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_005134 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +3 μC and +7 μC are separated by 7 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "38.57143 N",
"B": "39.57143 N",
"C": "40.57143 N",
"D": "57.85714 N"
} | A | 38.57143 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(3×10⁻⁶)(7×10⁻⁶)/(7×10⁻²)²=38.57143 N. | {
"A": "Correct. The worked calculation gives 38.57143 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 38.57143 N, not 39.57143 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 38.57143 N, not 40.57143 N.",
"D": "Incorrect. Substitutin... | [
"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_008839 | 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 1.89 mol L⁻¹ and initially B is absent. At equilibrium, 35% of A has dissociated at constant volume. Find K_c. | {
"A": "1.42477",
"B": "0.42477",
"C": "1.06858",
"D": "-1.42477"
} | A | 1.42477 | Let α=0.35. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.89)(0.35)²/(1−0.35)=1.42477. | {
"A": "Correct. The worked calculation gives 1.42477.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.42477, not 0.42477.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.42477, not 1.06858.",
"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_004754 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 5 Ω and 7 Ω are connected in series; this combination is connected in parallel with a 11 Ω resistor across a 30 V ideal battery. Find the total current supplied by the battery. | {
"A": "3.92045 A",
"B": "10.45455 A",
"C": "-5.22727 A",
"D": "5.22727 A"
} | D | 5.22727 A | Series branch resistance=5+7=12 Ω. Parallel equivalent R_eq=(12×11)/(12+11)=5.73913 Ω. Total current I=V/R_eq=30/5.73913=5.22727 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5.22727 A, not 3.92045 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5.22727 A, not 10.45455 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5.22727 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_009031 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.73 V, transfers 2 electrons, and has reaction quotient Q=10^6. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "3.5527 V",
"B": "1.5527 V",
"C": "-0.4473 V",
"D": "0.5527 V"
} | D | 0.5527 V | Apply the Nernst equation: E=0.73−(0.0591/2)(6)=0.5527 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.5527 V, not 3.5527 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.5527 V, not 1.5527 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.5527 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_000041 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 4x^2 -6x +4=0, let its roots be α and β. What is α²+β²? | {
"A": "-0.75",
"B": "1.25",
"C": "-2.75",
"D": "0.25"
} | D | 0.25 | By Vieta's relations, α+β = 6/4 and αβ = 4/4. Then α²+β²=(α+β)²−2αβ = (6/4)²−2(4/4) = 0.25. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.25, not -0.75.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.25, not 1.25.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.25, not -2.75.",
"D": "Co... | [
"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_001133 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=3x³ +2x² +7x+7. Evaluate f′(2). | {
"A": "6.1",
"B": "51",
"C": "76.5",
"D": "52"
} | B | 51 | Differentiate: f′(x)=3(3)x²+2(2)x+(7). At x=2, f′(2)=3(3)(2)²+2(2)(2)+(7)=51. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 51, not 6.1.",
"B": "Correct. The worked calculation gives 51.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 51, not 76.5.",
"D": "Incorrect. Substituting the stated values into the governin... | [
"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_007596 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 77 mL of 0.10 M HCl is mixed with 74 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "5.70186",
"B": "3.70186",
"C": "2.70186",
"D": "5.40371"
} | C | 2.70186 | Excess H⁺ moles=0.10×(77−74)/1000=0.000300 mol. Total volume=0.151 L, so [H⁺]=0.001987 M and pH=−log₁₀[H⁺]=2.70186. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.70186, not 5.70186.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.70186, not 3.70186.",
"C": "Correct. The worked calculation gives 2.70186.",
"D": "Incorrect. Substituting the stated va... | [
"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_004133 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 19 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.31. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "67.88 N",
"B": "50.16 N",
"C": "80.256 N",
"D": "66.88 N"
} | D | 66.88 N | Net force=mg(sinθ−μcosθ)=19×10×(0.6−0.31×0.8)=66.88 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 66.88 N, not 67.88 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 66.88 N, not 50.16 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 66.88 N, not 80.25... | [
"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_006858 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 18 mol H₂ and 27 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "21.6 mol",
"B": "14.4 mol",
"C": "19 mol",
"D": "18 mol"
} | D | 18 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 18 mol H₂O; O₂ can produce 54 mol H₂O. The limiting condition is H₂; product=18 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 18 mol, not 21.6 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 18 mol, not 14.4 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 18 mol, not 19 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_002056 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 11 red, 5 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": "-1.66304",
"B": "0.16848",
"C": "0.33696",
"D": "0.25272"
} | C | 0.33696 | Favourable pairs = C(11,2)+C(5,2)+C(8,2)=93. Total pairs=C(24,2)=276. Therefore P=93/276=0.33696. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.33696, not -1.66304.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.33696, not 0.16848.",
"C": "Correct. The worked calculation gives 0.33696.",
"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_004151 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 5 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.28. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "19.8 N",
"B": "20.8 N",
"C": "18.8 N",
"D": "16.8 N"
} | C | 18.8 N | Net force=mg(sinθ−μcosθ)=5×10×(0.6−0.28×0.8)=18.8 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 18.8 N, not 19.8 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 18.8 N, not 20.8 N.",
"C": "Correct. The worked calculation gives 18.8 N.",
"D": "Incorrect. Substituting the stated values ... | [
"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_005737 | 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 1.5 atm at 269 K. What is its pressure at 669 K? | {
"A": "-3.73048 atm",
"B": "2.79786 atm",
"C": "1.73048 atm",
"D": "3.73048 atm"
} | D | 3.73048 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(1.5)(669)/(269)=3.73048 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.73048 atm, not -3.73048 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.73048 atm, not 2.79786 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3... | [
"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_007394 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 61 mL of 0.10 M HCl is mixed with 84 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "11.20036",
