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 :

Options

A.
B.
C.
D.

Answer

A

Explanation

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.