Verified Solution • 2025–26 Session moderate

QJEEM24Ph52d7: JEE Main 2024 Physics - Class 12 JEEM24Ph52d7 Solution: To measure the temperature coefficient of resistivity of a…

To measure the temperature coefficient of resistivity α\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∘C25^{\circ} \mathrm{C} and resistance of the semiconductor arm is 3 mΩ3 \mathrm{~m} \Omega . Arm BC\mathrm{BC} is cooled at a constant rate of 2∘C/s2^{\circ} \mathrm{C} / \mathrm{s} . If the galvanometer G\mathrm{G} shows no deflection after 10 s10 \mathrm{~s} , then α\alpha is :

JEE Main 2024 (Online) 1st February Evening Shift Physics - Semiconductor Question 31 English

Multiple Choice Options

  • A −1×10−2∘C−1-1 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}
  • B −2×10−2∘C−1-2 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}
  • C −2.5×10−2∘C−1-2.5 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}
  • D −1.5×10−2∘C−1-1.5 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}

Step-by-Step Verified Solution

For no deflection 0.81=R3\frac{0.8}{1}=\frac{R}{3}

⇒R=2.4 mΩ

Temperature fall in 10 s=20∘C10 \mathrm{~s}=20^{\circ} \mathrm{C}

ΔR=RαΔtα=ΔRRΔt=−0.63×20=−10−2C−1

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring