A solid cube changes its volume such that its shape remains unchanged. For such a cube of unit volume, what will be the value of rate of change of volume?

The correct option is 3/2*(rate of change of area of any face of the cube).

To understand why this is correct, let us break down the mathematical relationship between the volume of a cube, the area of its faces, and their respective rates of change with respect to time.

Step 1: Define the variables
Let the side length of the cube at any time t be represented by:
$$x$$
The volume V of the cube is given by the formula:
$$V=x^3$$
The surface area A of any single face of the cube is given by:
$$A=x^2$$

Step 2: Differentiate with respect to time
Using the chain rule, we find the rate of change of volume with respect to time t:
$$\frac{dV}{dt}=\frac{d}{dt}\left(x^3\right)=3x^2\frac{dx}{dt}$$
Similarly, the rate of change of the area of any face with respect to time t is:
$$\frac{dA}{dt}=\frac{d}{dt}\left(x^2\right)=2x\frac{dx}{dt}$$

Step 3: Relate the two rates of change
From the equation for the rate of change of area, we can express the rate of change of the side length as:
$$\frac{dx}{dt}=\frac{1}{2x}\frac{dA}{dt}$$
Now, substitute this expression into the rate of change of volume equation:
$$\frac{dV}{dt}=3x^2\left(\frac{1}{2x}\frac{dA}{dt}\right)$$
Simplifying the expression yields:
$$\frac{dV}{dt}=\frac{3}{2}x\frac{dA}{dt}$$

Step 4: Evaluate for a unit volume
The question specifies that we are analyzing a cube of unit volume. Therefore:
$$V=1$$
Since $V=x^3=1$, the side length of this unit cube must be:
$$x=1$$
Substituting $x=1$ into our related rates equation:
$$\frac{dV}{dt}=\frac{3}{2}\left(1\right)\frac{dA}{dt}=\frac{3}{2}\frac{dA}{dt}$$

Thus, the rate of change of volume is exactly 3/2 times the rate of change of the area of any face of the cube.