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A body cools in a surrounding which is at a constant temperature of $\theta _0$ . Assuming that it obeys Newton's law of cooling, its temperature $\theta $ is plotted against time $t$ . Tangents are drawn to the curve at the points $A(\theta = \theta _1)$ and $B(\theta = \theta _2)$ . These tangents meet the time-axis at angles $\alpha _1$ and $\alpha _2$ as shown in the graph then
$\frac{{\tan \,{\alpha _1}}}{{\tan \,{\alpha _2}}} = \frac{{{\theta _2}}}{{{\theta _1}}}$
$\frac{{\tan \,{\alpha _1}}}{{\tan \,{\alpha _2}}} = \frac{{{\theta _1}}}{{{\theta _2}}}$
$\frac{{\tan \,{\alpha _1}}}{{\tan \,{\alpha _2}}} = \frac{{{\theta _1} - {\theta _0}}}{{{\theta _2} - {\theta _0}}}$
$\frac{{\tan \,{\alpha _1}}}{{\tan \,{\alpha _2}}} = \frac{{{\theta _2} - {\theta _0}}}{{{\theta _1} - {\theta _0}}}$
Solution
Rate of cooling $\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-\theta_{0}$
at ${\rm{A}},{\left( {\frac{{{\rm{d}}\theta }}{{{\rm{dt}}}}} \right)_{\theta = {\theta _1}}} = \tan \left( {{{180}^\circ } – {\alpha _1}} \right)$
$=\tan \alpha_{1}=\mathrm{K}\left(\theta_{1}-\theta_{0}\right)$
and at ${\rm{B}},{\left( {\frac{{{\rm{d}}\theta }}{{{\rm{dt}}}}} \right)_{\theta = {\theta _2}}} = \tan \left( {{{180}^\circ } – {\alpha _2}} \right) = \tan {\alpha _2}$
$=\mathrm{K}\left(\theta_{2}-\theta_{0}\right)$
where, $\mathrm{K}$ is proportionally constant
$\therefore $ $\frac{\tan \alpha_{1}}{\tan \alpha_{2}}=\frac{\theta_{1}-\theta_{0}}{\theta_{2}-\theta_{0}}$
