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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    Two bodies A and B having temperatures \[327{}^\circ C\] and \[427{}^\circ C\] are radiating heat to the surrounding. The surrounding temperature is \[27{}^\circ C\]. The ratio of rates of heat radiation of A to that of B is

    A)  0.52                                      

    B)  0.31

    C)  0.81                      

    D)         0.42

    Correct Answer: A

    Solution :

    If temperature of surrounding is considered, then net loss of energy of a body by radiation \[Q=Ae\sigma ({{T}^{4}}-T_{0}^{4})\Rightarrow \,\,Q\propto ({{T}^{4}}-T_{0}^{4})\] \[\therefore \]  \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{T_{1}^{4}-T_{0}^{4}}{T_{2}^{4}-T_{0}^{4}}\]                 \[=\frac{{{(273+327)}^{4}}-{{(273+27)}^{4}}}{{{(273+427)}^{4}}-{{(273+27)}^{4}}}\]                 \[=\frac{{{(600)}^{4}}-{{(300)}^{4}}}{{{(700)}^{4}}-{{(300)}^{4}}}=0.52\]


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