#moment of inertia

PHYSICS

A uniform rod of mass 5 kg and length 2 m is pivoted at its nter. What is its moment of inertia about the pivot?

Given: A uniform rod of mass 5 kg and length 2 m is pivoted at its nter. What is its moment of inertia about the pivot? Formula: For a rod pivoted at its nter: I = 1/12 M L². Substitution & Calculation: M = 5 kg, L = 2 m . I = 1/12 × 5 × (2)² = 20/12 = 5/3 approx 1.67 kg m² . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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Moment of Inertia and Radius of Gyration

What is the effect of doubling the radius of a rotating disk on its moment of inertia, assuming mass remains constant?

For a disk, I = 12MR2. Doubling R increases R2 by a factor of 4, so I increases by a factor of 4 if M is constant. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It quadruples. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A wheel with moment of inertia 8kg m2 rotates at 2rad/s. A torque of 16Nm acts for 3s. What is its final angular velocit

α = τI = 168 = 2rad/s2. Δω = αt = 2×3 = 6rad/s. ω = ω0+Δω = 2+6 = 8rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A solid sphere of mass 2kg and radius 0.6m rotates about its center at 5rad/s. What is its rotational kinetic energy?

I = 25MR2 = 25×2×(0.6)2 = 0.288kg m2. K = 12Iω2 = 12×0.288×(5)2 = 0.144×25 = 3.6J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.6 J. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A wheel with moment of inertia 5kg m2 rotates at 3rad/s. A torque of 10Nm acts for 5s. What is its final angular velocit

α = τI = 105 = 2rad/s2. Δω = αt = 2×5 = 10rad/s. ω = ω0+Δω = 3+10 = 13rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 13 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A torque of 25Nm acts on a wheel with moment of inertia 5kg m2 starting from rest. What is its angular speed after 2s?

α = τI = 255 = 5rad/s2. ω = ω0+αt = 0+5×2 = 10rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 rad/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A thin ring of mass 1kg and radius 0.4m has a kinetic energy of 8J. What is its angular speed?

I = MR2 = 1×(0.4)2 = 0.16kg m2. K = 12Iω2⇒8 = 12×0.16×ω2⇒16 = 0.16ω2⇒ω2 = 100⇒ω = 10rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A uniform rod of mass 5kg and length 2m is pivoted at its center. What is its moment of inertia about the pivot?

For a rod pivoted at its center: I = 112ML2. M = 5kg, L = 2m. I = 112×5×(2)2 = 2012 = 53≈1.67kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.67 kg m². This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A uniform rod of mass 2kg and length 1m is pivoted at one end. What is its moment of inertia about the pivot?

Moment of inertia of a rod about an end: I = 13ML2. M = 2kg, L = 1m. I = 13×2×(1)2 = 23≈0.67kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.67 kg m². This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Vector Product and Rotational Kinematics

A wheel with moment of inertia 6kg m2 rotates at 3rad/s. A torque of 12Nm acts for 3s. What is its final angular velocit

α = τI = 126 = 2rad/s2. Δω = αt = 2×3 = 6rad/s. ω = ω0+Δω = 3+6 = 9rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Vector Product and Rotational Kinematics

A solid sphere of mass 5kg and radius 0.2m has an angular momentum of 8kg m2/s. What is its angular velocity?

I = 25MR2 = 25×5×(0.2)2 = 0.08kg m2. ω = LI = 80.08 = 100rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 100 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Vector Product and Rotational Kinematics

A solid cylinder of mass 4kg and radius 0.2m rotates at 25rad/s. What is its angular momentum about its axis?

I = 12MR2 = 12×4×(0.2)2 = 0.08kg m2. L = Iω = 0.08×25 = 2kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 kg m²/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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