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પદાર્થને ગરમ કરીને $ {\theta _0} $ તાપમાનવાળા વાતાવરણમાં મૂકવામાં આવે,તો નીચેનામાંથી કયો સંબંધ આવશે?
$ \frac{{\tan \,{\varphi _2}}}{{\tan \,{\varphi _1}}} = \frac{{{\theta _1} - {\theta _0}}}{{{\theta _2} - {\theta _0}}} $
$ \frac{{\tan \,{\varphi _2}}}{{\tan \,{\varphi _1}}} = \frac{{{\theta _2} - {\theta _0}}}{{{\theta _1} - {\theta _0}}} $
$ \frac{{\tan \,{\varphi _1}}}{{\tan \,{\varphi _2}}} = \frac{{{\theta _1}}}{{{\theta _2}}} $
$ \frac{{\tan \,{\varphi _1}}}{{\tan \,{\varphi _2}}} = \frac{{{\theta _2}}}{{{\theta _1}}} $
Solution
(b) For $\theta-t$ plot, rate of cooling $ = \frac{{d\theta }}{{dt}} = $ slope of the curve.
At $P,\frac{{d\theta }}{{dt}} = \tan {\varphi _2} = k({\theta _2} – {\theta _0})$, where k = constant.
At $Q\frac{{d\theta }}{{dt}} = \tan {\varphi _1} = k({\theta _1} – {\theta _0})$
$\Rightarrow \,\,\,\frac{{\tan {\varphi _2}}}{{\tan {\varphi _1}}} = \frac{{{\theta _2} – {\theta _0}}}{{{\theta _1} – {\theta _0}}}$
