Question

$Thereare2cubesAandBeachofvolumeXc{m}^3$. $CubeBiscutintosmallercubeseachofvolumeYc{m}^3$. $Thesurfaceareaandvolumesoftheresultingcubesarecompared$

S1: The ratio of volumes of A and B = 1:1

S2: The ratio of surface areas of A and B = 1:1

• A. S1 is true but S2 is false
• B. S1 is false but S2 is true
• C. S1 and S2 are true
• D. S1 and S2 are false

Answer 1

The correct option is A

S1 is true but S2 is false

A volume is a space occupied by a body. So cutting an object does not change its volume. Therefore cube A and cube B after cutting will have the same volume.

The surface area of a body depends on a number of surfaces. So cutting a body creates new surfaces. Hence surface area of the newly formed cubes would be greater than that of cube A.

Let us try to understand this by converting variables to numbers.

Assume $X=64c{m}^3$ and $Y=1c{m}^3$.

$\therefore$ Number of cubes $=\frac{64}{1}=64$

Volume of cube A $=64c{m}^3$

Volume of cube B after cutting $=64\times 1=64c{m}^3$

Hence the volume is same.

Length of side of cube A $=\sqrt[3]{364}=4cm$

Surface area of cube A $=6\times 4^2=96c{m}^2$

Length of cube B after cutting $=1cm$

Surface area of 1 cube $=6\times 1^2=6c{m}^2$

Surface area of 64 such cubes $=64\times 6=384c{m}^2$

Hence surface areas are not same.

$\therefore$ S1 is true and S2 is false