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(A) Initial surface area of the cube $= 6 	imes (10 \ cm)^2 = 600 \ cm^2$. When the cube is chopped lengthwise into $5$ equal pieces,each piece is a

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increases by $2.33$ times

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(A) Initial surface area of the cube $= 6 	imes (10 \ cm)^2 = 600 \ cm^2$. When the cube is chopped lengthwise into $5$ equal pieces,each piece is a cuboid with dimensions $2 \ cm 	imes 10 \ cm 	imes 10 \ cm$. Surface area of one such cuboid $= 2 	imes (2 	imes 10 + 10 	imes 10 + 10 	imes 2) = 2 	imes (20 + 100 + 20) = 280 \ cm^2$. Total surface area of $5$ such cuboids $= 5 	imes 280 \ cm^2 = 1400 \ cm^2$. Since adsorption power is directly proportional to the surface area,the ratio of final to initial adsorption power $= \frac{1400}{600} = \frac{14}{6} = 2.33$. Thus,the adsorption power increases by $2.33$ times.

$A$ charcoal cube having side length $10 \ cm$ is chopped lengthwise into $5$ equal pieces. Then the effective adsorption power

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