50 questions on this topic of Rotational Motion (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
Free with a Google sign-in; NEET Daily needs no sign-in at all.
Answer: (A) (7/5)MR²
For a solid sphere, the moment of inertia about an axis passing through its center of mass is I_cm = (2/5)MR². Applying the theorem of parallel axes with distance d = R, I = I_cm + MR² = (2/5)MR² + MR² = (7/5)MR².
Answer: (A) (7/48) ML²
The centre of mass of the rod is at x = L/2. The axis passes through x = L/4, so the distance between the centre of mass and the given parallel axis is d = |L/2 - L/4| = L/4. By the parallel axis theorem, I = I_cm + Md² = (1/12)ML² + M(L/4)² = ML²(1/12 + 1/16) = ML²(4 + 3)/48 = (7/48)ML².
Answer: (A) (1/12)ML²
The bent rod consists of two arms, each of mass m = M/2 and length l = L/2. The moment of inertia of each arm about an axis through its end perpendicular to its length is (1/3)ml² = (1/3)(M/2)(L/2)² = (1/24)ML². The total moment of inertia about the vertex is 2 × (1/24)ML² = (1/12)ML².
Answer: (A) I₁ = I₂ = I₃
For a uniform thin square plate lying in the xy-plane, any axis passing through its center in the plane of the plate has the same moment of inertia by symmetry and the perpendicular axis theorem: I_x + I_y = I_z. Since both diagonals are mutually perpendicular and symmetric, I₁ + I₂ = I_z ⇒ 2 I₁ = I_z ⇒ I₁ = I₂ = I_z/2. Similarly, the axes through opposite midpoints are mutually perpendicular, giving I₃ = I_z/2. Therefore, I₁ = I₂ = I₃ = (1/12) M L².
Answer: (B) (5/2) I₀
The moment of inertia about the central perpendicular axis is I₀ = (1/2)MR², which means MR² = 2I₀. By perpendicular axis theorem, the moment of inertia about any diameter is I_dia = I₀/2 = (1/4)MR². Applying the parallel axis theorem for a tangent in the plane, I_tangent = I_dia + MR² = (1/4)MR² + MR² = (5/4)MR² = (5/4)(2I₀) = (5/2)I₀.
Answer: (B) (3/2)MR²
About a diameter, I = MR²/2 (perpendicular axis theorem from MR² about the central axis). By the parallel axis theorem, about a tangent in its plane, I = MR²/2 + MR² = (3/2)MR².
NCERT Class 11
Answer: (C) 2 M R²
The moment of inertia of a ring about an axis through its center and perpendicular to its plane is I_cm = M R². By the parallel axis theorem, the moment of inertia about a parallel axis passing through a point on its rim at distance d = R is I = I_cm + M R² = M R² + M R² = 2 M R².
Answer: (A) (6/5) I
The moment of inertia of a disc about its diameter is I_d = (1/4)MR². By the parallel axis theorem, about a tangent in its plane, I = I_d + MR² = (5/4)MR². About a tangent perpendicular to its plane, I' = I_cm + MR² = (1/2)MR² + MR² = (3/2)MR². Thus, I'/I = (3/2)/(5/4) = 6/5, giving I' = (6/5)I.
Answer: (A) Both Statement I and Statement II are correct
The parallel axis theorem holds for any rigid body of arbitrary 3D geometry provided one axis passes through the center of mass. The perpendicular axis theorem is derived specifically for thin, flat lamina where the z-coordinates of all mass elements are zero.
Answer: (B) Statement I is correct but Statement II is incorrect
Statement I is correct because the perpendicular axis theorem (I_z = I_x + I_y) holds for any two-dimensional laminar body lying in the xy-plane regardless of its mass distribution. Statement II is incorrect because the theorem of parallel axes strictly requires one of the parallel axes to pass through the centre of mass of the body (I = I_cm + Md²).
Answer: (D) (3/2) M R²
The y-axis is a diameter of the ring, so I_y = (1/2)MR². The line passing through (R, 0, 0) and parallel to the y-axis is parallel to the diameter and separated by a perpendicular distance d = R. By the parallel axis theorem, I = I_y + Md² = (1/2)MR² + MR² = (3/2)MR².
Answer: (D) M [(R²/4) + (L²/3)]
About a central transverse axis passing through the center of mass, I_cm = M (R²/4 + L²/12). The distance from the center of mass to the circular end face along the axis of symmetry is d = L/2. By the parallel axis theorem, I = I_cm + M (L/2)² = M (R²/4 + L²/12 + L²/4) = M [R²/4 + L²/3].