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NEET Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Mock 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 \[P(\theta ={{\theta }_{1}})\]and\[Q(\theta ={{\theta }_{2}})\]. These tangents meet the time axis at angles of \[{{\phi }_{2\,}}\] and \[{{\phi }_{1}}\], as shown

    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 \[Q\frac{d\theta }{dt}=\tan {{\phi }_{1}}=k({{\theta }_{1}}-{{\theta }_{0}})\] \[\Rightarrow \frac{\tan {{\phi }_{2}}}{\tan {{\phi }_{1}}}=\frac{{{\theta }_{2}}-{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}}\]


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