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NEET MOCK TEST 19

NEET Mock Test 19 is one test in the Examtube NEET practice series. It covers questions from across the NEET syllabus so you can sit a full attempt in one go and review your answers afterwards. Useful for timing practice and for checking which sections still slow you down.

180 questions

Which complex shows ionization isomerism with [Co(NH₃)5NO₂]Cl₂ ?

[Co(NH₃)5Cl]NO₂Cl swaps Cl^- and NO₂^- with [Co(NH₃)5NO₂]Cl₂, exhibiting ionization isomerism. 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)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

The dimensional formula of energy density is [M L^{-1 T^{-2] . What is the dimensional formula of energy flux (energy pe

Given: The dimensional formula of energy density is [M L^{-1 T^{-2] . What is the dimensional formula of energy flux (energy per unit area per unit time)? Formula: Energy flux = Energy / Area / Time. Substitution & Calculation: [M L² T^{-2] / [L²] / [T] = [M L² T^{-2] [L^{-2] [T^{-1] = [M T^{-3] . 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.

Which complex shows optical isomerism?

[PtCl₂(en)2]^{2+ (cis isomer) with two bidentate en ligands in a square planar geometry is chiral, showing optical isomerism (d and l forms). 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)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

What is the oxidation number of Mn in KMnO₄ ?

K is +1, O is -2. Sum is zero: +1 + x + 4(-2) = 0, 1 + x - 8 = 0, x = +7 . 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)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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

In CO, carbon has 1 triple bond (6 electrons) and 1 lone pair (2 electrons). Formal charge = 4 - 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

An electron moves at 6.5 × 10⁶m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 9.1 × 10⁻³

Given: An electron moves at 6.5 × 10⁶m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹kg, charge = 1.6 × 10⁻¹⁹C ) Formula: r = mv/qB. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 6.5 × 10⁶¹.6 × 10⁻¹⁹ × 0.2 = frac5.915 × 10⁻²⁴³.2 × 10⁻²⁰= 1.848 × 10⁻⁴m approx 0.0185 cm . 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.

For N₂(g) + 2O₂(g) 2NO₂(g), K_c = 0.01 at 500 K. What is K_c for the reverse reaction?

Given: For N₂(g) + 2O₂(g) 2NO₂(g), K_c = 0.01 at 500 K. What is K_c for the reverse reaction? Formula: For reverse reaction 2NO₂(g) N₂(g) + 2O₂(g), K_c' = 1/K_c = 1/0.01 = 100 .. 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 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.

A spring of k = 250 N/m has a 2.5 kg mass. If E = 1.25 J, what is the amplitude?

Given: A spring of k = 250 N/m has a 2.5 kg mass. If E = 1.25 J, what is the amplitude? Formula: Total energy: E = 1/2 k A². Substitution & Calculation: 1.25 = 0.5 × 250 × A² Rightarrow 1.25 = 125 A² Rightarrow A² = 0.01 Rightarrow A = 0.1 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.

A gas occupies 5.6 litres at STP. How many moles of the gas are present? (Molar volume at STP = 22.4 litres)

Given: A gas occupies 5.6 litres at STP. How many moles of the gas are present? (Molar volume at STP = 22.4 litres) Formula: Number of moles (μ) = fracVolumeMolar volume. Substitution & Calculation: μ = 5.6/22.4 = 0.25 mol. 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.

An electron is accelerated through 50 V . What is its de Broglie wavelength? (Take h = 6.63 × 10⁻³⁴J s, m_e = 9.11 × 10⁻

Given: An electron is accelerated through 50 V . What is its de Broglie wavelength? (Take h = 6.63 × 10⁻³⁴J s, m_e = 9.11 × 10⁻³¹kg, e = 1.6 × 10⁻¹⁹C ) Formula: K = e V = 1.6 × 10⁻¹⁹ × 50 = 8.0 × 10⁻¹⁸J. Substitution & Calculation: p = √2 m K = √2 × 9.11 × 10⁻³¹ × 8.0 × 10⁻¹⁸approx 3.816 × 10⁻²⁴kg m/s . lambda = h/p = frac6.63 × 10⁻³⁴³.816 × 10⁻²⁴approx 1.737 × 10⁻¹⁰m = 0.1737 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.

A square loop of side 26 cm rotates at 12 rad/s in a 0.3 T field. What is the maximum emf induced?

Given: A square loop of side 26 cm rotates at 12 rad/s in a 0.3 T field. What is the maximum emf induced? Formula: A = (0.26)² = 0.0676 m². Substitution & Calculation: ε_0 = N B A omega = 1 × 0.3 × 0.0676 × 12 = 0.24336 V approx 0.243 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), Chapter: Electromagnetic Induction (Latest NCERT 2026-27), Topic: Rotating coil, maximum emf ε₀ = NBAω, N = 300, B = 0.07 T, A = 0.012 m². The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI.

A wire of length 2.2 m carrying 3.5 A is at 60° to a magnetic field of 0.2 T . What is the force on the wire?

Given: A wire of length 2.2 m carrying 3.5 A is at 60° to a magnetic field of 0.2 T . What is the force on the wire? Formula: Force F = I l B sin θ. Substitution & Calculation: F = 3.5 × 2.2 × 0.2 × sin 60° = 7.7 × 0.2 × 0.866 = 1.3336 approx 1.33 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.