JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 17)
An electric toaster has resistance of $$60 \Omega$$ at room temperature $$\left(27^{\circ} \mathrm{C}\right)$$. The toaster is connected to a $$220 \mathrm{~V}$$ supply. If the current flowing through it reaches $$2.75 \mathrm{~A}$$, the temperature attained by toaster is around : ( if $$\alpha=2 \times 10^{-4}$$/$$^\circ \mathrm{C}$$)
1235 $$^\circ$$C
1667 $$^\circ$$C
694 $$^\circ$$C
1694 $$^\circ$$C
Explanation
$$\begin{aligned}
& \mathrm{R}_{\mathrm{T}-27}=60 \Omega, R_T=\frac{220}{2.75}=80 \Omega \\
& \mathrm{R}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T}) \\
& 80=60\left[1+2 \times 10^{-4}(\mathrm{~T}-27)\right] \\
& \mathrm{T} \approx 1694^{\circ} \mathrm{C}
\end{aligned}$$
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