A) \[\frac{\tan \,{{\varphi }_{2}}}{\tan \,{{\varphi }_{1}}}=\frac{{{\theta }_{1}}-{{\theta }_{0}}}{{{\theta }_{2}}-{{\theta }_{0}}}\]
B) \[\frac{\tan \,{{\varphi }_{2}}}{\tan \,{{\varphi }_{1}}}=\frac{{{\theta }_{2}}-{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}}\]
C) \[\frac{\tan \,{{\varphi }_{1}}}{\tan \,{{\varphi }_{2}}}=\frac{{{\theta }_{1}}}{{{\theta }_{2}}}\]
D) \[\frac{\tan \,{{\varphi }_{1}}}{\tan \,{{\varphi }_{2}}}=\frac{{{\theta }_{2}}}{{{\theta }_{1}}}\]
Correct Answer: B
Solution :
For q-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\[\frac{d\theta }{dt}=\tan {{\varphi }_{1}}=k({{\theta }_{1}}-{{\theta }_{0}})\] \[\Rightarrow \,\,\,\frac{\tan {{\varphi }_{2}}}{\tan {{\varphi }_{1}}}=\frac{{{\theta }_{2}}-{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}}\]You need to login to perform this action.
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