JEE Main 2025 · 22 January · morning · Single correct · +4 / −1

A uniform circular disc of radius ' R′ and mass ' M′ is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R / 2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.

JEE Main PYQ figure
  1. (a) 17/32 MR²
  2. (b) 13/32 MR²
  3. (c) 9/32 MR²
  4. (d) 7/32 MR²
Answer and solution

Answer (b) 13/32 MR²

Solution

Physics sir · 4 steps

Treat the hole as a negative disc: find its moment of inertia about O and subtract it from the full disc's.

  1. Mass of the hole. Mass is proportional to area. The hole has radius R/2, so its area is 1/4 of the full disc and its mass is m = M/4.
  2. Where the hole sits. In the figure the small disc runs from the centre O to the rim. Its diameter is R, so its centre is at d = R/2 from the axis.

    Iₕₒₗₑ = 1/2m(R/2)² + m(R/2)² = MR²/32 + 2MR²/32 = 3MR²/32

  3. Subtract. The full disc has I = 1/2MR² = 16/32MR². The remaining part has I = 16/32MR² - 3/32MR².
  4. So. The moment of inertia of the remaining part is 13/32MR², option (b).

The idea to keep. For a disc with a hole: full body minus the cut-out piece. Get the piece's mass from the area ratio, then shift its moment of inertia to the axis with the parallel axis theorem.

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

Study Rotational Motion

Asked about 180 times in JEE Main since 2020. This question is from 2025.