A) \[1.25\times {{10}^{-6}}N\]
B) \[2.50\times {{10}^{-6}}N\]
C) \[1.20\times {{10}^{-6}}N\]
D) \[3.0\times {{10}^{-6}}N\]
Correct Answer: A
Solution :
Energy flux = \[\frac{25\times {{10}^{4}}\text{J}}{s{{m}^{2}}}\] Force on unit area = momentum transferred in unit time on unit area \[=\frac{25\times {{10}^{4}}}{C}\] \[=\frac{25\times {{10}^{4}}}{3\times {{10}^{8}}}\] \[=\,8.3\times {{10}^{-4}}N/{{m}^{2}}\] Force on the \[15\times {{10}^{-4}}{{m}^{2}}\] area \[=8.3\times {{10}^{-4}}N/{{m}^{2}}\times 15\times {{10}^{-4}}{{m}^{2}}\] \[=124.5\times {{10}^{-8}}N\] \[1.25\times {{10}^{-6}}N\]
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