Given identical rings are arranged in a hexagonal plane pattern so as to touch each neighbouring ring as shown in figure. Each ring has mass M and radius R. The moment of inertia of the system of seven rings about an axis passing through the centre of central ring and normal to plane of all rings is
Show Hint
Use the parallel axis theorem for the six outer rings, each at distance 2R from the central axis.
Step 1: Count the contributions:
Total $I = I_{\text{central}} + 6\times I_{\text{outer}}$.
Step 2: Spin term and orbit term:
Each ring has a spin term $MR^2$ (rotation about its own centre) and an orbit term $Md^2$ (its centre moving around the axis). For all seven rings, the spin terms add up to $7MR^2$.
The orbit terms come only from the six outer rings: $6\times M(2R)^2 = 24MR^2$.
Step 3: Add:
$I = 7MR^2 + 24MR^2 = 31MR^2$.
Step 4: Check:
The distance 2R is the sum of two radii since the rings touch. This confirms the orbit term uses $d=2R$.
Final Answer:
$31MR^2$, option (A).
\[ \boxed{31MR^2 \text{ (A)}} \]
