id string | subject string | topic string | subtopic string | difficulty string | difficulty_labeling_method string | question_type string | question string | options dict | correct_answer string | correct_answer_value string | solution string | option_explanations dict | concepts_required list | reasoning_steps int64 | answer_type string | language string | origin string | generator string | generator_template_id string | verification_status string | verification_note string | difficulty_note string | source_type string | answer_validation string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
openstem_001899 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 7 elements can be selected from a set of 75 distinct elements? | {
"A": "1488622387.5",
"B": "1984829851",
"C": "-1984829850",
"D": "1984829850"
} | D | 1984829850 | Order does not matter, so the count is C(75,7)=75!/[7!(68)!]=1984829850. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1984829850, not 1488622387.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1984829850, not 1984829851.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1984... | [
"combinations",
"factorials"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_combinations | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003992 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 4 kg block initially moves at 1 m s⁻¹. A constant net force does 28 N of work over 2 m in the direction of motion. Find its final kinetic energy. | {
"A": "69.6 J",
"B": "87 J",
"C": "-58 J",
"D": "58 J"
} | D | 58 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(4)(1)²+(28)(2)=58 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 58 J, not 69.6 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 58 J, not 87 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 58 J, not -58 J.",
"D": "C... | [
"work–energy theorem",
"kinetic energy",
"work by constant force"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_work_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003360 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 47 m s⁻¹ at 45° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "218.9 m",
"B": "165.675 m",
"C": "220.9 m",
"D": "23.09 m"
} | C | 220.9 m | Range R=u²sin(2θ)/g. Substituting u=47, θ=45° and g=10 gives R=(47)²sin(90°)/10=220.9 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 220.9 m, not 218.9 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 220.9 m, not 165.675 m.",
"C": "Correct. The worked calculation gives 220.9 m.",
"D": "Incorrect. Substituting the stated ... | [
"projectile motion",
"range equation",
"trigonometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_projectile_range | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003392 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 33 m s⁻¹ at 20° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "83.99948 m",
"B": "69.99957 m",
"C": "55.99966 m",
"D": "52.49968 m"
} | B | 69.99957 m | Range R=u²sin(2θ)/g. Substituting u=33, θ=20° and g=10 gives R=(33)²sin(40°)/10=69.99957 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 69.99957 m, not 83.99948 m.",
"B": "Correct. The worked calculation gives 69.99957 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 69.99957 m, not 55.99966 m.",
"D": "Incorrect. Substitutin... | [
"projectile motion",
"range equation",
"trigonometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_projectile_range | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006380 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 24 kg is attached to an ideal spring of force constant 225 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "1.6532 s",
"B": "-0.3468 s",
"C": "3.6532 s",
"D": "0.6532 s"
} | D | 0.6532 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(24/225)=0.6532 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.6532 s, not 1.6532 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.6532 s, not -0.3468 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.6532 s, not... | [
"simple harmonic motion",
"mass–spring oscillator",
"time period"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_shm_period | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000187 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 1x^2 -16x +11=0, let its roots be α and β. What is α²+β²? | {
"A": "233",
"B": "175.5",
"C": "234",
"D": "232"
} | C | 234 | By Vieta's relations, α+β = 16/1 and αβ = 11/1. Then α²+β²=(α+β)²−2αβ = (16/1)²−2(11/1) = 234. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 234, not 233.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 234, not 175.5.",
"C": "Correct. The worked calculation gives 234.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000938 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 4 and common ratio 5. What is the sum of its first 12 terms? | {
"A": "244140626",
"B": "244140622",
"C": "366210936",
"D": "244140624"
} | D | 244140624 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=4(5^12−1)/(5−1)=244140624. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 244140624, not 244140626.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 244140624, not 244140622.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 244140624,... | [
"geometric progression",
"finite geometric sum"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_gp_sum | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006391 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 9 kg is attached to an ideal spring of force constant 1105 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "0.1805 s",
"B": "0.09025 s",
"C": "0.36099 s",
"D": "-1.8195 s"
} | A | 0.1805 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(9/1105)=0.1805 s. | {
"A": "Correct. The worked calculation gives 0.1805 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.1805 s, not 0.09025 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.1805 s, not 0.36099 s.",
"D": "Incorrect. Substituting the st... | [
"simple harmonic motion",
"mass–spring oscillator",
"time period"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_shm_period | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004708 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 4 Ω and 11 Ω are connected in series; this combination is connected in parallel with a 4 Ω resistor across a 18 V ideal battery. Find the total current supplied by the battery. | {
"A": "5.7 A",
"B": "8.55 A",
"C": "8.7 A",
"D": "2.85 A"
} | A | 5.7 A | Series branch resistance=4+11=15 Ω. Parallel equivalent R_eq=(15×4)/(15+4)=3.15789 Ω. Total current I=V/R_eq=18/3.15789=5.7 A. | {
"A": "Correct. The worked calculation gives 5.7 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5.7 A, not 8.55 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5.7 A, not 8.7 A.",
"D": "Incorrect. Substituting the stated values into... | [
"series resistors",
"parallel resistors",
"Ohm's law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_series_parallel | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009500 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_18H_27X_1. Treat X as the total number of monovalent halogen atoms. | {
"A": "5",
"B": "3",
"C": "7",
"D": "6"
} | A | 5 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×18+2+0−27−1)/2=5. | {
"A": "Correct. The worked calculation gives 5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5, not 3.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5, not 7.",
"D": "Incorrect. Substituting the stated values into the governing relati... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001395 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^7 (5x² -3x -5) dx. | {
"A": "463.16667",
"B": "370.53333",
"C": "347.375",
"D": "555.8"
} | A | 463.16667 | An antiderivative is (5/3)x³+(-3/2)x²+(-5)x. Evaluating between 0 and 7 gives (5/3)(7)³+(-3/2)(7)²+(-5)(7)=463.16667. | {
"A": "Correct. The worked calculation gives 463.16667.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 463.16667, not 370.53333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 463.16667, not 347.375.",
"D": "Incorrect. Substituting the s... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002250 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 6 red, 7 blue and 8 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "3.30476",
"B": "-1.69524",
"C": "0.30476",
"D": "-0.30476"
} | C | 0.30476 | Favourable pairs = C(6,2)+C(7,2)+C(8,2)=64. Total pairs=C(21,2)=210. Therefore P=64/210=0.30476. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.30476, not 3.30476.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.30476, not -1.69524.",
