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 2012]

A)  

B)      

C)  

D)      

Correct Answer: A

Solution :

[a]

.

So graph is straight line.