Question

Two identical solid cubes of volume \(8\text{ cm}^3\) are joined end to end. Then the length of diagonal of the resulting cuboid is :

• A. \(3\sqrt{6}\text{ cm}\)
• B. \(2\sqrt{6}\text{ cm}\)
• C. \(2\sqrt{2}\text{ cm}\)
• D. 4 cm

Hint:

For any cuboid formed by joining \(n\) identical cubes of side \(a\) end-to-end, the diagonal is always:
\[ D = a\sqrt{n^2 + 2} \]
Here, \(a = 2\) and \(n = 2\), so:
\[ D = 2\sqrt{2^2 + 2} = 2\sqrt{6}\text{ cm} \]
This formula saves calculation time for multi-cube questions.

Answer 1

The Correct Option is B

Step 1: Find the side of each cube. \[ a^3 = 8 \implies a = 2 \text{ cm} \]
Step 2: Set up the cuboid's dimensions as a vector from one corner to the opposite one.
Joining the two cubes end to end doubles only the length, so the cuboid measures $4 \times 2 \times 2$. We can think of the space diagonal as the vector \[ (l, b, h) = (4, 2, 2) \]
Step 3: Find the magnitude of this vector. \[ |D| = \sqrt{4^2 + 2^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} \]
Step 4: Simplify the surd and conclude. \[ \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6} \text{ cm} \] \[ \boxed{2\sqrt{6} \text{ cm}} \]