NEET MOCK TEST 14

CHEMISTRY

What is the oxidation state of iron in K₄[Fe(CN)6] ?

Given: What is the oxidation state of iron in K₄[Fe(CN)6] ? These values define the system as per NCERT data. Formula: Each CN^- is -1, 6 ligands = -6. This is standard NCERT relation. Substitution & Calculation: The complex [Fe(CN)6]^{4- is balanced by 4 K⁺, so Fe’s oxidation state is x - 6 = -4, x = +2 . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

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

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CHEMISTRY

What is the maximum oxidation state exhibited by vanadium in the 3d series?

Given: What is the maximum oxidation state exhibited by vanadium in the 3d series? These values define the system as per NCERT data. Formula: Vanadium (V, Z = 23) exhibits oxidation states from +2 to +5, with +5 being the maximum, as seen in VO₂^+ .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

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

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PHYSICS

The magnetic potential energy of a dipole with m = 0.9 A m² in a field B = 0.2 T at 90° is:

Given: The magnetic potential energy of a dipole with m = 0.9 A m² in a field B = 0.2 T at 90° is: These values define the system as per NCERT data. Formula: U_m = -m B cosθ. This is standard NCERT relation. Substitution & Calculation: Given: m = 0.9 A m², B = 0.2 T, θ = 90°, cos 90° = 0 . Substitute: U_m = -0.9 × 0.2 × 0 = 0 J . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Magnetism and Matter, Topic: Magnetic potential energy U = -m·B, dipole in magnetic field at 180°. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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CHEMISTRY

A gaseous reaction follows Rate = k p_A p_B with k = 0.04 atm^{-1 s^{-1, p_A = 0.5 atm, and p_B = 0.2 atm . What is the

Given: A gaseous reaction follows Rate = k p_A p_B with k = 0.04 atm^{-1 s^{-1, p_A = 0.5 atm, and p_B = 0.2 atm . What is the rate in atm s^{-1 ? These values define the system as per NCERT data. Formula: Rate = k p_A p_B = 0.04 × 0.5 × 0.2 = 0.004 .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

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

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CHEMISTRY

How many Faradays of electricity are required to deposit 0.635 g of copper from CuSO₄ solution? (Molar mass of Cu = 63.5

Given: How many Faradays of electricity are required to deposit 0.635 g of copper from CuSO₄ solution? (Molar mass of Cu = 63.5 g/mol) These values define the system as per NCERT data. Formula: Moles of Cu = 0.635/63.5 = 0.01 mol. This is standard NCERT relation. Substitution & Calculation: Reaction: Cu²⁺ + 2e⁻ → Cu(s), 2F deposits 63.5 g. . Faradays = 0.01 × 2 = 0.02 F . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

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

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PHYSICS

A 6 μF capacitor charged to 150 V is connected to an uncharged 9 μF capacitor. What is the energy lost?

Given: A 6 μF capacitor charged to 150 V is connected to an uncharged 9 μF capacitor. What is the energy lost? These values define the system as per NCERT data. Formula: Initial energy: U_i = 1/2 × 6 × 10⁻⁶ × (150)² = 0.0675 J. This is standard NCERT relation. Substitution & Calculation: Charge: Q = 6 × 10⁻⁶ × 150 = 9 × 10⁻⁴C . Total C = 6 + 9 = 15 μF, V = frac9 × 10⁻⁴¹⁵ × 10⁻⁶= 60 V . Final energy: U_f = 1/2 × 15 × 10⁻⁶ × (60)² = 0.027 J . Loss: U_i - U_f = 0.0675 - 0.027 = 0.0405 J . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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CHEMISTRY

The K_p for CS₂(g) + 4H₂(g) CH₄(g) + 2H₂S(g) is 0.25 at 900 K. What is K_c if R = 0.0821 L · atm · mol^{-1 · K^{-1 ?

Given: The K_p for CS₂(g) + 4H₂(g) CH₄(g) + 2H₂S(g) is 0.25 at 900 K. What is K_c if R = 0.0821 L · atm · mol^{-1 · K^{-1 ? These values define the system as per NCERT data. Formula: Δ n = (1 + 2) - (1 + 4) = -2, K_p = K_c (RT)^{Δ n. This is standard NCERT relation. Substitution & Calculation: RT = 0.0821 × 900 = 73.89, (RT)^{-2 = (73.89)^{-2 approx 1.83 × 10⁻⁴. 0.25 = K_c · 1.83 × 10⁻⁴, K_c approx 1365 . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

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

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PHYSICS

A rod of length 0.8 m moves at 1.5 m/s in a 0.4 T field perpendicular to its length. What is the induced emf?

Given: A rod of length 0.8 m moves at 1.5 m/s in a 0.4 T field perpendicular to its length. What is the induced emf? These values define the system as per NCERT data. Formula: varepsilon = B l v = 0.4 × 0.8 × 1.5 = 0.48 V .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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CHEMISTRY

A solution of two volatile liquids A and B has a total vapor pressure of 320 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 320 mm Hg. If the vapor pressure of pure A is 400 mm Hg and that of pure B is 200 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 standard NCERT relation. Substitution & Calculation: 320 = (1 - x_B) · 400 + x_B · 200 . 320 = 400 - 400 x_B + 200 x_B . 320 = 400 - 200 x_B, 200 x_B = 80, x_B = 0.4 . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

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

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PHYSICS

A 3.5 kg mass falls from 7 m onto a spring ( k = 1200 N/m ). What is the maximum compression? (Take g = 10 m/s² )

Given: A 3.5 kg mass falls from 7 m onto a spring ( k = 1200 N/m ). What is the maximum compression? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Potential energy mgh = 3.5 × 10 × 7 = 245 J. This is standard NCERT relation. Substitution & Calculation: Spring energy 1/2 k x_m² = 245 Rightarrow 600 x_m² = 245 Rightarrow x_m = sqrt0.4083 approx 0.639 m . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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CHEMISTRY

A solution contains 10 g of urea ( NH₂CONH₂ ) in 190 g of water. What is the molality of the solution? (Molar mass: NH₂C

Given: A solution contains 10 g of urea ( NH₂CONH₂ ) in 190 g of water. What is the molality of the solution? (Molar mass: NH₂CONH₂ = 60 g/mol ) These values define the system as per NCERT data. Formula: Moles of urea = 10/60 = 0.1667 mol. This is standard NCERT relation. Substitution & Calculation: Mass of solvent in kg = 190/1000 = 0.19 kg . Molality = 0.1667/0.19 = 0.877 mol/kg . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Boiling point elevation ΔT_b = K_b·m, urea NH₂CONH₂ example, colligative properties.

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PHYSICS

What is the magnetic field at a point on the equatorial line of a bar magnet with magnetic moment 2 A m² at a distance o

Given: What is the magnetic field at a point on the equatorial line of a bar magnet with magnetic moment 2 A m² at a distance of 10 cm from its nter? (Take μ_0 = 4π × 10⁻⁷T m A^{-1 ). These values define the system as per NCERT data. Formula: The magnetic field on the equatorial line is B = μ_0/4π m/r³. This is standard NCERT relation. Substitution & Calculation: Given: m = 2 A m², r = 0.1 m, μ_0/4π = 10⁻⁷T m A^{-1 . Substitute: B = 10⁻⁷ × 2/(0.1)³ = 10⁻⁷ × 2/0.001 = 2 × 10⁻⁴T . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Magnetism and Matter, Topic: Bar magnet cut transversely, magnetic moment halves, m' = m/2. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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