JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    A point source of electromagnetic radiation has an average power output of 1500 W. The maximum value of electric field at a distance of 3m from this source in \[V{{m}^{-1}}\] is

    A)  \[100\,V{{m}^{-1}}\]                         

    B)  \[100\,V{{m}^{-1}}\]

    C)  \[\frac{500}{3}V{{m}^{-1}}\]                        

    D)  \[\frac{250}{3}\,V{{m}^{-1}}\]

    Correct Answer: A

    Solution :

    Maximum value of electric field\[=\frac{\text{Average}\,\text{power}\,\text{output}}{\text{Distance}\,\text{at}\,\text{which}\,\text{electricd}\,\text{field}\,\text{is}\,\text{to}\,\text{be}\,\text{det}\,\text{re}\,\text{min}\,\text{ed}}\]          \[\Rightarrow \,\,E=\frac{1500}{3}\,=500\,V{{m}^{-1}}\]

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