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. `-2 xx 10^(-3)`
B. `-1 xx 10^(-3)`
C. `2 xx 10^(-3)`
D. `1 xx 10^(-3)`

Answer

To solve the problem step-by-step, we will analyze the effect of temperature change on the period of a bar pendulum made from a uniform metal rod.

### Step 1: Understand the formula for the period of a bar 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 pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Determine the change in length due to temperature change
The change in length \( \Delta L \) of the rod due to a temperature change \( \Delta T \) can be calculated using the formula for linear expansion:
\[ \Delta L = L \alpha \Delta T \]
where \( \alpha \) is the coefficient of linear expansion.

### Step 3: Substitute the values
Given:
- \( \Delta T = 10^\circ C \)
- \( \alpha = 2 \times 10^{-6} \, \text{per} \, ^\circ C \)

Substituting these values into the linear expansion formula:
\[ \Delta L = L \cdot (2 \times 10^{-6}) \cdot 10 = 2 \times 10^{-5} L \]

### Step 4: Find the new length of the rod
The new length \( L' \) of the rod after the temperature increase is:
\[ L' = L + \Delta L = L + 2 \times 10^{-5} L = L(1 + 2 \times 10^{-5}) \]

### Step 5: Calculate the new period \( T' \)
Now substituting \( L' \) into the period formula:
\[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{L(1 + 2 \times 10^{-5})}{g}} \]

### Step 6: Relate the new period to the old period
Using the original period \( T = 2\pi \sqrt{\frac{L}{g}} \):
\[ T' = T \sqrt{1 + 2 \times 10^{-5}} \]

### Step 7: Approximate the square root
Using the binomial approximation for small \( x \):
\[ \sqrt{1 + x} \approx 1 + \frac{x}{2} \]
we can write:
\[ \sqrt{1 + 2 \times 10^{-5}} \approx 1 + \frac{2 \times 10^{-5}}{2} = 1 + 10^{-5} \]

### Step 8: Calculate the percentage increase in the period
The change in period \( \Delta T \) is given by:
\[ \Delta T = T' - T = T(10^{-5}) \]
The percentage increase in the period is:
\[ \frac{\Delta T}{T} \times 100 = 10^{-5} \times 100 = 10^{-3} \% \]

### Final Answer
The percentage increase in the period of the pendulum is:
\[ \boxed{0.001\%} \]