"C": "Correct. The worked calculation gives 0.30476.",
"D": "Incorrect. Substituting the stated v... | [
"classical probability",
"combinations",
"without replacement"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_probability_colours | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003950 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 13 kg block initially moves at 7 m s⁻¹. A constant net force does 15 N of work over 4 m in the direction of motion. Find its final kinetic energy. | {
"A": "283.875 J",
"B": "378.5 J",
"C": "454.2 J",
"D": "379.5 J"
} | B | 378.5 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(13)(7)²+(15)(4)=378.5 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 378.5 J, not 283.875 J.",
"B": "Correct. The worked calculation gives 378.5 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 378.5 J, not 454.2 J.",
"D": "Incorrect. Substituting the stated ... | [
"work–energy theorem",
"kinetic energy",
"work by constant force"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_work_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005343 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 5 μF capacitor is charged to a potential difference of 2 V. Calculate the energy stored, expressed in μJ. | {
"A": "20 μJ",
"B": "10 μJ",
"C": "7.5 μJ",
"D": "9 μJ"
} | B | 10 μJ | U=½CV². Since C=5 μF, U in μJ is ½×5×(2)²=10 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 10 μJ, not 20 μJ.",
"B": "Correct. The worked calculation gives 10 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 10 μJ, not 7.5 μJ.",
"D": "Incorrect. Substituting the stated values into... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009379 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_3H_3N_3. Treat X as the total number of monovalent halogen atoms. | {
"A": "4",
"B": "3",
"C": "6",
"D": "7"
} | A | 4 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×3+2+3−3−0)/2=4. | {
"A": "Correct. The worked calculation gives 4.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4, not 3.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4, not 6.",
"D": "Incorrect. Substituting the stated values into the governing relati... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003406 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 63 m s⁻¹ at 75° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "200.45 m",
"B": "197.45 m",
"C": "198.45 m",
"D": "99.225 m"
} | C | 198.45 m | Range R=u²sin(2θ)/g. Substituting u=63, θ=75° and g=10 gives R=(63)²sin(150°)/10=198.45 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 198.45 m, not 200.45 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 198.45 m, not 197.45 m.",
"C": "Correct. The worked calculation gives 198.45 m.",
"D": "Incorrect. Substituting the stat... | [
"projectile motion",
"range equation",
"trigonometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_projectile_range | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008752 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 0.27 mol L⁻¹ and initially B is absent. At equilibrium, 74% of A has dissociated at constant volume. Find K_c. | {
"A": "-0.72535",
"B": "4.54929",
"C": "1.22746",
"D": "2.27465"
} | D | 2.27465 | Let α=0.74. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.27)(0.74)²/(1−0.74)=2.27465. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.27465, not -0.72535.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.27465, not 4.54929.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.27465, not 1.22... | [
"chemical equilibrium",
"equilibrium constant",
"degree of dissociation"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_equilibrium_alpha | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002695 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(-3, -5, -6) and v=(-6, -2, -5), calculate u·v. | {
"A": "58",
"B": "46.4",
"C": "60",
"D": "29"
} | A | 58 | The dot product is the sum of corresponding component products: (-3)(-6)+(-5)(-2)+(-6)(-5)=58. | {
"A": "Correct. The worked calculation gives 58.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 58, not 46.4.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 58, not 60.",
"D": "Incorrect. Substituting the stated values into the governing... | [
"three-dimensional vectors",
"scalar product"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_vector_dot | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001476 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^2 (-4x² +4x +1) dx. | {
"A": "-1",
"B": "-0.5",
"C": "-3.66667",
"D": "-0.66667"
} | D | -0.66667 | An antiderivative is (-4/3)x³+(4/2)x²+(1)x. Evaluating between 0 and 2 gives (-4/3)(2)³+(4/2)(2)²+(1)(2)=-0.66667. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -0.66667, not -1.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -0.66667, not -0.5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -0.66667, not -3.66667."... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003903 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 15 kg block initially moves at 6 m s⁻¹. A constant net force does 15 N of work over 16 m in the direction of motion. Find its final kinetic energy. | {
"A": "510 J",
"B": "508 J",
"C": "509 J",
"D": "-510 J"
} | A | 510 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(15)(6)²+(15)(16)=510 J. | {
"A": "Correct. The worked calculation gives 510 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 510 J, not 508 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 510 J, not 509 J.",
"D": "Incorrect. Substituting the stated values into ... | [
"work–energy theorem",
"kinetic energy",
"work by constant force"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_work_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000137 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 3x^2 -14x -10=0, let its roots be α and β. What is α²+β²? | {
"A": "3.84444",
"B": "30.44444",
"C": "22.75556",
"D": "28.44444"
} | D | 28.44444 | By Vieta's relations, α+β = 14/3 and αβ = -10/3. Then α²+β²=(α+β)²−2αβ = (14/3)²−2(-10/3) = 28.44444. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 28.44444, not 3.84444.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 28.44444, not 30.44444.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 28.44444, not 2... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009350 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_3H_4X_2. Treat X as the total number of monovalent halogen atoms. | {
"A": "1.1",
"B": "1",
"C": "0.75",
"D": "2"
} | B | 1 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×3+2+0−4−2)/2=1. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1, not 1.1.",
"B": "Correct. The worked calculation gives 1.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1, not 0.75.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005043 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +1 μC and +4 μC are separated by 4 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "18 N",
"B": "21.5 N",
"C": "22.5 N",
"D": "24.5 N"
} | C | 22.5 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(1×10⁻⁶)(4×10⁻⁶)/(4×10⁻²)²=22.5 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 22.5 N, not 18 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 22.5 N, not 21.5 N.",
"C": "Correct. The worked calculation gives 22.5 N.",
"D": "Incorrect. Substituting the stated values in... | [
"Coulomb's law",
"SI unit conversion",
"inverse-square law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_coulomb | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009802 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Co(CN)_6]^-2, each CN ligand has charge −1. Determine the formal oxidation state of Co. | {
"A": "4",
"B": "6",
"C": "1",
"D": "1.4"
} | A | 4 | Let the formal oxidation state of Co be x. Charge balance gives x+(6)(−1)=-2; hence x=-2+6=+4. | {
"A": "Correct. The worked calculation gives 4.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4, not 6.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4, not 1.",
"D": "Incorrect. Substituting the stated values into the governing relati... | [
"coordination compounds",
"oxidation state",
"charge balance"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_complex_oxidation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009211 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.18 V, transfers 2 electrons, and has reaction quotient Q=10^2. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "1.1209 V",
