NEET Mock Test 10 offers another full round of exam-style practice questions. Sit the test as you would the real paper, then go through the answers to check your accuracy and decide what to revise next.
A son has a 50% chance of inheriting an X-linked recessive disorder from a carrier mother, as he receives one X chromosome from his mother. 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.
Given: A gaseous reaction follows Rate = k p_A² . If k = 0.1 atm^{-1 s^{-1 and p_A = 0.4 atm, what is the rate in atm s^{-1 ? These values define the system as per NCERT data. Formula: Rate = k p_A² = 0.1 × (0.4)² = 0.1 × 0.16 = 0.016 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10â»âµ, 236 J kgâ»Â¹ Kâ»Â¹, CH₃CHâ‚‚NHâ‚‚ 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 kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ 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.
Given: The initial rate of a reaction is 2.0 × 10â»Â³ mol L^{-1 s^{-1 when [A] = 0.5 M and [B] = 0.2 M. If the rate law is Rate = k[A][B], what is k in L mol^{-1 s^{-1 ? These values define the system as per NCERT data. Formula: k = fracRate[A][B] = frac2.0 × 10â»Â³â°.5 × 0.2 = 0.02 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10â»âµ, 236 J kgâ»Â¹ Kâ»Â¹, CH₃CHâ‚‚NHâ‚‚ 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 kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ 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.
The expected 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: Principles of Inheritance and Molecular Basis of Inheritance, Topic: Genetics and molecular biology in plants.
Given: In an LCR circuit with R = 3 Ω, X_L = 8 Ω, X_C = 4 Ω, what is the power factor? These values define the system as per NCERT data. Formula: Impedance: Z = sqrtR² + (X_L - X_C)² = sqrt3² + (8 - 4)² = sqrt9 + 16 = 5 Ω. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Power factor: cos phi = R/Z = 3/5 = 0.6 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.
In CH₃CH=CHCH=CH₂, there are two double bonds, each contributing 1 π bond. Total π bonds = 2. 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.
Given: How many electrons are in the L shell of an atom with 10 electrons? These values define the system as per NCERT data. Formula: The L shell (n = 2) has 2s² 2pⶠ= 2 + 6 = 8 electrons.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 10 electrons (neon): 1s² 2s² 2pâ¶. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ 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.
Given: The K_a of phenol is 1.0 × 10â»Â¹â°. What is the pH of a 0.01 M solution? These values define the system as per NCERT data. Formula: K_a = x²/0.01, 1.0 × 10â»Â¹â°= x²/0.01, x² = 1.0 × 10â»Â¹Â². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: x = sqrt1.0 × 10â»Â¹Â²= 1.0 × 10â»â¶, pH = -log(1.0 × 10â»â¶) = 6.0 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.
Given: The rate law of a reaction is Rate = k[A][B] . If both [A] and [B] are increased by a factor of 3, by what factor does the rate increase? These values define the system as per NCERT data. Formula: Initial rate = k[A][B], new rate = k(3[A])(3[B]) = 9k[A][B]. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Factor = 9. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ 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.
Given: What is the pH of a 0.006 M KOH solution, assuming complete dissociation? These values define the system as per NCERT data. Formula: [OH-] = 0.006 M, pOH = -log(0.006) approx 2.22. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: pH = 14 - 2.22 = 11.78 approx 11.8 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.
[Ni(CN)4]^{2- (Ni²âº, d⸠) with strong field CN^- in a square planar geometry is low spin, with 0 unpaired electrons, making it diamagnetic. 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.
Given: A rectangular loop of 0.24 m × 0.38 m moves out of a 0.3 T field at 0.6 m/s along its shorter side. What is the emf? These values define the system as per NCERT data. Formula: varepsilon = B l v, l = 0.38 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 0.3 × 0.38 × 0.6 = 0.0684 V approx 0.068 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ 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.
