JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 13)
The resistance of an electrical toaster has a temperature dependence given by R(T) = R0 [1 + $$\alpha $$(T − T0)] in its range of operation. At T0 = 300 K, R = 100 $$\Omega $$ and at T = 500 K, R = 120 $$\Omega $$. 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 :
400 $$\ln \,{{1.5} \over {1.3}}\,J$$
200 $$\ln \,{{2} \over {3}}\,J$$
60000 $$\ln \,{{6} \over {5}}\,J$$
300 J
Explanation
Given,
R = R6 [1 + $$\alpha $$ (T $$-$$ t0)]
120 = 100 [1 + $$\alpha $$ (500 $$-$$ 300)]
$$ \Rightarrow $$ 200 $$\alpha $$ = $${1 \over 5}$$
$$ \Rightarrow $$ $$\alpha $$ = 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,
$$\Delta $$T = $${{500 - 300} \over {30}}t$$
= $${{200} \over 3}t$$
= $${{20} \over 3}t$$
Total work done in raising the temperature
= $$\int\limits_0^t {{{{V^2}} \over {R\left( t \right)}}\,dt} $$
= $$\int\limits_0^t {{{{V^2}} \over {{R_0}\left( {1 + \alpha \Delta t} \right)}}} \,dt$$
= $$\int\limits_0^{30} {{{{{\left( {200} \right)}^2}} \over {100\left( {1 + {{10}^{ - 3}} \times {{20} \over 3}t} \right)}}} \,dt$$
= $${{40000} \over {100}}\int\limits_0^{30} {{{dt} \over {\left( {1 + {t \over {150}}} \right)}}} $$
$$400 \times 150\left[ {\ln \left( {1 + {t \over {150}}} \right)} \right]_0^{30}$$
$$ = 60000\left[ {\ln \left( {1 + {{30} \over {150}}} \right) - \ln 1} \right]$$
$$ = 60000\ln \left( {{6 \over 5}} \right)\,J$$
R = R6 [1 + $$\alpha $$ (T $$-$$ t0)]
120 = 100 [1 + $$\alpha $$ (500 $$-$$ 300)]
$$ \Rightarrow $$ 200 $$\alpha $$ = $${1 \over 5}$$
$$ \Rightarrow $$ $$\alpha $$ = 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,
$$\Delta $$T = $${{500 - 300} \over {30}}t$$
= $${{200} \over 3}t$$
= $${{20} \over 3}t$$
Total work done in raising the temperature
= $$\int\limits_0^t {{{{V^2}} \over {R\left( t \right)}}\,dt} $$
= $$\int\limits_0^t {{{{V^2}} \over {{R_0}\left( {1 + \alpha \Delta t} \right)}}} \,dt$$
= $$\int\limits_0^{30} {{{{{\left( {200} \right)}^2}} \over {100\left( {1 + {{10}^{ - 3}} \times {{20} \over 3}t} \right)}}} \,dt$$
= $${{40000} \over {100}}\int\limits_0^{30} {{{dt} \over {\left( {1 + {t \over {150}}} \right)}}} $$
$$400 \times 150\left[ {\ln \left( {1 + {t \over {150}}} \right)} \right]_0^{30}$$
$$ = 60000\left[ {\ln \left( {1 + {{30} \over {150}}} \right) - \ln 1} \right]$$
$$ = 60000\ln \left( {{6 \over 5}} \right)\,J$$
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