NEET MOCK TEST 11

PHYSICS

What is the dimensional formula of electrical conductance ( G = 1/R ), where R is resistance with SI unit kg m² s^{-3 A^

Given: What is the dimensional formula of electrical conductance ( G = 1/R ), where R is resistance with SI unit kg m² s^{-3 A^{-2 ? These values define the system as per NCERT data. Formula: [R] = kg m² s^{-3 A^{-2 = [M L² T^{-3 A^{-2]. This is standard NCERT relation. Substitution & Calculation: [G] = 1 / [R] = [M^{-1 L^{-2 T³ A²] . 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

Which complex shows geometrical isomerism?

[PtCl₂(NH₃)2] is square planar and can form cis and trans isomers, unlike tetrahedral or fully symmetric octahedral complexes. This follows from NCERT principle where relation explains outcome clearly for students.

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 copper plate has an area of 0.8 m² at 60° C . What is the decrease in area when cooled to 10° C ? ( α_l = 1.7 × 10⁻⁵K^

Given: A copper plate has an area of 0.8 m² at 60° C . What is the decrease in area when cooled to 10° C ? ( α_l = 1.7 × 10⁻⁵K^{-1 ) These values define the system as per NCERT data. Formula: Given: A_0 = 0.8 m², Δ T = 10 - 60 = -50° C, α_l = 1.7 × 10⁻⁵K^{-1. This is standard NCERT relation. Substitution & Calculation: Δ A = A_0 × 2 α_l Δ T = 0.8 × 2 × 1.7 × 10⁻⁵ × (-50) . Δ A = 0.8 × 3.4 × 10⁻⁵ × (-50) = -0.00136 m² (decrease of 0.00136 m² ). 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 zero-order reaction has a half-life of 25 minutes with an initial concentration of 0.25 M. What is the rate constant i

Given: A zero-order reaction has a half-life of 25 minutes with an initial concentration of 0.25 M. What is the rate constant in mol L^{-1 min^{-1 ? These values define the system as per NCERT data. Formula: t_{1/2 = frac[R]_02k, 25 = 0.25/2k, k = 0.25/50 = 0.005 .. 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

A reaction follows the rate law Rate = k[A][B] . If k = 0.5 L mol^{-1 s^{-1, [A] = 0.2 M, and [B] = 0.3 M, what is the i

Given: A reaction follows the rate law Rate = k[A][B] . If k = 0.5 L mol^{-1 s^{-1, [A] = 0.2 M, and [B] = 0.3 M, what is the initial rate in mol L^{-1 s^{-1 ? These values define the system as per NCERT data. Formula: Rate = k[A][B] = 0.5 × 0.2 × 0.3 = 0.03 mol L^{-1 s^{-1 .. 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

For the reaction N₂(g) + O₂(g) 2NO(g), K_c = 0.1 at 2000 K. If 0.5 mol of N₂ and 0.5 mol of O₂ are placed in a 5 L vesse

Given: For the reaction N₂(g) + O₂(g) 2NO(g), K_c = 0.1 at 2000 K. If 0.5 mol of N₂ and 0.5 mol of O₂ are placed in a 5 L vessel, what is the equilibrium concentration of NO ? These values define the system as per NCERT data. Formula: Initial [N₂] = [O₂] = 0.5 / 5 = 0.1 M. This is standard NCERT relation. Substitution & Calculation: Let [NO] = 2x at equilibrium. Then, [N₂] = [O₂] = 0.1 - x . K_c = frac[NO]²[N₂][O₂] = (2x)²/(0.1 - x)² = 0.1 . 4x²/(0.1 - x)² = 0.1, 2x/0.1 - x = sqrt0.1 approx 0.316 . Solving: 2x = 0.0316 - 0.316x, 2.316x = 0.0316, x approx 0.0136, [NO] = 2x approx 0.027 M. 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

Calculate the boiling point elevation of a solution containing 17.1 g of sucrose ( C₁₂H₂₂O₁₁ ) in 450 g of water. ( K_b

Given: Calculate the boiling point elevation of a solution containing 17.1 g of sucrose ( C₁₂H₂₂O₁₁ ) in 450 g of water. ( K_b = 0.52 K kg/mol, Molar mass of sucrose = 342 g/mol ) These values define the system as per NCERT data. Formula: Moles of sucrose = 17.1/342 = 0.05 mol. This is standard NCERT relation. Substitution & Calculation: Molality = 0.05/0.45 = 0.111 mol/kg . Δ T_b = 0.52 × 0.111 = 0.058 K . 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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CHEMISTRY

What is the molality of a solution containing 10 g of sugar (molar mass = 342 g/mol) dissolved in 500 g of water?

Given: What is the molality of a solution containing 10 g of sugar (molar mass = 342 g/mol) dissolved in 500 g of water? These values define the system as per NCERT data. Formula: Moles = 10 / 342 ≈ 0.0292 mol. This is standard NCERT relation. Substitution & Calculation: Molality = 0.0292 / 0.5 ≈ 0.0584 m. 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 solenoid of 250 turns and length 0.8 m induces an emf of 0.75 V in a nearby coil when its current changes from 0 to 2.

Given: A solenoid of 250 turns and length 0.8 m induces an emf of 0.75 V in a nearby coil when its current changes from 0 to 2.5 A in 0.25 s. What is the mutual inductance? These values define the system as per NCERT data. Formula: varepsilon = M Δ I/Δ t. This is standard NCERT relation. Substitution & Calculation: Δ I = 2.5 - 0 = 2.5 A, Δ t = 0.25 s . M = fracvarepsilonΔ I/Δ t = frac0.752.5/0.25 = 0.75/10 = 0.075 H . 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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PHYSICS

The magnetic potential energy of a dipole with m = 0.8 A m² in a field B = 0.25 T at 180° is:

Given: The magnetic potential energy of a dipole with m = 0.8 A m² in a field B = 0.25 T at 180° 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 = 0.8 A m², B = 0.25 T, θ = 180°, cos 180° = -1 . Substitute: U_m = -0.8 × 0.25 × (-1) = 0.2 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

During the electrolysis of aqueous CuSO₄ using copper electrodes, what mass of copper is deposited at the cathode if a c

Given: During the electrolysis of aqueous CuSO₄ using copper electrodes, what mass of copper is deposited at the cathode if a current of 1.5 A flows for 10 minutes? (Molar mass of Cu = 63 g/mol, F = 96500 C/mol) These values define the system as per NCERT data. Formula: Charge, Q = I × t = 1.5 × 600 = 900 C. This is standard NCERT relation. Substitution & Calculation: For Cu²⁺ + 2e⁻ → Cu(s), 2F (2 × 96500 C) deposits 63 g of Cu. Mass = 63 × 900/2 × 96500 approx 0.294 g . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Electrochemistry, Topic: Faraday's first law, charge to reduce Al³⁺ to Al, 3F = 3 × 96500 C.

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PHYSICS

A long wire carries 20 A . At what distance from the wire is the magnetic field 4 × 10⁻⁶T ? ( μ_0 = 4 π × 10⁻⁷T m/A )

Given: A long wire carries 20 A . At what distance from the wire is the magnetic field 4 × 10⁻⁶T ? ( μ_0 = 4 π × 10⁻⁷T m/A ) These values define the system as per NCERT data. Formula: B = μ_0 I/2 π r, so r = μ_0 I/2 π B. This is standard NCERT relation. Substitution & Calculation: r = frac4 π × 10⁻⁷ × 202 π × 4 × 10⁻⁶= frac8 × 10⁻⁶⁸ × 10⁻⁶= 1 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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