Sumanta has made

Given that,

Sumanta has made many cubes of length, 1 cm.

This means, volume of each cube = length × length × length

Here, length = 1 cm

⇒ Volume of each cube = 1 × 1 × 1 cm³

⇒ Volume of each cube = 1 cm³

And a single cube out of those many cubes is made by Manami.

Let us assume that Sumanta has made ‘Z’ cubes of length 1 cm, where Z is some natural number.

So, if volume of 1 cube = 1 cm³

⇒ Volume of Z cubes = Z × 1 cm³

⇒ Volume of Z cubes = Z cm³

We have volumes of cube in the options.

(i). Let us check for 100.

We need to find cube root of 100. If we are able to find the cube root of 100, then we can say that a big cube is possible in this case.

100 = 2×2× 5× 5

We can see 100 cannot be written as a perfect cube.

∴ 100 ≠ (any integer)³

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 100 is not a natural number.

Thus, Manami won’t be able to make a big cube in this case.

(ii). Let us check for 1000.

We need to find the cube root of 1000. If we are able to find the cube root of 1000, then we can say that a big cube is possible in this case.

Cube root of 1000

1000 = 10× 10× 10

⇒ Cube root of 1000 = (1000)^1/3

= 10

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 1000 is a natural number.

Thus, Manami will be able to make a big cube in this case.

(iii). Let us check for 1331.

We need to find the cube root of 1331. If we are able to find the cube root of 1331, then we can say that a big cube is possible in this case.

Cube root of 1331

1331 = 11× 11× 11

⇒ Cube root of 1331 = (1331)^1/3

= 11

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 1331 is a natural number.

Thus, Manami will be able to make a big cube in this case.

(iv). Let us check for 1210.

We need to find the cube root of 1210. If we are able to find the cube root of 1210, then we can say that a big cube is possible in this case.

Cube root of 1210

1210 = 2× 5× 11× 11

We can see 1210 cannot be written as a perfect cube.

∴ 1210 ≠ (any integer)³

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 1210 is not a natural number.

Thus, Manami will not be able to make a big cube in this case.

(v). Let us check for 3375.

We need to find the cube root of 3375. If we are able to find the cube root of 3375, then we can say that a big cube is possible in this case.

Cube root of 3375

3375 = 15× 15× 15

⇒ Cube root of 3375 = (3375)^1/3

= 15

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 3375 is a natural number.

Thus, Manami will be able to make a big cube in this case.

(vi). Let us check for 2700.

We need to find the cube root of 2700. If we are able to find the cube root of 2700, we can say that a big cube is possible in this case.

Cube root of 2700

2700 = 3× 3× 3× 2× 2× 5× 5

We can see 2700 cannot be written as a perfect cube.

∴ 2700 ≠ (any integer)³

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 2700 is not a natural number.

Thus, Manami will not be able to make a big cube in this case.