NEET MOCK TEST 3

CHEMISTRY

What is the magnetic moment (in BM) of [Ni(NH₃)6]^{2+ ?

Given: What is the magnetic moment (in BM) of [Ni(NH₃)6]^{2+ ? These values define the system as per NCERT data. Formula: Magnetic moment = sqrt2(2+2) = sqrt8 approx 2.83 BM.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Ni²⁺ ( d⁸ ) with NH₃ (moderate field) in an octahedral field is high spin ( t_{2g⁶ e_g² ), with 2 unpaired electrons. 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.

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PHYSICS

A gas mixture contains equal numbers of neon and argon molecules at 300 K. What is the ratio of their average kinetic en

Given: A gas mixture contains equal numbers of neon and argon molecules at 300 K. What is the ratio of their average kinetic energies? These values define the system as per NCERT data. Formula: Average KE per molecule = 3/2 k_B T, independent of mass. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Since T is same, fracKE_{NeKE_{Ar = 1:1. 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.

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CHEMISTRY

Calculate the mole fraction of methanol ( CH₃OH ) in a solution containing 32 g of methanol and 144 g of water. (Molar

Given: Calculate the mole fraction of methanol ( CH₃OH ) in a solution containing 32 g of methanol and 144 g of water. (Molar mass: CH₃OH = 32 g/mol, H₂O = 18 g/mol ) These values define the system as per NCERT data. Formula: Moles of methanol = 32/32 = 1 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles of water = 144/18 = 8 mol . Total moles = 1 + 8 = 9. Mole fraction of methanol = 1/9 = 0.111 . 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.

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BOTANY

How many ATP molecules are produced through substrate-level phosphorylation during glycolysis?

In glycolysis, substrate-level phosphorylation directly generates 4 ATP molecules. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Biology Textbook for Class XI and XII (Botany section), Chapter: Morphology and Anatomy of Flowering Plants, Topic: Plant structure and tissue systems.

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CHEMISTRY

The K_a of chloroacetic acid is 1.35 × 10⁻³. What is the pH of a 0.05 M solution?

Given: The K_a of chloroacetic acid is 1.35 × 10⁻³. What is the pH of a 0.05 M solution? These values define the system as per NCERT data. Formula: K_a = x²/0.05, 1.35 × 10⁻³= x²/0.05, x² = 6.75 × 10⁻⁵. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: x = sqrt6.75 × 10⁻⁵ approx 8.22 × 10⁻³, pH = -log(8.22 × 10⁻³) approx 2.09 . 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: Solutions, Topic: Concentration terms and colligative properties.

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CHEMISTRY

What is the maximum number of emission lines when an electron drops from n = 5 to the ground state in a hydrogen atom?

Given: What is the maximum number of emission lines when an electron drops from n = 5 to the ground state in a hydrogen atom? These values define the system as per NCERT data. Formula: Number of emission lines = n(n-1)/2, where n = 5. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: So, 5 × 4/2 = 10. 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.

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CHEMISTRY

The rate constant of a reaction is 5.0 × 10⁻⁵ s^{-1 at 290 K and 1.5 × 10⁻⁴ s^{-1 at 300 K. What is the activa

Given: The rate constant of a reaction is 5.0 × 10⁻⁵ s^{-1 at 290 K and 1.5 × 10⁻⁴ s^{-1 at 300 K. What is the activation energy in kJ mol^{-1 ? (R = 8.314 J/mol · K) These values define the system as per NCERT data. Formula: log k_2/k_1 = E_a/2.303R ( T_2 - T_1/T_1 T_2 ). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: log frac1.5 × 10⁻⁴⁵.0 × 10⁻⁵= log 3 = 0.477 . 0.477 = E_a/2.303 × 8.314 ( 10/290 × 300 ), E_a approx 51.5 kJ mol^{-1 . 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.

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CHEMISTRY

Which of the following amines forms a salt soluble in excess NaOH when treated with benzenesulphonyl chloride?

Primary amines like CH₃NH₂ form sulphonamides with benzenesulphonyl chloride that have an acidic hydrogen, making them soluble in excess NaOH in the Hinsberg test.

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

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PHYSICS

Two small charged spheres with charges 5 × 10⁻⁷ C and 7 × 10⁻⁷ C are placed 50 cm apart in air. What is the fo

Given: Two small charged spheres with charges 5 × 10⁻⁷ C and 7 × 10⁻⁷ C are placed 50 cm apart in air. What is the force between them? These values define the system as per NCERT data. Formula: Using Coulomb’s law: F = k |q_1 q_2|/r². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k = 9 × 10⁹ Nm²/C², q_1 = 5 × 10⁻⁷ C, q_2 = 7 × 10⁻⁷ C, r = 0.5 m . |q_1 q_2| = 5 × 7 × 10⁻¹⁴= 35 × 10⁻¹⁴ C² . r² = (0.5)² = 0.25 m² . F = 9 × 10⁹ × frac35 × 10⁻¹⁴⁰.25 = 9 × 10⁹ × 1.4 × 10⁻¹²= 0.0126 N . 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.

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PHYSICS

A square loop of side 25 cm rotates at 8 rad/s in a 0.15 T field. What is the maximum emf induced?

Given: A square loop of side 25 cm rotates at 8 rad/s in a 0.15 T field. What is the maximum emf induced? These values define the system as per NCERT data. Formula: A = (0.25)² = 0.0625 m². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon_0 = N B A omega = 1 × 0.15 × 0.0625 × 8 = 0.075 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.

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PHYSICS

In a new system, the unit of mass is 2 kg, length is 0.5 m, and time is 3 s . What is the value of 1 J ( kg m² s^{-2 )

Given: In a new system, the unit of mass is 2 kg, length is 0.5 m, and time is 3 s . What is the value of 1 J ( kg m² s^{-2 ) in this system? These values define the system as per NCERT data. Formula: 1 J = 1 kg m² s^{-2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: New units: kg = 2 α, m = 0.5 β, s = 3 γ . 1 J = (2 α) (0.5 β)² (3 γ)^{-2 = 2 × 0.25 × 1/9 = 0.5/9 = 0.0556 α β² γ^{-2 . 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.

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CHEMISTRY

What is the ll constant if the resistance of a conductivity ll with 0.01 M KCl solution is 200 Ω and its conductivity i

Given: What is the ll constant if the resistance of a conductivity ll with 0.01 M KCl solution is 200 Ω and its conductivity is 0.14 × 10⁻² S cm⁻¹? These values define the system as per NCERT data. Formula: Cell constant, G^* = kappa × R = 0.14 × 10⁻² × 200 = 0.28 cm^{-1 .. 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: Solutions, Topic: Concentration terms and colligative properties.

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