JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Self Evaluation Test - Thermal Properties of Matter

  • question_answer
    A spherical black body with a radius of 12 cm radiates 459 W power at 599 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

    A) 225      

    B)        459  

    C) 999      

    D)        1899

    Correct Answer: D

    Solution :

    [d] The energy radiated per second by a black body is given by Stefan's Law \[\frac{E}{t}=\sigma {{T}^{4}}\times A\], where A is the surface area. \[\frac{E}{t}=\sigma {{T}^{4}}\times A\pi {{r}^{2}}(\because \text{ }\operatorname{For}\,a\,sphere,=4\pi {{r}^{2}})\] Dividing (ii) and (i), we get \[\frac{E/t}{450}=\frac{{{\left( 1000 \right)}^{4}}{{\left( 0.06 \right)}^{2}}}{{{\left( 500 \right)}^{4}}{{\left( 0.12 \right)}^{2}}}=\frac{{{2}^{4}}}{{{2}^{2}}}=4\] \[\Rightarrow \frac{E}{t}=450\times 4= 1800 W\]


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