PHYSICS

PHYSICS

A material with susceptibility chi = 3 × 10⁻⁴ has a relative permeability μ_r of:

Given: A material with susceptibility chi = 3 × 10⁻⁴ has a relative permeability μ_r of: These values define the system as per NCERT data. Formula: μ_r = 1 + chi. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: chi = 3 × 10⁻⁴. Substitute: μ_r = 1 + 3 × 10⁻⁴= 1.0003 . 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 hydrogen atom absorbs a photon of energy 12.09 eV from the ground state. To which energy level does it jump? (Use E_n

Given: A hydrogen atom absorbs a photon of energy 12.09 eV from the ground state. To which energy level does it jump? (Use E_n = -13.6/n² eV ) These values define the system as per NCERT data. Formula: E_1 = -13.6 eV. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_n = -13.6 + 12.09 = -1.51 eV . -1.51 = -13.6/n² Rightarrow n² = 9 Rightarrow n = 3 . 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

The dimensional formula of momentum is [M L T^{-1] . What is the dimensional formula of impulse?

Given: The dimensional formula of momentum is [M L T^{-1] . What is the dimensional formula of impulse? These values define the system as per NCERT data. Formula: Impulse = Force × Time. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: [F] = [M L T^{-2], [t] = [T] . [Impulse] = [M L T^{-2] [T] = [M L T^{-1], same as momentum. 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 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

An electron moves at 8 × 10⁶ m/s perpendicular to a field of 0.15 T . What is the radius of its path? (Mass = 9.1 ×

Given: An electron moves at 8 × 10⁶ m/s perpendicular to a field of 0.15 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹ kg, charge = 1.6 × 10⁻¹⁹ C ) These values define the system as per NCERT data. Formula: r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 8 × 10⁶¹.6 × 10⁻¹⁹ × 0.15 = frac7.28 × 10⁻²⁴².4 × 10⁻²⁰= 3.033 × 10⁻⁴ m approx 0.0303 cm . 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 wheel with 6 spokes of 0.3 m each rotates at 40 rpm in a 0.6 T field. What is the induced emf?

Given: A wheel with 6 spokes of 0.3 m each rotates at 40 rpm in a 0.6 T field. What is the induced emf? These values define the system as per NCERT data. Formula: omega = 2π × 40/60 = 4π/3 rad/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 1/2 B omega R² = 1/2 × 0.6 × 4π/3 × (0.3)² = 0.07536 V approx 0.075 V . 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: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

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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

The magnetic potential energy of a dipole with m = 0.3 A m² in a field B = 0.6 T at 45° is:

Given: The magnetic potential energy of a dipole with m = 0.3 A m² in a field B = 0.6 T at 45° is: These values define the system as per NCERT data. Formula: U_m = -m B cosθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.3 A m², B = 0.6 T, θ = 45°, cos 45° = frac1sqrt2 approx 0.707 . Substitute: U_m = -0.3 × 0.6 × 0.707 approx -0.12726 J approx -0.13 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: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

A magnetic dipole of moment 0.6 A m² is in a uniform field of 0.4 T at 60° . What is the torque on it?

Given: A magnetic dipole of moment 0.6 A m² is in a uniform field of 0.4 T at 60° . What is the torque on it? These values define the system as per NCERT data. Formula: Torque is tau = m B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.6 A m², B = 0.4 T, θ = 60°, sin 60° = fracsqrt32 approx 0.866 . Substitute: tau = 0.6 × 0.4 × 0.866 approx 0.20784 N m approx 0.21 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: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

Two lls in parallel have emf 10 V and 5 V with internal resistances 2 Ω and 1 Ω . What is the equivalent internal resi

Given: Two lls in parallel have emf 10 V and 5 V with internal resistances 2 Ω and 1 Ω . What is the equivalent internal resistance? These values define the system as per NCERT data. Formula: For parallel: frac1r_{eq = 1/r_1 + 1/r_2 = 1/2 + 1/1 = 1 + 2/2 = 3/2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r_{eq = 2/3 approx 0.67 Ω . 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: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

A string vibrates with a stationary wave y = 0.1 sin (2Ï€ x/3) cos (150Ï€ t) . What is the distance between a node and t

Given: A string vibrates with a stationary wave y = 0.1 sin (2π x/3) cos (150π t) . What is the distance between a node and the next antinode? These values define the system as per NCERT data. Formula: k = 2π/3, lambda = 2π/k = 3 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Distance between node and antinode: lambda/4 = 3/4 = 0.75 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: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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PHYSICS

A bar magnet with original m = 2.0 A m² is cut transversely into two equal parts. What is m of each part?

Given: A bar magnet with original m = 2.0 A m² is cut transversely into two equal parts. What is m of each part? These values define the system as per NCERT data. Formula: Given: m = 2.0 A m². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: When cut transversely, each part has half the original magnetic moment. . Each part: m' = 2.0/2 = 1.0 A m² . 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: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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