Answer:
\[F=10\text{ }N\] \[A=0.01c{{m}^{2}}=\frac{0.01}{10000}{{m}^{2}}\] \[=\frac{1}{10,000\times 100}{{m}^{2}}=\frac{1}{{{10}^{6}}}{{m}^{2}}={{10}^{-6}}{{m}^{2}}\] \[\therefore \]\[P=\frac{F}{A}=\frac{10}{{{10}^{-6}}}Pa=10\times {{10}^{6}}Pa={{10}^{7}}Pa\]
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