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Physics Question 28 – JEE-MAIN 2025

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :

Recall the moment of inertia of a circular ring about an axis passing through its center and perpendicular to its plane, and then about a diameter.

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Ninja StrategyCareful Variable Interpretation

The key to quickly eliminating options is to correctly interpret 'r' as the diameter, not the radius, and to distinguish between moment of inertia about a diameter versus a tangential axis.

Step 1: Moment of Inertia about a Diameter✦ Active

For a circular ring of mass M and radius R, the moment of inertia about an axis passing through its center and perpendicular to its plane is ICM,=MR2. By the perpendicular axis theorem, the moment of inertia about a diameter (an axis lying in the plane of the ring and passing through its center) is Idiameter=12MR2.

Step 2: Apply Parallel Axis Theorem○ Expand

The tangential axis lying in the plane of the ring is parallel to a diameter and is at a distance R from it. Using the parallel axis theorem, Itangential=Idiameter+MR2.

Itangential=12MR2+MR2=32MR2
💡 Teacher's Secret Hint

Remember that the distance 'd' in the parallel axis theorem is the radius 'R' in this case.

Step 3: Substitute Radius in terms of Diameter○ Expand

The problem states that the diameter is r. Therefore, the radius of the ring is R=r2. Substitute this into the expression for Itangential.

Itangential=32M(r2)2=32Mr24=38Mr2
💡 Teacher's Secret Hint

A common mistake is to assume 'r' is the radius. Always double-check the definition of variables in the problem statement.

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