"B": "6.10018",
"C": "18.30054",
"D": "12.20036"
} | D | 12.20036 | Excess OH⁻ moles=0.10×(84−61)/1000=0.002300 mol. Total volume=0.145 L, so [OH⁻]=0.015862 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.20036. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12.20036, not 11.20036.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.20036, not 6.10018.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12.20036, not 1... | [
"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_009793 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Fe(Cl)_5]^-1, each Cl ligand has charge −1. Determine the formal oxidation state of Fe. | {
"A": "4",
"B": "1",
"C": "2",
"D": "6"
} | A | 4 | Let the formal oxidation state of Fe be x. Charge balance gives x+(5)(−1)=-1; hence x=-1+5=+4. | {
"A": "Correct. The worked calculation gives 4.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4, not 1.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4, not 2.",
"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_001144 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-4x³ +9x² -1x+7. Evaluate f′(5). | {
"A": "-253.2",
"B": "-20.1",
"C": "-211",
"D": "211"
} | C | -211 | Differentiate: f′(x)=3(-4)x²+2(9)x+(-1). At x=5, f′(5)=3(-4)(5)²+2(9)(5)+(-1)=-211. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -211, not -253.2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -211, not -20.1.",
"C": "Correct. The worked calculation gives -211.",
"D": "Incorrect. Substituting the stated values into th... | [
"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_009815 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Pd(CN)_3]^+3, each CN ligand has charge −1. Determine the formal oxidation state of Pd. | {
"A": "8",
"B": "1.6",
"C": "6",
"D": "-6"
} | C | 6 | Let the formal oxidation state of Pd be x. Charge balance gives x+(3)(−1)=+3; hence x=3+3=+6. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6, not 8.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6, not 1.6.",
"C": "Correct. The worked calculation gives 6.",
"D": "Incorrect. Substituting the stated values into the governing rela... | [
"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_001381 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^4 (2x² -5x +6) dx. | {
"A": "-26.66667",
"B": "27.66667",
"C": "32",
"D": "26.66667"
} | D | 26.66667 | An antiderivative is (2/3)x³+(-5/2)x²+(6)x. Evaluating between 0 and 4 gives (2/3)(4)³+(-5/2)(4)²+(6)(4)=26.66667. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 26.66667, not -26.66667.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 26.66667, not 27.66667.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 26.66667, not... | [
"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_000281 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 5x^2 -9x -10=0, let its roots be α and β. What is α²+β²? | {
"A": "5.24",
"B": "10.86",
"C": "8.24",
"D": "7.24"
} | D | 7.24 | By Vieta's relations, α+β = 9/5 and αβ = -10/5. Then α²+β²=(α+β)²−2αβ = (9/5)²−2(-10/5) = 7.24. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7.24, not 5.24.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 7.24, not 10.86.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7.24, not 8.24.",
"D": "Cor... | [
"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_002319 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 10 red, 5 blue and 7 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "1.329",
"B": "-0.329",
"C": "-0.671",
"D": "0.329"
} | D | 0.329 | Favourable pairs = C(10,2)+C(5,2)+C(7,2)=76. Total pairs=C(22,2)=231. Therefore P=76/231=0.329. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.329, not 1.329.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.329, not -0.329.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.329, not -0.671.",
"D... | [
"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_005739 | 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 4.2 atm at 352 K. What is its pressure at 458 K? | {
"A": "5.46477 atm",
"B": "6.46477 atm",
"C": "4.46477 atm",
"D": "7.46477 atm"
} | A | 5.46477 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(4.2)(458)/(352)=5.46477 atm. | {
"A": "Correct. The worked calculation gives 5.46477 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5.46477 atm, not 6.46477 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5.46477 atm, not 4.46477 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_005775 | 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 1.9 atm at 424 K. What is its pressure at 580 K? | {
"A": "3.59906 atm",
"B": "2.59906 atm",
"C": "1.59906 atm",
"D": "5.59906 atm"
} | B | 2.59906 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(1.9)(580)/(424)=2.59906 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.59906 atm, not 3.59906 atm.",
"B": "Correct. The worked calculation gives 2.59906 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.59906 atm, not 1.59906 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_003932 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 3 kg block initially moves at 7 m s⁻¹. A constant net force does 16 N of work over 10 m in the direction of motion. Find its final kinetic energy. | {
"A": "234.5 J",
"B": "233.5 J",
"C": "175.125 J",
"D": "-233.5 J"
} | B | 233.5 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(3)(7)²+(16)(10)=233.5 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 233.5 J, not 234.5 J.",
"B": "Correct. The worked calculation gives 233.5 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 233.5 J, not 175.125 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_003311 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(9+13i)/(10+3i). What is Re(z), the real part of z? | {
"A": "1.11835",
"B": "1.18349",
"C": "1.77523",
"D": "2.36697"
} | B | 1.18349 | Multiply numerator and denominator by the conjugate (10-3i). Then Re(z)=(ac+bd)/(c²+d²)=((9)(10)+(13)(3))/(10²+(3)²)=1.18349. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.18349, not 1.11835.",
"B": "Correct. The worked calculation gives 1.18349.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.18349, not 1.77523.",
"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_009249 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.22 V, transfers 3 electrons, and has reaction quotient Q=10^4. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "4.1412 V",
"B": "1.1412 V",
"C": "1.11412 V",
"D": "3.1412 V"
} | B | 1.1412 V | Apply the Nernst equation: E=1.22−(0.0591/3)(4)=1.1412 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.1412 V, not 4.1412 V.",
"B": "Correct. The worked calculation gives 1.1412 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.1412 V, not 1.11412 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_004019 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 24 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.42. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "76.032 N",
"B": "63.36 N",
"C": "65.36 N",
"D": "62.36 N"
} | B | 63.36 N | Net force=mg(sinθ−μcosθ)=24×10×(0.6−0.42×0.8)=63.36 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 63.36 N, not 76.032 N.",
"B": "Correct. The worked calculation gives 63.36 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 63.36 N, not 65.36 N.",
"D": "Incorrect. Substituting the stated v... | [
