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_004672 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 10 kg and radius 1 m is acted on by a constant torque of 62 N m about its central axis. Find its angular acceleration. | {
"A": "10.4 rad s⁻²",
"B": "6.2 rad s⁻²",
"C": "14.4 rad s⁻²",
"D": "12.4 rad s⁻²"
} | D | 12.4 rad s⁻² | For a solid disc, I=½MR²=½(10)(1)²=5 kg m². Using τ=Iα, α=62/5=12.4 rad s⁻². | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12.4 rad s⁻², not 10.4 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.4 rad s⁻², not 6.2 rad s⁻².",
"C": "Incorrect. Substituting the stated values into the governing relation gives... | [
"moment of inertia",
"rotational dynamics",
"torque"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_disc_torque | 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_008144 | 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 6390 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": "6.30723 g",
"B": "4.20482 g",
"C": "1.42048 g",
"D": "1.20482 g"
} | B | 4.20482 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×2×6390/(2×96500)=4.20482 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.20482 g, not 6.30723 g.",
"B": "Correct. The worked calculation gives 4.20482 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4.20482 g, not 1.42048 g.",
"D": "Incorrect. Substituting the... | [
"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_000941 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 11 and common ratio 5. What is the sum of its first 7 terms? | {
"A": "214843",
"B": "214842",
"C": "214840",
"D": "214841"
} | D | 214841 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=11(5^7−1)/(5−1)=214841. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 214841, not 214843.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 214841, not 214842.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 214841, not 214840.",
... | [
"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_009476 | 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_4H_11N_3. Treat X as the total number of monovalent halogen atoms. | {
"A": "1",
"B": "1.5",
"C": "3",
"D": "4"
} | A | 1 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×4+2+3−11−0)/2=1. | {
"A": "Correct. The worked calculation gives 1.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1, not 1.5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1, not 3.",
"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_006694 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 30 mol H₂ and 20 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "30 mol",
"B": "-30 mol",
"C": "32 mol",
"D": "36 mol"
} | A | 30 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 30 mol H₂O; O₂ can produce 40 mol H₂O. The limiting condition is H₂; product=30 mol. | {
"A": "Correct. The worked calculation gives 30 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 30 mol, not -30 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 30 mol, not 32 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_008512 | 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 0.91 mol kg⁻¹. If its van't Hoff factor is 3 and K_b for water is 0.512 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "4.39776 K",
"B": "0.69888 K",
"C": "1.39776 K",
"D": "2.39776 K"
} | C | 1.39776 K | ΔT_b=iK_bm. Substitution gives ΔT_b=3×0.512×0.91=1.39776 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.39776 K, not 4.39776 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.39776 K, not 0.69888 K.",
"C": "Correct. The worked calculation gives 1.39776 K.",
"D": "Incorrect. Substituting the... | [
"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_009873 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Pd(Cl)_5]^-2, each Cl ligand has charge −1. Determine the formal oxidation state of Pd. | {
"A": "4.5",
"B": "3",
"C": "1.5",
"D": "2.25"
} | B | 3 | Let the formal oxidation state of Pd be x. Charge balance gives x+(5)(−1)=-2; hence x=-2+5=+3. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3, not 4.5.",
"B": "Correct. The worked calculation gives 3.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3, not 1.5.",
"D": "Incorrect. Substituting the stated values into the governing re... | [
"coordination compounds",
"oxidation state",
"charge balance"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_complex_oxidation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002082 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 11 red, 6 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": "0.32971",
"B": "-0.67029",
"C": "0.49457",
"D": "1.32971"
} | A | 0.32971 | Favourable pairs = C(11,2)+C(6,2)+C(7,2)=91. Total pairs=C(24,2)=276. Therefore P=91/276=0.32971. | {
"A": "Correct. The worked calculation gives 0.32971.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.32971, not -0.67029.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.32971, not 0.49457.",
"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_004138 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 47 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": "17.544 N",
"B": "248.16 N",
"C": "82.72 N",
"D": "165.44 N"
} | D | 165.44 N | Net force=mg(sinθ−μcosθ)=47×10×(0.6−0.31×0.8)=165.44 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 165.44 N, not 17.544 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 165.44 N, not 248.16 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 165.44 N, not ... | [
"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_008887 | 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 1.3 mol L⁻¹ and initially B is absent. At equilibrium, 49% of A has dissociated at constant volume. Find K_c. | {
"A": "1.83606",
"B": "1.22404",
"C": "2.44808",
"D": "5.44808"
} | C | 2.44808 | Let α=0.49. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.3)(0.49)²/(1−0.49)=2.44808. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.44808, not 1.83606.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.44808, not 1.22404.",
"C": "Correct. The worked calculation gives 2.44808.",
"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_008186 | 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 7 A for 7762 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": "19.87673 g",
"B": "-17.87673 g",
"C": "17.87673 g",
"D": "16.87673 g"
} | C | 17.87673 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×7×7762/(2×96500)=17.87673 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 17.87673 g, not 19.87673 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 17.87673 g, not -17.87673 g.",