"B": "4.1209 V",
"C": "2.1209 V",
"D": "0.56045 V"
} | A | 1.1209 V | Apply the Nernst equation: E=1.18−(0.0591/2)(2)=1.1209 V. | {
"A": "Correct. The worked calculation gives 1.1209 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.1209 V, not 4.1209 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.1209 V, not 2.1209 V.",
"D": "Incorrect. Substituting the stat... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005809 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 3.6 atm at 381 K. What is its pressure at 570 K? | {
"A": "8.38583 atm",
"B": "10.77165 atm",
"C": "-5.38583 atm",
"D": "5.38583 atm"
} | D | 5.38583 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(3.6)(570)/(381)=5.38583 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5.38583 atm, not 8.38583 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5.38583 atm, not 10.77165 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5... | [
"ideal gas law",
"isochoric process",
"absolute temperature"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_isochoric_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003059 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(10+6i)/(10+4i). What is Re(z), the real part of z? | {
"A": "0.80172",
"B": "1.06897",
"C": "-0.93103",
"D": "2.13793"
} | B | 1.06897 | Multiply numerator and denominator by the conjugate (10-4i). Then Re(z)=(ac+bd)/(c²+d²)=((10)(10)+(6)(4))/(10²+(4)²)=1.06897. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.06897, not 0.80172.",
"B": "Correct. The worked calculation gives 1.06897.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.06897, not -0.93103.",
"D": "Incorrect. Substituting the stated v... | [
"complex conjugate",
"complex division",
"real and imaginary parts"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_complex_quotient | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007720 | chemistry | physical_chemistry | first_order_half_life | hard | provisional_template_level_label | single_correct_mcq | A first-order reaction has a half-life of 20 minutes. If the initial reactant concentration is 60 mmol L⁻¹, what concentration remains after 80 minutes? | {
"A": "3.75 mmol L⁻¹",
"B": "1.875 mmol L⁻¹",
"C": "6.75 mmol L⁻¹",
"D": "2.75 mmol L⁻¹"
} | A | 3.75 mmol L⁻¹ | 80 minutes corresponds to 4 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=60/2^4=3.75 mmol L⁻¹. | {
"A": "Correct. The worked calculation gives 3.75 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.75 mmol L⁻¹, not 1.875 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.75 mmol L⁻¹, not 6.75 mmol L⁻¹.",
"D": "Incorr... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002492 | mathematics | linear_algebra | 2x2_determinant | hard | provisional_template_level_label | single_correct_mcq | Find the determinant of the matrix [[8, -7], [9, 2]]. | {
"A": "158",
"B": "79",
"C": "94.8",
"D": "59.25"
} | B | 79 | For a 2×2 matrix [[a,b],[c,d]], det=ad−bc. Here it is (8)(2)−(-7)(9)=79. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 79, not 158.",
"B": "Correct. The worked calculation gives 79.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 79, not 94.8.",
"D": "Incorrect. Substituting the stated values into the governin... | [
"determinants",
"matrix operations"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_determinant_2x2 | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000025 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 6x^2 -13x +6=0, let its roots be α and β. What is α²+β²? | {
"A": "3.69444",
"B": "2.02083",
"C": "2.69444",
"D": "4.04167"
} | C | 2.69444 | By Vieta's relations, α+β = 13/6 and αβ = 6/6. Then α²+β²=(α+β)²−2αβ = (13/6)²−2(6/6) = 2.69444. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.69444, not 3.69444.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.69444, not 2.02083.",
"C": "Correct. The worked calculation gives 2.69444.",
"D": "Incorrect. Substituting the stated va... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005051 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +5 μC and +8 μC are separated by 7 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "110.20408 N",
"B": "58.77551 N",
"C": "71.46939 N",
"D": "73.46939 N"
} | D | 73.46939 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(5×10⁻⁶)(8×10⁻⁶)/(7×10⁻²)²=73.46939 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 73.46939 N, not 110.20408 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 73.46939 N, not 58.77551 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 73.46... | [
"Coulomb's law",
"SI unit conversion",
"inverse-square law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_coulomb | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006974 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 6 mol H₂ and 29 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "3 mol",
"B": "7 mol",
"C": "6 mol",
"D": "9 mol"
} | C | 6 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 6 mol H₂O; O₂ can produce 58 mol H₂O. The limiting condition is H₂; product=6 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6 mol, not 3 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6 mol, not 7 mol.",
"C": "Correct. The worked calculation gives 6 mol.",
"D": "Incorrect. Substituting the stated values into ... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006824 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 2 mol H₂ and 5 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "-2 mol",
"B": "3 mol",
"C": "4 mol",
"D": "2 mol"
} | D | 2 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 2 mol H₂O; O₂ can produce 10 mol H₂O. The limiting condition is H₂; product=2 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2 mol, not -2 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2 mol, not 3 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2 mol, not 4 mol.",
"D"... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003439 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 107 m s⁻¹ at 65° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "1315.56642 m",
"B": "1052.45314 m",
"C": "701.63543 m",
"D": "877.04428 m"
} | D | 877.04428 m | Range R=u²sin(2θ)/g. Substituting u=107, θ=65° and g=10 gives R=(107)²sin(130°)/10=877.04428 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 877.04428 m, not 1315.56642 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 877.04428 m, not 1052.45314 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives ... | [
"projectile motion",
"range equation",
"trigonometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_projectile_range | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003236 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(12+13i)/(12+2i). What is Re(z), the real part of z? | {
"A": "0.57432",
"B": "1.14865",
"C": "1.11486",
"D": "-0.85135"
} | B | 1.14865 | Multiply numerator and denominator by the conjugate (12-2i). Then Re(z)=(ac+bd)/(c²+d²)=((12)(12)+(13)(2))/(12²+(2)²)=1.14865. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.14865, not 0.57432.",
"B": "Correct. The worked calculation gives 1.14865.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.14865, not 1.11486.",
"D": "Incorrect. Substituting the stated va... | [
"complex conjugate",
"complex division",
"real and imaginary parts"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_complex_quotient | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007276 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 14 mol of gas at 426 K and 1.79 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "273.40109 L",
"B": "410.10163 L",
"C": "-273.40109 L",
"D": "275.40109 L"
} | A | 273.40109 L | PV=nRT, so V=nRT/P=(14)(0.082057)(426)/(1.79)=273.40109 L. | {
"A": "Correct. The worked calculation gives 273.40109 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 273.40109 L, not 410.10163 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 273.40109 L, not -273.40109 L.",
"D": "Incorrect. Subst... | [
"ideal gas equation",
"unit consistency"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_ideal_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004041 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 46 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.26. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "179.32 N",
"B": "180.32 N",
"C": "360.64 N",
"D": "216.384 N"
} | B | 180.32 N | Net force=mg(sinθ−μcosθ)=46×10×(0.6−0.26×0.8)=180.32 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 180.32 N, not 179.32 N.",