"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_005077 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +9 μC and +9 μC are separated by 3 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "810 N",
"B": "-810 N",
"C": "607.5 N",
"D": "82 N"
} | A | 810 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(9×10⁻⁶)(9×10⁻⁶)/(3×10⁻²)²=810 N. | {
"A": "Correct. The worked calculation gives 810 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 810 N, not -810 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 810 N, not 607.5 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_009510 | 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_10H_10N_3X_3. Treat X as the total number of monovalent halogen atoms. | {
"A": "3",
"B": "-6",
"C": "6",
"D": "7"
} | C | 6 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×10+2+3−10−3)/2=6. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6, not 3.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6, not -6.",
"C": "Correct. The worked calculation gives 6.",
"D": "Incorrect. Substituting the stated values into the governing relat... | [
"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_009672 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Fe(Cl)_2]^-4, each Cl ligand has charge −1. Determine the formal oxidation state of Fe. | {
"A": "-2",
"B": "-1.5",
"C": "-3",
"D": "0.8"
} | A | -2 | Let the formal oxidation state of Fe be x. Charge balance gives x+(2)(−1)=-4; hence x=-4+2=-2. | {
"A": "Correct. The worked calculation gives -2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -2, not -1.5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -2, not -3.",
"D": "Incorrect. Substituting the stated values into the governing... | [
"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_009667 | 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_13H_14. Treat X as the total number of monovalent halogen atoms. | {
"A": "-7",
"B": "10",
"C": "7",
"D": "10.5"
} | C | 7 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×13+2+0−14−0)/2=7. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7, not -7.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 7, not 10.",
"C": "Correct. The worked calculation gives 7.",
"D": "Incorrect. Substituting the stated values into the governing rela... | [
"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_006284 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 45 cm in front of a thin convex lens of focal length 12 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "0.36364",
"B": "0.54545",
"C": "1.36364",
"D": "0.27273"
} | A | 0.36364 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(45)(12)/(45−12)=16.36364 cm. Thus |m|=v/u=16.36364/45=0.36364. | {
"A": "Correct. The worked calculation gives 0.36364.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.36364, not 0.54545.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.36364, not 1.36364.",
"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_007482 | 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 27 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "1.1421",
"B": "3.42101",
"C": "-0.57899",
"D": "1.42101"
} | D | 1.42101 | Excess H⁺ moles=0.10×(60−27)/1000=0.003300 mol. Total volume=0.087 L, so [H⁺]=0.037931 M and pH=−log₁₀[H⁺]=1.42101. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.42101, not 1.1421.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.42101, not 3.42101.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.42101, not -0.578... | [
"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_008242 | 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 2 A for 22458 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": "12.77806 g",
"B": "-14.77806 g",
"C": "15.77806 g",
"D": "14.77806 g"
} | D | 14.77806 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×2×22458/(2×96500)=14.77806 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 14.77806 g, not 12.77806 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 14.77806 g, not -14.77806 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 14.77... | [
"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_006277 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 59 cm in front of a thin convex lens of focal length 16 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "2.37209",
"B": "0.37209",
"C": "1.37209",
"D": "0.55814"
} | B | 0.37209 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(59)(16)/(59−16)=21.95349 cm. Thus |m|=v/u=21.95349/59=0.37209. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.37209, not 2.37209.",
"B": "Correct. The worked calculation gives 0.37209.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.37209, not 1.37209.",
"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_009872 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Cr(Cl)_2]^+3, each Cl ligand has charge −1. Determine the formal oxidation state of Cr. | {
"A": "5",
"B": "2",
"C": "3.75",
"D": "6"
} | A | 5 | Let the formal oxidation state of Cr be x. Charge balance gives x+(2)(−1)=+3; hence x=3+2=+5. | {
"A": "Correct. The worked calculation gives 5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5, not 2.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5, not 3.75.",
"D": "Incorrect. Substituting the stated values into the governing rel... | [
"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_004312 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 12 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.37. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "4.648 N",
"B": "18.24 N",
"C": "35.48 N",
"D": "36.48 N"
} | D | 36.48 N | Net force=mg(sinθ−μcosθ)=12×10×(0.6−0.37×0.8)=36.48 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 36.48 N, not 4.648 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 36.48 N, not 18.24 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 36.48 N, not 35.48... | [
"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_006099 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 47 cm in front of a thin convex lens of focal length 13 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "2.38235",
"B": "1.38235",
"C": "0.57353",
"D": "0.38235"
} | D | 0.38235 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(47)(13)/(47−13)=17.97059 cm. Thus |m|=v/u=17.97059/47=0.38235. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.38235, not 2.38235.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.38235, not 1.38235.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.38235, not 0.573... | [
"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_008799 | 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 0.15 mol L⁻¹ and initially B is absent. At equilibrium, 13% of A has dissociated at constant volume. Find K_c. | {
"A": "0.01748",
"B": "-1.98834",
"C": "1.01166",
"D": "0.01166"
} | D | 0.01166 | Let α=0.13. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.15)(0.13)²/(1−0.13)=0.01166. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.01166, not 0.01748.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.01166, not -1.98834.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.01166, not 1.01... | [
"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_000797 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 7 and common ratio 2. What is the sum of its first 10 terms? | {
"A": "5728.8",
"B": "7161",
"C": "-7161",
"D": "7159"
} | B | 7161 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=7(2^10−1)/(2−1)=7161. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7161, not 5728.8.",
