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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    \[50W/{{m}^{2}}\]energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on \[1{{m}^{2}}\]surface area will be close to \[\left( c=3\times {{10}^{8}}m/s \right)\]:- [JEE Main 9-4-2019 Afternoon]

    A) \[15\times {{10}^{8}}N\]                

    B) \[35\times {{10}^{8}}N\]

    C) \[10\times {{10}^{8}}N\]                

    D) \[20\times {{10}^{8}}N\]

    Correct Answer: D

    Solution :

    Force on the surface (25% reflecting and rest absorbing) \[F=\frac{25}{100}\left( \frac{2I}{C} \right)+\frac{75}{100}\left( \frac{I}{C} \right)=\frac{125}{100}\left( \frac{I}{C} \right)\] \[=\frac{25}{100}\times \left( \frac{50}{3\times {{10}^{8}}} \right)=20.83\times {{10}^{-8}}N.\]


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