Skip to content
Login Register

NEET MOCK TEST 6

NEET Mock Test 6 is a practice test built along the lines of the NEET pattern, with questions drawn from Physics, Chemistry and Biology. Attempting it in one sitting gives you a sense of how you manage time across sections and where accuracy drops. Review the answers afterwards to see which topics need more work.

180 questions

CHEMISTRY

For Nâ‚‚(g) + Oâ‚‚(g) 2NO(g), K_c = 0.02 at 1500 K. What is K_c for NO(g) 1/2 Nâ‚‚(g) + 1/2 Oâ‚‚(g) ?

Given: For N₂(g) + O₂(g) 2NO(g), K_c = 0.02 at 1500 K. What is K_c for NO(g) 1/2 N₂(g) + 1/2 O₂(g) ? These values define the system as per NCERT data. Formula: For the reverse halved reaction, K_c' = frac1sqrtK_c = frac1sqrt0.02 approx 7.07 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

View Details
CHEMISTRY

A solution of two volatile liquids A and B has a total vapor pressure of 280 mm Hg. If the vapor pressure of pure A is 4

Given: A solution of two volatile liquids A and B has a total vapor pressure of 280 mm Hg. If the vapor pressure of pure A is 400 mm Hg and that of pure B is 160 mm Hg, what is the mole fraction of B in the solution? These values define the system as per NCERT data. Formula: Using Raoult's law: p_{total = x_A p_A⁰ + x_B p_B⁰, where x_A + x_B = 1. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 280 = (1 - x_B) · 400 + x_B · 160 . 280 = 400 - 400 x_B + 160 x_B . 280 = 400 - 240 x_B, 240 x_B = 120, x_B = 0.5 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

View Details
CHEMISTRY

A first-order reaction has a rate constant of 0.01386 min^{-1 . What percentage of the reactant remains after 100 minute

Given: A first-order reaction has a rate constant of 0.01386 min^{-1 . What percentage of the reactant remains after 100 minutes? These values define the system as per NCERT data. Formula: ln frac[R]_0[R] = k t = 0.01386 × 100 = 1.386. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: frac[R][R]_0 = e^{-1.386 = 0.25, so 25% remains. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

View Details
CHEMISTRY

How many optical isomers are possible for [Co(NH₃)2(en)2]^{3+ ?

[Co(NH₃)2(en)2]^{3+ is octahedral with two bidentate en ligands and lacks symmetry in its cis form, forming 2 optical isomers (d and l). This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

View Details
CHEMISTRY

How many moles of electrons are required to reduce 1 mol of Cr³⁺ to Cr(s)?

Given: How many moles of electrons are required to reduce 1 mol of Cr³⁺ to Cr(s)? These values define the system as per NCERT data. Formula: Reaction: Cr³+ + 3e- -> Cr(s). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 3 moles of electrons are required per mole of Cr³⁺. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

View Details
PHYSICS

A hydrogen atom absorbs a photon of energy 12.75 eV from the ground state. To which energy level does it jump? (Use E_n

Given: A hydrogen atom absorbs a photon of energy 12.75 eV from the ground state. To which energy level does it jump? (Use E_n = -13.6/n² eV ) These values define the system as per NCERT data. Formula: E_1 = -13.6 eV. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_n = -13.6 + 12.75 = -0.85 eV . -0.85 = -13.6/n² Rightarrow n² = 16 Rightarrow n = 4 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

View Details
PHYSICS

The number of valence electrons in Si and Ge atoms is:

Si (third orbit) and Ge (fourth orbit) are group IV elements, each with four valence electrons (2s and 2p for Si, 4s and 4p for Ge), forming covalent bonds in their lattice.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

View Details
CHEMISTRY

A first-order reaction has a rate constant of 0.01732 min^{-1 . What is the half-life in minutes?

Given: A first-order reaction has a rate constant of 0.01732 min^{-1 . What is the half-life in minutes? These values define the system as per NCERT data. Formula: t_{1/2 = 0.693/k = 0.693/0.01732 = 40 min .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

View Details
PHYSICS

A compound microscope has an objective of focal length 2 cm and eyepiece of focal length 5 cm with a tube length of 18 c

Given: A compound microscope has an objective of focal length 2 cm and eyepiece of focal length 5 cm with a tube length of 18 cm . What is the magnification at infinity? These values define the system as per NCERT data. Formula: Objective magnification: m_o = L/f_o = 18/2 = 9. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Eyepiece magnification: m_e = D/f_e = 25/5 = 5 . Total magnification: m = m_o × m_e = 9 × 5 = 45 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

View Details
CHEMISTRY

Which complex is diamagnetic?

[Pt(CN)4]^{2- (Pt²⁺, d⁸ ) with strong field CN^- in a square planar geometry is low spin, with 0 unpaired electrons, making it diamagnetic. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

View Details
PHYSICS

A circular loop of radius 18 cm is deformed into a straight wire in a 0.14 T field in 0.7 s. What is the induced emf?

Given: A circular loop of radius 18 cm is deformed into a straight wire in a 0.14 T field in 0.7 s. What is the induced emf? These values define the system as per NCERT data. Formula: Initial flux: Phi = B A = 0.14 × π × (0.18)² = 0.01425 Wb. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Final flux = 0. varepsilon = Δ Phi/Δ t = 0.01425/0.7 = 0.02036 V approx 0.02 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

View Details