A body cools down from temperature $9{\theta }_{0}to3{\theta }_0$ in time $t_{1}and3{\theta }_{0}to2{\theta }_{0}intime{t}_2$ in a surrounding ambience of constant temperature ${\theta }_0$. Following Newton’s law of cooling, the value of $t_1/t_2=$………. (nearest integer value)

By using Newton’s law of cooling , $\frac{d\theta }{dt}=-k(\theta -{\theta }_s)$

Solving this differential equation, we have ${\int }_{{\theta }_0}^{\theta }\frac{d\theta }{\theta -{\theta }_s}=-k{\int }_{{\theta }_{ }}^{t}dt$

This gives, $kt=\ln \frac{{\theta }_0-{\theta }_s}{\theta -{\theta }_s}$

$puttingt=t_1,{\theta }_0={\theta }_{},{\theta }_0={\theta }_2$ we have $k{t}_1=\ln \frac{{\theta }_1-{\theta }_s}{{\theta }_2-{\theta }_s}$

$puttingt=t_2,{\theta }_0={{\theta }_2}_{},\theta ={\theta }_3 we have k{t}_2=\ln \frac{{\theta }_2-{\theta }_s}{{\theta }_3-{\theta }_s}$

By using equation (i) and (ii) $\frac{t_1}{t_2}=\frac{\ln [({\theta }_1-{\theta }_s)/({\theta }_2-{\theta }_s)]}{\ln [({\theta }_2-{\theta }_s)/({\theta }_3-{\theta }_s)]}$