PHYSICS

PHYSICS

A rod of length 0.25 m moves at 3 m/s in a 0.4 T field perpendicular to its length. What is the induced emf?

Given: A rod of length 0.25 m moves at 3 m/s in a 0.4 T field perpendicular to its length. What is the induced emf? Formula: ε = B l v = 0.4 × 0.25 × 3 = 0.3 V .. 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 square loop of side 0.18 m with 30 turns carries 2 A in a magnetic field of 0.4 T . The plane of the loop is at 60° to

Given: A square loop of side 0.18 m with 30 turns carries 2 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 = 0.18 × 0.18 = 0.0324 m². Substitution & Calculation: tau = 30 × 2 × 0.0324 × 0.4 × sin 60° = 0.7776 × 0.866 = 0.6734 approx 0.67 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 uniform disk of mass 3 kg and radius 0.6 m rotates about its nter. What is its moment of inertia?

Given: A uniform disk of mass 3 kg and radius 0.6 m rotates about its nter. What is its moment of inertia? Formula: For a uniform disk: I = 1/2 M R². Substitution & Calculation: M = 3 kg, R = 0.6 m . I = 1/2 × 3 × (0.6)² = 1.5 × 0.36 = 0.54 kg 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),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 2 kg block falls from 5 m onto a spring ( k = 500 N/m ). What is the maximum compression? (Take g = 10 m/s² )

Given: A 2 kg block falls from 5 m onto a spring ( k = 500 N/m ). What is the maximum compression? (Take g = 10 m/s² ) Formula: Potential energy mgh = 2 × 10 × 5 = 100 J. Substitution & Calculation: Spring energy 1/2 k x_m² = 100 Rightarrow 250 x_m² = 100 Rightarrow x_m = √0.4 approx 0.63 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), Chapter: Work, Energy and Power (Latest NCERT 2026-27), Topic: Energy conservation, kinetic energy ½mv² to spring potential ½kx² and compression. 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 pendulum of length 0.49 m oscillates with g = 9.8 m/s² . What is its frequency?

Given: A pendulum of length 0.49 m oscillates with g = 9.8 m/s² . What is its frequency? Formula: Period: T = 2π √L/g = 2π √0.49/9.8 = 2π √0.05 approx 1.404 s. Substitution & Calculation: Frequency: v = 1/T = 1/1.404 approx 0.712 Hz . 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

Light of wavelength 450 nm is incident on a metal with work function 1.9 eV . What is the maximum kinetic energy in eV?

Given: Light of wavelength 450 nm is incident on a metal with work function 1.9 eV . What is the maximum kinetic energy in eV? (Take h c = 1240 eV nm ) Formula: E = h c/lambda = 1240/450 approx 2.756 eV. Substitution & Calculation: K_{max = E - phi_0 = 2.756 - 1.9 approx 0.856 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 gas has a C_v of 20.8 J mol^{-1 K^{-1. What is the ratio of specific heats (γ)? (R = 8.31 J mol^{-1 K^{-1)

Given: A gas has a C_v of 20.8 J mol^{-1 K^{-1. What is the ratio of specific heats (γ)? (R = 8.31 J mol^{-1 K^{-1) Formula: C_p = C_v + R = 20.8 + 8.31 = 29.11 J mol^{-1 K^{-1. Substitution & Calculation: γ = C_p/C_v = 29.11/20.8 approx 1.40. 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

In Rutherford’s scattering experiment, what percentage of alpha-particles scatter by more than 1°?

Given: In Rutherford’s scattering experiment, what percentage of alpha-particles scatter by more than 1°? Formula: Percentage = 0.14%.. Substitution & Calculation: Given: 0.14% of alpha-particles scatter by more than 1°. 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 energy equivalent of 2 kg of matter in Joules? (Given c = 3 × 10⁸m/s )

Given: What is the energy equivalent of 2 kg of matter in Joules? (Given c = 3 × 10⁸m/s ) Formula: E = m c². Substitution & Calculation: m = 2 kg, c² = 9 × 10¹⁶m²/s² . E = 2 × 9 × 10¹⁶= 1.8 × 10¹⁷J . 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 coil of self-inductance 1.2 H has its current increased from 2 A to 6 A in 0.4 s. What is the magnitude of the induced

Given: A coil of self-inductance 1.2 H has its current increased from 2 A to 6 A in 0.4 s. What is the magnitude of the induced emf? Formula: ε = L Δ I/Δ t. Substitution & Calculation: Δ I = 6 - 2 = 4 A, Δ t = 0.4 s . ε = 1.2 × 4/0.4 = 1.2 × 10 = 12 V . 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 2 kg block slides down a frictionless incline from 9 m height. What is its speed at the bottom? (Take g = 10 m/s² )

Given: A 2 kg block slides down a frictionless incline from 9 m height. What is its speed at the bottom? (Take g = 10 m/s² ) Formula: Potential energy mgh = 2 × 10 × 9 = 180 J. Substitution & Calculation: Kinetic energy 1/2 m v² = 180 Rightarrow v² = 180 Rightarrow v = √180 approx 13.42 m/s . 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: Work, Energy and Power (Latest NCERT 2026-27), Topic: Work W = mgh, power P = W/t, lifting at constant speed and power. The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI units and.

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PHYSICS

A manometer with water ( rho = 1000 kg/m³ ) shows a height difference of 0.15 m . What is the pressure difference? (Take

Given: A manometer with water ( rho = 1000 kg/m³ ) shows a height difference of 0.15 m . What is the pressure difference? (Take g = 9.8 m/s² ) Formula: Δ P = rho g h. Substitution & Calculation: rho = 1000 kg/m³, g = 9.8 m/s², h = 0.15 m . Δ P = 1000 × 9.8 × 0.15 = 1470 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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