A) \[2\times {{10}^{3}}\,N/{{m}^{2}}\]
B) \[2\times {{10}^{5}}\,N/{{m}^{2}}\]
C) \[{{10}^{7}}\,N/{{m}^{2}}\]
D) \[2\times {{10}^{7}}\,N/{{m}^{2}}\]
Correct Answer: D
Solution :
\[P=\frac{F}{A}=\frac{n[mv-(-mv)]}{A}=\frac{2mnv}{A}\] \[=\frac{2\times {{10}^{-3}}\times {{10}^{4}}\times {{10}^{2}}}{{{10}^{-4}}}\]\[=2\times {{10}^{7}}N/{{m}^{2}}\]
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