Temperature Dependence Of Resistance
A metallic coil has a resistance of 50 (\Omega) at 20 (^\circ)C and a temperature coefficient of resistance of (4 \times 10^{-3}) per (^\circ)C. What is its resistance when heated to 120 (^\circ)C?
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Solution
70 \(\Omega\)
In metals, lattice vibrations intensify with temperature and scatter conduction electrons more frequently, so resistance rises almost linearly over moderate temperature ranges. This behaviour is modelled by (R_T = R_0[1 + \alpha(T - T_0)]), where (\alpha) is the temperature coefficient of resistance and (T_0) the reference temperature. Here (R_0 = 50;\Omega), (\alpha = 4 \times 10^{-3},^\circ\text{C}^{-1}), and the temperature rise is (\Delta T = 120 - 20 = 100,^\circ)C. Therefore (R_T = 50[1 + 4 \times 10^{-3} \times 100] = 50[1 + 0.4] = 50 \times 1.4 = 70;\Omega). The value 60 (\Omega) wrongly uses a 50 (^\circ)C rise instead of 100 (^\circ)C. The value 65 (\Omega) results from halving the correct increment. The value 75 (\Omega) comes from adding an extra ten percent through a miscalculated product. This is the standard NCERT linear-resistance relation valid for metallic conductors. A sanity check confirms the trend: resistance must increase for a metal on heating, and a 40 percent rise over a 100 (^\circ)C span is consistent with the given coefficient. It is also worth noting that this linear model is an approximation valid only over moderate temperature ranges; at very low temperatures the resistance of metals deviates and approaches a small residual value set by impurities and lattice defects rather than thermal vibrations.
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About This Question
- Subject
- physics
- Chapter
- current electricity
- Topic
- temperature dependence of resistance
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
70 \(\Omega\)
In metals, lattice vibrations intensify with temperature and scatter conduction electrons more frequently, so resistance rises almost linearly over moderate temperature ranges. This behaviour is modelled by (R_T = R_0[1 + \alpha(T - T_0)]), where (\alpha) is the temperature coefficient of resistance and (T_0) the reference temperature. Here (R_0 = 50;\Omega), (\alpha = 4 \times 10^{-3},^\circ\text{C}^{-1}), and the temperature rise is (\Delta T = 120 - 20 = 100,^\circ)C. Therefore (R_T = 50[1 + 4 \times 10^{-3} \times 100] = 50[1 + 0.4] = 50 \times 1.4 = 70;\Omega). The value 60 (\Omega) wrongly uses a 50 (^\circ)C rise instead of 100 (^\circ)C. The value 65 (\Omega) results from halving the correct increment. The value 75 (\Omega) comes from adding an extra ten percent through a miscalculated product. This is the standard NCERT linear-resistance relation valid for metallic conductors. A sanity check confirms the trend: resistance must increase for a metal on heating, and a 40 percent rise over a 100 (^\circ)C span is consistent with the given coefficient. It is also worth noting that this linear model is an approximation valid only over moderate temperature ranges; at very low temperatures the resistance of metals deviates and approaches a small residual value set by impurities and lattice defects rather than thermal vibrations.
This medium difficulty physics question is from the chapter current electricity, covering the topic of temperature dependence of resistance. It appeared in the 2025 exam.
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