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JEE Main 2016 (10 Apr, Shift 1) · Ohm's Law, Resistivity & Temperature

Ohm's Law, Resistivity & Temperature question from JEE Main 2016 (10 Apr, Shift 1)

Q55JEE Main 2016 · 10 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature

The resistance of an electrical toaster has a temperature dependence given by R(T) = R0 [1 + α\alpha(T − T0)] in its range of operation. At T0 = 300 K, R = 100 Ω\Omega and at T = 500 K, R = 120 Ω\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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Answer: (c)

  1. 120=100[1+α(200)]⇒α=10−3120 = 100[1+\alpha(200)] \Rightarrow \alpha = 10^{-3} K−1^{-1}.
  2. T−T0=20030tT - T_0 = \tfrac{200}{30}t, so R(t)=100(1+t150)R(t) = 100\left(1 + \tfrac{t}{150}\right).
  3. W=∫030V2Rdt=40000100∫030dt1+t/150=400×150ln⁡180150W = \int_0^{30}\dfrac{V^2}{R}dt = \dfrac{40000}{100}\int_0^{30}\dfrac{dt}{1+t/150} = 400\times150\ln\dfrac{180}{150}
  4. W=60000ln⁡65W = 60000\ln\dfrac{6}{5} J. The answer is (c).
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