BCECE Engineering BCECE Engineering Solved Paper-2014

  • question_answer
    Light with an energy flux of \[\text{18W}/\text{c}{{\text{m}}^{\text{2}}}\]falls on a non-reflecting surface at normal incidence, The surface has an area of\[\text{2}0\text{ c}{{\text{m}}^{\text{2}}}\], then the total momentum delivered on the surface during a  span of 30 mm is    

    A)                           \[2.16\times {{10}^{-3}}kg-m/s\]

    B) \[1.52\times {{10}^{-5}}kg-m/s\]

    C)  \[8.31\times {{10}^{-8}}kg-m/s\]

    D) \[18.2\times {{10}^{-6}}kg-m/s\]

    Correct Answer: A

    Solution :

    Total energy falling \[U=(18W/c{{m}^{2}})(30\times 60s)(20c{{m}^{2}})\]                 \[=6.48\times {{10}^{5}}J\] Total, momentum delivered is \[P=\frac{U}{c}=\frac{6.48\times {{10}^{5}}}{3\times {{10}^{8}}}=2.16\times {{10}^{-3}}kg-m/s\]


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