$(a)\;0.1cm^2\qquad(b)\;0.2cm^2\qquad(c)\;0.3cm^2\qquad(d)\;0.4cm^2$

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Meq of Al=Meq of HCl

$\qquad\quad\;\;=12\times 0.05=0.6$

Weight of Al=$\large\frac{0.6\times 9}{1000}$$=0.0054g$

Volume of Al foil=$\large\frac{0.0054}{2.7}$$cm^3$

$\qquad\qquad\qquad=0.002cm^3$

Area$\times$ thickness=volume

Area=$\large\frac{0.002}{0.01}$$=0.02cm^2$

Hence (b) is the correct answer.

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