Moment of Inertia – The moment of inertia: NEET MCQs with Answers

Explanation: Moment of inertia is smallest about the axis from which the mass distribution is least spread out. EG is the axis parallel to the longer sides and passing through the centre, so the material extends only over the shorter dimension on either side of it — the distances are smallest. By contrast, the diagonal BD and the axis HF have the mass distributed much farther out. Hence I is minimum about EG.

Explanation: The moment of inertia of a uniform triangular plate about one of its sides depends on the altitude drawn to that side: the larger the altitude, the farther the mass lies from the axis. With sides BC = 3, AB = 4 and CA = 5, the altitude on BC is 4, the altitude on AB is 3 and the altitude on CA is (3 × 4)/5 = 2.4. Hence I_BC > I_AB > I_CA, so I_BC > I_AC is the correct relation. The relation I_AB + I_BC = I_CA cannot hold, because the perpendicular-axis theorem applies only to an axis perpendicular to the plate, not to three axes lying in its plane; and I_CA is the smallest of the three, not the largest.

Explanation: For a plate of fixed mass, the moment of inertia about an axis grows with the average squared distance of the mass from that axis. The mass distribution lies farthest from the shortest side BC, so I₂ is the largest. The hypotenuse AC passes closest to the bulk of the material, so I₃ is the smallest. Hence I₂ > I₁ > I₃.

Explanation: I ∝ R² for equal mass. I₁/I₂ = (1/2)² = 1/4.

Explanation: About a tangent I = (7/5)MR² = MK² ⇒ K² = (7/5)R² = (7/5)(25x²/7) = 5x². With K = 5: 25 = 5x² ⇒ x = √5.

Explanation: Ring about its central normal axis: I_A = MR². Disc about a diameter: I_B = MR²/4. Hollow cylinder about its axis: I_C = MR². Solid cylinder about its axis: I_D = MR²/2. Hence I_A = I_C and 2I_B = I_D.

Explanation: Each rod makes 45° with the diagonal and its centre is at distance L/(2√2) from it. MI of a rod about an axis through its centre at 45°: (ML²/12)sin²45° = ML²/24; parallel shift adds M(L²/8). Each rod: ML²/24 + ML²/8 = ML²/6. Four rods: 2ML²/3.

Explanation: Shell with XX′ as diameter: (2/3)mr². Each shell touched tangentially: (2/3)mr² + mr² = (5/3)mr². Total = 2/3 + 10/3 = 4mr².

Explanation: Each rod rotates about a perpendicular axis through one end: ML²/3. Two rods: 2ML²/3.

Explanation: W = ½Iω², and for equal M and R the moments of inertia rank as I_ring (MR²) > I_disc (½MR²) > I_sphere ((2/5)MR²). Hence W_C > W_B > W_A.

Explanation: The removed disc has radius R/2 and mass M/4, and its centre is R/2 from the axis, so its contribution is ½(M/4)(R/2)² + (M/4)(R/2)² = 3MR²/32. Subtracting from ½MR² = 16MR²/32 gives 13MR²/32.

Explanation: About a diameter, I_d = MR²/4. A tangent in the plane of the disc is parallel to a diameter and a distance R from it, so by the parallel-axis theorem: I = MR²/4 + MR² = (5/4)MR².

Explanation: Moment of inertia depends on how the mass is distributed relative to the axis — mass placed farther out contributes far more, since the contribution goes as r². Iron is much denser than aluminium, so the iron should be placed as far from the axis as possible. Hence aluminium should form the interior and iron should surround it.

Explanation: Disc, tangent in its own plane: I_d = MR²/4 + MR² = (5/4)MR² ⟹ K_d = √(5/4) R. Ring, tangent in its own plane: I_r = MR²/2 + MR² = (3/2)MR² ⟹ K_r = √(3/2) R. K_d/K_r = √(5/4) / √(3/2) = √(5/4 × 2/3) = √(5/6) = √5 : √6.

Explanation: About the central normal axis, I_CM = ½MR². The new axis is parallel to it and a distance R away, so by the parallel-axis theorem: I = ½MR² + MR² = (3/2)MR².