Formula
Temperature dependence of R
R(T) = R₀ [1 + α (T - T₀)], where α is the temperature coefficient (1/°C). For metals α > 0; for semiconductors α < 0.
-- NCERT Class 12 Physics, Ch. 3, p. 90Temperature Dependence Resistance
8 MCQs9-step worked example
Source: NCERT Current ElectricityPYQ coverage: NEET 2020, 2022, 2023, 2026Official key: NTA-verifiedLast updated: 21 Sep 2026
Temperature Dependence Resistance, explained for NEET
The trap: treating every material like a metal. Metals have a positive temperature coefficient α — resistance rises when they are heated. Semiconductors and insulators behave the opposite way: their resistivity falls as temperature rises. Giving a semiconductor a positive α is a common confusion, and NEET options are built around it.
The relation. NCERT Class 12 Physics Part I, Chapter 3, Section 3.8 (page 90, Eq. 3.26) gives, over a limited range of temperature:
ρ_T = ρ₀[1 + α(T − T₀)]
Here ρ₀ is the resistivity at the reference temperature T₀, and α has the dimension (temperature)⁻¹. For metals α is positive. When the wire's dimensions change negligibly, the same form is used for resistance, R_T = R₀[1 + α(T − T₀)] — this is how NCERT's toaster (Example 3.3) and platinum thermometer (Example 3.4) problems are solved.
Why the sign differs (NCERT page 91, Eq. 3.27): ρ = m/(ne²τ).
- Metals: the free-electron density n is nearly independent of temperature, but the relaxation time τ decreases as temperature rises, so ρ increases.
- Insulators and semiconductors: n increases with temperature, more than compensating for any decrease in τ, so ρ decreases. Silicon and germanium are the standard semiconductor examples.
Alloys. Nichrome, manganin and constantan have resistivity with a very weak dependence on temperature, which is why they are used in standard resistors (NCERT page 90).
Bridge to NEET. Questions come in two forms: (1) sign/concept — "which solids have a negative temperature coefficient?" (insulators and semiconductors); (2) two-temperature calculation — find α from R at two temperatures, or find a temperature from a measured resistance.
Watch-out. α = (R_T − R₀)/[R₀(T − T₀)]. The denominator uses the resistance at the reference temperature, not the heated value, and uses the temperature difference, not the final temperature. Check the power of ten at the end — 3×10⁻² and 3×10⁻³ both appear as options.
Can you answer these Temperature Dependence Resistance MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
For a metallic conductor, which statement about the temperature coefficient of resistivity α is correct?
Show answer and why every option is right or wrong
Answer: A. A is correct. NCERT Class 12 Physics Part I, Chapter 3, page 90 states that α is positive for metals; with ρ_T = ρ₀[1 + α(T − T₀)], a positive α means resistivity increases with temperature.
Why B is wrong: B is wrong because a negative α describes semiconductors and insulators, not metals (trap: swapping the metal and semiconductor behaviour).
Why C is wrong: C is wrong because metals do show temperature dependence; only certain alloys such as nichrome, manganin and constantan show a very weak dependence, and even that is not zero.
Why D is wrong: D is wrong because it contradicts itself: with a positive α the term α(T − T₀) is positive when T > T₀, so resistivity must rise, not fall.
MCQ 2Easy RecallPractice
Which solids have a negative temperature coefficient of resistance?
Show answer and why every option is right or wrong
Answer: B. B is correct. NCERT Class 12 Physics Part I, Chapter 3, page 91 explains that in insulators and semiconductors the carrier density n increases with temperature, so resistivity decreases — a negative coefficient.
Why A is wrong: A is wrong because metals have a positive α; their resistivity rises with temperature since τ decreases while n stays nearly constant.
Why C is wrong: C is wrong because it includes metals, whose coefficient is positive (trap: treating all solids alike).
Why D is wrong: D is wrong because it includes metals; semiconductors have a negative coefficient but metals do not (trap: giving metals and semiconductors the same sign).
MCQ 3Easy RecallPractice
Nichrome, manganin and constantan are used to make standard resistors mainly because:
Show answer and why every option is right or wrong
Answer: A. A is correct. NCERT Class 12 Physics Part I, Chapter 3, page 90 notes that these alloys show a very weak dependence of resistivity on temperature, so their resistance value stays nearly fixed — the property a standard resistor needs.
Why B is wrong: B is wrong because a large α would make the resistance change noticeably with temperature, which is the opposite of what a standard resistor needs.
Why C is wrong: C is wrong because these are metallic alloys, not semiconductors; a large negative α would also make the resistance drift with temperature (trap: attaching semiconductor behaviour to a metal).
Why D is wrong: D is wrong because a rapid rise in carrier density with temperature is the mechanism in semiconductors and insulators, and it would make resistance strongly temperature dependent.
