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A liquid in a beaker has temperature the...

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 :

A

B

C

D

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The correct Answer is:
A
`(d theta)/(dt) =- k (theta-theta_(0)), underset(theta_(0))overset(theta)int(d theta)/(theta-theta_(0))=-kunderset(0)overset(t)intdt`
In `(theta -theta_(0) = - kt+C` so graph is a straight line .
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