JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    The graph as shown represents the variation of temperature \[(T)\] of the bodies \[X\] and \[Y\]having same surface area with time \[(t)\] due to emission of radiation. Find the correct relation between emissivity and absorptivity power of the two bodies.

    A)  \[{{E}_{X}}>{{E}_{y}}\]and \[{{a}_{X}}<{{a}_{y}}\]

    B)  \[{{E}_{X}}<{{E}_{y}}\]and \[{{a}_{X}}>{{a}_{y}}\]

    C)  \[{{E}_{X}}>{{E}_{y}}\]and \[{{a}_{X}}<{{a}_{y}}\]

    D)  \[{{E}_{X}}<{{E}_{y}}\]and \[{{a}_{X}}<{{a}_{y}}\]

    Correct Answer: C

    Solution :

    Emissivity (e) \[\propto \] Rate of cooling \[\left( -\frac{dT}{dt} \right)\] The graph shows \[\left( -\frac{dT}{dt} \right)\,>{{\left( \frac{dT}{dt} \right)}_{y}}\] Emissivity (e) \[\propto \] absorptive power So, \[{{a}_{x}}>{{a}_{y}}\]


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