JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Self Evaluation Test - Thermal Properties of Matter

  • question_answer
    A body cools in a surrounding which is at a constant temperature of \[{{\theta }_{0}}\]. Assume 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\[P\left( \theta ={{\theta }_{2}} \right)\]and\[Q\left( \theta ={{\theta }_{1}} \right)\]. These tangents meet the time axis at angle of\[{{\phi }_{2}}\] and \[{{\phi }_{1}}\], as shown, then 

    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, \[\frac{\operatorname{dQ}}{dt} =| tan\,(180{}^\circ -{{\varphi }_{1}}) |\,= tan{{\varphi }_{1}}, = k({{\theta }_{1}}-{{\theta }_{0}})\]\[\therefore \,\,\frac{\tan {{\phi }_{2}}}{\tan {{\phi }_{1}}}=\frac{{{\theta }_{2}}-{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}}\]


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