NEET MOCK TEST 18

CHEMISTRY

What is the formula for hexaamminecobalt(III) sulphate?

Co³⁺ with 6 NH₃ forms [Co(NH₃)6]^{3+ . Sulphate ( SO₄^{2- ) requires 3 units to balance two +3 complexes, giving [Co(NH₃)6]2(SO₄)3 . 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.

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PHYSICS

The SI unit of energy is kg m² s^{-2 . What is its dimensional formula?

Given: The SI unit of energy is kg m² s^{-2 . What is its dimensional formula? Formula: kg = [M], m² = [L²], s^{-2 = [T^{-2]. Substitution & Calculation: kg m² s^{-2 = [M L² T^{-2] . 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.

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CHEMISTRY

What is the total number of lone pairs on the nitrogen atom in the HNO₃ molecule?

In HNO₃, nitrogen forms 3 bonds (1 double to O, 1 single to O, 1 single to OH), using all 5 valence electrons, leaving 0 lone pairs. 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: Chemical Bonding (Latest NCERT 2026-27), Topic: Hybridization of nitrogen in CH₃NH₂, sp³ hybridization and pyramidal geometry

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CHEMISTRY

A first-order reaction is 25% complete in 15 minutes. What is the rate constant in min^{-1 ?

Given: A first-order reaction is 25% complete in 15 minutes. What is the rate constant in min^{-1 ? Formula: 25% complete, 75% remains: frac[R]_0[R] = 100/75 = 1.333. Substitution & Calculation: k = 2.303/t log frac[R]_0[R] = 2.303/15 log 1.333 = 0.0192 min^{-1 . 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), Chapter: Chemical Kinetics (Latest NCERT 2026-27), Topic: Arrhenius equation, k = A e^(-E_a/RT), activation energy and rate constant

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PHYSICS

A soap film supports 3.0 × 10⁻²N over a 50 cm slider. What is the surface tension?

Given: A soap film supports 3.0 × 10⁻²N over a 50 cm slider. What is the surface tension? Formula: S = F/2 l. Substitution & Calculation: F = 3.0 × 10⁻²N, l = 0.5 m . S = frac3.0 × 10⁻²² × 0.5 = 0.03 N/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.

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CHEMISTRY

How much current (in amperes) is required to deposit 0.108 g of silver from AgNO₃ solution in 965 seconds? (Molar mass o

Given: How much current (in amperes) is required to deposit 0.108 g of silver from AgNO₃ solution in 965 seconds? (Molar mass of Ag = 108 g/mol, F = 96500 C/mol) Formula: Moles of Ag = 0.108/108 = 0.001 mol. Substitution & Calculation: Reaction: Ag⁺ + e⁻ → Ag(s), 1F deposits 108 g. . Charge = 0.001 × 96500 = 96.5 C . Current = Q/t = 96.5/965 = 0.1 A . 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), Chapter: Electrochemistry (Latest NCERT 2026-27), Topic: Faraday's laws, charge required to reduce Al³⁺, 3F = 3 × 96500 C

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PHYSICS

A 4 μF capacitor charged to 300 V is connected to an uncharged 8 μF capacitor. What is the energy lost?

Given: A 4 μF capacitor charged to 300 V is connected to an uncharged 8 μF capacitor. What is the energy lost? Formula: Initial energy: U_i = 1/2 × 4 × 10⁻⁶ × (300)² = 0.18 J. Substitution & Calculation: Charge: Q = 4 × 10⁻⁶ × 300 = 1.2 × 10⁻³C . Total C = 4 + 8 = 12 μF, V = frac1.2 × 10⁻³¹² × 10⁻⁶= 100 V . Final energy: U_f = 1/2 × 12 × 10⁻⁶ × (100)² = 0.06 J . Loss: U_i - U_f = 0.18 - 0.06 = 0.12 J . 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.

