#particles

Vector Product and Rotational Kinematics

Two particles of masses 3kg and 7kg are at (1,4) and (9,2) respectively. What is the distance of their center of mass fr

X = (3×1)+(7×9)3+7 = 3+6310 = 6.6. Y = (3×4)+(7×2)3+7 = 12+1410 = 2.6. Distance = (6.6)2+(2.6)2 = 43.56+6.76 = 50.32≈7.09m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.09 m. 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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Vector Product and Rotational Kinematics

For a body rotating about a fixed axis, what is true about particles on the axis?

Particles on the axis of rotation have a perpendicular distance r = 0 from the axis, so their linear velocity v = ωr = 0, meaning they remain stationary. As per NCERT, applying relevant law/formula with correct units and sign convention leads to They remain stationary. 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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Rotational Motion Applications

Three particles of masses 3kg, 6kg, and 9kg are located at (0,1), (2,2), and (4,0) respectively. What is the x-coordinat

Formula: X = m1x1+m2x2+m3x3m1+m2+m3. Masses: 3,6,9kg; x-coordinates: 0,2,4. X = (3×0)+(6×2)+(9×4)3+6+9 = 0+12+3618 = 4818 = 2.67m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.67 m. 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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Rotational Motion Applications

Two particles of masses 8kg and 2kg are at (0,5) and (10,0) respectively. What is the distance of their center of mass f

X = (8×0)+(2×10)8+2 = 2010 = 2. Y = (8×5)+(2×0)8+2 = 4010 = 4. Distance = (2)2+(4)2 = 4+16 = 20≈4.47m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.47 m. 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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Rotational Motion Applications

Three particles of masses 5kg, 10kg, and 15kg are at (0,0), (2,4), and (6,2) respectively. What is the x-coordinate of t

Formula: X = m1x1+m2x2+m3x3m1+m2+m3. Masses: 5,10,15kg; x-coordinates: 0,2,6. X = (5×0)+(10×2)+(15×6)5+10+15 = 0+20+9030 = 11030≈3.67m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.67 m. 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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Conservation of Angular Momentum

Two particles of masses 6kg and 4kg are at (2,3) and (8,7) respectively. What is the distance of their center of mass fr

X = (6×2)+(4×8)6+4 = 12+3210 = 4.4. Y = (6×3)+(4×7)6+4 = 18+2810 = 4.6. Distance = (4.4)2+(4.6)2 = 19.36+21.16 = 40.52≈6.36m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.36 m. 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

Four particles of masses 1kg, 3kg, 5kg, and 7kg are at (0,0), (2,0), (0,3), and (2,3) respectively. What is the y-coordi

Formula: Y = m1y1+m2y2+m3y3+m4y4m1+m2+m3+m4. Masses: 1,3,5,7kg; y-coordinates: 0,0,3,3. Y = (1×0)+(3×0)+(5×3)+(7×3)1+3+5+7 = 0+0+15+2116 = 3616 = 2.25m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.25 m. 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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