Physics · Current Electricity

JEE Main 2024 — 9 April, Shift 2 — Question 55

At room temperature $\left(27^{\circ} \mathrm{C}\right)$ , the resistance of a heating element is $50 \Omega$ . The temperature coefficient of the material is $2.4 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$ . The temperature of the element, when its resistance is $62 \Omega$ , is $\qquad$ ${ }^{\circ} \mathrm{C}$ .

Answer: 1027

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})$

$62=50\left[1+2.4 \times 10^{-4} \Delta \mathrm{~T}\right]$

$\Delta \mathrm{T}=1000^{\circ} \mathrm{C}$

$\Rightarrow \mathrm{T}-27^{\circ}=1000^{\circ} \mathrm{C}$

$\mathrm{T}=1027^{\circ} \mathrm{C}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Current Electricity
Topic: Ohm's Law and Calculation of Resistance

To determine the resistance (R) of a wire, a circuit is designed below, The V-I characteristic curve for this circuit is plotted for the…

Answer: 1027

Step-by-step solution

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