"B": "Correct. The worked calculation gives 7161.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7161, not -7161.",
"D": "Incorrect. Substituting the stated values into th... | [
"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_005947 | 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 0.4 atm at 302 K. What is its pressure at 464 K? | {
"A": "-0.38543 atm",
"B": "1.61457 atm",
"C": "-2.38543 atm",
"D": "0.61457 atm"
} | D | 0.61457 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(0.4)(464)/(302)=0.61457 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.61457 atm, not -0.38543 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.61457 atm, not 1.61457 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0... | [
"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_009685 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Pt(OH)_4]^+1, each OH ligand has charge −1. Determine the formal oxidation state of Pt. | {
"A": "5",
"B": "7",
"C": "1.5",
"D": "10"
} | A | 5 | Let the formal oxidation state of Pt be x. Charge balance gives x+(4)(−1)=+1; hence x=1+4=+5. | {
"A": "Correct. The worked calculation gives 5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5, not 7.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5, not 1.5.",
"D": "Incorrect. Substituting the stated values into the governing rela... | [
"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_009356 | 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_17H_26N_2X_4. Treat X as the total number of monovalent halogen atoms. | {
"A": "5",
"B": "4",
"C": "6",
"D": "1.4"
} | B | 4 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×17+2+2−26−4)/2=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 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_008686 | 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 0.8 mol L⁻¹ and initially B is absent. At equilibrium, 19% of A has dissociated at constant volume. Find K_c. | {
"A": "0.21393",
"B": "0.14262",
"C": "-0.85738",
"D": "0.28523"
} | B | 0.14262 | Let α=0.19. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.8)(0.19)²/(1−0.19)=0.14262. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.14262, not 0.21393.",
"B": "Correct. The worked calculation gives 0.14262.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.14262, not -0.85738.",
"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_009415 | 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_4N_2X_2. Treat X as the total number of monovalent halogen atoms. | {
"A": "11",
"B": "1.8",
"C": "8",
"D": "6"
} | C | 8 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×9+2+2−4−2)/2=8. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 8, not 11.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 8, not 1.8.",
"C": "Correct. The worked calculation gives 8.",
"D": "Incorrect. Substituting the stated values into the governing rel... | [
"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_005712 | 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 3.5 atm at 399 K. What is its pressure at 532 K? | {
"A": "1.66667 atm",
"B": "4.66667 atm",
"C": "7.66667 atm",
"D": "-4.66667 atm"
} | B | 4.66667 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(3.5)(532)/(399)=4.66667 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.66667 atm, not 1.66667 atm.",
"B": "Correct. The worked calculation gives 4.66667 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4.66667 atm, not 7.66667 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_007485 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 51 mL of 0.10 M HCl is mixed with 68 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "13.1549",
"B": "12.1549",
"C": "10.1549",
"D": "14.1549"
} | B | 12.1549 | Excess OH⁻ moles=0.10×(68−51)/1000=0.001700 mol. Total volume=0.119 L, so [OH⁻]=0.014286 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.1549. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12.1549, not 13.1549.",
"B": "Correct. The worked calculation gives 12.1549.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12.1549, not 10.1549.",
"D": "Incorrect. Substituting the stated va... | [
"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_006439 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 19 kg is attached to an ideal spring of force constant 111 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "0.82746 s",
"B": "1.24119 s",
"C": "1.65491 s",
"D": "2.82746 s"
} | A | 0.82746 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(19/111)=0.82746 s. | {
"A": "Correct. The worked calculation gives 0.82746 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.82746 s, not 1.24119 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.82746 s, not 1.65491 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_005213 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +8 μC and +1 μC are separated by 8 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "14.25 N",
"B": "8.25 N",
"C": "11.25 N",
"D": "8.4375 N"
} | C | 11.25 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(8×10⁻⁶)(1×10⁻⁶)/(8×10⁻²)²=11.25 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 11.25 N, not 14.25 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 11.25 N, not 8.25 N.",
"C": "Correct. The worked calculation gives 11.25 N.",
"D": "Incorrect. Substituting the stated val... | [
"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_007540 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 80 mL of 0.10 M HCl is mixed with 23 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "0.25696",
"B": "-1.25696",
"C": "1.88544",
"D": "1.25696"
} | D | 1.25696 | Excess H⁺ moles=0.10×(80−23)/1000=0.005700 mol. Total volume=0.103 L, so [H⁺]=0.055340 M and pH=−log₁₀[H⁺]=1.25696. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.25696, not 0.25696.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.25696, not -1.25696.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.25696, not 1.88... | [
"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_008995 | 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 1.92 mol L⁻¹ and initially B is absent. At equilibrium, 38% of A has dissociated at constant volume. Find K_c. | {
"A": "1.7887",
"B": "-1.7887",
"C": "0.89435",
"D": "1.34152"
} | A | 1.7887 | Let α=0.38. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.92)(0.38)²/(1−0.38)=1.7887. | {
"A": "Correct. The worked calculation gives 1.7887.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.7887, not -1.7887.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.7887, not 0.89435.",
"D": "Incorrect. Substituting the stated value... | [
"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_001008 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-3x³ -5x² -10x+7. Evaluate f′(4). | {
"A": "-194",
"B": "-195",
"C": "-18.4",
"D": "-291"
} | A | -194 | Differentiate: f′(x)=3(-3)x²+2(-5)x+(-10). At x=4, f′(4)=3(-3)(4)²+2(-5)(4)+(-10)=-194. | {
"A": "Correct. The worked calculation gives -194.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -194, not -195.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -194, not -18.4.",
"D": "Incorrect. Substituting the stated values into the ... | [
"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_004150 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 38 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.31. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "134.76 N",
