JEE MAIN - Physics (2024 - 1st February Evening Shift - No. 16)
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 $\mathrm{BC}$ is cooled at a constant rate of $2^{\circ} \mathrm{C} / \mathrm{s}$. If the galvanometer $\mathrm{G}$ shows no deflection after $10 \mathrm{~s}$, then $\alpha$ is :
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$-1 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}$
$-2 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}$
$-2.5 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}$
$-1.5 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}$
Explanation
For no deflection $\frac{0.8}{1}=\frac{R}{3}$
$$ \Rightarrow \mathrm{R}=2.4 \mathrm{~m} \Omega $$
Temperature fall in $10 \mathrm{~s}=20^{\circ} \mathrm{C}$
$$ \begin{aligned} & \Delta \mathrm{R}=\mathrm{R} \alpha \Delta \mathrm{t} \\\\ & \alpha=\frac{\Delta \mathrm{R}}{\mathrm{R} \Delta \mathrm{t}}=\frac{-0.6}{3 \times 20} \\\\ & =-10^{-2} \mathrm{C}^{-1} \end{aligned} $$
$$ \Rightarrow \mathrm{R}=2.4 \mathrm{~m} \Omega $$
Temperature fall in $10 \mathrm{~s}=20^{\circ} \mathrm{C}$
$$ \begin{aligned} & \Delta \mathrm{R}=\mathrm{R} \alpha \Delta \mathrm{t} \\\\ & \alpha=\frac{\Delta \mathrm{R}}{\mathrm{R} \Delta \mathrm{t}}=\frac{-0.6}{3 \times 20} \\\\ & =-10^{-2} \mathrm{C}^{-1} \end{aligned} $$
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