Two identical solid spheres each of mass 2 kg and radii 10 cm are fixed at the ends of a light rod. The separation between the centres of the spheres is 40 cm. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is ______ $$\times 10^{-3}$$ kg m$$^2$$.

We have two identical solid spheres, each of mass $$m = 2\;\text{kg}$$ and radius $$r = 10\;\text{cm} = 0.1\;\text{m}$$, fixed at the ends of a light rod with separation between centres = 40 cm. The axis passes through the middle point of the rod, perpendicular to it, so the distance of each sphere's centre from the axis is $$d = 20\;\text{cm} = 0.2\;\text{m}$$.

Using the parallel axis theorem, the moment of inertia of each sphere about the given axis is:

$$I_{each} = \frac{2}{5}mr^2 + md^2$$

Substituting values:

$$I_{each} = \frac{2}{5}(2)(0.1)^2 + 2(0.2)^2 = \frac{2}{5}(2)(0.01) + 2(0.04)$$

$$I_{each} = 0.008 + 0.08 = 0.088\;\text{kg m}^2$$

Now the total moment of inertia is:

$$I = 2 \times I_{each} = 2 \times 0.088 = 0.176\;\text{kg m}^2 = 176 \times 10^{-3}\;\text{kg m}^2$$

So, the answer is $$176$$.