Initial volume of cube \(=(10)^3=1000 \mathrm{~cm}^3\)
\(V\) when cube is chopped along length then it will form cuboid in which length \(=6 \mathrm{~cm}\)
\(\begin{aligned}
\text {breadth } & =10 \mathrm{~cm} \\
\text {height } & =10 \mathrm{~cm}
\end{aligned}\)
Adsorption power \(\propto\) more surface area \(\propto 6 \mathrm{~cm}\)
\(\propto 6 \times(10)^2 \propto 600 \mathrm{~cm}^2\)
Final surface area of the cuboid
\(\begin{aligned}
& =2(l \times b+b \times h+h \times l) \text { Area of cuboid } \\
& =2(2 \times 10+100+20) \\
& =280 \mathrm{~cm}^2
\end{aligned}\)
For final cuboid area \(=280 \times 5=1400 \mathrm{~cm}^2\)
So, final adsorption power \(\propto 1400 \mathrm{~cm}^2\)
\(\begin{aligned}
\frac{\text { Final adsorption power }}{\text { Initial adsorption power }} & =\frac{1400}{600}=\frac{14}{6}=\frac{7}{3} \\
& =2.33 \text { times }
\end{aligned}\)