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[" 30) A liquid in a beaker has temperat...

[" 30) A liquid in a beaker has temperature "theta(t)" at "],[" time "^(6)t'" and "^(6)theta" ' is temperature of "],[" surroundings,then according to Newton's law "],[" of cooling the correct graph between loge "],[(theta-theta_(0))" and "t" is "]

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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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In Newton's law of cooling (d theta)/(dt)=-k(theta-theta_(0)) the constant k is proportional to .

In Newton's law of cooling (d theta)/(dt)=-k(theta-theta_(0)) the constant k is proportional to .

In Newton's law of cooling (d theta)/(dt)=-k(theta-theta_(0)) the constant k is proportional to .