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JEE Main & Advanced Physics Transmission of Heat Question Bank Graphical Questions

  • 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 \[{{\varphi }_{2}}\]and \[{{\varphi }_{1}}\], as shown

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


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