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JCECE Medical JCECE Medical 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 Newtons 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 :

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


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