If a cube has surface area S and volume V, then the volume of the cube of surface area 2S is :

1. $\sqrt{2}V$

2. 2V

3. $2\sqrt{2}V$

4. $\dfrac{V}{\sqrt{2}}$

Let the side of the first cube be a units and volume be V cubic units.

We know that,

Surface area of the cube = 6a²

⇒ S = 6a²

⇒ a² = $\dfrac{S}{6}$

⇒ a = $\sqrt{\dfrac{S}{6}}$

Calculating the volume of first cube,

Volume of cube (V) = a³

Substituting the value of a, we get,

V = $\Big(\sqrt{\dfrac{S}{6}}\Big)^3$

V = $\Big(\dfrac{S}{6}\Big)^\dfrac{3}{2}$ …….(1)

Let the side of second cube be b units and volume be V' cubic units.

Surface area = 6b²

⇒ 2S = 6b²

⇒ S = 3b²

⇒ b² = $\dfrac{S}{3}$

⇒ b = $\sqrt{\dfrac{S}{3}}$

Volume of second cube (V') = b³

Substituting the value of b, we get

V' = $\Big(\sqrt{\dfrac{S}{3}}\Big)^3$

V' = $\Big(\dfrac{S}{3}\Big)^\dfrac{3}{2}$…….(2)

Taking the ratio of equation (1) and (2) we get,

$\Rightarrow \dfrac{V}{V'} = \dfrac{\Big(\dfrac{S}{6}\Big)^\dfrac{3}{2}}{\Big(\dfrac{S}{3}\Big)^\dfrac{3}{2}} \\[1em] \Rightarrow \dfrac{V}{V'} = \Big(\dfrac{\dfrac{S}{6}}{\dfrac{S}{3}}\Big)^{\dfrac{3}{2}} \\[1em] \Rightarrow \dfrac{V}{V'} = \Big(\dfrac{3}{6}\Big)^\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{V}{V'} = \Big(\dfrac{1}{2}\Big)^\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{V}{V'} = \dfrac{1}{2\sqrt{2}} \\[1em] \Rightarrow V' = 2 \sqrt{2}V$

Hence, option 3 is the correct option.