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JEE Main 20241 FebruaryEvening Shift
Exam
JEE Main
Year
2024
Date
1 February
Shift
Evening Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2024 · 1 February · Evening Shift · Physics · Current Electricity

To measure the temperature coefficient of resistivity α 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 BCBC is made up of the semiconductor. The experiment is being conducted at 25°C25 ° C and resistance of the semiconductor arm is 3mΩ3mΩ. Arm BCBC is cooled at a constant rate of 2°Cs−12 ° C{s}^{- 1}. If the galvanometer GG shows no deflection after 10s,10s, then α\alpha is:

To measure the temperature coefficient of resistivity α of a semiconductor, an electrical arrangement shown in the

Options

  1. A−2×10−2C−1∘- 2 \times {10}^{- 2} C - 1 ∘
  2. B−1.5×10−2C−1∘- 1 . 5 \times {10}^{- 2} C - 1 ∘
  3. C−1×10−2C−1∘- 1 \times {10}^{- 2} C - 1 ∘
  4. D−2.5×10−2C−1∘- 2 . 5 \times {10}^{- 2} C - 1 ∘

Correct answer

C. −1×10−2C−1∘- 1 \times {10}^{- 2} C - 1 ∘

Step-by-step solution

To measure the temperature coefficient of resistivity α of a semiconductor, an electrical arrangement shown in the

For no deflection(balanced condition) 0.81=R3\dfrac{0 . 8}{1} = \dfrac{R}{3}

⇒R=2.4mΩ⇒ R = 2 . 4mΩ

Temperature fall in 10s=20°C10s = 20 ° C

\DeltaR=RαΔt\DeltaR = R\alpha \Delta t

⇒α=\DeltaRRΔt=2.4−33×20⇒ \alpha = \dfrac{\DeltaR}{R\Delta t} = \dfrac{2 . 4 - 3}{3 \times 20}

=−1×10−2°C−1= - 1 \times {10}^{- 2} ° {C}^{- 1}

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