MCQ 4Direct ApplicationPractice
A metal wire has resistance 10.0 Ω at 0 °C. Its temperature coefficient of resistance is 4.0×10⁻³ °C⁻¹. Assuming the linear relation holds and the wire's dimensions change negligibly, what is its resistance at 50.0 °C?
Show answer and why every option is right or wrong
Answer: D. D is correct. R_T = R₀[1 + α(T − T₀)] = 10.0 × [1 + (4.0×10⁻³)(50.0)] = 10.0 × 1.20 = 12.0 Ω (NCERT Class 12 Physics Part I, Chapter 3, page 90).
Why A is wrong: A is wrong because it adds αΔT = 0.20 directly to 10.0 Ω instead of multiplying R₀ by (1 + αΔT); the increase is R₀αΔT = 2.0 Ω.
Why B is wrong: B is wrong because it uses 1 − αΔT = 0.80, as if the metal had a negative coefficient (trap: treating a metal like a semiconductor).
Why C is wrong: C is wrong because it misreads α as 4.0×10⁻² °C⁻¹, giving αΔT = 2.0 and R = 10.0 × 3.0 = 30.0 Ω — a power-of-ten slip.
MCQ 5Direct ApplicationPYQ Pattern
A platinum resistance thermometer reads 4.00 Ω at the ice point (0 °C) and 4.40 Ω at the steam point (100 °C). When placed in a bath it reads 4.60 Ω. Assuming a linear variation, the bath temperature is:
Show answer and why every option is right or wrong
Answer: C. C is correct. t = [(R_t − R₀)/(R₁₀₀ − R₀)] × 100 °C = (0.60/0.40) × 100 °C = 1.50×10² °C. This is the same method as NCERT Class 12 Physics Part I, Chapter 3, Example 3.4 (page 92).
Why A is wrong: A is wrong because it divides the change 0.60 Ω by R₀ = 4.00 Ω instead of by the 0–100 °C change of 0.40 Ω, giving 0.15 × 100 = 15.0 °C.
Why B is wrong: B is wrong because it measures the rise from the steam-point reading, (4.60 − 4.40)/0.40 × 100 = 50.0 °C, and then forgets to add the 100 °C of that reference point.
Why D is wrong: D is wrong because it takes the plain ratio 4.60/4.40 × 100 ≈ 105 °C, ignoring that the relation is linear in the resistance change from R₀, not proportional to R itself.
MCQ 6Direct ApplicationPractice
A nichrome heating element has resistance 75.3 Ω at 27.0 °C. At its steady operating temperature its resistance is 85.8 Ω. Taking α = 1.70×10⁻⁴ °C⁻¹ (averaged over the range) and using room temperature as the reference, the steady temperature of the element is:
Show answer and why every option is right or wrong
Answer: C. C is correct. T₂ − T₁ = (R₂ − R₁)/(R₁α) = 10.5/(75.3 × 1.70×10⁻⁴) ≈ 820 °C, so T₂ = 820 + 27.0 = 847 °C. This matches NCERT Class 12 Physics Part I, Chapter 3, Example 3.3 (page 91).
Why A is wrong: A is wrong because it puts the hot resistance 85.8 Ω in the denominator: 10.5/(85.8 × 1.70×10⁻⁴) ≈ 720 °C, then adds 27.0 °C. The reference resistance belongs in the denominator.
Why B is wrong: B is wrong because 820 °C is only the temperature rise T₂ − T₁; the reference temperature 27.0 °C must be added to get T₂.
Why D is wrong: D is wrong because it subtracts 27.0 °C from the rise (820 − 27.0) instead of adding it.
MCQ 7Concept TrapPractice
Using ρ = m/(ne²τ), which explanation correctly accounts for the fall in resistivity of a semiconductor as its temperature rises?
Show answer and why every option is right or wrong
Answer: B. B is correct. NCERT Class 12 Physics Part I, Chapter 3, page 91: in insulators and semiconductors n increases with temperature, and this increase more than compensates any decrease in τ, so ρ decreases.
Why A is wrong: A is wrong because τ does not increase with temperature; NCERT describes τ as decreasing as temperature rises.
Why C is wrong: C is wrong because constant n with falling τ is the mechanism for metals, and it makes ρ increase, not decrease (trap: applying the metal explanation to a semiconductor).
Why D is wrong: D is wrong because the electron mass m is a constant in this relation; the temperature effect enters through n and τ.
MCQ 8CalculationPractice
A metal wire has resistance 2.50 Ω at 20.0 °C and 3.00 Ω at 70.0 °C. Taking 20.0 °C as the reference temperature and assuming the linear relation holds, what is its resistance at 120.0 °C?
