Moment of Inertia & Radius of Gyration — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Moment of Inertia & Radius of Gyration MCQs with step-by-step solutions (22 questions). Part of Rotational Dynamics. Practise online on Prepizo — no login needed.
▶ Practise Moment of Inertia & Radius of Gyration online (free)
Questions with solutions
Q1 — Moment of Inertia & Radius of Gyration · easy · theory
The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is:
A. $MR^2$
B. $\dfrac{1}{2}MR^2$
C. $\dfrac{2}{3}MR^2$
D. $\dfrac{2}{5}MR^2$ ✓ Correct
Solution: For a uniform solid sphere rotating about any diameter, $I = \dfrac{2}{5}MR^2$.
Q2 — Moment of Inertia & Radius of Gyration · easy · theory
The moment of inertia of a solid cylinder of mass $M$ and radius $R$ about its own geometric axis is:
A. $MR^2$
B. $\dfrac{1}{2}MR^2$ ✓ Correct
C. $\dfrac{2}{3}MR^2$
D. $\dfrac{1}{4}MR^2$
Solution: A solid cylinder (or disc) about its central longitudinal axis has $I = \dfrac{1}{2}MR^2$, independent of its length.
Q3 — Moment of Inertia & Radius of Gyration · easy · theory
The moment of inertia of a thin circular ring of mass $M$ and radius $R$ about an axis perpendicular to its plane through its centre is:
A. $\dfrac{1}{2}MR^2$
B. $\dfrac{2}{5}MR^2$
C. $\dfrac{2}{3}MR^2$
D. $MR^2$ ✓ Correct
Solution: Every element of a thin ring lies at the same distance $R$ from that axis, so $I = \sum m_i R^2 = MR^2$.
Q4 — Moment of Inertia & Radius of Gyration · medium · theory
The moment of inertia of a thin hollow (spherical shell) sphere of mass $M$ and radius $R$ about its diameter is:
A. $\dfrac{2}{3}MR^2$ ✓ Correct
B. $\dfrac{2}{5}MR^2$
C. $\dfrac{1}{2}MR^2$
D. $MR^2$
Solution: For a thin spherical shell, $I = \dfrac{2}{3}MR^2$ — larger than the solid sphere's $\dfrac{2}{5}MR^2$ because all the mass sits at the outer radius.
Q5 — Moment of Inertia & Radius of Gyration · medium · numerical
A uniform rod of mass $4\text{ kg}$ and length $2\text{ m}$ rotates about an axis through its centre, perpendicular to its length. Its moment of inertia is:
A. $5.33\text{ kg}\cdot\text{m}^2$
B. $1.33\text{ kg}\cdot\text{m}^2$ ✓ Correct
C. $0.33\text{ kg}\cdot\text{m}^2$
D. $16\text{ kg}\cdot\text{m}^2$
Solution: $I = \dfrac{ML^2}{12} = \dfrac{4 \times 2^2}{12} = \dfrac{16}{12} \approx 1.33\text{ kg}\cdot\text{m}^2$.
Q6 — Moment of Inertia & Radius of Gyration · medium · theory
A uniform rod of mass $M$ and length $L$ is rotated about an axis perpendicular to its length passing through one end. Its moment of inertia is:
A. $\dfrac{2}{3}ML^2$
B. $\dfrac{1}{3}ML^2$ ✓ Correct
C. $\dfrac{1}{12}ML^2$
D. $\dfrac{1}{2}ML^2$
Solution: By the parallel axis theorem, $I = \dfrac{ML^2}{12} + M\left(\dfrac{L}{2}\right)^2 = \dfrac{ML^2}{12} + \dfrac{ML^2}{4} = \dfrac{ML^2}{3}$.
Q7 — Moment of Inertia & Radius of Gyration · easy · numerical
A disc of mass $10\text{ kg}$ and radius $0.5\text{ m}$ rotates about its own axis. Its moment of inertia is:
A. $5\text{ kg}\cdot\text{m}^2$
B. $2.5\text{ kg}\cdot\text{m}^2$
C. $0.625\text{ kg}\cdot\text{m}^2$
D. $1.25\text{ kg}\cdot\text{m}^2$ ✓ Correct
Solution: $I = \dfrac{1}{2}MR^2 = \dfrac{1}{2}(10)(0.5)^2 = 5 \times 0.25 = 1.25\text{ kg}\cdot\text{m}^2$.
Q8 — Moment of Inertia & Radius of Gyration · easy · numerical
Two discs have the same mass but the radius of the second is twice that of the first. The ratio of their moments of inertia about their own axes is:
A. $4 : 1$
B. $1 : 2$
C. $1 : 4$ ✓ Correct
D. $2 : 1$
Solution: $I = \dfrac{1}{2}MR^2 \propto R^2$ for equal masses. Hence $I_1 : I_2 = R^2 : (2R)^2 = 1 : 4$.
Q9 — Moment of Inertia & Radius of Gyration · easy · theory
The radius of gyration $k$ of a body of mass $M$ and moment of inertia $I$ is defined by the relation:
A. $I = Mk^2$ ✓ Correct
B. $I = Mk$
C. $k^2 = \dfrac{M}{I}$
D. $k = IM$
Solution: The radius of gyration is the distance from the axis at which the whole mass could be concentrated without changing the moment of inertia, i.e. $I = Mk^2$, so $k = \sqrt{I/M}$.
Q10 — Moment of Inertia & Radius of Gyration · medium · numerical
The radius of gyration of a uniform disc of radius $R$ about its own axis is:
A. $\dfrac{R}{2}$
B. $\dfrac{R}{\sqrt{2}}$ ✓ Correct
C. $\dfrac{R}{2\sqrt{2}}$
D. $R$
Solution: $I = \dfrac{1}{2}MR^2 = Mk^2 \Rightarrow k^2 = \dfrac{R^2}{2} \Rightarrow k = \dfrac{R}{\sqrt{2}}$.
Q11 — Moment of Inertia & Radius of Gyration · medium · numerical
The radius of gyration of a uniform disc of mass $M$ and radius $R$ about its transverse diameter is:
A. $\dfrac{R}{2}$ ✓ Correct
B. $\dfrac{R}{2\sqrt{2}}$
C. $R$
D. $\dfrac{R}{\sqrt{2}}$
Solution: About a diameter, $I = \dfrac{1}{4}MR^2 = Mk^2 \Rightarrow k = \dfrac{R}{2}$.
Q12 — Moment of Inertia & Radius of Gyration · easy · theory
The theorem of parallel axes relates the moment of inertia $I$ about an axis at distance $d$ from a parallel axis through the centre of mass by:
A. $I = I_{cm} + Md^2$ ✓ Correct
B. $I = Md^2$
C. $I = I_{cm} + Md$
D. $I = I_{cm} - Md^2$
Solution: The parallel axis theorem states $I = I_{cm} + Md^2$, valid for any rigid body and any pair of parallel axes.
Q13 — Moment of Inertia & Radius of Gyration · medium · theory
The theorem of perpendicular axes, $I_z = I_x + I_y$, is applicable only to:
A. Any three-dimensional rigid body
B. A plane lamina (two-dimensional body) ✓ Correct
C. A hollow cylinder only
D. A solid sphere
Solution: The perpendicular axes theorem holds only for a planar body, with $x$ and $y$ axes in the plane of the lamina and $z$ perpendicular to it through their intersection.
Q14 — Moment of Inertia & Radius of Gyration · hard · theory
The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about a tangent lying in its own plane is:
A. $\dfrac{3}{2}MR^2$
B. $\dfrac{5}{4}MR^2$ ✓ Correct
C. $\dfrac{1}{2}MR^2$
D. $\dfrac{1}{4}MR^2$
Solution: About a diameter $I_d = \dfrac{1}{4}MR^2$. Shifting to a parallel tangent at distance $R$: $I = \dfrac{1}{4}MR^2 + MR^2 = \dfrac{5}{4}MR^2$.
Q15 — Moment of Inertia & Radius of Gyration · hard · theory
The moment of inertia of a uniform disc of mass $M$ and radius $R$ about a tangent perpendicular to its plane is:
A. $\dfrac{5}{4}MR^2$
B. $2MR^2$
C. $\dfrac{1}{2}MR^2$
D. $\dfrac{3}{2}MR^2$ ✓ Correct
Solution: About the central perpendicular axis $I = \dfrac{1}{2}MR^2$. By the parallel axis theorem with $d = R$: $I = \dfrac{1}{2}MR^2 + MR^2 = \dfrac{3}{2}MR^2$.
