NEET MOCK TEST 5

CHEMISTRY

Calculate the mole fraction of ethanol ( Câ‚‚Hâ‚…OH ) in a solution containing 69 g of ethanol and 198 g of water. (Mola

Given: Calculate the mole fraction of ethanol ( C₂H₅OH ) in a solution containing 69 g of ethanol and 198 g of water. (Molar mass: C₂H₅OH = 46 g/mol, H₂O = 18 g/mol ) These values define the system as per NCERT data. Formula: Moles of ethanol = 69/46 = 1.5 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles of water = 198/18 = 11 mol . Total moles = 1.5 + 11 = 12.5. Mole fraction of ethanol = 1.5/12.5 = 0.12 . 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

Which of the following represents the correct genotypic ratio in a monohybrid cross (F2 generation)?

The genotypic ratio in a monohybrid cross (F2 generation) is 1:2:1 (homozygous dominant: heterozygous: homozygous recessive). 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: Biology - Botany portion covering relevant concept.

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CHEMISTRY

In the Carius method, the percentage of sulphur in a 0.3 g compound yielding 0.699 g BaSOâ‚„ is (Atomic masses: Ba = 137

Given: In the Carius method, the percentage of sulphur in a 0.3 g compound yielding 0.699 g BaSO₄ is (Atomic masses: Ba = 137, S = 32, O = 16) These values define the system as per NCERT data. Formula: Molar mass of BaSO₄ = 137 + 32 + 64 = 233 g/mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Mass of S = 32 × 0.699/233 = 0.096 g. Percentage = 0.096 × 100/0.3 = 32% . 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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BOTANY

What is the estimated percentage of animal species among all recorded species?

More than 70% of all recorded species are animals, with insects being the most numerous. 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

What is the magnetic moment (in BM) of [CoClâ‚„]^{2- ?

Given: What is the magnetic moment (in BM) of [CoCl₄]^{2- ? These values define the system as per NCERT data. Formula: Magnetic moment = sqrt3(3+2) = sqrt15 approx 3.87 BM.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Co²⁺ ( d⁷ ) in tetrahedral [CoCl₄]^{2- with weak field Cl^- is high spin ( e⁴ t_2³ ), with 3 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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CHEMISTRY

What is the bond order of O₂²- according to molecular orbital theory?

For O₂²- : 18 electrons, (sigma 1s)² (sigma^* 1s)² (sigma 2s)² (sigma^* 2s)² (sigma 2p_z)² (π 2p_x)² (π 2p_y)² (π^* 2p_x)² (π^* 2p_y)² . Bonding = 10, antibonding = 8. Bond order = 1/2 (10 - 8) = 1 .

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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CHEMISTRY

How many geometrical isomers are possible for [Cr(NH₃)2Cl₄]^{- ?

For octahedral [Ma₂b4], 2 geometrical isomers exist: cis (two NH₃ adjacent) and trans (opposite). 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: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

What is the minimum number of electrons required to fill the M shell completely?

The M shell (n = 3) has a maximum capacity of 2n² = 2 × 3² = 18 electrons, assuming all subshells (3s, 3p, 3d) are filled. 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.

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CHEMISTRY

The rate law for a reaction is Rate = k[A]² . If the concentration of A is tripled, by what factor does the rate increa

Initial rate = k[A]², new rate = k(3[A])² = 9k[A]² . Factor = frac9k[A]²k[A]² = 9 . 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.

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PHYSICS

What is the approximate mass number of a nucleus with radius 4.8 × 10⁻¹⁵ m ? (Given R_0 = 1.2 × 10⁻¹⁵ m )

Given: What is the approximate mass number of a nucleus with radius 4.8 × 10⁻¹⁵ m ? (Given R_0 = 1.2 × 10⁻¹⁵ m ) These values define the system as per NCERT data. Formula: R = R_0 A^{1/3. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 4.8 × 10⁻¹⁵= 1.2 × 10⁻¹⁵ × A^{1/3 . A^{1/3 = 4.8/1.2 = 4 . A = 4³ = 64 . 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: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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PHYSICS

A square loop of side 30 cm rotates at 12 rad/s in a 0.1 T field. What is the maximum emf induced?

Given: A square loop of side 30 cm rotates at 12 rad/s in a 0.1 T field. What is the maximum emf induced? These values define the system as per NCERT data. Formula: A = (0.3)² = 0.09 m². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon_0 = N B A omega = 1 × 0.1 × 0.09 × 12 = 0.108 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

The de Broglie wavelength of a particle with momentum 4.0 × 10⁻²⁴ kg m/s is:

Given: The de Broglie wavelength of a particle with momentum 4.0 × 10⁻²⁴ kg m/s is: These values define the system as per NCERT data. Formula: lambda = h/p = frac6.63 × 10⁻³⁴⁴.0 × 10⁻²⁴= 1.6575 × 10⁻¹⁰ m = 0.16575 nm .. 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 Physics Textbook for Class XI and XII, Chapter: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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