A) \[\frac{\tan {{\phi }_{2}}}{\tan {{\theta }_{1}}}=\frac{{{\theta }_{1}}-{{\theta }_{0}}}{{{\theta }_{2}}-{{\theta }_{0}}}\]
B) \[\frac{\tan {{\phi }_{2}}}{\tan {{\theta }_{1}}}=\frac{{{\theta }_{2}}-{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}}\]
C) \[\frac{\tan {{\phi }_{1}}}{\tan {{\theta }_{2}}}=\frac{{{\theta }_{1}}}{{{\theta }_{2}}}\]
D) \[\frac{\tan {{\phi }_{1}}}{\tan {{\theta }_{2}}}=\frac{{{\theta }_{2}}}{{{\theta }_{1}}}\]
Correct Answer: B
Solution :
[b] For \[\theta -t\] plot, rate of cooling \[=\frac{d\theta }{dt}=\] slope of the curve. At \[P,\frac{d\theta }{dt}=\tan {{\phi }_{2}}=k({{\theta }_{2}}-{{\theta }_{0}})\], where k=constant. At \[Q\frac{d\theta }{dt}=\tan {{\phi }_{1}}=k({{\theta }_{1}}-{{\theta }_{0}})\] \[\Rightarrow \frac{\tan {{\phi }_{2}}}{\tan {{\phi }_{1}}}=\frac{{{\theta }_{2}}-{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}}\]You need to login to perform this action.
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