"B": "Correct. The worked calculation gives 180.32 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 180.32 N, not 360.64 N.",
"D": "Incorrect. Substituting the stat... | [
"Newton's laws",
"inclined plane",
"kinetic friction"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_incline_friction | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008481 | chemistry | physical_chemistry | boiling_point_elevation | hard | provisional_template_level_label | single_correct_mcq | A non-volatile solute is dissolved in water to produce a solution of molality 1.6 mol kg⁻¹. If its van't Hoff factor is 2 and K_b for water is 0.512 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "2.4576 K",
"B": "0.6384 K",
"C": "1.16384 K",
"D": "1.6384 K"
} | D | 1.6384 K | ΔT_b=iK_bm. Substitution gives ΔT_b=2×0.512×1.6=1.6384 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.6384 K, not 2.4576 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.6384 K, not 0.6384 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.6384 K, not ... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005511 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 75 μF capacitor is charged to a potential difference of 6 V. Calculate the energy stored, expressed in μJ. | {
"A": "136 μJ",
"B": "-1350 μJ",
"C": "1350 μJ",
"D": "1349 μJ"
} | C | 1350 μJ | U=½CV². Since C=75 μF, U in μJ is ½×75×(6)²=1350 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1350 μJ, not 136 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1350 μJ, not -1350 μJ.",
"C": "Correct. The worked calculation gives 1350 μJ.",
"D": "Incorrect. Substituting the stated va... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006571 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 3 kg is attached to an ideal spring of force constant 606 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "0.28144 s",
"B": "1.14072 s",
"C": "0.14072 s",
"D": "0.21108 s"
} | C | 0.14072 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(3/606)=0.14072 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.14072 s, not 0.28144 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.14072 s, not 1.14072 s.",
"C": "Correct. The worked calculation gives 0.14072 s.",
"D": "Incorrect. Substituting the... | [
"simple harmonic motion",
"mass–spring oscillator",
"time period"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_shm_period | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001089 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-1x³ -6x² +4x+7. Evaluate f′(-1). | {
"A": "15",
"B": "14",
"C": "13",
"D": "9.75"
} | C | 13 | Differentiate: f′(x)=3(-1)x²+2(-6)x+(4). At x=-1, f′(-1)=3(-1)(-1)²+2(-6)(-1)+(4)=13. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 13, not 15.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 13, not 14.",
"C": "Correct. The worked calculation gives 13.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009105 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.47 V, transfers 3 electrons, and has reaction quotient Q=10^1. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "2.4503 V",
"B": "-1.5497 V",
"C": "-0.5497 V",
"D": "1.4503 V"
} | D | 1.4503 V | Apply the Nernst equation: E=1.47−(0.0591/3)(1)=1.4503 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.4503 V, not 2.4503 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.4503 V, not -1.5497 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.4503 V, not... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007695 | chemistry | physical_chemistry | first_order_half_life | hard | provisional_template_level_label | single_correct_mcq | A first-order reaction has a half-life of 20 minutes. If the initial reactant concentration is 51 mmol L⁻¹, what concentration remains after 80 minutes? | {
"A": "6.375 mmol L⁻¹",
"B": "1.31875 mmol L⁻¹",
"C": "3.1875 mmol L⁻¹",
"D": "2.1875 mmol L⁻¹"
} | C | 3.1875 mmol L⁻¹ | 80 minutes corresponds to 4 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=51/2^4=3.1875 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.1875 mmol L⁻¹, not 6.375 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.1875 mmol L⁻¹, not 1.31875 mmol L⁻¹.",
"C": "Correct. The worked calculation gives 3.1875 mmol L⁻¹.",
"D"... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001443 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^6 (3x² -5x +3) dx. | {
"A": "115.2",
"B": "172.8",
"C": "-144",
"D": "144"
} | D | 144 | An antiderivative is (3/3)x³+(-5/2)x²+(3)x. Evaluating between 0 and 6 gives (3/3)(6)³+(-5/2)(6)²+(3)(6)=144. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 144, not 115.2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 144, not 172.8.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 144, not -144.",
"D": "Corre... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007527 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 58 mL of 0.10 M HCl is mixed with 32 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "1.53927",
"B": "-0.46073",
"C": "4.53927",
"D": "1.15445"
} | A | 1.53927 | Excess H⁺ moles=0.10×(58−32)/1000=0.002600 mol. Total volume=0.090 L, so [H⁺]=0.028889 M and pH=−log₁₀[H⁺]=1.53927. | {
"A": "Correct. The worked calculation gives 1.53927.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.53927, not -0.46073.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.53927, not 4.53927.",
"D": "Incorrect. Substituting the stated v... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000149 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 3x^2 -10x +1=0, let its roots be α and β. What is α²+β²? | {
"A": "8.44444",
"B": "10.44444",
"C": "2.04444",
"D": "9.44444"
} | B | 10.44444 | By Vieta's relations, α+β = 10/3 and αβ = 1/3. Then α²+β²=(α+β)²−2αβ = (10/3)²−2(1/3) = 10.44444. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 10.44444, not 8.44444.",
"B": "Correct. The worked calculation gives 10.44444.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 10.44444, not 2.04444.",
"D": "Incorrect. Substituting the stated... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007791 | chemistry | physical_chemistry | first_order_half_life | hard | provisional_template_level_label | single_correct_mcq | A first-order reaction has a half-life of 20 minutes. If the initial reactant concentration is 61 mmol L⁻¹, what concentration remains after 140 minutes? | {
"A": "-2.52344 mmol L⁻¹",
"B": "-0.52344 mmol L⁻¹",
"C": "-0.47656 mmol L⁻¹",
"D": "0.47656 mmol L⁻¹"
} | D | 0.47656 mmol L⁻¹ | 140 minutes corresponds to 7 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=61/2^7=0.47656 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.47656 mmol L⁻¹, not -2.52344 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.47656 mmol L⁻¹, not -0.52344 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the gover... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001015 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=4x³ -9x² +4x+7. Evaluate f′(-4). | {
"A": "266",
"B": "268",
"C": "214.4",
"D": "402"
} | B | 268 | Differentiate: f′(x)=3(4)x²+2(-9)x+(4). At x=-4, f′(-4)=3(4)(-4)²+2(-9)(-4)+(4)=268. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 268, not 266.",
"B": "Correct. The worked calculation gives 268.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 268, not 214.4.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005669 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 68 μF capacitor is charged to a potential difference of 16 V. Calculate the energy stored, expressed in μJ. | {
"A": "8704 μJ",
"B": "4352 μJ",
"C": "10444.8 μJ",
"D": "-8704 μJ"
} | A | 8704 μJ | U=½CV². Since C=68 μF, U in μJ is ½×68×(16)²=8704 μJ. | {
"A": "Correct. The worked calculation gives 8704 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 8704 μJ, not 4352 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 8704 μJ, not 10444.8 μJ.",
"D": "Incorrect. Substituting the stated... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009575 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_12H_6N_2X_2. Treat X as the total number of monovalent halogen atoms. | {
"A": "10",
"B": "12",
"C": "2",
"D": "15"
} | A | 10 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×12+2+2−6−2)/2=10. | {
"A": "Correct. The worked calculation gives 10.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 10, not 12.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 10, not 2.",
