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How many cubes of \[{\text{10 cm}}\] edge can be put in the cubic box of \[{\text{1m}}\] edge ?
A. \[10\]
B. \[100\]
C. \[1000\]
D. \[10000\]

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Last updated date: 21st Jun 2024
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Answer
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Hint: First of all considering the unit conversion that \[{\text{1m = 100cm}}\]and calculating the volume of both the cubes and then diving the volume of a bigger cube by smaller cube to get an exact number of cubes that can be inserted into the bigger one.

Complete step by step answer:

As per the given that one cube is of \[{\text{10cm}}\] edge and another is of \[{\text{1m}}\] edge
Volume of cube (side length as a) \[{\text{ = }}{{\text{a}}^{\text{3}}}\]
So, now converting all into similar units.
As \[{\text{1m = 100cm}}\],
So the volume of the larger cube is
\[
  {\text{V = }}{{\text{a}}^{\text{3}}} \\
  {\text{ = (100)(100)(100)}} \\
  {\text{ = 1000000c}}{{\text{m}}^{\text{3}}} \\
 \]
And the volume of a smaller cube is
\[
  {\text{V = }}{{\text{a}}^{\text{3}}} \\
  {\text{ = (10)(10)(10)}} \\
  {\text{ = 1000c}}{{\text{m}}^{\text{3}}} \\
 \]
Now, to calculate the number of the smaller cube that can be inserted into the larger cube is given by
\[
  {\text{1000n = 1000000}} \\
  \therefore {\text{n = }}\dfrac{{1000000}}{{1000}} \\
  {\text{n = 1000}} \\
 \]
Hence, option (c) is our required correct answer.

Note: In geometry, a cube is a three-dimensional solid object bounded by six square faces, facets or sides, with three meetings at each vertex. The cube is the only regular hexahedron and is one of the five Platonic solids. The volume of a cube is found by multiplying the length of any edge by itself thrice.