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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

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The correct Answer is:
B
Rate of cooling = `-(d theta)/(dt) = k(theta-theta_(0)) (d theta)/(theta- theta_(0)) = -kdt implies int(d"theta)/(theta- theta_(0)) = - intkdt`
`implies ln(theta- theta_(0))= -kt + C implies` correct answer is(`2`)
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ALLEN -GEOMETRICAL OPTICS-EXERCISE - 05 (A)
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