Question

Calculate the temperature (in K) at which the resistance of a metal becomes 20% more than its resistance at 300 K. The value of the temperature coefficient of resistance for this metal is \(2.0 \times  10^{-4}\)/K.

Answer 1

Correct Answer: 1300

The resistance \( R_T \) of a metal at temperature \( T \) can be related to its resistance at a reference temperature \( R_0 \) (300 K in this case) using the formula: \(R_T = R_0(1 + \alpha(T - 300))\). Here, \(\alpha = 2.0 \times 10^{-4}\) is the temperature coefficient of resistance.

1. We are given that the resistance at temperature \( T \), \( R_T \), is 20% more than \( R_0 \). Therefore, we can express \( R_T \) as: \(R_T = 1.2R_0\).

2. Substitute the expression for \( R_T \) into the resistance formula: \(1.2R_0 = R_0(1 + 2.0 \times 10^{-4}(T - 300))\).

3. Simplify and solve for \( T \):

• Cancel \( R_0 \) from both sides: \(1.2 = 1 + 2.0 \times 10^{-4}(T - 300)\).
• Rearrange to find: \(0.2 = 2.0 \times 10^{-4}(T - 300)\).
• Solve for \( T - 300 \):
  \(T - 300 = \frac{0.2}{2.0 \times 10^{-4}} = 1000\).
• Thus,  \(T = 1300 \)

 

Therefore, the temperature at which the resistance becomes 20% more than its resistance at 300 K is 1300 K.