JEE Main 2016 — Physics Electromagnetism

2016mcqhard

The resistance of an electrical toaster has a temperature dependence given by $R(T)={R}_{0}[1+\alpha (T-{T}_{0})]$ in its range of operation. At ${T}_{0}=300 K, R=100\Omega \text{and} at T=500 K, R=120 \Omega$ . The toaster is connected to a voltage source at $200V$ and its temperature is raised at a constant rate from $300\text{to}500K$ in $30s$ . 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.

Official previous-year question

Held on 10 Apr 2016 · Verified 6 Jul 2026.

Options

A
$60000\mathrm{ln}(\frac{6}{5})J$
B
$200 \mathrm{ln}\frac{2}{3} J$
C
$300 J$
D
$400 \mathrm{ln}(\frac{1.5}{1.3}) J$

Did you get this right?

Solution

$P= \frac{ {V}^{2} }{R} =\frac{{(200)}^{2}}{{R}_{0}[1+\alpha (T-{T}_{0})]}$

T $\rightarrow$ temperature at 't'

${\text{T}}_{0}=300K \rightarrow$ temperature at t = 0

$\text{T} - {\text{T}}_{0} = \frac{ 5 0 0 - 3 0 0 }{ 3 0 } ( \text{t} )$

$\text{T} - {\text{T}}_{0} = \frac{ 2 0 0 }{ 3 0 } \text{t}$

$\text{T} - {\text{T}}_{0}= \frac{ 2 0 \text{t} }{ 3 }$

$\alpha =\frac{1}{{R}_{0}}(\frac{\Delta R}{\Delta T})=\frac{1}{100}(\frac{20}{200})=\frac{{10}^{-3}}{K}$

$H=\int Pdt=$

$\int _{0}^{ 3 0 } \frac{ { ( 2 0 0 ) }^{2} }{ 1 0 0 ( 1 + \alpha \frac{ 2 0t }{ 3 } ) } \text{dt} = \frac{ 2 0 0 \times 2 0 0 }{ 1 0 0 } \int _{0}^{ 3 0 } \frac{ \text{dt} }{ 1 + \frac{ 2 0 \alpha }{ 3 } \text{t} }$

$= \frac{ 4 0 0 \times 3 }{ 2 0 \alpha }\ell \text{n} ( \frac{ 1+\frac{ 20\propto }{3} \times 30}{1} )$

$= 6 0 0 0 0 \ell \text{n } ( \frac{ 6 }{ 5 } )$

Your note

JEE Main Electromagnetism in other years

Electromagnetism 2026 Electromagnetism 2025 Electromagnetism 2024 Electromagnetism 2023 Electromagnetism 2022 Electromagnetism 2021 Electromagnetism 2020 Electromagnetism 2019 Electromagnetism 2018 Electromagnetism 2017 Electromagnetism 2015 Electromagnetism 2014

Keep practicing Electromagnetism

Work through every JEE Main Electromagnetism PYQ, year by year.

Browse all Physics topics

Solution

$P= \frac{ {V}^{2} }{R} =\frac{{(200)}^{2}}{{R}_{0}[1+\alpha (T-{T}_{0})]}$

T $\rightarrow$ temperature at 't'

${\text{T}}_{0}=300K \rightarrow$ temperature at t = 0

$\text{T} - {\text{T}}_{0} = \frac{ 5 0 0 - 3 0 0 }{ 3 0 } ( \text{t} )$

$\text{T} - {\text{T}}_{0} = \frac{ 2 0 0 }{ 3 0 } \text{t}$

$\text{T} - {\text{T}}_{0}= \frac{ 2 0 \text{t} }{ 3 }$

$\alpha =\frac{1}{{R}_{0}}(\frac{\Delta R}{\Delta T})=\frac{1}{100}(\frac{20}{200})=\frac{{10}^{-3}}{K}$

$H=\int Pdt=$

$= \frac{ 4 0 0 \times 3 }{ 2 0 \alpha }\ell \text{n} ( \frac{ 1+\frac{ 20\propto }{3} \times 30}{1} )$

$= 6 0 0 0 0 \ell \text{n } ( \frac{ 6 }{ 5 } )$

JEE Main 2016 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism

JEE Main 2016 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism