PHYSICS

PHYSICS

A Wheatstone bridge with R_1 = 4 Ω, R_2 = 8 Ω, R_3 = 6 Ω, R_4 = 12 Ω has a 12 V battery across AC and a galvanometer ( 2

Given: A Wheatstone bridge with R_1 = 4 Ω, R_2 = 8 Ω, R_3 = 6 Ω, R_4 = 12 Ω has a 12 V battery across AC and a galvanometer ( 2 Ω ) across BD. What is the current through the galvanometer? Formula: Check balance: R_1/R_2 = 4/8 = 0.5, R_3/R_4 = 6/12 = 0.5. Substitution & Calculation: Bridge is balanced. Since balanced, I_g = 0 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

A silicon diode has a threshold voltage of approximately:

The threshold or cut-in voltage for a silicon diode is about 0.7 V, beyond which the forward current increases significantly. This follows from latest NCERT 2026-27 principle explaining the concept clearly for NEET students in simple steps as per rationalized syllabus.

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 magnetic field contribution B_m due to a material with M = 2.5 × 10⁵A m^{-1 is: (Take μ_0 = 4π × 10⁻⁷T m A^{-1 ).

Given: The magnetic field contribution B_m due to a material with M = 2.5 × 10⁵A m^{-1 is: (Take μ_0 = 4π × 10⁻⁷T m A^{-1 ). Formula: B_m = μ_0 M. Substitution & Calculation: Given: M = 2.5 × 10⁵A m^{-1, μ_0 = 4π × 10⁻⁷. B_m = 4π × 10⁻⁷ × 2.5 × 10⁵= 0.314 T approx 0.31 T . 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 wheel with 5 spokes of 0.65 m each rotates at 42 rpm in a 0.7 T field. What is the induced emf?

Given: A wheel with 5 spokes of 0.65 m each rotates at 42 rpm in a 0.7 T field. What is the induced emf? Formula: omega = 2π × 42/60 = 1.4π rad/s. Substitution & Calculation: ε = 1/2 B omega R² = 1/2 × 0.7 × 1.4π × (0.65)² = 0.623 V approx 0.62 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 wire of length 0.8 m carrying 5 A is at 30° to a magnetic field of 0.8 T . What is the force on the wire?

Given: A wire of length 0.8 m carrying 5 A is at 30° to a magnetic field of 0.8 T . What is the force on the wire? Formula: Force F = I l B sin θ. Substitution & Calculation: F = 5 × 0.8 × 0.8 × sin 30° = 4 × 0.8 × 0.5 = 1.6 N . 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 12 μF capacitor is connected to a 220 V, 50 Hz AC source. What is the rms current?

Given: A 12 μF capacitor is connected to a 220 V, 50 Hz AC source. What is the rms current? Formula: X_C = 1/omega C, omega = 2π × 50 = 314 rad/s. Substitution & Calculation: C = 12 × 10⁻⁶F . X_C = frac1314 × 12 × 10⁻⁶approx 265.3 Ω . RMS current: I = V/X_C = 220/265.3 approx 0.83 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

How much energy is required to move a 200 kg satellite from 6 R_E to 12 R_E from Earth’s nter? ( M_E = 6 × 10²⁴kg, R_E =

Given: How much energy is required to move a 200 kg satellite from 6 R_E to 12 R_E from Earth’s nter? ( M_E = 6 × 10²⁴kg, R_E = 6.4 × 10⁶m, G = 6.67 × 10⁻¹¹N m²/kg² ) Formula: Δ E = -G M_E m (1/r_2 - 1/r_1). Substitution & Calculation: r_1 = 3.84 × 10⁷m, r_2 = 7.68 × 10⁷m . Δ E = -6.67 × 10⁻¹¹ × 6 × 10²⁴ × 200 (1/7.68 × 10⁷- 1/3.84 × 10⁷) . Δ E = -8.004 × 10¹⁶(-1.302 × 10⁻⁸) approx 1.04 × 10⁹J . 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 5 V and 7 V with internal resistances 2 Ω and 4 Ω are connected in series with a 6 Ω resistor. What is th

Given: Two lls of emf 5 V and 7 V with internal resistances 2 Ω and 4 Ω are connected in series with a 6 Ω resistor. What is the current through the circuit? Formula: Equivalent emf: ε_{eq = 5 + 7 = 12 V. Substitution & Calculation: Total resistance: R_{total = 2 + 4 + 6 = 12 Ω . Current: I = fracε_{eqR_{total = 12/12 = 1 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

A force F = 6x² N acts from x = 0 to x = 1 m . What is the work done?

Given: A force F = 6x² N acts from x = 0 to x = 1 m . What is the work done? Formula: Work W = int_0¹ 6x² dx = [ 2x³ ]_0¹ = 2 × 1 - 0 = 2 J .. 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 gas mixture has equal masses of hydrogen and argon at 300 K. What is the ratio of their rms speeds? (Molecular mass: H

Given: A gas mixture has equal masses of hydrogen and argon at 300 K. What is the ratio of their rms speeds? (Molecular mass: H_2 = 2 u, Ar = 39.9 u) Formula: v_{rms ∝ frac1√m, fracv_{H_2v_{Ar = √fracm_{Arm_{H_2. Substitution & Calculation: fracv_{H_2v_{Ar = √39.9/2 approx √19.95 approx 4.47. 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: Kinetic Theory (Latest NCERT 2026-27), Topic: RMS speed v_rms ∝ √T, temperature dependence, ratio v₂/v₁ = √(T₂/T₁) and calculation. 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 6 kg particle moves with velocity v = 3 j m/s at r = -4 i m . What is the magnitude of its angular momentum about the

Given: A 6 kg particle moves with velocity v = 3 j m/s at r = -4 i m . What is the magnitude of its angular momentum about the origin? Formula: L = r × p = beginvmatrix i & j & k -4 & 0 & 0 0 & 3 & 0 endvmatrix = k ((-4) × 3 - 0 × 0) = -12 k kg m²/s. Substitution & Calculation: Magnitude = 12 kg m²/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),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 convex mirror has a radius of curvature of 50 cm . An object is placed 25 cm from it. What is the image distance?

Given: A convex mirror has a radius of curvature of 50 cm . An object is placed 25 cm from it. What is the image distance? Formula: Focal length: f = R/2 = 50/2 = 25 cm (positive for convex). Substitution & Calculation: Object distance: u = -25 cm . Mirror equation: 1/v + 1/u = 1/f . 1/v + 1/-25 = 1/25 Rightarrow 1/v = 1/25 + 1/25 = 2/25 . v = 25/2 = 12.5 cm (virtual image). 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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