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AMU Medical AMU Solved Paper-2005

  • question_answer
    A thin square plate with each side equal to 10 cm, is heated by a blacksmith The rate of radiated energy by the heated plate is 1134 W. The temperature of hot square plate is (Stefans constant \[\sigma =\text{5}.\text{67}\times \text{l}{{0}^{-8}}\text{ W}-{{m}^{2}}\text{ }{{\text{K}}^{-4}},\]emissivity of plate = 1)

    A)  1000 K                                 

    B)  1189 K

    C)   2000 K                                

    D)  2378 K

    Correct Answer: A

    Solution :

                     From Stefans law, the radiant energy emitted is given by \[E=e\sigma A{{T}^{4}}\]where \[\sigma \] is Stefans constant, e the emissivity, A the area and T the absolute temperature. Total surface area of plate is \[=2{{a}^{2}}\]where a is length of side of square plate. \[\therefore \]  \[A=2{{(10\times {{10}^{-2}})}^{2}}=2\times {{10}^{-2}}{{m}^{2}}\]                 e = 1,                 \[\sigma =5.67\times {{10}^{-8}}W-{{m}^{2}}{{K}^{-4}}\],                 E = 1134 W \[\therefore \]  \[{{T}^{4}}=\frac{1134}{1\times 5.67\times {{10}^{-8}}\times 2\times {{10}^{-2}}}\] \[\Rightarrow \]               \[T={{10}^{3}}K=1000\,K\].


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