"D": "Incorrect. Substituting the stated values into the governing re... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009846 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Cr(CN)_2]^-2, each CN ligand has charge −1. Determine the formal oxidation state of Cr. | {
"A": "-2",
"B": "0",
"C": "2",
"D": "-1"
} | B | 0 | Let the formal oxidation state of Cr be x. Charge balance gives x+(2)(−1)=-2; hence x=-2+2=+0. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0, not -2.",
"B": "Correct. The worked calculation gives 0.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0, not 2.",
"D": "Incorrect. Substituting the stated values into the governing relat... | [
"coordination compounds",
"oxidation state",
"charge balance"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_complex_oxidation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001841 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 4 elements can be selected from a set of 18 distinct elements? | {
"A": "2295",
"B": "3061",
"C": "4590",
"D": "3060"
} | D | 3060 | Order does not matter, so the count is C(18,4)=18!/[4!(14)!]=3060. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3060, not 2295.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3060, not 3061.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3060, not 4590.",
"D": "Corr... | [
"combinations",
"factorials"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_combinations | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005493 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 37 μF capacitor is charged to a potential difference of 14 V. Calculate the energy stored, expressed in μJ. | {
"A": "3627 μJ",
"B": "4351.2 μJ",
"C": "7252 μJ",
"D": "3626 μJ"
} | D | 3626 μJ | U=½CV². Since C=37 μF, U in μJ is ½×37×(14)²=3626 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3626 μJ, not 3627 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3626 μJ, not 4351.2 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3626 μJ, not 725... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000053 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 6x^2 -6x +4=0, let its roots be α and β. What is α²+β²? | {
"A": "-0.25",
"B": "-0.33333",
"C": "-0.16667",
"D": "-0.66667"
} | B | -0.33333 | By Vieta's relations, α+β = 6/6 and αβ = 4/6. Then α²+β²=(α+β)²−2αβ = (6/6)²−2(4/6) = -0.33333. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -0.33333, not -0.25.",
"B": "Correct. The worked calculation gives -0.33333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -0.33333, not -0.16667.",
"D": "Incorrect. Substituting the stated ... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004782 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 12 Ω and 5 Ω are connected in series; this combination is connected in parallel with a 12 Ω resistor across a 30 V ideal battery. Find the total current supplied by the battery. | {
"A": "-4.26471 A",
"B": "2.26471 A",
"C": "7.26471 A",
"D": "4.26471 A"
} | D | 4.26471 A | Series branch resistance=12+5=17 Ω. Parallel equivalent R_eq=(17×12)/(17+12)=7.03448 Ω. Total current I=V/R_eq=30/7.03448=4.26471 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.26471 A, not -4.26471 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4.26471 A, not 2.26471 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4.26471 A... | [
"series resistors",
"parallel resistors",
"Ohm's law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_series_parallel | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001890 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 3 elements can be selected from a set of 14 distinct elements? | {
"A": "364",
"B": "728",
"C": "362",
"D": "546"
} | A | 364 | Order does not matter, so the count is C(14,3)=14!/[3!(11)!]=364. | {
"A": "Correct. The worked calculation gives 364.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 364, not 728.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 364, not 362.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"combinations",
"factorials"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_combinations | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008927 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 1.27 mol L⁻¹ and initially B is absent. At equilibrium, 35% of A has dissociated at constant volume. Find K_c. | {
"A": "1.95738",
"B": "-1.04262",
"C": "1.43608",
"D": "0.95738"
} | D | 0.95738 | Let α=0.35. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.27)(0.35)²/(1−0.35)=0.95738. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.95738, not 1.95738.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.95738, not -1.04262.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.95738, not 1.43... | [
"chemical equilibrium",
"equilibrium constant",
"degree of dissociation"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_equilibrium_alpha | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009985 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Cr(OH)_4]^+0, each OH ligand has charge −1. Determine the formal oxidation state of Cr. | {
"A": "5",
"B": "4",
"C": "3",
"D": "-4"
} | B | 4 | Let the formal oxidation state of Cr be x. Charge balance gives x+(4)(−1)=+0; hence x=0+4=+4. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4, not 5.",
"B": "Correct. The worked calculation gives 4.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4, not 3.",
"D": "Incorrect. Substituting the stated values into the governing relati... | [
"coordination compounds",
"oxidation state",
"charge balance"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_complex_oxidation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006832 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 5 mol H₂ and 8 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "2.5 mol",
"B": "3.75 mol",
"C": "8 mol",
"D": "5 mol"
} | D | 5 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 5 mol H₂O; O₂ can produce 16 mol H₂O. The limiting condition is H₂; product=5 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not 2.5 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not 3.75 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not 8 mol.",
... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004308 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 30 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.29. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "109.4 N",
"B": "110.4 N",
"C": "12.04 N",
"D": "82.8 N"
} | B | 110.4 N | Net force=mg(sinθ−μcosθ)=30×10×(0.6−0.29×0.8)=110.4 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 110.4 N, not 109.4 N.",
"B": "Correct. The worked calculation gives 110.4 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 110.4 N, not 12.04 N.",
"D": "Incorrect. Substituting the stated va... | [
"Newton's laws",
"inclined plane",
"kinetic friction"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_incline_friction | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007859 | chemistry | physical_chemistry | first_order_half_life | hard | provisional_template_level_label | single_correct_mcq | A first-order reaction has a half-life of 20 minutes. If the initial reactant concentration is 94 mmol L⁻¹, what concentration remains after 160 minutes? | {
"A": "0.18359 mmol L⁻¹",
"B": "0.55078 mmol L⁻¹",
"C": "3.36719 mmol L⁻¹",
"D": "0.36719 mmol L⁻¹"
} | D | 0.36719 mmol L⁻¹ | 160 minutes corresponds to 8 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=94/2^8=0.36719 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.36719 mmol L⁻¹, not 0.18359 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.36719 mmol L⁻¹, not 0.55078 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governi... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007578 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 34 mL of 0.10 M HCl is mixed with 44 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "10.10791",
"B": "2.21079",
"C": "12.10791",
"D": "9.08093"
} | C | 12.10791 | Excess OH⁻ moles=0.10×(44−34)/1000=0.001000 mol. Total volume=0.078 L, so [OH⁻]=0.012821 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.10791. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12.10791, not 10.10791.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.10791, not 2.21079.",
"C": "Correct. The worked calculation gives 12.10791.",
"D": "Incorrect. Substituting the state... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007487 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 60 mL of 0.10 M HCl is mixed with 59 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "6.15109",
"B": "0.07555",
"C": "5.07555",
"D": "3.07555"
