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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    Light is incident normally on a completely absorbing surface with an energy flux of \[25Wc{{m}^{-2}}.\]if the surface has an area of \[25c{{m}^{-2}},\]the momentum transferred to the surface in 40 min time duration will be : [JEE Main 10-4-2019 Afternoon]

    A) \[5.0\times {{10}^{3}}Ns\]

    B) \[3.5\times {{10}^{6}}Ns\]

    C) \[1.4\times {{10}^{6}}Ns\]

    D) \[6.3\times {{10}^{4}}Ns\]

    Correct Answer: A

    Solution :

                \[Pressure=\frac{I}{C}\] \[Force=Pressure\times Area=\frac{I}{C}.Area\] Momentum transferred = Force .\[\Delta t\] \[=\frac{I}{C}.Area.\Delta t\] \[=\frac{25\times {{10}^{4}}}{3\times {{10}^{8}}}\times 25\times {{10}^{-4}}\times 40\times 60\] \[=5\times {{10}^{-3}}N-s\]


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