8 public questions tagged with this topic.
For a rod pivoted at its center: I = 112ML2. M = 5kg, L = 2m. I = 112×5×(2)2 = 2012 = 53≈1.67kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.67 kg m². This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Moment of inertia of a rod about an end: I = 13ML2. M = 2kg, L = 1m. I = 13×2×(1)2 = 23≈0.67kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.67 kg m². This satisfies dimensional consistency and physical conditions given, so option C 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.
For a rod pivoted at one end: I = 13ML2. M = 6kg, L = 3m. I = 13×6×(3)2 = 2×9 = 18kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 18 kg m². This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
For a rod pivoted at its center: I = 112ML2. M = 5kg, L = 1.2m. I = 112×5×(1.2)2 = 5×1.4412 = 0.6kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.6 kg m². 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 rod pivoted at one end: I = 13ML2. M = 3kg, L = 1.5m. I = 13×3×(1.5)2 = 1×2.25 = 2.25kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.25 kg m². 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 rod pivoted at its center: I = 112ML2. M = 4kg, L = 2.5m. I = 112×4×(2.5)2 = 4×6.2512 = 2512≈2.08kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.08 kg m². 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 rod pivoted at one end: I = 13ML2. M = 7kg, L = 1.5m. I = 13×7×(1.5)2 = 7×2.253 = 5.25kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.25 kg m². This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
