11 public questions tagged with this topic.
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.
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.
For a uniform rod, the center of mass is at its geometric center. Length = 2m, so the center is at 1m from the origin along the x-axis. Position vector: R = 1i^+0j^+0k^ = 1i^m.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
τ = r×F = |i^j^k^250−3−40| = k^(2×(−4)−5×(−3)) = k^(−8+15) = 7k^Nm. Magnitude = 7Nm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7 Nm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
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.
τ = r×F = |i^j^k^120−230| = k^(1×3−2×(−2)) = k^(3+4) = 7k^Nm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7 Nm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
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.
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.
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.
Angular momentum: L = r×p, where p = mv = 1×(3i^+4j^). L = |i^j^k^020340| = k^(0×4−2×3) = −6k^kgm2/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 B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Torque τ = r×F is perpendicular to both the position vector r and force F, following the right-hand rule. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Perpendicular to both force and position vector. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
