A liquid in a beaker has temperature phi(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 liquid in a beaker has temperature phi(t) at time t and theta_(0) is

1.	A liquid in a beaker has temperature phi(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
Answer» Solution :`(d theta)/(dt)=-K(theta-theta_(0))` `int_(theta_(0))^(theta)(d theta)/(theta-theta_(0))=-Kint_(0)^(t)dt` `lntheta_(0)-theta_(0)=-Kt+C` The above EQUATION shows that graph is a STRAIGHT line Correct CHOICE is (b).

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A liquid in a beaker has temperature phi(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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