Explanation

Moment of inertia of a solid sphere about its centre is:

$$ I_{\text{cm}} = \frac{2}{5}MR^2 $$

The axis is a tangent passing through the point of contact, so we must apply the parallel axis theorem:

$$ I = I_{\text{cm}} + Md^2 $$

Sphere 1:

Mass $M_1 = 5\ \text{kg}$
Radius $R_1 = 10\ \text{cm} = 0.1\ \text{m}$

Moment of inertia about its centre:

$$ I_{\text{cm1}} = \frac{2}{5}\times 5 \times (0.1)^2 = 2 \times 0.01 = 0.02 $$

Distance of centre from tangent $d_1 = R_1 = 0.1\ \text{m}$

Using parallel axis theorem:

$$ I_1 = 0.02 + 5(0.1)^2 = 0.02 + 0.05 = 0.07\ \text{kg·m}^2 $$

Sphere 2:

Mass $M_2 = 10\ \text{kg}$
Radius $R_2 = 20\ \text{cm} = 0.2\ \text{m}$

Moment of inertia about its centre:

$$ I_{\text{cm2}} = \frac{2}{5}\times 10 \times (0.2)^2 = 4 \times 0.04 = 0.16 $$

Distance of centre from tangent $d_2 = R_2 = 0.2\ \text{m}$

Using parallel axis theorem:

$$ I_2 = 0.16 + 10(0.2)^2 = 0.16 + 0.40 = 0.56\ \text{kg·m}^2 $$

Total moment of inertia of the system:

$$ I = I_1 + I_2 $$

$$ I = 0.07 + 0.56 = 0.63\ \text{kg·m}^2 $$

Therefore, the moment of inertia of the pair of spheres is 0.63 kg·m².

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