Verified Solution • 2025–26 Session moderate

QJEEM16Pha71a: JEE Main 2016 Physics - Class 12 JEEM16Pha71a Solution: The resistance of an electrical toaster has a temperature…

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

Multiple Choice Options

  • A 400 ln⁡ 1.51.3 J\ln \,{{1.5} \over {1.3}}\,J
  • B 200 ln⁡ 23 J\ln \,{{2} \over {3}}\,J
  • C 60000 ln⁡ 65 J\ln \,{{6} \over {5}}\,J
  • D 300 J

Step-by-Step Verified Solution

Given,

R = R6 [1 + α\alpha (T −- t0)]

120 = 100 [1 + α\alpha (500 −- 300)]

⇒\Rightarrow    200 α\alpha = 15{1 \over 5}

⇒\Rightarrow     α\alpha = 10 −- 3   oC −- 1

Temperature of the toaster raised from 300 K to 500 K in 30 s.

∴\therefore    Increment in the temperature in time t,

Δ\Delta T = 500−30030t{{500 - 300} \over {30}}t

= 2003t{{200} \over 3}t

= 203t{{20} \over 3}t

Total work done in raising the temperature

= ∫0tV2R(t) dt\int\limits_0^t {{{{V^2}} \over {R\left( t \right)}}\,dt}

= ∫0tV2R0(1+αΔt) dt\int\limits_0^t {{{{V^2}} \over {{R_0}\left( {1 + \alpha \Delta t} \right)}}} \,dt

= ∫030(200)2100(1+10−3×203t) dt\int\limits_0^{30} {{{{{\left( {200} \right)}^2}} \over {100\left( {1 + {{10}^{ - 3}} \times {{20} \over 3}t} \right)}}} \,dt

= 40000100∫030dt(1+t150){{40000} \over {100}}\int\limits_0^{30} {{{dt} \over {\left( {1 + {t \over {150}}} \right)}}}

400×150[ln⁡(1+t150)]030400 \times 150\left[ {\ln \left( {1 + {t \over {150}}} \right)} \right]_0^{30}

=60000[ln⁡(1+30150)−ln⁡1]= 60000\left[ {\ln \left( {1 + {{30} \over {150}}} \right) - \ln 1} \right]

=60000ln⁡(65) J= 60000\ln \left( {{6 \over 5}} \right)\,J

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