Question

There are 2 cubes A and B each of volume X $c{m}^3$. Cube B is cut into smaller cubes each of volume Y $c{m}^3$. The surface area and volumes of the resulting cubes are compared.

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

The volume is the 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.

Surface area of a body depends on number of surfaces. So cutting a body creates new surfaces. Hence sum of the TSA of the new cubes would be greater than the TSA of cube A.

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

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

Therefore number of cubes = $\frac{64}{1}=64$

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

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 $A=6\times 4^2$ = 96$c{m}^2$

Length of cube B after cutting = 1cm

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

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