Four solid spheres each of diameter √5 cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N × 10⁻⁸ kgm², then N is:

MI of the system
I=I₁+I₂+I₃+I₄
where Iᵢ is the MI of the ith sphere about the diagonal XY where i goes from 1 to 4.
I₁=I₃ by symmetry
I₂=I₄ by symmetry
I₁ = (2/5)mR² + m(√2a/2)² using Parallel Axis Theorem since (2/5)mR² is the MI of the sphere about the Center of Mass of the sphere.
√2a/2 is the distance between the center of the square and sphere.
I₂ = (2/5)mR²
∴I = 2I₁ + 2I₂ = 2 × ((2/5)mR² + m(√2a/2)²) + 2 × ((2/5)mR²) = (8/5)mR² + ma²
= (8/5) × 0.5 × (√5/2)² × 10⁻⁸ + 0.5 × 16 × 10⁻⁸
= 9 × 10⁻⁸ kgm²
∴comparing I with N × 10⁻⁸ kgm², we get N=9