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PHYSICS

In a Wheatstone bridge with R_1 = 30 Ω, R_2 = 60 Ω, R_3 = 15 Ω, and R_4 = 32 Ω, a 10 V battery is connected across AC. W

Given: In a Wheatstone bridge with R_1 = 30 Ω, R_2 = 60 Ω, R_3 = 15 Ω, and R_4 = 32 Ω, a 10 V battery is connected across AC. What is the current through the galvanometer ( R_g = 10 Ω )? Formula: Junction B: I_1 = I_g + I_4, Junction D: I_2 = I_g + I_3. Substitution & Calculation: Apply Kirchhoff’s rules. Let currents be I_1 (AB), I_2 (AD), I_g (BD). . Loop BADB: 30 I_1 + 10 I_g - 60 I_2 = 0 Rightarrow 3 I_1 + I_g - 6 I_2 = 0 . Loop BCDB: 60 (I_1 - I_g) - 10 I_g - 15 (I_2 + I_g) = 0 Rightarrow 4 I_1 - 2 I_2 - 2 I_g = 0 . Loop ADCEA: 60 I_2 + 15 (I_2 + I_g) = 10 Rightarrow 75 I_2 + 15 I_g = 10 Rightarrow 5 I_2 + I_g = 2/3 . Solve: From (3) I_g = 2/3 - 5 I_2, substitute in (1): 3 I_1 + 2/3 - 5 I_2 - 6 I_2 = 0 Rightarrow 3 I_1 - 11 I_2 = -2/3 . From (2): 4 I_1 - 2 I_2 - 2 (2/3 - 5 I_2) = 0 Rightarrow 4 I_1 - 2 I_2 - 4/3 + 10 I_2 = 0 Rightarrow 4 I_1 + 8 I_2 = 4/3 . Solve: 12 I_1 - 33 I_2 = -2, 12 I_1 + 24 I_2 = 4 . Subtract: -57 I_2 = -6 Rightarrow I_2 = 6/57 = 2/19 A . I_g = 2/3 - 5 × 2/19 = 38/57 - 30/57 = 8/57 approx 0.14 A . 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.

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PHYSICS

A dipole with m = 0.7 A m² in B = 0.6 T at 30° has torque:

Given: A dipole with m = 0.7 A m² in B = 0.6 T at 30° has torque: Formula: tau = m B sinθ. Substitution & Calculation: Given: m = 0.7 A m², B = 0.6 T, θ = 30°, sin 30° = 0.5 . tau = 0.7 × 0.6 × 0.5 = 0.21 N 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.

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CHEMISTRY

The K_p for NH₃(g) 1/2 N₂(g) + 3/2 H₂(g) is 0.2 atm at 500 K. What is K_c if R = 0.0821 L · atm · mol^{-1 · K^{-1 ?

Given: The K_p for NH₃(g) 1/2 N₂(g) + 3/2 H₂(g) is 0.2 atm at 500 K. What is K_c if R = 0.0821 L · atm · mol^{-1 · K^{-1 ? Formula: Δ n = (1/2 + 3/2) - 1 = 1, K_p = K_c (RT)^{Δ n. Substitution & Calculation: RT = 0.0821 × 500 = 41.05, 0.2 = K_c · 41.05, K_c = 0.2 / 41.05 approx 0.00487 . 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.

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PHYSICS

A neutron ( 1 u ) at 10⁶m/s collides elastically with a carbon ( 12 u ). What fraction of its kinetic energy is transfer

Given: A neutron ( 1 u ) at 10⁶m/s collides elastically with a carbon ( 12 u ). What fraction of its kinetic energy is transferred? Formula: Fraction transferred f_2 = 4 m_1 m_2/(m_1 + m_2)² = 4 × 1 × 12/(1 + 12)² = 48/169 approx 0.284 .. 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 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.

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CHEMISTRY

What is the molality of a solution made by dissolving 5 g of urea (molar mass = 60 g/mol) in 200 g of water?

Given: What is the molality of a solution made by dissolving 5 g of urea (molar mass = 60 g/mol) in 200 g of water? Formula: Moles = 5 / 60 = 0.0833 mol. Substitution & Calculation: Mass of solvent = 0.2 kg. Molality = 0.0833 / 0.2 = 0.4165 ≈ 0.417 m. 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), Chapter: Solutions (Latest NCERT 2026-27), Topic: Colligative properties, ΔT_b = K_b·m, urea NH₂CONH₂ example

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