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#center of mass

43 public questions tagged with this topic.

PHYSICS

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

Given: Three particles of masses 2 kg, 4 kg, and 6 kg are at (0, 2), (3, 0), and (1, 4) respectively. What is the x-coordinate of their nter of mass? Formula: Formula: X = m_1 x_1 + m_2 x_2 + m_3 x_3/m_1 + m_2 + m_3. Substitution & Calculation: Masses: 2, 4, 6 kg ; x-coordinates: 0, 3, 1 . X = (2 × 0) + (4 × 3) + (6 × 1)/2 + 4 + 6 = 0 + 12 + 6/12 = 18/12 = 1.5 m . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

Four particles of masses 1 kg, 2 kg, 3 kg, and 4 kg are placed at (0, 0), (3, 0), (0, 4), and (3, 4) respectively. What

Given: Four particles of masses 1 kg, 2 kg, 3 kg, and 4 kg are placed at (0, 0), (3, 0), (0, 4), and (3, 4) respectively. What is the x-coordinate of their nter of mass? These values define the system as per NCERT data. Formula: Formula: X = m_1 x_1 + m_2 x_2 + m_3 x_3 + m_4 x_4/m_1 + m_2 + m_3 + m_4. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Masses: 1, 2, 3, 4 kg ; x-coordinates: 0, 3, 0, 3 . X = (1 × 0) + (2 × 3) + (3 × 0) + (4 × 3)/1 + 2 + 3 + 4 = 0 + 6 + 0 + 12/10 = 18/10 = 1.8 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A system of two particles, 3 kg at (1, 2) and 1 kg at (5, 6), moves as a rigid body. What is the distance of the nter of

Given: A system of two particles, 3 kg at (1, 2) and 1 kg at (5, 6), moves as a rigid body. What is the distance of the nter of mass from the origin? These values define the system as per NCERT data. Formula: X = (3 × 1) + (1 × 5)/3 + 1 = 3 + 5/4 = 2. This is standard NCERT relation. Substitution & Calculation: Y = (3 × 2) + (1 × 6)/3 + 1 = 6 + 6/4 = 3 . Distance = sqrtX² + Y² = sqrt2² + 3² = sqrt4 + 9 = sqrt13 approx 3.6 m . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

Two particles of masses 4 kg and 6 kg are at (0, 7) and (8, 3) respectively. What is the distance of their nter of mass

Given: Two particles of masses 4 kg and 6 kg are at (0, 7) and (8, 3) respectively. What is the distance of their nter of mass from the origin? These values define the system as per NCERT data. Formula: X = (4 × 0) + (6 × 8)/4 + 6 = 48/10 = 4.8. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Y = (4 × 7) + (6 × 3)/4 + 6 = 28 + 18/10 = 4.6 . Distance = sqrt(4.8)² + (4.6)² = sqrt23.04 + 21.16 = sqrt44.2 approx 6.65 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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Moment of Inertia and Radius of Gyration

A uniform square lamina of side 4m and mass 8kg has one corner at the origin and lies along the x- and y-axes. What is t

For a uniform square, the center of mass is at the centroid. Vertices: (0,0),(4,0),(0,4),(4,4). CM: X = 0+42 = 2, Y = 0+42 = 2. Position: (2,2)m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to (2,2). 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

A 7kg object moves with a velocity of 3i^−4j^m/s. What is the magnitude of the velocity of its center of mass?

For a single object, the center of mass velocity equals the object’s velocity. Magnitude = (3)2+(−4)2 = 9+16 = 25 = 5m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.0 m/s. 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

A uniform triangular lamina has vertices at (0,0), (5,0), and (0,6). What is the x-coordinate of its center of mass?

For a uniform triangular lamina, the center of mass is at the centroid, the average of the vertices. X = 0+5+03 = 53≈1.67m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.67 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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Vector Product and Rotational Kinematics

A 4kg block moves with a velocity of −2i^+3j^m/s. What is the magnitude of the velocity of its center of mass?

For a single object, the center of mass velocity equals the object’s velocity. Magnitude = (−2)2+(3)2 = 4+9 = 13≈3.6m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.6 m/s. 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

A uniform rectangular plate of dimensions 3m×5m and mass 12kg has one corner at (1,1) along the x- and y-axes. What is t

For a uniform rectangle, the center of mass is at the centroid. Vertices: (1,1),(4,1),(1,6),(4,6). CM: X = 1+32 = 2.5, Y = 1+52 = 3.5. Position: (2.5,3.5)m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to (2.5,3.5). 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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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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Rotational Motion Applications

Three particles of masses 1kg, 5kg, and 10kg are at (0,2), (3,0), and (4,6) respectively. What is the x-coordinate of th

Formula: X = m1x1+m2x2+m3x3m1+m2+m3. Masses: 1,5,10kg; x-coordinates: 0,3,4. X = (1×0)+(5×3)+(10×4)1+5+10 = 0+15+4016 = 5516≈3.44m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.44 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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