Current Electricity

JEE Main level, single correct, medium

84 of 380

Question 84 of 380

The resistance of an electrical toaster has a temperature dependence given by R(T)=R0[1+α(T−T0)]R\left(T\right)={R}_{0}\left[1+\alpha \left(T-{T}_{0}\right)\right] in its range of operation. At T0=300K,R=100ΩandatT=500K,R=120Ω{T}_{0}=300K,R=100\Omega \text{and}atT=500K,R=120\Omega. The toaster is connected to a voltage source at 200V200V and its temperature is raised at a constant rate from 300to500K300\text{to}500K in 30s30s. 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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Answer: (A) 60000ln⁡(65)J60000\ln \left(\dfrac{6}{5}\right)J

P=V2R=(200)2R0[1+α(T−T0)]P=\dfrac{{V}^{2}}{R}=\dfrac{{\left(200\right)}^{2}}{{R}_{0}[1+\alpha (T-{T}_{0})]}

T →\rightarrow temperature at 't'

T0=300K→{\text{T}}_{0}=300K\rightarrow temperature at t = 0

T−T0=500−30030(t)\text{T}-{\text{T}}_{0}=\dfrac{500-300}{30}\left(\text{t}\right)

T−T0=20030t\text{T}-{\text{T}}_{0}=\dfrac{200}{30}\text{t}

T−T0=20t3\text{T}-{\text{T}}_{0}=\dfrac{20\text{t}}{3}

α=1R0(∆R∆T)=1100(20200)=10−3K\alpha =\dfrac{1}{{R}_{0}}\left(\dfrac{∆R}{∆T}\right)=\dfrac{1}{100}\left(\dfrac{20}{200}\right)=\dfrac{{10}^{-3}}{K}

H=∫Pdt=H=\int Pdt=

∫030(200)2100(1+α20t3)dt=200×200100∫030dt1+20α3t\int _{0}^{30}\dfrac{{\left(200\right)}^{2}}{100\left(1+\alpha \dfrac{20t}{3}\right)}\text{dt}=\dfrac{200\times 200}{100}\int _{0}^{30}\dfrac{\text{dt}}{1+\dfrac{20\alpha }{3}\text{t}}

=400×320αℓn(1+20∝3×301)=\dfrac{400\times 3}{20\alpha }\ell \text{n}(\dfrac{1+\dfrac{20\propto }{3}\times 30}{1})

=60000ℓn(65)=60000\ell \text{n}\left(\dfrac{6}{5}\right)

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