Show answer and why every option is right or wrong
Answer: D. D is correct. Step 1: α = (3.00 − 2.50)/(2.50 × 50.0) = 4.00×10⁻³ °C⁻¹. Step 2: R = 2.50 × [1 + (4.00×10⁻³)(100.0)] = 2.50 × 1.40 = 3.50 Ω (NCERT Class 12 Physics Part I, Chapter 3, page 90).
Why A is wrong: A is wrong because it uses the final temperature 120.0 °C instead of the difference T − T₀ = 100.0 °C: 2.50 × (1 + 0.480) = 3.70 Ω.
Why B is wrong: B is wrong because it uses 1 − αΔT = 0.600, as if the metal had a negative coefficient (trap: treating a metal like a semiconductor).
Why C is wrong: C is wrong because α was found with the hot resistance 3.00 Ω in the denominator (≈ 3.33×10⁻³ °C⁻¹), giving 2.50 × 1.333 ≈ 3.33 Ω; the reference resistance 2.50 Ω belongs in the denominator.
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How do you solve a Temperature Dependence Resistance question? A worked example
- 1
Given
A platinum wire has resistance R₀ = 2 Ω at T₀ = 0 °C and R_T = 6.8 Ω at T = 80 °C. The values are used as stated in the question and treated as exact.
- 2
Required
The temperature coefficient of resistance α of platinum.
- 3
Concept
For a metal, resistance varies linearly with temperature over a limited range, with α positive (NCERT Class 12 Physics Part I, Chapter 3, Section 3.8, page 90). Since resistance rose on heating, α must come out positive.
- 4
Formula
R_T = R₀[1 + α(T − T₀)], rearranged: α = (R_T − R₀)/[R₀(T − T₀)]
- 5
Substitution
α = (6.8 − 2)/[2 × (80 − 0)]
- 6
Calculation
Numerator: 6.8 − 2 = 4.8 Ω. Denominator: 2 × 80 = 160 Ω °C. α = 4.8/160 = 0.03 °C⁻¹. The 0 °C reference and the values as given in the question are treated as exact, so they do not limit the significant figures of the result.
- 7
Final answer
α = 3×10⁻² °C⁻¹
- 8
Common trap
Writing 3×10⁻³ °C⁻¹ — a decimal slip in 4.8/160 — appears as a distractor. Other errors: dividing by the hot resistance 6.8 Ω instead of R₀ = 2 Ω (giving about 8.8×10⁻³ °C⁻¹), or reporting a negative α, which would describe a semiconductor, not platinum. Treat 3×10⁻² °C⁻¹ as the answer this question's data gives, not as a value of platinum to memorise — for comparison, NCERT's nichrome example uses α = 1.70×10⁻⁴ °C⁻¹.
- 9
Similar NEET-style question
"A metal wire has resistance 3.0 Ω at 0 °C and 3.6 Ω at 50.0 °C. Find its temperature coefficient of resistance."
Strategy: α = (3.6 − 3.0)/(3.0 × 50.0) = 0.6/150 = 4×10⁻³ °C⁻¹. Check the power of ten, and confirm the sign is positive for a metal.
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What to remember before solving Temperature Dependence Resistance questions
Where do students lose marks on Temperature Dependence Resistance?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Root cause: concept gap
Correction
Metals: α > 0 (R increases with T). Semiconductors: α < 0 (R decreases with T) because more thermal carriers. Carbon, silicon, germanium have negative α.
More in Current Electricity: 2 exam traps and mistakes · 7 formulas · 5 question patterns from its other lessons.
Temperature Dependence Resistance questions from past NEET papers
5 questions from NEET 2020, 2022, 2023, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
1I_A remains constant while I_B increases
2I_A decreases while I_B increases
3I_A increases while I_B decreases
4both I_A and I_B remain constant
NTA Answer: Option 1(final)
NEET 2023
13 × 10⁻³ °C⁻¹
23 × 10⁻² °C⁻¹
33 × 10⁻¹ °C⁻¹
43 × 10⁻⁴ °C⁻¹
NTA Answer: Option 2(final)
NEET 2022
As the temperature increases, the electrical resistance
1Decreases for conductors but increases for semiconductors
2Increases for both conductors and semiconductors
3Decreases for both conductors and semiconductors
4Increases for conductors but decreases for semiconductors
NTA Answer: Option 4(final)
NEET 2020
The solids which have the negative temperature coefficient of resistance are :
1insulators and semiconductors
2metals
3insulators only
4semiconductors only
NTA Answer: Option 1(final)
NEET 2020
1
2
3
4
NTA Answer: Option 4(final)
How does NEET ask about Temperature Dependence Resistance?
1 recurring pattern from past papers — click to collapse
Temperature coefficient α: R(T) = R0(1 + α(T-T0)). Compare at two temperatures.
InterpretationEasy
Common distractors
treats semiconductor like metal
Uses α > 0 for semiconductors
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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