#Mechanics

PHYSICS

A uniform square plate of side 6 m and mass 9 kg has one corner at (2, 2) along the x- and y-axes. What is the position

Given: A uniform square plate of side 6 m and mass 9 kg has one corner at (2, 2) along the x- and y-axes. What is the position of its nter of mass? These values define the system as per NCERT data. Formula: CM: X = 2 + 6/2 = 5, Y = 2 + 6/2 = 5. This is standard NCERT relation. Substitution & Calculation: For a uniform square, the nter of mass is at the ntroid. Vertices: (2, 2), (8, 2), (2, 8), (8, 8) . . Position: (5, 5) 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

A 3 kg mass moving at 12 m/s collides elastically with an identical stationary mass. What is the speed of the first mass

Given: A 3 kg mass moving at 12 m/s collides elastically with an identical stationary mass. What is the speed of the first mass after collision? These values define the system as per NCERT data. Formula: For equal masses in elastic collision, v_{1f = 0 m/s (first mass stops).. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. 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 soap film supports 2.0 × 10⁻² N over a 40 cm slider. What is the surface tension?

Given: A soap film supports 2.0 × 10⁻² N over a 40 cm slider. What is the surface tension? These values define the system as per NCERT data. Formula: S = F/2 l. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: F = 2.0 × 10⁻² N, l = 0.4 m . S = frac2.0 × 10⁻²² × 0.4 = 0.025 N/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 soap film supports 3.0 × 10⁻²N over a 50 cm slider. What is the surface tension?

Given: A soap film supports 3.0 × 10⁻²N over a 50 cm slider. What is the surface tension? Formula: S = F/2 l. Substitution & Calculation: F = 3.0 × 10⁻²N, l = 0.5 m . S = frac3.0 × 10⁻²² × 0.5 = 0.03 N/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

A spring of k = 250 N/m has a 2.5 kg mass. If E = 1.25 J, what is the amplitude?

Given: A spring of k = 250 N/m has a 2.5 kg mass. If E = 1.25 J, what is the amplitude? Formula: Total energy: E = 1/2 k A². Substitution & Calculation: 1.25 = 0.5 × 250 × A² Rightarrow 1.25 = 125 A² Rightarrow A² = 0.01 Rightarrow A = 0.1 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

A spring system has m = 0.1 kg, k = 40 N/m, A = 10 cm . What is the kinetic energy at x = 5 cm ?

Given: A spring system has m = 0.1 kg, k = 40 N/m, A = 10 cm . What is the kinetic energy at x = 5 cm ? These values define the system as per NCERT data. Formula: Total energy: E = 1/2 k A² = 0.5 × 40 × (0.1)² = 0.2 J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Potential energy: U = 1/2 k x² = 0.5 × 40 × (0.05)² = 0.05 J . Kinetic energy: K = E - U = 0.2 - 0.05 = 0.15 J . 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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

A dipole with m = 0.7 A m² in B = 0.6 T at 30° has torque:

Given: A dipole with m = 0.7 A m² in B = 0.6 T at 30° has torque: Formula: tau = m B sinθ. Substitution & Calculation: Given: m = 0.7 A m², B = 0.6 T, θ = 30°, sin 30° = 0.5 . tau = 0.7 × 0.6 × 0.5 = 0.21 N 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

A spring-mass system has m = 2.25 kg, k = 900 N/m . What is its period?

Given: A spring-mass system has m = 2.25 kg, k = 900 N/m . What is its period? Formula: Period: T = 2π √m/k = 2π √2.25/900 = 2π √0.0025 = 2π × 0.05 = 0.314 s .. Substitution: Substituting given values into formula as per NCERT 2026-27 method. Calculation: Simplifying step by step with proper SI units like m/s², J kg⁻¹ K⁻¹, 10⁻⁵, A m⁻¹ etc. 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

A neutron ( 1 u ) at 10⁶m/s collides elastically with a carbon ( 12 u ). What fraction of its kinetic energy is transfer

Given: A neutron ( 1 u ) at 10⁶m/s collides elastically with a carbon ( 12 u ). What fraction of its kinetic energy is transferred? Formula: Fraction transferred f_2 = 4 m_1 m_2/(m_1 + m_2)² = 4 × 1 × 12/(1 + 12)² = 48/169 approx 0.284 .. Substitution: Substituting given values into formula as per NCERT 2026-27 method. Calculation: Simplifying step by step with proper SI units like m/s², J kg⁻¹ K⁻¹, 10⁻⁵, A m⁻¹ etc. 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

A block of mass 1 kg oscillates with a spring of k = 100 N/m . If the amplitude is 10 cm, what is the total energy?

Given: A block of mass 1 kg oscillates with a spring of k = 100 N/m . If the amplitude is 10 cm, what is the total energy? These values define the system as per NCERT data. Formula: Total energy: E = 1/2 k A². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: A = 0.1 m, k = 100 N/m . E = 1/2 × 100 × (0.1)² = 0.5 J . 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: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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PHYSICS

A spring system has m = 3 kg, k = 1200 N/m, A = 5 cm . What is the potential energy at x = 2.5 cm ?

Given: A spring system has m = 3 kg, k = 1200 N/m, A = 5 cm . What is the potential energy at x = 2.5 cm ? These values define the system as per NCERT data. Formula: Potential energy: U = 1/2 k x². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k = 1200 N/m, x = 0.025 m . U = 0.5 × 1200 × (0.025)² = 0.5 × 1200 × 0.000625 = 0.375 J . 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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

A steel rod of radius 0.005 m and length 1 m is compressed by a force producing a stress of 2 × 10 ⁷ N/m ² . If the

Given: A steel rod of radius 0.005 m and length 1 m is compressed by a force producing a stress of 2 × 10 ⁷ N/m ² . If the Young's modulus of steel is 2 × 10 ¹¹ N/m ², what is the strain? These values define the system as per NCERT data. Formula: Young's modulus: Y = Stress / Strain. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Strain: Strain = Stress / Y = (2 × 10 ⁷ ) / (2 × 10 ¹¹ ) = 1 × 10 ⁻⁴ . 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: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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