Three solid spheres each of mass m and diameter d are stuck together such that the lines connecting the centres form an equilateral triangle of side of length d. The ratio I₀/I A of moment of inertia I₀ of the system about an axis passing the centroid and about center of any of the spheres I A and perpendicular to the plane of the triangle is :

Solution

Physics sir · 4 steps

Each sphere spins about its own centre, then gets shifted to the axis using the parallel axis theorem.

1. One sphere alone. Each sphere has radius d/2. About a diameter through its own centre, I_(cm) = 2/5m(d/2)² = md²/10.
2. Shift the axis. For an axis parallel to that diameter at distance x, each sphere gives md²/10 + mx². The centroid of an equilateral triangle of side d is d/√3 from each vertex, so x² = d²/3 for all three spheres.

   I₀ = 3(md²/10 + md²/3) = 3md²/10 + md² = 13/10md²

3. About centre A. Sphere A sits on the axis, so it adds only md²/10. Spheres B and C are each a distance d away, so each adds md²/10 + md². That gives I_A = 3md²/10 + 2md² = 23/10md².
4. So. I₀/I_A = (13/10)/(23/10) = 13/23, which is option (a).

The idea to keep. For a system of rigid bodies, add each body's own I_(cm) and then mx² for its distance from the axis. Never treat a sphere as a point mass when its size is given.

Answer from the NTA final answer key. Solution by PrepWiser.