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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
(1) `log_(e) (theta = theta_(0)) = log_(e) (theta_(i) = theta_(0)) - kt`
`log_(e) (theta - theta_(0)) v//s t` will be straight line of negative slope and positive intercept
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