"B": "133.76 N",
"C": "107.008 N",
"D": "160.512 N"
} | B | 133.76 N | Net force=mg(sinθ−μcosθ)=38×10×(0.6−0.31×0.8)=133.76 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 133.76 N, not 134.76 N.",
"B": "Correct. The worked calculation gives 133.76 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 133.76 N, not 107.008 N.",
"D": "Incorrect. Substituting the sta... | [
"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_004986 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 2 Ω and 7 Ω are connected in series; this combination is connected in parallel with a 5 Ω resistor across a 24 V ideal battery. Find the total current supplied by the battery. | {
"A": "7.46667 A",
"B": "10.46667 A",
"C": "4.46667 A",
"D": "8.46667 A"
} | A | 7.46667 A | Series branch resistance=2+7=9 Ω. Parallel equivalent R_eq=(9×5)/(9+5)=3.21429 Ω. Total current I=V/R_eq=24/3.21429=7.46667 A. | {
"A": "Correct. The worked calculation gives 7.46667 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 7.46667 A, not 10.46667 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7.46667 A, not 4.46667 A.",
"D": "Incorrect. Substituting th... | [
"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_002953 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(-6, 6, -3) and v=(-2, 5, -6), calculate u·v. | {
"A": "90",
"B": "30",
"C": "60",
"D": "58"
} | C | 60 | The dot product is the sum of corresponding component products: (-6)(-2)+(6)(5)+(-3)(-6)=60. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 60, not 90.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 60, not 30.",
"C": "Correct. The worked calculation gives 60.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"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_001152 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=5x³ +1x² -2x+7. Evaluate f′(-4). | {
"A": "460",
"B": "229",
"C": "230",
"D": "24"
} | C | 230 | Differentiate: f′(x)=3(5)x²+2(1)x+(-2). At x=-4, f′(-4)=3(5)(-4)²+2(1)(-4)+(-2)=230. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 230, not 460.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 230, not 229.",
"C": "Correct. The worked calculation gives 230.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"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_001257 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=4x³ -7x² -11x+7. Evaluate f′(2). | {
"A": "4.5",
"B": "11",
"C": "9",
"D": "7"
} | C | 9 | Differentiate: f′(x)=3(4)x²+2(-7)x+(-11). At x=2, f′(2)=3(4)(2)²+2(-7)(2)+(-11)=9. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 9, not 4.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 9, not 11.",
"C": "Correct. The worked calculation gives 9.",
"D": "Incorrect. Substituting the stated values into the governing rel... | [
"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_000098 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 3x^2 +9x +6=0, let its roots be α and β. What is α²+β²? | {
"A": "1.5",
"B": "7",
"C": "8",
"D": "5"
} | D | 5 | By Vieta's relations, α+β = -9/3 and αβ = 6/3. Then α²+β²=(α+β)²−2αβ = (-9/3)²−2(6/3) = 5. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5, not 1.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5, not 7.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5, not 8.",
"D": "Correct. The worked ... | [
"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_002724 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(8, -2, 2) and v=(1, 3, 7), calculate u·v. | {
"A": "16",
"B": "18",
"C": "-16",
"D": "24"
} | A | 16 | The dot product is the sum of corresponding component products: (8)(1)+(-2)(3)+(2)(7)=16. | {
"A": "Correct. The worked calculation gives 16.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 16, not 18.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 16, not -16.",
"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_003892 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 12 kg block initially moves at 7 m s⁻¹. A constant net force does 17 N of work over 2 m in the direction of motion. Find its final kinetic energy. | {
"A": "327 J",
"B": "328 J",
"C": "246 J",
"D": "262.4 J"
} | B | 328 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(12)(7)²+(17)(2)=328 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 328 J, not 327 J.",
"B": "Correct. The worked calculation gives 328 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 328 J, not 246 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_003534 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 30 m s⁻¹ at 25° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "70.944 m",
"B": "82.7328 m",
"C": "68.944 m",
"D": "55.1552 m"
} | C | 68.944 m | Range R=u²sin(2θ)/g. Substituting u=30, θ=25° and g=10 gives R=(30)²sin(50°)/10=68.944 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 68.944 m, not 70.944 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 68.944 m, not 82.7328 m.",
"C": "Correct. The worked calculation gives 68.944 m.",
"D": "Incorrect. Substituting the sta... | [
"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_005485 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 74 μF capacitor is charged to a potential difference of 27 V. Calculate the energy stored, expressed in μJ. | {
"A": "26975 μJ",
"B": "26971 μJ",
"C": "20229.75 μJ",
"D": "26973 μJ"
} | D | 26973 μJ | U=½CV². Since C=74 μF, U in μJ is ½×74×(27)²=26973 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 26973 μJ, not 26975 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 26973 μJ, not 26971 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 26973 μJ, not ... | [
"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_008142 | 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 9 A for 9379 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": "13.88627 g",
"B": "41.6588 g",
"C": "27.77253 g",
"D": "55.54506 g"
} | C | 27.77253 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×9×9379/(2×96500)=27.77253 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 27.77253 g, not 13.88627 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 27.77253 g, not 41.6588 g.",
"C": "Correct. The worked calculation gives 27.77253 g.",
"D": "Incorrect. Substituting... | [
"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_001440 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^2 (6x² +0x -4) dx. | {
"A": "6",
"B": "1.8",
"C": "8",
"D": "-8"
} | C | 8 | An antiderivative is (6/3)x³+(0/2)x²+(-4)x. Evaluating between 0 and 2 gives (6/3)(2)³+(0/2)(2)²+(-4)(2)=8. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 8, not 6.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 8, not 1.8.",
"C": "Correct. The worked calculation gives 8.",
"D": "Incorrect. Substituting the stated values into the governing rela... | [
"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_007233 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 20 mol of gas at 421 K and 2.04 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "169.34312 L",
"B": "338.68625 L",
"C": "34.86862 L",
"D": "254.01468 L"
} | B | 338.68625 L | PV=nRT, so V=nRT/P=(20)(0.082057)(421)/(2.04)=338.68625 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 338.68625 L, not 169.34312 L.",
"B": "Correct. The worked calculation gives 338.68625 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 338.68625 L, not 34.86862 L.",
