#angular momentum

PHYSICS

A 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the n

Given: A 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the nter? These values define the system as per NCERT data. Formula: L = m v r. This is standard NCERT relation. Substitution & Calculation: m = 4 kg, v = 2 m/s, r = 0.25 m . L = 4 × 2 × 0.25 = 2 kg m²/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: System of Particles and Rotational Motion, Topic: Angular momentum L = m v r, rotating mass and moment of momentum. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

View Details

PHYSICS

What is the effect of increasing the mass of a rotating body on its angular momentum, if angular velocity remains consta

Angular momentum L = I omega, and I increases with mass. If omega is constant, L increases proportionally with mass. This follows from latest NCERT 2026-27 principle explaining the concept clearly for NEET students in simple steps as per rationalized syllabus.

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.

View Details

Moment of Inertia and Radius of Gyration

A 6kg particle moves with velocity v\=3j^m/s at r\=−4i^m. What is the magnitude of its angular momentum about the origin

L = r×p = |i^j^k^−400030| = k^((−4)×3−0×0) = −12k^kg m2/s. Magnitude = 12kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 12 kg m²/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

View Details

Moment of Inertia and Radius of Gyration

A 4kg particle moves with velocity v\=3i^+5j^m/s at r\=−2i^m. What is the z-component of its angular momentum?

L = r×p = |i^j^k^−200350| = k^((−2)×5−0×3) = −10k^kg m2/s. Z-component = −10kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -10 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.

View Details

Moment of Inertia and Radius of Gyration

A 3kg particle moves with velocity v\=6i^−4j^m/s at r\=2j^m. What is the z-component of its angular momentum?

L = r×p = |i^j^k^0206−40| = k^(0×(−4)−2×6) = −12k^kg m2/s. Z-component = −12kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -12 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.

View Details

Moment of Inertia and Radius of Gyration

A 3kg particle has a velocity v\=2i^m/s at position r\=4j^m. What is the magnitude of its angular momentum about the ori

L = r×p = r×(mv) = |i^j^k^040200| = −8k^kg m2/s. Magnitude = 8kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8 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.

View Details

Moment of Inertia and Radius of Gyration

A solid sphere of mass 6kg and radius 0.4m has an angular momentum of 11.52kg m2/s. What is its angular velocity?

I = 25MR2 = 25×6×(0.4)2 = 0.384kg m2. ω = LI = 11.520.384 = 30rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 30 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.

View Details

Vector Product and Rotational Kinematics

A 3kg particle moves with velocity v\=6i^m/s at r\=2j^m. What is the magnitude of its angular momentum about the origin?

L = r×p = |i^j^k^020600| = k^(0×0−2×6) = −12k^kg m2/s. Magnitude = 12kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 12 kg m²/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

View Details

Vector Product and Rotational Kinematics

A 4kg particle moves with velocity v\=−5i^+2j^m/s at r\=3i^m. What is the z-component of its angular momentum?

L = r×p = |i^j^k^300−520| = k^(3×2−0×(−5)) = 6k^kg m2/s. Z-component = 6kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6 kg m²/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

View Details

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.

View Details

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.

View Details

Vector Product and Rotational Kinematics

A solid sphere of mass 5kg and radius 0.3m rotates about an axis through its center. If its angular momentum is 9kg m2/s

L = Iω. For a sphere: I = 25MR2 = 25×5×(0.3)2 = 0.18kg m2. ω = LI = 90.18 = 50rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 50 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.

View Details