} | D | 3.07555 | Excess H⁺ moles=0.10×(60−59)/1000=0.000100 mol. Total volume=0.119 L, so [H⁺]=0.000840 M and pH=−log₁₀[H⁺]=3.07555. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.07555, not 6.15109.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.07555, not 0.07555.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.07555, not 5.075... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008542 | chemistry | physical_chemistry | boiling_point_elevation | hard | provisional_template_level_label | single_correct_mcq | A non-volatile solute is dissolved in water to produce a solution of molality 0.33 mol kg⁻¹. If its van't Hoff factor is 1 and K_b for water is 0.6 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "0.198 K",
"B": "-2.802 K",
"C": "-0.198 K",
"D": "3.198 K"
} | A | 0.198 K | ΔT_b=iK_bm. Substitution gives ΔT_b=1×0.6×0.33=0.198 K. | {
"A": "Correct. The worked calculation gives 0.198 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.198 K, not -2.802 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.198 K, not -0.198 K.",
"D": "Incorrect. Substituting the stated ... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002889 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(7, -2, 8) and v=(-2, 8, -5), calculate u·v. | {
"A": "-105",
"B": "70",
"C": "-69",
"D": "-70"
} | D | -70 | The dot product is the sum of corresponding component products: (7)(-2)+(-2)(8)+(8)(-5)=-70. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -70, not -105.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -70, not 70.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -70, not -69.",
"D": "Correct. T... | [
"three-dimensional vectors",
"scalar product"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_vector_dot | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009009 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.53 V, transfers 1 electrons, and has reaction quotient Q=10^-3. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "0.7073 V",
"B": "-2.2927 V",
"C": "0.35365 V",
"D": "3.7073 V"
} | A | 0.7073 V | Apply the Nernst equation: E=0.53−(0.0591/1)(-3)=0.7073 V. | {
"A": "Correct. The worked calculation gives 0.7073 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.7073 V, not -2.2927 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.7073 V, not 0.35365 V.",
"D": "Incorrect. Substituting the st... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008574 | chemistry | physical_chemistry | boiling_point_elevation | hard | provisional_template_level_label | single_correct_mcq | A non-volatile solute is dissolved in water to produce a solution of molality 0.51 mol kg⁻¹. If its van't Hoff factor is 3 and K_b for water is 1.0 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "0.765 K",
"B": "1.53 K",
"C": "1.153 K",
"D": "1.1475 K"
} | B | 1.53 K | ΔT_b=iK_bm. Substitution gives ΔT_b=3×1.0×0.51=1.53 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.53 K, not 0.765 K.",
"B": "Correct. The worked calculation gives 1.53 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.53 K, not 1.153 K.",
"D": "Incorrect. Substituting the stated value... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001660 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^3 (-2x² +0x -2) dx. | {
"A": "-26",
"B": "-23",
"C": "-24",
"D": "24"
} | C | -24 | An antiderivative is (-2/3)x³+(0/2)x²+(-2)x. Evaluating between 0 and 3 gives (-2/3)(3)³+(0/2)(3)²+(-2)(3)=-24. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -24, not -26.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -24, not -23.",
"C": "Correct. The worked calculation gives -24.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008308 | chemistry | physical_chemistry | faradays_law_copper | very_hard | provisional_template_level_label | single_correct_mcq | An aqueous Cu²⁺ solution is electrolysed using a current of 3 A for 4973 s. Assuming 100% current efficiency, calculate the mass of copper deposited. Use M(Cu)=63.5 g mol⁻¹ and F=96500 C mol⁻¹. | {
"A": "7.90858 g",
"B": "6.90858 g",
"C": "1.90858 g",
"D": "4.90858 g"
} | D | 4.90858 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×3×4973/(2×96500)=4.90858 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.90858 g, not 7.90858 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4.90858 g, not 6.90858 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4.90858 g,... | [
"electrolysis",
"Faraday's laws",
"redox stoichiometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_faraday_copper | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009010 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.23 V, transfers 1 electrons, and has reaction quotient Q=10^-6. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "0.5846 V",
"B": "1.5846 V",
"C": "-1.5846 V",
"D": "-0.4154 V"
} | B | 1.5846 V | Apply the Nernst equation: E=1.23−(0.0591/1)(-6)=1.5846 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.5846 V, not 0.5846 V.",
"B": "Correct. The worked calculation gives 1.5846 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.5846 V, not -1.5846 V.",
"D": "Incorrect. Substituting the sta... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005097 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +1 μC and +3 μC are separated by 4 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "25.3125 N",
"B": "16.875 N",
"C": "2.6875 N",
"D": "15.875 N"
} | B | 16.875 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(1×10⁻⁶)(3×10⁻⁶)/(4×10⁻²)²=16.875 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 16.875 N, not 25.3125 N.",
"B": "Correct. The worked calculation gives 16.875 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 16.875 N, not 2.6875 N.",
"D": "Incorrect. Substituting the sta... | [
"Coulomb's law",
"SI unit conversion",
"inverse-square law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_coulomb | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009109 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.55 V, transfers 3 electrons, and has reaction quotient Q=10^3. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "2.4909 V",
"B": "1.4909 V",
"C": "0.4909 V",
"D": "0.36818 V"
} | C | 0.4909 V | Apply the Nernst equation: E=0.55−(0.0591/3)(3)=0.4909 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.4909 V, not 2.4909 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.4909 V, not 1.4909 V.",
"C": "Correct. The worked calculation gives 0.4909 V.",
"D": "Incorrect. Substituting the stat... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007602 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 93 mL of 0.10 M HCl is mixed with 86 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "0.40775",
"B": "1.24078",
"C": "3.61163",
"D": "2.40775"
} | D | 2.40775 | Excess H⁺ moles=0.10×(93−86)/1000=0.000700 mol. Total volume=0.179 L, so [H⁺]=0.003911 M and pH=−log₁₀[H⁺]=2.40775. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.40775, not 0.40775.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.40775, not 1.24078.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.40775, not 3.611... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001247 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-5x³ -2x² +9x+7. Evaluate f′(2). | {
"A": "-58",
"B": "-29.5",
"C": "-59",
"D": "-70.8"
} | C | -59 | Differentiate: f′(x)=3(-5)x²+2(-2)x+(9). At x=2, f′(2)=3(-5)(2)²+2(-2)(2)+(9)=-59. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -59, not -58.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -59, not -29.5.",
"C": "Correct. The worked calculation gives -59.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007509 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 53 mL of 0.10 M HCl is mixed with 45 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "2.08814",
"B": "4.17627",
"C": "3.1322",
"D": "0.08814"
} | A | 2.08814 | Excess H⁺ moles=0.10×(53−45)/1000=0.000800 mol. Total volume=0.098 L, so [H⁺]=0.008163 M and pH=−log₁₀[H⁺]=2.08814. | {
"A": "Correct. The worked calculation gives 2.08814.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.08814, not 4.17627.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.08814, not 3.1322.",
"D": "Incorrect. Substituting the stated val... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006366 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 21 kg is attached to an ideal spring of force constant 732 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "2.33875 s",
"B": "0.67751 s",
"C": "-0.66125 s",
"D": "0.33875 s"
