#position

Moment of Inertia and Radius of Gyration

A force F\=7i^+3j^N acts at r\=−2i^+1j^m. What is the magnitude of the torque about the origin?

τ = r×F = |i^j^k^−210730| = k^((−2)×3−1×7) = k^(−6−7) = −13k^Nm. Magnitude = 13Nm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 13 Nm. 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 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.

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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.

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Torque and Angular Momentum

A 5kg particle moves with velocity v\=4i^m/s at r\=−3j^m. What is the magnitude of its angular momentum about the origin

L = r×p = |i^j^k^0−30400| = k^(0×0−(−3)×4) = 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 B is scientifically correct.

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

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Torque and Angular Momentum

A force F\=4i^−3j^N acts at r\=2i^+1j^m. What is the magnitude of the torque about the origin?

τ = r×F = |i^j^k^2104−30| = k^(2×(−3)−1×4) = k^(−6−4) = −10k^Nm. Magnitude = 10Nm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 Nm. 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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