Physics · Thermal Properties of Matter

JEE Main 2024 — 30 January, Shift 1 — Question 39

An electric toaster has resistance of $60 \Omega$ at room temperature $\left(27^{\circ} \mathrm{C}\right)$ . The toaster is connected to a 220 V supply. If the current flowing through it reaches 2.75 A , the temperature attained by toaster is around : (if $\alpha=2 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ )

Option A:
$694^{\circ} \mathrm{C}$
Option B:
$1235^{\circ} \mathrm{C}$
Option C:
$1694^{\circ} \mathrm{C}$
Correct
Option D:
$1667^{\circ} \mathrm{C}$

Answer: C

Step-by-step solution

$\mathrm{R}_{\mathrm{T}=27}=60 \Omega, R_{T}=\frac{220}{2.75}=80 \Omega$

$\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})$ $80=60\left[1+2 \times 10^{-4}(\mathrm{~T}-27)\right]$

$\mathrm{T} \approx 1694^{\circ} \mathrm{C}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Physics
Chapter: Thermal Properties of Matter
Topic: Thermal Expansion of Solids and its Applications