Physics · Thermal Properties of Matter

JEE Main 2024 — 1 February, Shift 2 — Question 32

To measure the temperature coefficient of resistivity $\alpha$ 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 $25^\circ\text{C}$ and resistance of the semiconductor arm is $3\,\text{m}\Omega$ . Arm BC is cooled at a constant rate of $2^\circ\text{C/s}$ . If the galvanometer $G$ shows no deflection after $10\,\text{s}$ , then $\alpha$ is:

Option A:
$\ -2 \times 10^{-2}\,^{\circ}\mathrm{C}^{-1}$
Option B:
$\ -1.5 \times 10^{-2}\,^{\circ}\mathrm{C}^{-1}$
Option C:
$\ -1 \times 10^{-2}\,^{\circ}\mathrm{C}^{-1}$
Correct
Option D:
$\ -2.5 \times 10^{-2}\,^{\circ}\mathrm{C}^{-1}$

Answer: C

Step-by-step solution

$\text{For no deflection, } \frac{0.8}{1} = \frac{R}{3}$

$\Rightarrow R = 2.4\,\text{m}\Omega$

$\text{Temperature fall in } 10\,\text{s} = 20^\circ\text{C}$

$\Delta R = R \alpha \Delta t$

$\alpha = \frac{\Delta R}{R \Delta t}$ $= \frac{-0.6}{3 \times 20}$

$= -10^{-2}\,^\circ\text{C}^{-1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Physics
Chapter: Thermal Properties of Matter
Topic: Thermal Expansion of Solids and its Applications