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NEET MOCK TEST 20

NEET Mock Test 20 is a full practice set for NEET aspirants, with questions drawn from physics, chemistry and biology. Attempt it under timed conditions to get a feel for the paper, then go through the solutions to find the topics worth revising again.

180 questions

CHEMISTRY

What is the formal charge on the nitrogen atom in the NO₂^- ion in one of its resonance structures?

In one resonance structure of NO₂^-, N has 1 double bond (4 electrons), 1 single bond (2 electrons), and 1 lone pair (2 electrons). Formal charge = 5 - 2 - 1/2 × 6 = 0.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII), Chapter: Chemical Bonding (Latest NCERT 2026-27), Topic: Hybridization of nitrogen in CH₃NH₂, sp³ hybridization and pyramidal geometry

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CHEMISTRY

What is the formal charge on the oxygen atom in the CO molecule?

In CO, oxygen has 1 triple bond (6 electrons) and 1 lone pair (2 electrons). Formal charge = 6 - 2 - 1/2 × 6 = +1. This follows from latest NCERT 2026-27 principle explaining the concept clearly for NEET students in simple steps as per rationalized syllabus.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII), Chapter: Electrochemistry (Latest NCERT 2026-27), Topic: Faraday's laws, charge required to reduce Al³⁺, 3F = 3 × 96500 C

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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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CHEMISTRY

A solution contains 10 g of a non-volatile solute (molar mass 200 g/mol) in 190 g of water. What is the molality of the

Given: A solution contains 10 g of a non-volatile solute (molar mass 200 g/mol) in 190 g of water. What is the molality of the solution? Formula: Moles of solute = 10/200 = 0.05 mol. Substitution & Calculation: Mass of solvent in kg = 190/1000 = 0.19 kg . Molality = 0.05/0.19 = 0.263 mol/kg . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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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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CHEMISTRY

A first-order reaction has a half-life of 35 minutes. What fraction of the reactant remains after 70 minutes?

Given: A first-order reaction has a half-life of 35 minutes. What fraction of the reactant remains after 70 minutes? Formula: k = 0.693/35 = 0.0198 min^{-1. Substitution & Calculation: ln frac[R]_0[R] = 0.0198 × 70 = 1.386, frac[R][R]_0 = e^{-1.386 = 0.25 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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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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CHEMISTRY

How much current (in amperes) is required to deposit 0.159 g of copper from CuSO₄ solution in 482.5 seconds? (Molar mass

Given: How much current (in amperes) is required to deposit 0.159 g of copper from CuSO₄ solution in 482.5 seconds? (Molar mass of Cu = 63.5 g/mol, F = 96500 C/mol) Formula: Moles of Cu = 0.159/63.5 = 0.0025 mol. Substitution & Calculation: Reaction: Cu²⁺ + 2e⁻ → Cu(s), 2F deposits 63.5 g. . Charge = 0.0025 × 2 × 96500 = 482.5 C . Current = Q/t = 482.5/482.5 = 1 A . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII), Chapter: Electrochemistry (Latest NCERT 2026-27), Topic: Faraday's laws, charge required to reduce Al³⁺, 3F = 3 × 96500 C

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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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