19 public questions tagged with this topic.
a×b = |i^j^k^630−420| = k^(6×2−3×(−4)) = k^(12+12) = 24k^. Magnitude = 24. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 24. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
The equation v = ω×r gives the linear velocity of a particle in rotational motion, where ω is the angular velocity and r is the position vector from the axis.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
a×b = |i^j^k^−430250| = k^((−4)×5−3×2) = k^(−20−6) = −26k^. Magnitude = 26. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 26. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
a×b = |i^j^k^8−30270| = k^(8×7−(−3)×2) = k^(56+6) = 62k^. Magnitude = 62. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 62. 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^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.
a×b = |i^j^k^720−340| = k^(7×4−2×(−3)) = k^(28+6) = 34k^. Magnitude = 34. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 34. 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^−300050| = k^((−3)×5−0×0) = −15k^kg m2/s. Magnitude = 15kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 15 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.
a×b = |i^j^k^3−50240| = k^(3×4−(−5)×2) = k^(12+10) = 22k^. Magnitude = 22. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 22. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
a×b = |i^j^k^4−20350| = k^(4×5−(−2)×3) = k^(20+6) = 26k^. Direction is along the positive z-axis (k^). As per NCERT, applying relevant law/formula with correct units and sign convention leads to k^. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
The vector product (cross product) of two vectors results in a vector perpendicular to the plane of the two vectors, with magnitude absinθ, and is not commutative (a×b = −b×a).
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
a×b = |i^j^k^5−40360| = k^(5×6−(−4)×3) = k^(30+12) = 42k^. Magnitude = 42. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 42. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
