The resistance of an electrical toaster has a temperature dependence given by R ( T) = R0 [ 1 + α ( T - T0) ] in its range of operation. At T0 = 300,K,R = 100, and at T = 500,K,,R = 120,. The toaster is connected to a voltage source at 200, V and its temperature is raised at a constant rate from 300 to 500, K in 30, s. The total work done in raising the temperature is

(A) 400\,\ln \,\frac{1.5}{1.3}\,J

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JEE Mains · Physics · STD 12 - Current electricity

JEE Mains 2016
Hard
MCQ

The resistance of an electrical toaster has a temperature dependence given by \(R\left( T \right) = {R_0}\left[ {1 + \alpha \left( {T - {T_0}} \right)} \right]\) in its range of operation. At \({T_0} = 300\,K,R = 100\,\Omega \) and at \(T = 500\,K,\,R = 120\,\Omega \). The toaster is connected to a voltage source at \(200\, V\) and its temperature is raised at a constant rate from \(300\) to \(500\, K\) in \(30\, s\). The total work done in raising the temperature is

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(D) \(400\,\ln \,\frac{5}{6}\,J\)

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\(R(T)=R_{o}\left(1+\alpha\left(T-T_{o}\right)\right)\)

Applying boundary conditions, \(120=100(1+200 \alpha)\)

\(\alpha=10^{-3}\, K^{-1}\)

It is given that temperature increases at a constant rate from \(300\, \mathrm{K}\) to \(500\, \mathrm{K}\) in \(30 \,\mathrm{s}\). Hence, \(T(t)=300+20 t / 3\)

By Joule's Law, heat dissipated in a resistor is given by:

\(W=\int_{0}^{30} \frac{V^{2}}{R} d t\)

\(=\int_{0}^{30} \frac{V^{2}}{R_{0}\left(1+\alpha\left(T-T_{0}\right)\right)} d t\)

\(=\frac{V^{2}}{R_{o}} \int_{0}^{30} \frac{1}{(1+20 \alpha t / 3)} d t\)

Solving, \(W=400 \ln (6 / 5)\)

Work done on resistor \(=-W=400 \ln (5 / 6) \,J\)

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