A) \[\frac{\tan {{\phi }_{2}}}{\tan {{\phi }_{1}}}=\frac{{{\theta }_{1}}-{{\theta }_{0}}}{{{\theta }_{2}}-{{\theta }_{0}}}\]
B) \[\frac{\tan {{\phi }_{2}}}{\tan {{\phi }_{1}}}=\frac{{{\theta }_{2}}-{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}}\]
C) \[\frac{\tan {{\phi }_{1}}}{\tan {{\phi }_{2}}}=\frac{{{\theta }_{1}}}{{{\theta }_{2}}}\]
D) \[\frac{\tan {{\phi }_{1}}}{\tan {{\phi }_{2}}}=\frac{{{\theta }_{2}}}{{{\theta }_{1}}}\]
Correct Answer: B
Solution :
[b] For 0-t plot, rate of cooling \[=\frac{dQ}{dt}=\]slope of the curve. \[\operatorname{AT} P,\frac{dQ}{dt}=\left| tan\left( 180{}^\circ -{{\phi }_{2}} \right) \right|\] \[=tan{{\phi }_{2}}=k\left( {{\theta }_{2}}-{{\theta }_{1}} \right)\] where k= constant. At Q, \[\frac{\operatorname{dQ}}{dt} =| tan\,(180{}^\circ -{{\varphi }_{1}}) |\,= tan{{\varphi }_{1}}, = k({{\theta }_{1}}-{{\theta }_{0}})\]\[\therefore \,\,\frac{\tan {{\phi }_{2}}}{\tan {{\phi }_{1}}}=\frac{{{\theta }_{2}}-{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}}\]You need to login to perform this action.
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