"D": "Incorrect. Substit... | [
"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_000688 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 11 and common ratio -2. What is the sum of its first 12 terms? | {
"A": "-15015",
"B": "-15013",
"C": "-22522.5",
"D": "-1500.5"
} | A | -15015 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=11(-2^12−1)/(-2−1)=-15015. | {
"A": "Correct. The worked calculation gives -15015.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -15015, not -15013.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -15015, not -22522.5.",
"D": "Incorrect. Substituting the stated value... | [
"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_007408 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 23 mL of 0.10 M HCl is mixed with 82 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "12.74966",
"B": "13.74966",
"C": "9.74966",
"D": "19.12449"
} | A | 12.74966 | Excess OH⁻ moles=0.10×(82−23)/1000=0.005900 mol. Total volume=0.105 L, so [OH⁻]=0.056190 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.74966. | {
"A": "Correct. The worked calculation gives 12.74966.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.74966, not 13.74966.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12.74966, not 9.74966.",
"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_006433 | 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 1151 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "3.17685 s",
"B": "0.13264 s",
"C": "0.17685 s",
"D": "1.01769 s"
} | C | 0.17685 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(9/1151)=0.17685 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.17685 s, not 3.17685 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.17685 s, not 0.13264 s.",
"C": "Correct. The worked calculation gives 0.17685 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_009232 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.06 V, transfers 2 electrons, and has reaction quotient Q=10^4. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "0.70635 V",
"B": "0.9418 V",
"C": "3.9418 V",
"D": "-0.9418 V"
} | B | 0.9418 V | Apply the Nernst equation: E=1.06−(0.0591/2)(4)=0.9418 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.9418 V, not 0.70635 V.",
"B": "Correct. The worked calculation gives 0.9418 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.9418 V, not 3.9418 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_003370 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 27 m s⁻¹ at 70° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "5.68592 m",
"B": "48.85922 m",
"C": "93.71843 m",
"D": "46.85922 m"
} | D | 46.85922 m | Range R=u²sin(2θ)/g. Substituting u=27, θ=70° and g=10 gives R=(27)²sin(140°)/10=46.85922 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 46.85922 m, not 5.68592 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 46.85922 m, not 48.85922 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 46.8592... | [
"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_003398 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 109 m s⁻¹ at 75° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "-594.05 m",
"B": "297.025 m",
"C": "712.86 m",
"D": "594.05 m"
} | D | 594.05 m | Range R=u²sin(2θ)/g. Substituting u=109, θ=75° and g=10 gives R=(109)²sin(150°)/10=594.05 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 594.05 m, not -594.05 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 594.05 m, not 297.025 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 594.05 m, no... | [
"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_002127 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 12 red, 6 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": "0.33538",
"B": "1.03354",
"C": "0.67077",
"D": "0.25154"
} | A | 0.33538 | Favourable pairs = C(12,2)+C(6,2)+C(8,2)=109. Total pairs=C(26,2)=325. Therefore P=109/325=0.33538. | {
"A": "Correct. The worked calculation gives 0.33538.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.33538, not 1.03354.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.33538, not 0.67077.",
"D": "Incorrect. Substituting the stated va... | [
"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_002257 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 7 red, 12 blue and 7 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "1.33231",
"B": "3.33231",
"C": "0.33231",
"D": "-0.33231"
} | C | 0.33231 | Favourable pairs = C(7,2)+C(12,2)+C(7,2)=108. Total pairs=C(26,2)=325. Therefore P=108/325=0.33231. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.33231, not 1.33231.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.33231, not 3.33231.",
"C": "Correct. The worked calculation gives 0.33231.",
"D": "Incorrect. Substituting the stated va... | [
"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_003726 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 3 kg block initially moves at 7 m s⁻¹. A constant net force does 26 N of work over 17 m in the direction of motion. Find its final kinetic energy. | {
"A": "1031 J",
"B": "516.5 J",
"C": "515.5 J",
"D": "773.25 J"
} | C | 515.5 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(3)(7)²+(26)(17)=515.5 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 515.5 J, not 1031 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 515.5 J, not 516.5 J.",
"C": "Correct. The worked calculation gives 515.5 J.",
"D": "Incorrect. Substituting the stated val... | [
"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_008247 | 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 13 A for 22562 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": "77.20179 g",
"B": "96.50223 g",
"C": "-96.50223 g",
"D": "10.65022 g"
} | B | 96.50223 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×13×22562/(2×96500)=96.50223 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 96.50223 g, not 77.20179 g.",
"B": "Correct. The worked calculation gives 96.50223 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 96.50223 g, not -96.50223 g.",
"D": "Incorrect. Substituti... | [
"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_009563 | 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_11H_2X_4. Treat X as the total number of monovalent halogen atoms. | {
"A": "13.5",
"B": "9",
"C": "4.5",
"D": "7"
} | B | 9 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×11+2+0−2−4)/2=9. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 9, not 13.5.",
"B": "Correct. The worked calculation gives 9.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 9, not 4.5.",
"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_006913 | 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 6 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "12 mol",
"B": "14 mol",
"C": "18 mol",
"D": "13 mol"
} | A | 12 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 47 mol H₂O; O₂ can produce 12 mol H₂O. The limiting condition is O₂; product=12 mol. | {
"A": "Correct. The worked calculation gives 12 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12 mol, not 14 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12 mol, not 18 mol.",
"D": "Incorrect. Substituting the stated values ... | [
"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_004136 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 33 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.35. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "105.6 N",
