A) \[36 \times 1{{0}^{-5}} kg m/s.\]
B) \[36 \times 1{{0}^{-4}} kg m/s.\]
C) \[108 \times 1{{0}^{4}} kg m/s.\]
D) \[1.08 \times 1{{0}^{7}} kg m/s.\]
Correct Answer: B
Solution :
| Option [b] is correct. |
| Explanation: Energy flux, \[\phi =20\operatorname{W}/c{{m}^{2}}\] |
| Area A = 30\[c{{m}^{2}}\], time = 30 \[\times \] 60 sec |
| U = Total energy falling in t sec= Energy flux |
| \[\times \] Area \[\times \] time =\[\phi \]At |
| U= 20\[\times \]30\[\times \]30\[\times \]60 J |
| Momentum of the incident light |
| \[=\frac{u}{c}=\frac{20\times 30\times 30\times 60}{3\times {{10}^{8}}}=36\times {{10}^{-4}}\operatorname{kg}-m{{s}^{-1}}\] |
| As no reflection from the surface and for complete absorption, momentum of reflected radiation is zero. |
| Momentum delivered to surface = Change in momentum |
| \[={{p}_{f}}-{{p}_{i}}=0-36\times {{10}^{-4}}\] \[kgm\,/\,s\] |
| =\[-36\times {{10}^{-4}}\] \[kg\text{ }m\,/\,s\] |
| \[(-)\] Sign shows the direction of momentum. |
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