} | D | 0.33875 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(21/732)=0.33875 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.33875 s, not 2.33875 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.33875 s, not 0.67751 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.33875 s,... | [
"simple harmonic motion",
"mass–spring oscillator",
"time period"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_shm_period | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008834 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 0.66 mol L⁻¹ and initially B is absent. At equilibrium, 24% of A has dissociated at constant volume. Find K_c. | {
"A": "1.20008",
"B": "0.20008",
"C": "0.40017",
"D": "0.30013"
} | B | 0.20008 | Let α=0.24. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.66)(0.24)²/(1−0.24)=0.20008. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.20008, not 1.20008.",
"B": "Correct. The worked calculation gives 0.20008.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.20008, not 0.40017.",
"D": "Incorrect. Substituting the stated va... | [
"chemical equilibrium",
"equilibrium constant",
"degree of dissociation"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_equilibrium_alpha | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005358 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 53 μF capacitor is charged to a potential difference of 38 V. Calculate the energy stored, expressed in μJ. | {
"A": "38266 μJ",
"B": "19133 μJ",
"C": "38265 μJ",
"D": "38268 μJ"
} | A | 38266 μJ | U=½CV². Since C=53 μF, U in μJ is ½×53×(38)²=38266 μJ. | {
"A": "Correct. The worked calculation gives 38266 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 38266 μJ, not 19133 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 38266 μJ, not 38265 μJ.",
"D": "Incorrect. Substituting the stat... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007400 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 40 mL of 0.10 M HCl is mixed with 45 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "11.76955",
"B": "8.76955",
"C": "-11.76955",
"D": "17.65433"
} | A | 11.76955 | Excess OH⁻ moles=0.10×(45−40)/1000=0.000500 mol. Total volume=0.085 L, so [OH⁻]=0.005882 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=11.76955. | {
"A": "Correct. The worked calculation gives 11.76955.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 11.76955, not 8.76955.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 11.76955, not -11.76955.",
"D": "Incorrect. Substituting the stat... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006756 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 47 mol H₂ and 1 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "2 mol",
"B": "3 mol",
"C": "4 mol",
"D": "-2 mol"
} | A | 2 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 47 mol H₂O; O₂ can produce 2 mol H₂O. The limiting condition is O₂; product=2 mol. | {
"A": "Correct. The worked calculation gives 2 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2 mol, not 3 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2 mol, not 4 mol.",
"D": "Incorrect. Substituting the stated values into ... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006094 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 71 cm in front of a thin convex lens of focal length 29 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "1.06905",
"B": "-1.30952",
"C": "-2.30952",
"D": "0.69048"
} | D | 0.69048 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(71)(29)/(71−29)=49.02381 cm. Thus |m|=v/u=49.02381/71=0.69048. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.69048, not 1.06905.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.69048, not -1.30952.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.69048, not -2.3... | [
"thin lens equation",
"linear magnification",
"real image"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_lens_magnification | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009229 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.50 V, transfers 3 electrons, and has reaction quotient Q=10^-4. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "0.5788 V",
"B": "3.5788 V",
"C": "-1.4212 V",
"D": "1.5788 V"
} | A | 0.5788 V | Apply the Nernst equation: E=0.50−(0.0591/3)(-4)=0.5788 V. | {
"A": "Correct. The worked calculation gives 0.5788 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.5788 V, not 3.5788 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.5788 V, not -1.4212 V.",
"D": "Incorrect. Substituting the sta... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009730 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Mn(Cl)_2]^-1, each Cl ligand has charge −1. Determine the formal oxidation state of Mn. | {
"A": "1",
"B": "0.75",
"C": "-1",
"D": "-2"
} | A | 1 | Let the formal oxidation state of Mn be x. Charge balance gives x+(2)(−1)=-1; hence x=-1+2=+1. | {
"A": "Correct. The worked calculation gives 1.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1, not 0.75.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1, not -1.",
"D": "Incorrect. Substituting the stated values into the governing re... | [
"coordination compounds",
"oxidation state",
"charge balance"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_complex_oxidation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005672 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 5 atm at 193 K. What is its pressure at 317 K? | {
"A": "8.21244 atm",
"B": "9.21244 atm",
"C": "10.21244 atm",
"D": "12.31865 atm"
} | A | 8.21244 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(5)(317)/(193)=8.21244 atm. | {
"A": "Correct. The worked calculation gives 8.21244 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 8.21244 atm, not 9.21244 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 8.21244 atm, not 10.21244 atm.",
"D": "Incorrect. Subst... | [
"ideal gas law",
"isochoric process",
"absolute temperature"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_isochoric_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007721 | chemistry | physical_chemistry | first_order_half_life | hard | provisional_template_level_label | single_correct_mcq | A first-order reaction has a half-life of 20 minutes. If the initial reactant concentration is 63 mmol L⁻¹, what concentration remains after 120 minutes? | {
"A": "1.47656 mmol L⁻¹",
"B": "0.98438 mmol L⁻¹",
"C": "0.49219 mmol L⁻¹",
"D": "-1.01562 mmol L⁻¹"
} | B | 0.98438 mmol L⁻¹ | 120 minutes corresponds to 6 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=63/2^6=0.98438 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.98438 mmol L⁻¹, not 1.47656 mmol L⁻¹.",
"B": "Correct. The worked calculation gives 0.98438 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.98438 mmol L⁻¹, not 0.49219 mmol L⁻¹.",
... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005961 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 2.8 atm at 397 K. What is its pressure at 728 K? | {
"A": "7.13451 atm",
"B": "2.13451 atm",
"C": "5.13451 atm",
"D": "7.70176 atm"
} | C | 5.13451 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(2.8)(728)/(397)=5.13451 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5.13451 atm, not 7.13451 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5.13451 atm, not 2.13451 atm.",
"C": "Correct. The worked calculation gives 5.13451 atm.",
"D": "Incorrect. Substi... | [
"ideal gas law",
"isochoric process",
"absolute temperature"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_isochoric_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007511 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 11 mL of 0.10 M HCl is mixed with 17 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "9.33099",
"B": "10.33099",
"C": "12.33099",
"D": "14.33099"
} | C | 12.33099 | Excess OH⁻ moles=0.10×(17−11)/1000=0.000600 mol. Total volume=0.028 L, so [OH⁻]=0.021429 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.33099. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12.33099, not 9.33099.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.33099, not 10.33099.",
"C": "Correct. The worked calculation gives 12.33099.",
"D": "Incorrect. Substituting the state... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008814 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 0.3 mol L⁻¹ and initially B is absent. At equilibrium, 61% of A has dissociated at constant volume. Find K_c. | {
"A": "-0.85508",
"B": "1.71738",
"C": "1.14492",
"D": "2.28985"
