If four atoms of same radius are placed at the alternate corner of a cube touching each other, than the length of body diagonal of the cube is equal to $\sqrt{x} \times R,$ where R is the radius of atom.
Find the value of x?

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answer is 6.

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Detailed Solution

Face diagonal of cube $(AB) = 2R$
Edge length of cube $(B E) = \frac{2 R}{\sqrt{2}}$
Body diagonal of cube $(A D) = \sqrt{3} \times \frac{2 R}{\sqrt{2}} = \sqrt{6} \times R$

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