Four identical uniform solid spheres each of same mass ' $M$ ' and radius ' $R$ ' are placed touching each…

Four identical uniform solid spheres each of same mass ' $M$ ' and radius ' $R$ ' are placed touching each other as shown in figure with centres A, B, C, D. $\mathrm{I}_{\mathrm{A}}, \mathrm{I}_{\mathrm{B}}, \mathrm{I}_{\mathrm{C}}, \mathrm{I}_{\mathrm{D}}$ are the moment of inertia of these spheres respectively about an axis passing through centre and perpendicular to the plane, then
  1. $\mathrm{I}_{\mathrm{A}}>\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{C}}>\mathrm{I}_{\mathrm{D}}$
  2. $\mathrm{I}_{\mathrm{D}}>\mathrm{I}_{\mathrm{C}}>\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{A}}$
  3. $\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{D}}>\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}}$
  4. $\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{D}} < \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}}$

Solution

$\therefore \quad$ The moment of inertia of $\mathrm{A}$ and $\mathrm{D}$ will be equal as the mass, distance, and position are similar. Similarly, the moment of inertia of B and C are equal. $\therefore \quad$ But the moment of inertia of A and D is greater than $\mathrm{B}$ and $\mathrm{C}$ as they are at the ends.

Asked in: MHT CET 2023 (13 May Shift 2)

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