Q16 — Moment of Inertia & Radius of Gyration · medium · theory
The moment of inertia of a thin ring of mass $M$ and radius $R$ about one of its diameters is:
A. $MR^2$
B. $\dfrac{1}{4}MR^2$
C. $\dfrac{1}{2}MR^2$ ✓ Correct
D. $\dfrac{3}{2}MR^2$
Solution: By the perpendicular axes theorem, $I_z = I_x + I_y$. For a ring $I_z = MR^2$ and by symmetry $I_x = I_y$, so each diameter gives $I = \dfrac{1}{2}MR^2$.
Q17 — Moment of Inertia & Radius of Gyration · easy · theory
The moment of inertia of a rigid body depends on:
A. The mass, its distribution, and the position of the axis of rotation ✓ Correct
B. Only the torque applied to the body
C. Only the angular velocity of the body
D. Only the total mass of the body
Solution: Since $I = \sum m_i r_i^2$, it depends on how much mass there is and how far that mass sits from the chosen axis. The same body has different $I$ about different axes.
Q18 — Moment of Inertia & Radius of Gyration · medium · numerical
The moment of inertia of a solid sphere of mass $5\text{ kg}$ and radius $0.2\text{ m}$ about its diameter is:
A. $0.02\text{ kg}\cdot\text{m}^2$
B. $0.2\text{ kg}\cdot\text{m}^2$
C. $0.4\text{ kg}\cdot\text{m}^2$
D. $0.08\text{ kg}\cdot\text{m}^2$ ✓ Correct
Solution: $I = \dfrac{2}{5}MR^2 = \dfrac{2}{5}(5)(0.2)^2 = 2 \times 0.04 = 0.08\text{ kg}\cdot\text{m}^2$.
Q19 — Moment of Inertia & Radius of Gyration · easy · numerical
A thin ring of mass $3\text{ kg}$ and radius $0.4\text{ m}$ rotates about an axis through its centre perpendicular to its plane. Its moment of inertia is:
A. $0.48\text{ kg}\cdot\text{m}^2$ ✓ Correct
B. $1.2\text{ kg}\cdot\text{m}^2$
C. $0.96\text{ kg}\cdot\text{m}^2$
D. $0.24\text{ kg}\cdot\text{m}^2$
Solution: $I = MR^2 = 3 \times (0.4)^2 = 3 \times 0.16 = 0.48\text{ kg}\cdot\text{m}^2$.
Q20 — Moment of Inertia & Radius of Gyration · medium · numerical
A uniform rod of mass $6\text{ kg}$ and length $3\text{ m}$ rotates about an axis through one end perpendicular to its length. Its moment of inertia is:
A. $18\text{ kg}\cdot\text{m}^2$ ✓ Correct
B. $4.5\text{ kg}\cdot\text{m}^2$
C. $9\text{ kg}\cdot\text{m}^2$
D. $54\text{ kg}\cdot\text{m}^2$
Solution: $I = \dfrac{ML^2}{3} = \dfrac{6 \times 9}{3} = 18\text{ kg}\cdot\text{m}^2$.
Q21 — Moment of Inertia & Radius of Gyration · hard · numerical
The radius of gyration of a solid sphere of radius $0.5\text{ m}$ about its diameter is approximately:
A. $0.316\text{ m}$ ✓ Correct
B. $0.2\text{ m}$
C. $0.354\text{ m}$
D. $0.5\text{ m}$
Solution: $I = \dfrac{2}{5}MR^2 = Mk^2 \Rightarrow k = R\sqrt{\dfrac{2}{5}} = 0.5 \times 0.632 \approx 0.316\text{ m}$.
Q22 — Moment of Inertia & Radius of Gyration · medium · numerical
Two point masses of $2\text{ kg}$ each are fixed at the ends of a light rod of length $2\text{ m}$. The moment of inertia about a perpendicular axis through the centre of the rod is:
A. $4\text{ kg}\cdot\text{m}^2$ ✓ Correct
B. $8\text{ kg}\cdot\text{m}^2$
C. $2\text{ kg}\cdot\text{m}^2$
D. $16\text{ kg}\cdot\text{m}^2$
Solution: Each mass sits $1\text{ m}$ from the axis, so $I = \sum mr^2 = 2(2)(1)^2 = 4\text{ kg}\cdot\text{m}^2$. The rod is light and contributes nothing.