Class 12 Physics Chapter 3: Current Electricity case study question with answers
One case-based question of the kind the board asks: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
| Temperature (°C) | Resistance of M (Ω) | Resistance of N (Ω) |
|---|---|---|
| 0 | 10.0 | 500 |
| 50 | 12.0 | 180 |
| 100 | 14.0 | 70 |
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Answer: (a) 4.0 × 10⁻³ °C⁻¹
RT = R₀[1 + α(T − T₀)]: 12.0 = 10.0(1 + 50α), α = 4.0 × 10⁻³ °C⁻¹. The 3.3 × 10⁻³ option refers the change to 12.0 Ω instead of R₀.
Answer and marking scheme
N: its resistance falls as the temperature rises, while a metal's rises.
- N, because its resistance decreases with rising temperature (1)
Answer and marking scheme
In both, the faster thermal motion makes collisions more frequent, so τ falls. In the metal M, n hardly changes, so ρ rises. In the semiconductor N, n rises with temperature, more than compensating the fall in τ, so ρ falls.
- τ decreases in both with rising temperature (1)
- n nearly constant in M (ρ rises); n increases in N and outweighs τ (ρ falls) (1)
Answer and marking scheme
R = 10.0(1 + 4.0 × 10⁻³ × 150) = 16.0 Ω. 13.0 = 10.0(1 + 4.0 × 10⁻³T) gives T = 75 °C.
- 16.0 Ω at 150 °C (1)
- 75 °C for 13.0 Ω (1)
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