A)
B)
C)
D)
Correct Answer: B
Solution :
According to Newton's law of cooling Rate of cooling µ Temperature difference Þ \[-\frac{d\theta }{dt}\propto (\theta -{{\theta }_{0}})\]Þ\[-\frac{d\theta }{dt}\]=\[\alpha \ (\theta -{{\theta }_{0}})\] (a= constant) Þ \[\int\limits_{{{\theta }_{i}}}^{\theta }{\frac{d\theta }{(\theta -{{\theta }_{0}})}=-\alpha \int\limits_{0}^{t}{dt}}\] Þ \[\theta ={{\theta }_{0}}+({{\theta }_{i}}-{{\theta }_{0}}){{e}^{-\alpha \,t}}\] This relation tells us that, temperature of the body varies exponentially with time from \[{{\theta }_{i}}\] to \[{{\theta }_{0}}\] Hence graph (b) is correct.You need to login to perform this action.
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