JEE Main 2024 · 1 February · evening · Single correct · +4 / −1
To measure the temperature coefficient of resistivity α of a semiconductor, an electrical arrangement shown in the figure is prepared. The arm BC is made up of the semiconductor. The experiment is being conducted at 25° C and resistance of the semiconductor arm is 3 m Ω. Arm BC is cooled at a constant rate of 2° C / s. If the galvanometer G shows no deflection after 10 s, then α is :
- (a) -1 × 10⁻² ° C⁻¹
- (b) -2 × 10⁻² ° C⁻¹
- (c) -2.5 × 10⁻² ° C⁻¹
- (d) -1.5 × 10⁻² ° C⁻¹
Answer and solution
Answer (a) -1 × 10⁻² ° C⁻¹
Solution
Physics sir · 4 steps
Balance the bridge to find what BC must become, then compare that change with the 20 degree drop.
- Balance condition. No deflection means the products of opposite arms are equal: R_(AB) × R_(DC) = R_(BC) × R_(AD). So 0.8 × 3 = R_(BC) × 1, which gives R_(BC) = 2.4 mΩ.
- Temperature change. BC cools at 2 °C s⁻¹ for 10 s, so its temperature changes by 20 °C. Over that change its resistance must move from 3 mΩ to 2.4 mΩ, a change of 0.6 mΩ.
|α| = (|Δ R|)/(R₀ |Δ T|) = 0.6/(3 × 20) = 1 × 10⁻² °C⁻¹
- Sign. A semiconductor has a negative temperature coefficient of resistivity, and all four options carry a minus sign. The size of α is what picks the option.
- So. α = -1 × 10⁻² °C⁻¹, which is option (a).
The idea to keep. In a bridge problem, first get the balancing value of the unknown arm. Then Δ R = R₀ α Δ T, with R₀ the starting resistance and Δ T = rate × time.
Answer from the NTA final answer key. Solution by PrepWiser.