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Light with an energy flux of 20 \mathrm{~W} / \mathrm{cm}^2 falls on a non-reflecting surface at normal incidence. If the surface has an area of \mathrm{30\: cm^{2}}, then the total momentum delivered (for complete absorption) during 30 \mathrm{~min} is:
 

Option: 1

36 \times 10^{-5} \mathrm{~kg}-\mathrm{m} / \mathrm{s}



 


Option: 2

36 \times 10^{-4} \mathrm{~kg}-\mathrm{m} / \mathrm{s}


Option: 3

108 \times 10^{-4} \mathrm{~kg}-\mathrm{m} / \mathrm{s}


Option: 4

1.08 \times 10^7 \mathrm{~kg}-\mathrm{m} / \mathrm{s}


Answers (1)

best_answer

Given, energy flux, \mathrm{\phi=20 \mathrm{~W} / \mathrm{cm}^2}

Area, \mathrm{A=30 \mathrm{~cm}^2}

Time, \mathrm{\mathrm{t}=30 \mathrm{~min}=30 \times 60 \mathrm{~s}}

Now, total energy falling on the surface in time \mathrm{t\: is,\, u=\phi A t=20 \times 30 \times(30 \times 60) \mathrm{J}}
Momentum of the reflected light =0

\therefore \quad Momentum delivered to the surface=36 \times 10^{-4}-0=36 \times 10^{-4} \mathrm{~kg}-\mathrm{m} / \mathrm{s}

Hence option 2 is correct.

Posted by

Deependra Verma

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