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AMU Medical AMU Solved Paper-2001

  • question_answer
    Calculate the energy radiated per minute the filament of an incandescent lamp at 2000 K. If the surface area is \[\text{5}\times \text{1}{{0}^{-5}}\text{ }{{\text{m}}^{\text{2}}}\]and its relative emittance is \[0.\text{85},\sigma =\text{ 5}.\text{7}\times {{10}^{-8}}\] MKS units

    A)  1230 J                                  

    B)  2225 J

    C)  2115 J                                  

    D)   2315 J

    Correct Answer: B

    Solution :

                     From Stefans law of radiation, the energy radiated (E) is given by                 \[E=e\,\sigma {{T}^{4}}A\,t\] Given,   \[e=0.85,\,\,\,\sigma =5.7\times {{10}^{-8}}\],                 T = 2000 K,                 \[A=5\times {{10}^{-5}}{{m}^{2}},\,\,\,t=60\,s\] \[E=0.85\times 5.7\times {{10}^{-8}}\times {{(2000)}^{4}}\times 5\times {{10}^{-5}}\times 60\] E = 2325 J


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