The ratio of surface areas of 1 cube of volume $64 c m^{3}$ and 64 cubes of volume $1 c m^{3}$ is

1: 1

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1: 2

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1: 3

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1: 4

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Solution

The correct option is D
1: 4

Let us call the cube of volume $64 c m^{3}$ as cube A and the cube with volume $1 c m^{3}$ as cube B.

Length of side of cube A $= \sqrt[3]{64} = 4 c m$

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

Length of side of cube B $= 1 c m$

Surface area of a cube of side 1 cm $6 l^{2} = 6 (1)^{2} = 6 c m^{2}$

Therefore surface area of 64 such cubes $= 64 \times 6 = 384^{2}$

Ratio of surface areas = 96:384

⇒1:4

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The ratio of surface areas of $1$ cube of volume $64 c m^{3}$ and 64 cubes of volume $1 c m^{3}$ is:

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Q. The ratio of surface area of a cube of volume 64 cm

^{3}

to that of 64 cubes of volume 1 cm

^{3}

each = .

Q. The ratio of surface areas or 1 cube of volume 64 cubic cm and 64 cubes of volume 1 cubic cm is ________________
Fill in the blank.

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

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The ratio of surface areas of 1 cube of volume 64 cm3 and 64 cubes of volume 1 cm3 is

The ratio of surface areas of 1 cube of volume $64 c m^{3}$ and 64 cubes of volume $1 c m^{3}$ is