to `10^(@)C` is [`alpha` of brass `= 18 xx 10^(-6//@)C`]

Options

A. `18 xx 10^(-6) sec`
B. `18 xx 10^(-5) sec`
C. `0.0018 sec`
D. `0.018 sec`

Explanation

To solve the problem of how many seconds a clock gains or loses per second when the temperature falls from \(30^\circ C\) to \(10^\circ C\), we can follow these steps:

### Step 1: Understand the Problem
The clock keeps correct time at \(30^\circ C\) with a pendulum made of brass. When the temperature decreases to \(10^\circ C\), we need to determine how this change in temperature affects the timekeeping of the clock.

### Step 2: Determine the Change in Temperature
Calculate the change in temperature (\(\Delta T\)):
\[ \Delta T = T_{\text{initial}} - T_{\text{final}} = 30^\circ C - 10^\circ C = 20^\circ C \]

### Step 3: Use the Coefficient of Linear Expansion
The coefficient of linear expansion (\(\alpha\)) for brass is given as \(18 \times 10^{-6} \, \text{per} \, ^\circ C\). This coefficient indicates how much the length of the pendulum will change with temperature.

### Step 4: Calculate the Change in Length
The change in length (\(\Delta L\)) of the pendulum can be expressed as:
\[ \Delta L = L_0 \alpha \Delta T \]
However, we do not need the initial length (\(L_0\)) to find the time gained or lost per second.

### Step 5: Relate Time Gain to Temperature Change
The relation between the time gained or lost (\(\frac{\Delta T}{T}\)) and the temperature change is given by:
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta T \]
Here, \(T\) is the temperature at which the clock keeps correct time (in this case, \(30^\circ C\)).

### Step 6: Substitute Values into the Formula
Substituting the values into the formula:
\[ \frac{\Delta T}{T} = \frac{1}{2} \times (18 \times 10^{-6}) \times (20) \]

### Step 7: Calculate the Result
Calculating the above expression:
\[ \frac{\Delta T}{T} = \frac{1}{2} \times 18 \times 10^{-6} \times 20 = 180 \times 10^{-6} = 18 \times 10^{-5} \]

### Step 8: Interpret the Result
This result indicates that the clock gains \(18 \times 10^{-5}\) seconds for every second when the temperature drops from \(30^\circ C\) to \(10^\circ C\).

### Final Answer
The clock gains \(18 \times 10^{-5}\) seconds per second.

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