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JCECE Engineering JCECE Engineering Solved Paper-2009

  • question_answer
    Two solid spheres \[A\] and \[B\] made of the same material have radii \[{{r}_{A}}\] and \[{{r}_{B}}\] respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature \[A\] and \[B\] is

    A) \[\frac{{{r}_{A}}}{{{r}_{B}}}\]                                     

    B) \[\frac{{{r}_{B}}}{{{r}_{A}}}\]

    C) \[\frac{r_{A}^{2}}{r_{B}^{2}}\]                                  

    D) \[\frac{r_{B}^{2}}{r_{A}^{2}}\]

    Correct Answer: B

    Solution :

    \[\frac{4\pi }{3}{{r}^{3}}pc\left( -\frac{dT}{dt} \right)=\sigma 4\pi {{r}^{2}}({{T}^{4}}-T_{0}^{4})\] \[\therefore \]  \[\left( -\frac{dT}{dt} \right)=\frac{3\sigma }{\rho rc}({{T}^{4}}-T_{0}^{4})=H\]     (say) Ratio of rates of fall of temperature                 \[\frac{{{H}_{A}}}{{{H}_{B}}}=\frac{{{r}_{B}}}{{{r}_{A}}}\]


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