"B": "52.8 N",
"C": "-105.6 N",
"D": "11.56 N"
} | A | 105.6 N | Net force=mg(sinθ−μcosθ)=33×10×(0.6−0.35×0.8)=105.6 N. | {
"A": "Correct. The worked calculation gives 105.6 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 105.6 N, not 52.8 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 105.6 N, not -105.6 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_005747 | 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 187 K. What is its pressure at 514 K? | {
"A": "1.98952 atm",
"B": "11.89519 atm",
"C": "9.89519 atm",
"D": "14.84278 atm"
} | C | 9.89519 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(3.6)(514)/(187)=9.89519 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 9.89519 atm, not 1.98952 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 9.89519 atm, not 11.89519 atm.",
"C": "Correct. The worked calculation gives 9.89519 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_008444 | 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 0.25 mol kg⁻¹. If its van't Hoff factor is 4 and K_b for water is 0.512 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "-2.488 K",
"B": "0.512 K",
"C": "0.384 K",
"D": "-0.512 K"
} | B | 0.512 K | ΔT_b=iK_bm. Substitution gives ΔT_b=4×0.512×0.25=0.512 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.512 K, not -2.488 K.",
"B": "Correct. The worked calculation gives 0.512 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.512 K, not 0.384 K.",
"D": "Incorrect. Substituting the stated v... | [
"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_004916 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 13 Ω and 2 Ω are connected in series; this combination is connected in parallel with a 12 Ω resistor across a 9 V ideal battery. Find the total current supplied by the battery. | {
"A": "1.35 A",
"B": "3.35 A",
"C": "2.35 A",
"D": "4.35 A"
} | A | 1.35 A | Series branch resistance=13+2=15 Ω. Parallel equivalent R_eq=(15×12)/(15+12)=6.66667 Ω. Total current I=V/R_eq=9/6.66667=1.35 A. | {
"A": "Correct. The worked calculation gives 1.35 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.35 A, not 3.35 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.35 A, not 2.35 A.",
"D": "Incorrect. Substituting the stated values ... | [
"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_006944 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 23 mol H₂ and 30 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "46 mol",
"B": "23 mol",
"C": "22 mol",
"D": "34.5 mol"
} | B | 23 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 23 mol H₂O; O₂ can produce 60 mol H₂O. The limiting condition is H₂; product=23 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 23 mol, not 46 mol.",
"B": "Correct. The worked calculation gives 23 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 23 mol, not 22 mol.",
"D": "Incorrect. Substituting the stated values ... | [
"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_004027 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 26 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.30. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "70.2 N",
"B": "10.36 N",
"C": "95.6 N",
"D": "93.6 N"
} | D | 93.6 N | Net force=mg(sinθ−μcosθ)=26×10×(0.6−0.30×0.8)=93.6 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 93.6 N, not 70.2 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 93.6 N, not 10.36 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 93.6 N, not 95.6 N.",... | [
"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_007161 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 8 mol of gas at 520 K and 1.05 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "390.12242 L",
"B": "326.10202 L",
"C": "325.10202 L",
"D": "162.55101 L"
} | C | 325.10202 L | PV=nRT, so V=nRT/P=(8)(0.082057)(520)/(1.05)=325.10202 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 325.10202 L, not 390.12242 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 325.10202 L, not 326.10202 L.",
"C": "Correct. The worked calculation gives 325.10202 L.",
"D": "Incorrect. Substi... | [
"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_004185 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 40 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.16. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "226.56 N",
"B": "188.8 N",
"C": "187.8 N",
"D": "19.88 N"
} | B | 188.8 N | Net force=mg(sinθ−μcosθ)=40×10×(0.6−0.16×0.8)=188.8 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 188.8 N, not 226.56 N.",
"B": "Correct. The worked calculation gives 188.8 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 188.8 N, not 187.8 N.",
"D": "Incorrect. Substituting the stated v... | [
"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_004767 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 9 Ω and 15 Ω are connected in series; this combination is connected in parallel with a 12 Ω resistor across a 6 V ideal battery. Find the total current supplied by the battery. | {
"A": "0.75 A",
"B": "2.75 A",
"C": "-2.25 A",
"D": "-0.75 A"
} | A | 0.75 A | Series branch resistance=9+15=24 Ω. Parallel equivalent R_eq=(24×12)/(24+12)=8 Ω. Total current I=V/R_eq=6/8=0.75 A. | {
"A": "Correct. The worked calculation gives 0.75 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.75 A, not 2.75 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.75 A, not -2.25 A.",
"D": "Incorrect. Substituting the stated values... | [
"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_006628 | 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 21 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "1.75593 s",
"B": "0.75593 s",
"C": "1.51186 s",
"D": "3.75593 s"
} | B | 0.75593 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(3/21)=0.75593 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.75593 s, not 1.75593 s.",
"B": "Correct. The worked calculation gives 0.75593 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.75593 s, not 1.51186 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_000463 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 30 and common difference 6. Find the sum of its first 9 terms. | {
"A": "583.2",
"B": "487",
"C": "486",
"D": "243"
} | C | 486 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=30, d=6, n=9: Sₙ=9/2[2(30)+(9−1)(6)]=486. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 486, not 583.2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 486, not 487.",
"C": "Correct. The worked calculation gives 486.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"arithmetic progression",
"finite series"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_ap_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_001244 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=2x³ +2x² -12x+7. Evaluate f′(0). | {
"A": "-10",
"B": "-18",
"C": "-12",
"D": "-15"
} | C | -12 | Differentiate: f′(x)=3(2)x²+2(2)x+(-12). At x=0, f′(0)=3(2)(0)²+2(2)(0)+(-12)=-12. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -12, not -10.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -12, not -18.",
"C": "Correct. The worked calculation gives -12.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"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_008638 | 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 1.1 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": "4.4 K",
"B": "3.3 K",
"C": "2.4 K",
"D": "2.2 K"
