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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    At temperature\[227{}^\circ C,\]the power radiated by a body is\[Q\text{ }cal/c{{m}^{^{2}}}\]. At the temperature\[727{}^\circ C\] the power radiated by it will be

    A)  2Q          

    B)  4 Q

    C)  16 Q            

    D)  32 Q

    Correct Answer: C

    Solution :

     According to Stefan law \[Q\propto {{T}^{4}}\] \[\Rightarrow \] \[\frac{{{Q}_{2}}}{{{Q}_{1}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}\] Or \[\frac{{{Q}_{2}}}{{{Q}_{1}}}={{\left( \frac{727+273}{227+273} \right)}^{4}}=16\] or         \[{{Q}_{2}}=16Q\]


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