} | C | 1.14492 | Let α=0.61. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.3)(0.61)²/(1−0.61)=1.14492. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.14492, not -0.85508.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.14492, not 1.71738.",
"C": "Correct. The worked calculation gives 1.14492.",
"D": "Incorrect. Substituting the stated v... | [
"chemical equilibrium",
"equilibrium constant",
"degree of dissociation"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_equilibrium_alpha | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009100 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.89 V, transfers 3 electrons, and has reaction quotient Q=10^2. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "-1.1494 V",
"B": "0.8506 V",
"C": "1.08506 V",
"D": "2.8506 V"
} | B | 0.8506 V | Apply the Nernst equation: E=0.89−(0.0591/3)(2)=0.8506 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.8506 V, not -1.1494 V.",
"B": "Correct. The worked calculation gives 0.8506 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.8506 V, not 1.08506 V.",
"D": "Incorrect. Substituting the st... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000191 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 5x^2 +13x -1=0, let its roots be α and β. What is α²+β²? | {
"A": "8.16",
"B": "7.16",
"C": "-7.16",
"D": "5.37"
} | B | 7.16 | By Vieta's relations, α+β = -13/5 and αβ = -1/5. Then α²+β²=(α+β)²−2αβ = (-13/5)²−2(-1/5) = 7.16. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7.16, not 8.16.",
"B": "Correct. The worked calculation gives 7.16.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7.16, not -7.16.",
"D": "Incorrect. Substituting the stated values into the ... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001664 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^8 (6x² +3x +6) dx. | {
"A": "1166",
"B": "934.4",
"C": "1168",
"D": "117.8"
} | C | 1168 | An antiderivative is (6/3)x³+(3/2)x²+(6)x. Evaluating between 0 and 8 gives (6/3)(8)³+(3/2)(8)²+(6)(8)=1168. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1168, not 1166.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1168, not 934.4.",
"C": "Correct. The worked calculation gives 1168.",
"D": "Incorrect. Substituting the stated values into the ... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008432 | chemistry | physical_chemistry | boiling_point_elevation | hard | provisional_template_level_label | single_correct_mcq | A non-volatile solute is dissolved in water to produce a solution of molality 0.67 mol kg⁻¹. If its van't Hoff factor is 4 and K_b for water is 1.0 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "1.68 K",
"B": "3.68 K",
"C": "2.68 K",
"D": "1.268 K"
} | C | 2.68 K | ΔT_b=iK_bm. Substitution gives ΔT_b=4×1.0×0.67=2.68 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.68 K, not 1.68 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.68 K, not 3.68 K.",
"C": "Correct. The worked calculation gives 2.68 K.",
"D": "Incorrect. Substituting the stated values ... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006035 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 17 cm in front of a thin convex lens of focal length 11 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "1.83333",
"B": "1.18333",
"C": "2.83333",
"D": "-0.16667"
} | A | 1.83333 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(17)(11)/(17−11)=31.16667 cm. Thus |m|=v/u=31.16667/17=1.83333. | {
"A": "Correct. The worked calculation gives 1.83333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.83333, not 1.18333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.83333, not 2.83333.",
"D": "Incorrect. Substituting the stated va... | [
"thin lens equation",
"linear magnification",
"real image"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_lens_magnification | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009516 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_9H_14N_1X_3. Treat X as the total number of monovalent halogen atoms. | {
"A": "1",
"B": "2",
"C": "4",
"D": "5"
} | B | 2 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×9+2+1−14−3)/2=2. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2, not 1.",
"B": "Correct. The worked calculation gives 2.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2, not 4.",
"D": "Incorrect. Substituting the stated values into the governing relati... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000766 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 1 and common ratio -2. What is the sum of its first 11 terms? | {
"A": "1024.5",
"B": "1366",
"C": "684",
"D": "683"
} | D | 683 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=1(-2^11−1)/(-2−1)=683. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 683, not 1024.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 683, not 1366.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 683, not 684.",
"D": "Correc... | [
"geometric progression",
"finite geometric sum"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_gp_sum | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002760 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(4, 0, 9) and v=(5, 0, 4), calculate u·v. | {
"A": "28",
"B": "56",
"C": "44.8",
"D": "54"
} | B | 56 | The dot product is the sum of corresponding component products: (4)(5)+(0)(0)+(9)(4)=56. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 56, not 28.",
"B": "Correct. The worked calculation gives 56.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 56, not 44.8.",
"D": "Incorrect. Substituting the stated values into the governing... | [
"three-dimensional vectors",
"scalar product"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_vector_dot | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004742 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 3 Ω and 15 Ω are connected in series; this combination is connected in parallel with a 7 Ω resistor across a 18 V ideal battery. Find the total current supplied by the battery. | {
"A": "1.35714 A",
"B": "2.67857 A",
"C": "4.57143 A",
"D": "3.57143 A"
} | D | 3.57143 A | Series branch resistance=3+15=18 Ω. Parallel equivalent R_eq=(18×7)/(18+7)=5.04 Ω. Total current I=V/R_eq=18/5.04=3.57143 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.57143 A, not 1.35714 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.57143 A, not 2.67857 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.57143 A,... | [
"series resistors",
"parallel resistors",
"Ohm's law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_series_parallel | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000079 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 2x^2 -7x -11=0, let its roots be α and β. What is α²+β²? | {
"A": "18.6",
"B": "34.875",
"C": "46.5",
"D": "23.25"
} | D | 23.25 | By Vieta's relations, α+β = 7/2 and αβ = -11/2. Then α²+β²=(α+β)²−2αβ = (7/2)²−2(-11/2) = 23.25. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 23.25, not 18.6.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 23.25, not 34.875.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 23.25, not 46.5.",
"D": ... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005204 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +8 μC and +5 μC are separated by 9 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "46.44444 N",
"B": "44.44444 N",
"C": "5.44444 N",
"D": "45.44444 N"
} | B | 44.44444 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(8×10⁻⁶)(5×10⁻⁶)/(9×10⁻²)²=44.44444 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 44.44444 N, not 46.44444 N.",
"B": "Correct. The worked calculation gives 44.44444 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 44.44444 N, not 5.44444 N.",
"D": "Incorrect. Substituting... | [
"Coulomb's law",
"SI unit conversion",
"inverse-square law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_coulomb | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000830 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 6 and common ratio 3. What is the sum of its first 7 terms? | {
"A": "9837",
"B": "6559",
"C": "6558",
"D": "6556"
} | C | 6558 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=6(3^7−1)/(3−1)=6558. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6558, not 9837.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6558, not 6559.",
"C": "Correct. The worked calculation gives 6558.",
"D": "Incorrect. Substituting the stated values into the g... | [
"geometric progression",
"finite geometric sum"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_gp_sum | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
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