Question

To increase the length of a metal rod by 0.4%, the temperature of the rod is to be increased by (Coefficient of linear expansion of the metal = \(20 \times 10^{-6}\) °C⁻¹)

• A. 373 K
• B. 473 K
• C. 200 K
• D. 100 K

Hint:

Be careful to distinguish between a change in temperature (\(\Delta T\)) and an absolute temperature (\(T\)). The question asks for the increase, which is \(\Delta T\). A temperature change of \(X^\circ C\) is always equal to a temperature change of \(X\) K.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept: Thermal expansion causes a change in the dimensions of a body when its temperature changes. For linear expansion, the fractional change in length is directly proportional to the change in temperature.
Step 2: Key Formula: The formula for linear expansion is: \[ \frac{\Delta L}{L} = \alpha \Delta T \] Where: - \( \frac{\Delta L}{L} \) is the fractional change in length. - \( \alpha \) is the coefficient of linear expansion. - \( \Delta T \) is the rise in temperature.
Step 3: Detailed Calculation: Given: Percentage increase in length = \( 0.4% \). \[ \frac{\Delta L}{L} = \frac{0.4}{100} = 4 \times 10^{-3} \] Coefficient of linear expansion, \( \alpha = 20 \times 10^{-6} \, ^\circ\text{C}^{-1} \). Substitute these values into the formula: \[ 4 \times 10^{-3} = (20 \times 10^{-6}) \times \Delta T \] \[ \Delta T = \frac{4 \times 10^{-3}}{20 \times 10^{-6}} \] \[ \Delta T = \frac{4}{20} \times 10^{3} \] \[ \Delta T = 0.2 \times 1000 \] \[ \Delta T = 200 \, ^\circ\text{C} \] Since the magnitude of temperature difference is the same in Celsius and Kelvin scales (\( \Delta T_{^\circ\text{C}} = \Delta T_{\text{K}} \)): \[ \Delta T = 200 \text{ K} \]
Step 4: Final Answer: The temperature must be increased by 200 K.