PHYSICS

PHYSICS

A uniform field E = 5 × 10³N/C is along the x-axis. What is the net flux through a cube of side 30 cm with faces paralle

Given: A uniform field E = 5 × 10³N/C is along the x-axis. What is the net flux through a cube of side 30 cm with faces parallel to coordinate planes? Formula: Flux through face at x = 0 : phi = E × A = 5 × 10³ × (0.3)² = 450 Nm²/C (inward). Substitution & Calculation: Flux through face at x = 0.3 : 450 Nm²/C (outward). Net flux: 450 - 450 = 0 Nm²/C (no charge enclosed). 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

What is the mass defect of a nucleus with binding energy 102.3 MeV ? (Given 1 u = 931.5 MeV/c² )

Given: What is the mass defect of a nucleus with binding energy 102.3 MeV ? (Given 1 u = 931.5 MeV/c² ) Formula: Δ M = E_b/c². Substitution & Calculation: Δ M = 102.3/931.5 approx 0.11 u . 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

The de Broglie wavelength of a particle of mass 2.0 × 10⁻³⁰kg moving at 1.5 × 10⁶m/s is:

Given: The de Broglie wavelength of a particle of mass 2.0 × 10⁻³⁰kg moving at 1.5 × 10⁶m/s is: Formula: p = m v = 2.0 × 10⁻³⁰ × 1.5 × 10⁶= 3.0 × 10⁻²⁴kg m/s. Substitution & Calculation: lambda = h/p = frac6.63 × 10⁻³⁴³.0 × 10⁻²⁴= 2.21 × 10⁻¹⁰m = 0.221 nm . 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

What is the excess pressure inside a water drop of radius 2.5 mm at 100° C ? (Surface tension = 0.3 × 10⁻³N/m )

Given: What is the excess pressure inside a water drop of radius 2.5 mm at 100° C ? (Surface tension = 0.3 × 10⁻³N/m ) Formula: Excess pressure: Δ P = 2 S/r. Substitution & Calculation: S = 0.3 × 10⁻³N/m, r = 2.5 × 10⁻³m . Δ P = frac2 × 0.3 × 10⁻³².5 × 10⁻³= 0.24 Pa . 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 square loop of side 0.2 m with 30 turns carries 1.5 A in a magnetic field of 0.4 T . The plane of the loop is at 60° t

Given: A square loop of side 0.2 m with 30 turns carries 1.5 A in a magnetic field of 0.4 T . The plane of the loop is at 60° to the field. What is the torque? Formula: Torque tau = N I A B sin θ, where A = 0.2 × 0.2 = 0.04 m². Substitution & Calculation: tau = 30 × 1.5 × 0.04 × 0.4 × sin 60° = 1.8 × 0.4 × 0.866 = 0.6235 approx 0.62 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 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? Formula: Given: m = 2.0 A m². Substitution & Calculation: When cut transversely, each part has half the original magnetic moment. . Each part: m' = 2.0/2 = 1.0 A m² . 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), Chapter: Magnetism and Matter (Latest NCERT 2026-27), Topic: Magnetic dipole moment, bar magnet cut transversely, moment halves m' = m/2. The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI units and illustrative examples. Page.

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PHYSICS

A steel wire of length 2 m and cross-sectional area 2 × 10 ⁻⁶ m ² is stretched by a force of 200 N. If the Young's modul

Given: A steel wire of length 2 m and cross-sectional area 2 × 10 ⁻⁶ m ² is stretched by a force of 200 N. If the Young's modulus of steel is 2 × 10 ¹¹ N/m ², what is the elongation of the wire? Formula: Using Young's modulus formula: Y = (F L) / (A ΔL). Substitution & Calculation: Rearrange for elongation: ΔL = (F L) / (A Y). Substitute values: ΔL = (200 × 2) / (2 × 10 ⁻⁶ × 2 × 10 ¹¹ ). Calculate: ΔL = 400 / (4 × 10 ⁵ ) = 10 ⁻³ m = 1 mm. 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 material with susceptibility chi = 5 × 10⁻³has a relative permeability μ_r of:

Given: A material with susceptibility chi = 5 × 10⁻³has a relative permeability μ_r of: Formula: μ_r = 1 + chi. Substitution & Calculation: Given: chi = 5 × 10⁻³. Substitute: μ_r = 1 + 5 × 10⁻³= 1.005 . 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

Two lls of emf 4 V and 5 V with internal resistances 0.5 Ω and 1 Ω are connected in series with a 5.5 Ω resistor. What i

Given: Two lls of emf 4 V and 5 V with internal resistances 0.5 Ω and 1 Ω are connected in series with a 5.5 Ω resistor. What is the current through the circuit? Formula: Equivalent emf: ε_{eq = 4 + 5 = 9 V. Substitution & Calculation: Total resistance: R_{total = 0.5 + 1 + 5.5 = 7 Ω . Current: I = fracε_{eqR_{total = 9/7 approx 1.29 A . 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

The work function of a metal is 1.5 eV . Light of wavelength 400 nm is incident on it. What is the maximum kinetic energ

Given: The work function of a metal is 1.5 eV . Light of wavelength 400 nm is incident on it. What is the maximum kinetic energy of emitted electrons in eV? (Take h c = 1240 eV nm ) Formula: E = h c/lambda = 1240/400 = 3.1 eV. Substitution & Calculation: K_{max = E - phi_0 = 3.1 - 1.5 = 1.6 eV . 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 hydrogen atom is excited to the n = 4 state and returns to the ground state. What is the maximum energy of the emitted

Given: A hydrogen atom is excited to the n = 4 state and returns to the ground state. What is the maximum energy of the emitted photon? (Use E_n = -13.6/n² eV ) Formula: E_4 = -0.85 eV, E_1 = -13.6 eV. Substitution & Calculation: Δ E = -0.85 - (-13.6) = 12.75 eV . 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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