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:
This problem deals with linear thermal expansion, which describes how the length of an object changes with a change in temperature. The change in length is proportional to the original length, the change in temperature, and the coefficient of linear expansion.
Step 2: Key Formula or Approach:
The formula for linear expansion is:
\[ \Delta L = \alpha L_0 \Delta T \] where \(\Delta L\) is the change in length, \(L_0\) is the original length, \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the change in temperature.
The fractional change in length is \(\frac{\Delta L}{L_0} = \alpha \Delta T\).
The percentage change in length is \(\left(\frac{\Delta L}{L_0}\right) \times 100\).
Step 3: Detailed Explanation:
We are given that the percentage increase in length is 0.4%.
\[ \left(\frac{\Delta L}{L_0}\right) \times 100 = 0.4 \] This means the fractional increase in length is:
\[ \frac{\Delta L}{L_0} = \frac{0.4}{100} = 0.004 \] We are also given the coefficient of linear expansion, \(\alpha = 20 \times 10^{-6}\) °C⁻¹.
Now, we can use the formula to find the required change in temperature, \(\Delta T\).
\[ \alpha \Delta T = \frac{\Delta L}{L_0} \] \[ (20 \times 10^{-6}) \Delta T = 0.004 \] \[ \Delta T = \frac{0.004}{20 \times 10^{-6}} = \frac{4 \times 10^{-3}}{20 \times 10^{-6}} = \frac{4}{20} \times 10^{3} = 0.2 \times 1000 = 200 \] The required temperature increase is 200 °C.
A change in temperature has the same magnitude in both Celsius and Kelvin scales. That is, \(\Delta T(K) = \Delta T(^\circ C)\).
Therefore, the temperature of the rod must be increased by 200 K.
Step 4: Final Answer:
The required increase in temperature is 200 K. Option (C) is correct.