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

(C) $-1 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}$

(A) $-2 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}$

(B) $-1.5 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}$

(D) $-2.5 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}$

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Current Electricity

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

2 5^{\circ} C

and resistance of the semiconductor arm is

3 m Ω

. Arm BC is cooled at a constant rate of

2^{\circ} C / s

. If the galvanometer G shows no deflection after 10 s , then

α

is :

(A) $- 2 \times 1 0^{- 2}^{\circ} C^{- 1}$
(B) $- 1.5 \times 1 0^{- 2}^{\circ} C^{- 1}$
(C) $- 1 \times 1 0^{- 2}^{\circ} C^{- 1}$ ✓ Correct
(D) $- 2.5 \times 1 0^{- 2}^{\circ} C^{- 1}$

Answer: (C)

Solution

The bridge is balanced, so

\frac{0.8}{1} = \frac{R}{3}

, giving the semiconductor's resistance at the moment of balance as

R = 2.4 m Ω

. Cooling at

2^{\circ} C/s

for 10 s produces a temperature drop of

2 0^{\circ} C

. Starting from

3 m Ω

2 5^{\circ} C

, the change in resistance is

Δ R = 2.4 - 3 = - 0.6 m Ω

. Using

Δ R = R α Δ T

with the original resistance

R = 3 m Ω

α = \frac{Δ R}{R Δ T} = \frac{- 0.6}{3 \times 20} = - 1 \times 1 0^{- 2}^{\circ} C^{- 1}

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Current Electricity

Solution

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