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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

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The correct Answer is:
A
According to Newton’s law of cooling
`(d theta)/(dt)-=k(theta-theta_(0))`
`int(d theta)/(theta-theta_(0))=-kintdt`
In `(theta-theta_(0))=-kt+C`
So graph is straight line.
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