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Haryana PMT Haryana PMT Solved Paper-2006

  • question_answer
    A body at a temperature of \[728{}^\circ C\] and has surface area \[5\text{ }c{{m}^{2}}\], radiates 300 J of energy                 each minute. The emissivity is :                 (Given: Boltzmann constant \[\text{=5}\text{.67 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-8}}}\text{W}{{\text{m}}^{\text{2}}}{{\text{K}}^{\text{4}}}\]

    A)                  e=0.18                                

    B)  e=0.02

    C)                  e=0.2                                   

    D)  e=0.15

    Correct Answer: A

    Solution :

                    \[Q=EAt=e\,\sigma ({{T}^{4}}-T_{0}^{4})At\] Where, t = time \[{{T}_{o}}\]= temperature of surrounding When,        \[T>{{T}_{o}}\]                 \[Q=e\,\sigma {{T}^{4}}\]  At                 \[300=e\times (5.67\times {{10}^{-8}}){{(1000)}^{4}}\]                                                 \[(5.00\times {{10}^{-4}})(60)\]                 \[e=0.18\]


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