Question

A metal rod of length 1 m at 20°C is made up of a material of coefficient of linear expansion \( 2 \times 10^{-5} \, \text{°C}^{-1} \). The temperature at which its length is increased by 1 mm is:

• A. 45°C
• B. 70°C
• C. 65°C
• D. 60°C
• E. 50°C

Hint:

To calculate the temperature change for length expansion, use the formula \( \Delta L = \alpha L \Delta T \) and solve for \( \Delta T \).

Answer 1

The Correct Option is B

Step 1: The formula for linear expansion is: \[ \Delta L = \alpha L \Delta T \] where:
- \( \Delta L \) is the change in length,
- \( \alpha \) is the coefficient of linear expansion,
- \( L \) is the original length,
- \( \Delta T \) is the change in temperature.
Step 2: Given:
- \( \Delta L = 1 \, \text{mm} = 0.001 \, \text{m} \),
- \( \alpha = 2 \times 10^{-5} \, \text{°C}^{-1} \),
- \( L = 1 \, \text{m} \).
Substitute the values into the formula: \[ 0.001 = 2 \times 10^{-5} \times 1 \times \Delta T \] \[ \Delta T = \frac{0.001}{2 \times 10^{-5}} = 50^\circ \text{C} \] Step 3: Thus, the final temperature is: \[ T = 20 + 50 = 70^\circ \text{C} \]