} | A | 4.4 K | ΔT_b=iK_bm. Substitution gives ΔT_b=4×1.0×1.1=4.4 K. | {
"A": "Correct. The worked calculation gives 4.4 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4.4 K, not 3.3 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4.4 K, not 2.4 K.",
"D": "Incorrect. Substituting the stated values into ... | [
"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_004079 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 43 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": "17.856 N",
"B": "202.272 N",
"C": "168.56 N",
"D": "337.12 N"
} | C | 168.56 N | Net force=mg(sinθ−μcosθ)=43×10×(0.6−0.26×0.8)=168.56 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 168.56 N, not 17.856 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 168.56 N, not 202.272 N.",
"C": "Correct. The worked calculation gives 168.56 N.",
"D": "Incorrect. Substituting the sta... | [
"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_007951 | 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 73 mmol L⁻¹, what concentration remains after 100 minutes? | {
"A": "1.14062 mmol L⁻¹",
"B": "2.28125 mmol L⁻¹",
"C": "1.71094 mmol L⁻¹",
"D": "1.22812 mmol L⁻¹"
} | B | 2.28125 mmol L⁻¹ | 100 minutes corresponds to 5 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=73/2^5=2.28125 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.28125 mmol L⁻¹, not 1.14062 mmol L⁻¹.",
"B": "Correct. The worked calculation gives 2.28125 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.28125 mmol L⁻¹, not 1.71094 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_009497 | 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_15X_1. Treat X as the total number of monovalent halogen atoms. | {
"A": "2",
"B": "5",
"C": "0",
"D": "1"
} | A | 2 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×9+2+0−15−1)/2=2. | {
"A": "Correct. The worked calculation gives 2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2, not 5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2, not 0.",
"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_003908 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 4 kg block initially moves at 12 m s⁻¹. A constant net force does 17 N of work over 3 m in the direction of motion. Find its final kinetic energy. | {
"A": "169.5 J",
"B": "508.5 J",
"C": "339 J",
"D": "271.2 J"
} | C | 339 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(4)(12)²+(17)(3)=339 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 339 J, not 169.5 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 339 J, not 508.5 J.",
"C": "Correct. The worked calculation gives 339 J.",
"D": "Incorrect. Substituting the stated values i... | [
"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_003185 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(-8+9i)/(7+10i). What is Re(z), the real part of z? | {
"A": "0.34228",
"B": "-0.77181",
"C": "0.22819",
"D": "1.22819"
} | C | 0.22819 | Multiply numerator and denominator by the conjugate (7-10i). Then Re(z)=(ac+bd)/(c²+d²)=((-8)(7)+(9)(10))/(7²+(10)²)=0.22819. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.22819, not 0.34228.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.22819, not -0.77181.",
"C": "Correct. The worked calculation gives 0.22819.",
"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_008434 | 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 0.88 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": "3.52 K",
"B": "5.52 K",
"C": "1.52 K",
"D": "5.28 K"
} | A | 3.52 K | ΔT_b=iK_bm. Substitution gives ΔT_b=4×1.0×0.88=3.52 K. | {
"A": "Correct. The worked calculation gives 3.52 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.52 K, not 5.52 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.52 K, not 1.52 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_005135 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +9 μC and +2 μC are separated by 10 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "16.2 N",
"B": "12.15 N",
"C": "8.1 N",
"D": "14.2 N"
} | A | 16.2 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(9×10⁻⁶)(2×10⁻⁶)/(10×10⁻²)²=16.2 N. | {
"A": "Correct. The worked calculation gives 16.2 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 16.2 N, not 12.15 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 16.2 N, not 8.1 N.",
"D": "Incorrect. Substituting the stated values ... | [
"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_000831 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 6 and common ratio 5. What is the sum of its first 8 terms? | {
"A": "439452",
"B": "585936",
"C": "585937",
"D": "703123.2"
} | B | 585936 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=6(5^8−1)/(5−1)=585936. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 585936, not 439452.",
"B": "Correct. The worked calculation gives 585936.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 585936, not 585937.",
"D": "Incorrect. Substituting the stated values ... | [
"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_003204 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(13-4i)/(4+7i). What is Re(z), the real part of z? | {
"A": "0.36923",
"B": "0.55385",
"C": "-0.63077",
"D": "1.36923"
} | A | 0.36923 | Multiply numerator and denominator by the conjugate (4-7i). Then Re(z)=(ac+bd)/(c²+d²)=((13)(4)+(-4)(7))/(4²+(7)²)=0.36923. | {
"A": "Correct. The worked calculation gives 0.36923.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.36923, not 0.55385.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.36923, not -0.63077.",
"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_003364 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 15 m s⁻¹ at 55° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "10.57154 m",
"B": "23.14308 m",
"C": "31.71463 m",
"D": "21.14308 m"
} | D | 21.14308 m | Range R=u²sin(2θ)/g. Substituting u=15, θ=55° and g=10 gives R=(15)²sin(110°)/10=21.14308 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 21.14308 m, not 10.57154 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 21.14308 m, not 23.14308 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 21.143... | [
"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_008854 | 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 0.4 mol L⁻¹ and initially B is absent. At equilibrium, 62% of A has dissociated at constant volume. Find K_c. | {
"A": "4.61853",
"B": "1.61853",
"C": "-1.61853",
"D": "2.61853"
} | B | 1.61853 | Let α=0.62. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.4)(0.62)²/(1−0.62)=1.61853. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.61853, not 4.61853.",
"B": "Correct. The worked calculation gives 1.61853.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.61853, not -1.61853.",
"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_003218 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(-10-11i)/(12+7i). What is Re(z), the real part of z? | {
"A": "-2.04145",
"B": "-0.02073",
"C": "-4.02073",
"D": "-1.02073"
} | D | -1.02073 | Multiply numerator and denominator by the conjugate (12-7i). Then Re(z)=(ac+bd)/(c²+d²)=((-10)(12)+(-11)(7))/(12²+(7)²)=-1.02073. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -1.02073, not -2.04145.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -1.02073, not -0.02073.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -1.02073, not ... | [
"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 |
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