JEE Main 2016PhysicsCurrent ElectricityHardMCQ

JEE Main 2016 — Current Electricity Question with Solution

JEE Main 2016 (10 Apr Online)

Question

The resistance of an electrical toaster has a temperature dependence given by RT=R01+αT-T0 in its range of operation. At T0=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 :

Note: This question was awarded as the bonus since all options were incorrect in the exam. 

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Show full solutionCorrect option: A
Correct answer
A60000 ln65  J

Step-by-step explanation

P= V 2 R =2002R0[1+α(T-T0)]

T → temperature at 't'

T 0 =300K → temperature at t = 0  

T - T 0 = 5 0 0 - 3 0 0 3 0 t

T - T 0 = 2 0 0 3 0 t

T - T 0 = 2 0 t 3

α=1R0∆R∆T=110020200=10-3K    

H=∫Pdt=

∫ 0 3 0 2 0 0 2 1 0 0 1 + α 2 0t 3 dt = 2 0 0 × 2 0 0 1 0 0 ∫ 0 3 0 dt 1 + 2 0 α 3 t

= 4 0 0 × 3 2 0 α ℓ n ( 1+ 20∝ 3 ×30 1 )

= 6 0 0 0 0  ℓ n  6 5

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About this question

This is a previous-year question from JEE Main 2016, covering the Current Electricity chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.