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`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the formula for the period of a bar pendulum
The time period \( T \) of a bar pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgd}} \]
where:
- \( I \) is the moment of inertia,
- \( m \) is the mass of the pendulum,
- \( g \) is the acceleration due to gravity,
- \( d \) is the distance from the pivot to the center of mass.

For a uniform rod of length \( L \) pivoted at one end, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{3} m L^2 \]

### Step 2: Determine the effect of temperature on the length of the rod
When the temperature increases, the length of the rod changes due to thermal expansion. The change in length \( \Delta L \) can be expressed as:
\[ \Delta L = L \alpha \Delta T \]
where:
- \( \alpha \) is the coefficient of linear expansion,
- \( \Delta T \) is the change in temperature.

Given:
- \( \alpha = 2 \times 10^{-6} \, \text{per} \, ^\circ C \)
- \( \Delta T = 10 \, ^\circ C \)

The new length \( L' \) of the rod after the temperature increase is:
\[ L' = L + \Delta L = L + L \alpha \Delta T = L (1 + \alpha \Delta T) \]

### Step 3: Substitute the new length into the formula for the period
The new period \( T' \) can be expressed as:
\[ T' = 2\pi \sqrt{\frac{I}{mgd}} = 2\pi \sqrt{\frac{\frac{1}{3} m (L')^2}{mgd}} = 2\pi \sqrt{\frac{\frac{1}{3} m (L(1 + \alpha \Delta T))^2}{mgd}} \]

### Step 4: Calculate the percentage change in the period
The percentage change in the period \( \Delta T \) can be calculated using the formula:
\[ \frac{\Delta T}{T} = \frac{T' - T}{T} = \frac{1}{2} \frac{\Delta L}{L} \]
Substituting \( \Delta L = L \alpha \Delta T \):
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta T \]

### Step 5: Plug in the values and calculate
Substituting the values:
\[ \frac{\Delta T}{T} = \frac{1}{2} \times (2 \times 10^{-6}) \times 10 = 10^{-5} \]
To convert this to a percentage:
\[ \text{Percentage Change} = \frac{\Delta T}{T} \times 100 = 10^{-5} \times 100 = 10^{-3} \% \]

### Final Answer
The percentage increase in the period of the pendulum is:
\[ \text{Percentage Increase} = 0.001\% \text{ or } 10^{-3} \% \]

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