MHT CET · Physics · Rotational Motion
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
- A \(\mathrm{I}_{\mathrm{A}}>\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{C}}>\mathrm{I}_{\mathrm{D}}\)
- B \(\mathrm{I}_{\mathrm{D}}>\mathrm{I}_{\mathrm{C}}>\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{A}}\)
- C \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{D}}>\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}}\)
- D \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{D}} < \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{D}}>\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}}\)
Step-by-step Solution
Detailed explanation
\(\therefore \quad\) 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.
\(\therefore \quad\) But the moment of inertia of A and D is greater than \(\mathrm{B}\) and \(\mathrm{C}\) as they are at the ends.
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