Given data

• Mass of each sphere: $m = 1.5\,\text{kg}$
• Radius of each sphere: $R = 50\,\text{cm} = 0.5\,\text{m}$
• Distance between centres: $d = 5\,\text{m}$

The axis is perpendicular to the rod and passes through its midpoint.

Distance of each sphere’s centre from the axis Since the axis passes through the midpoint of the line joining the centres, $a = \frac{d}{2} = \frac{5}{2} = 2.5\,\text{m}$

Moment of inertia of one sphere about the given axis For a solid sphere, moment of inertia about an axis through its own centre is $I_{\text{cm}} = \frac{2}{5}mR^2$

Using parallel axis theorem, for the axis at distance $a$ from the centre: $I_1 = I_{\text{cm}} + ma^2$ $I_1 = \frac{2}{5}(1.5)(0.5)^2 + (1.5)(2.5)^2$

Now calculate: $\frac{2}{5}(1.5)(0.25) = 0.15$ $1.5(6.25) = 9.375$

Therefore, $I_1 = 0.15 + 9.375 = 9.525\,\text{kg m}^2$

Moment of inertia of two spheres Since there are two identical spheres, $I = 2I_1 = 2(9.525) = 19.05\,\text{kg m}^2$

Match with options $I = 19.05\,\text{kg m}^2$ This corresponds to Option C.

Comparison with stored answer Stored correct answer: C

Our derived answer also gives C, so they agree.