JEE Main 2016 — Current Electricity Question with Solution
From: JEE Main 2016 (Online) 10th April Morning Slot
Question
The resistance of an electrical toaster has a temperature dependence given by R(T) = R0 [1 + (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 :
Choose an option
Show full solutionCorrect option: C
Correct answer
C60000
Step-by-step explanation
Given,
R = R6 [1 + (T t0)]
120 = 100 [1 + (500 300)]
200 =
= 103 oC1
Temperature of the toaster raised from 300 K to 500 K in 30 s.
Increment in the temperature in time t,
T =
=
=
Total work done in raising the temperature
=
=
=
=
R = R6 [1 + (T t0)]
120 = 100 [1 + (500 300)]
200 =
= 103 oC1
Temperature of the toaster raised from 300 K to 500 K in 30 s.
Increment in the temperature in time t,
T =
=
=
Total work done in raising the temperature
=
=
=
=
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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.