question_answer

A liquid in a beaker has temperature \[\theta (t)\] at time t and \[{{\theta }_{0}}\] is temperature of surroundings, then according to Newton's law of cooling the correct graph between \[{{\log }_{e}}(\theta -{{\theta }_{0}})\] and t is :   AIEEE  Solved  Paper-2012

A)             

B)  

C)          

D)  

Correct Answer: A

Solution :

             \[\frac{d\theta }{dt}=-k(\theta -{{\theta }_{0}})\] \[\int\limits_{{{\theta }_{0}}}^{\theta }{\frac{d\theta }{\theta -{{\theta }_{0}}}=-k}\int\limits_{0}^{t}{dt}\]                 ln \[(\theta -{{\theta }_{0}})=-kt+C\] So graph is straight line.