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← JEE Main 2016 · 10 Apr · Shift 1
PhysicsSingle correct+4 / -1Question 14 of 90

JEE Main 2016 · 10 Apr · Shift 1, Physics question 14

The resistance of an electrical toaster has a temperature dependence given by R(T)=R0[1+α(T−T0)]R(T)=R_0[1+\alpha(T-T_0)] in its range of operation. At T0=300T_0=300 K, R = 100 Ω and at T = 500 K, R = 120 Ω. 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 :

Choose one option
Answer and worked solution

Official answer: Bonus: full marks to everyone

From the data, α=10−3\alpha=10^{-3} K⁻¹ and R rises linearly from 100 Ω to 120 Ω over 30 s, R(t)=100+23tR(t)=100+\frac23t. The electrical work is
W=∫0302002100+23tdt=40000×32ln⁡120100=60000ln⁡1.2≈1.09×104 J.W=\int_0^{30}\frac{200^2}{100+\frac23t}dt=40000\times\frac32\ln\frac{120}{100}=60000\ln1.2\approx1.09\times10^4\text{ J}.
None of the options gives this value, so this derived key treats the question as having no correct option (as a bonus).

Topics:Electrical power and heatingOhm's law and resistivity
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