A uniform metal rod is used as a bar pendulum. If the room temperature rises by `10^(@)C`, and the coefficient of linear expansion of the metal of the rod is `2 xx 10^(-6) per^(@)C`, the period of the pendulum will have percentage increase of

Options

A. `-2xx10^(-3)`
B. `-1xx10^(-3)`
C. `2xx10^(-3)`
D. `1xx10^(-3)`

Explanation

To solve the problem, we need to determine the percentage increase in the period of a bar pendulum when the temperature of the room rises by \(10^\circ C\). The coefficient of linear expansion of the metal rod is given as \(2 \times 10^{-6} \, \text{per} \, ^\circ C\).

### Step-by-Step Solution:

1. **Understand the Formula for the Period of a Pendulum:**
The period \(T\) of a bar pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \(L\) is the length of the rod and \(g\) is the acceleration due to gravity.

2. **Determine the Change in Length Due to Temperature Change:**
When the temperature increases, the length \(L\) of the rod changes due to thermal expansion. The change in length \(\Delta L\) can be expressed as:
\[ \Delta L = L \alpha \Delta \theta \]
where \(\alpha\) is the coefficient of linear expansion and \(\Delta \theta\) is the change in temperature.

3. **Calculate the Relative Change in Length:**
The relative change in length \(\frac{\Delta L}{L}\) is:
\[ \frac{\Delta L}{L} = \alpha \Delta \theta \]
Substituting the given values:
\[ \alpha = 2 \times 10^{-6} \, \text{per} \, ^\circ C, \quad \Delta \theta = 10^\circ C \]
Therefore,
\[ \frac{\Delta L}{L} = (2 \times 10^{-6}) \times (10) = 2 \times 10^{-5} \]

4. **Determine the Change in Period:**
The change in period \(\Delta T\) can be related to the change in length. The change in period relative to the original period is:
\[ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} \]
Substituting the value of \(\frac{\Delta L}{L}\):
\[ \frac{\Delta T}{T} = \frac{1}{2} \times (2 \times 10^{-5}) = 1 \times 10^{-5} \]

5. **Calculate the Percentage Increase in Period:**
To find the percentage increase in the period, we multiply the relative change by 100:
\[ \text{Percentage Increase} = \left(\frac{\Delta T}{T}\right) \times 100 = (1 \times 10^{-5}) \times 100 = 1 \times 10^{-3} \% \]

6. **Conclusion:**
The percentage increase in the period of the pendulum is:
\[ \text{Percentage Increase} = 1 \times 10^{-3} \% \]

### Final Answer:
The period of the pendulum will have a percentage increase of \(1 \times 10^{-3} \%\).

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