#force vector

Conservation of Angular Momentum

A force F\=−2i^+5j^N acts at r\=4i^−3j^m. What is the magnitude of the torque about the origin?

τ = r×F = |i^j^k^4−30−250| = k^(4×5−(−3)×(−2)) = k^(20−6) = 14k^Nm. Magnitude = 14Nm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 14 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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Conservation of Angular Momentum

A force F\=5i^+4j^N acts at r\=−3i^−2j^m. What is the magnitude of the torque about the origin?

τ = r×F = |i^j^k^−3−20540| = k^((−3)×4−(−2)×5) = k^(−12+10) = −2k^Nm. Magnitude = 2Nm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2 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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Center of Mass and System of Particles

A force F\=4i^−6j^N acts at r\=5i^+2j^m. What is the magnitude of the torque about the origin?

τ = r×F = |i^j^k^5204−60| = k^(5×(−6)−2×4) = k^(−30−8) = −38k^Nm. Magnitude = 38Nm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 38 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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Torque and Angular Momentum

A force F\=5i^+2j^N acts at r\=−1i^+3j^m. What is the torque about the origin?

τ = r×F = |i^j^k^−130520| = k^((−1)×2−3×5) = k^(−2−15) = −17k^Nm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -17 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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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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Energy Conservation and Friction

A force F\=4i^+6j^N acts on a particle moving along d\=−3i^+2j^m. What is the work done?

Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives 0 J. Work-energy theorem W=ΔK confirms. So option B is correct.

Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Energy Conservation and Friction

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