"C": "Correct. The worked calculation gives 17.87673 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_006669 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 13 kg is attached to an ideal spring of force constant 897 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "0.24077 s",
"B": "-1.75923 s",
"C": "0.18058 s",
"D": "-0.75923 s"
} | A | 0.24077 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(13/897)=0.24077 s. | {
"A": "Correct. The worked calculation gives 0.24077 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.24077 s, not -1.75923 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.24077 s, not 0.18058 s.",
"D": "Incorrect. Substituting th... | [
"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_004759 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 9 Ω and 10 Ω are connected in series; this combination is connected in parallel with a 12 Ω resistor across a 36 V ideal battery. Find the total current supplied by the battery. | {
"A": "-4.89474 A",
"B": "6.89474 A",
"C": "4.89474 A",
"D": "1.48947 A"
} | C | 4.89474 A | Series branch resistance=9+10=19 Ω. Parallel equivalent R_eq=(19×12)/(19+12)=7.35484 Ω. Total current I=V/R_eq=36/7.35484=4.89474 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.89474 A, not -4.89474 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4.89474 A, not 6.89474 A.",
"C": "Correct. The worked calculation gives 4.89474 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_001989 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 7 elements can be selected from a set of 61 distinct elements? | {
"A": "523524936",
"B": "872541560",
"C": "-436270780",
"D": "436270780"
} | D | 436270780 | Order does not matter, so the count is C(61,7)=61!/[7!(54)!]=436270780. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 436270780, not 523524936.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 436270780, not 872541560.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 436270780,... | [
"combinations",
"factorials"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_combinations | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006964 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 32 mol H₂ and 29 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "24 mol",
"B": "25.6 mol",
"C": "32 mol",
"D": "30 mol"
} | C | 32 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 32 mol H₂O; O₂ can produce 58 mol H₂O. The limiting condition is H₂; product=32 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 32 mol, not 24 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 32 mol, not 25.6 mol.",
"C": "Correct. The worked calculation gives 32 mol.",
"D": "Incorrect. Substituting the stated value... | [
"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_007179 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 1 mol of gas at 287 K and 1.72 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "12.69207 L",
"B": "10.69207 L",
"C": "13.69207 L",
"D": "15.69207 L"
} | C | 13.69207 L | PV=nRT, so V=nRT/P=(1)(0.082057)(287)/(1.72)=13.69207 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 13.69207 L, not 12.69207 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 13.69207 L, not 10.69207 L.",
"C": "Correct. The worked calculation gives 13.69207 L.",
"D": "Incorrect. Substitutin... | [
"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_001248 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=4x³ -4x² -2x+7. Evaluate f′(-3). | {
"A": "260",
"B": "130",
"C": "104",
"D": "14"
} | B | 130 | Differentiate: f′(x)=3(4)x²+2(-4)x+(-2). At x=-3, f′(-3)=3(4)(-3)²+2(-4)(-3)+(-2)=130. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 130, not 260.",
"B": "Correct. The worked calculation gives 130.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 130, not 104.",
"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_004966 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 11 Ω and 5 Ω are connected in series; this combination is connected in parallel with a 3 Ω resistor across a 6 V ideal battery. Find the total current supplied by the battery. | {
"A": "1.2375 A",
"B": "2.375 A",
"C": "1.78125 A",
"D": "5.375 A"
} | B | 2.375 A | Series branch resistance=11+5=16 Ω. Parallel equivalent R_eq=(16×3)/(16+3)=2.52632 Ω. Total current I=V/R_eq=6/2.52632=2.375 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.375 A, not 1.2375 A.",
"B": "Correct. The worked calculation gives 2.375 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.375 A, not 1.78125 A.",
"D": "Incorrect. Substituting the stated... | [
"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_005626 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 6 μF capacitor is charged to a potential difference of 15 V. Calculate the energy stored, expressed in μJ. | {
"A": "337.5 μJ",
"B": "540 μJ",
"C": "1350 μJ",
"D": "675 μJ"
} | D | 675 μJ | U=½CV². Since C=6 μF, U in μJ is ½×6×(15)²=675 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 675 μJ, not 337.5 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 675 μJ, not 540 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 675 μJ, not 1350 μJ.... | [
"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_005657 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 24 μF capacitor is charged to a potential difference of 40 V. Calculate the energy stored, expressed in μJ. | {
"A": "19201 μJ",
"B": "19202 μJ",
"C": "19200 μJ",
"D": "23040 μJ"
} | C | 19200 μJ | U=½CV². Since C=24 μF, U in μJ is ½×24×(40)²=19200 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 19200 μJ, not 19201 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 19200 μJ, not 19202 μJ.",
"C": "Correct. The worked calculation gives 19200 μJ.",
"D": "Incorrect. Substituting the stat... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002537 | mathematics | linear_algebra | 2x2_determinant | hard | provisional_template_level_label | single_correct_mcq | Find the determinant of the matrix [[9, 6], [5, -4]]. | {
"A": "-5.6",
"B": "-65",
"C": "-66",
"D": "-99"
} | C | -66 | For a 2×2 matrix [[a,b],[c,d]], det=ad−bc. Here it is (9)(-4)−(6)(5)=-66. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -66, not -5.6.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -66, not -65.",
"C": "Correct. The worked calculation gives -66.",
"D": "Incorrect. Substituting the stated values into the gover... | [
"determinants",
"matrix operations"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_determinant_2x2 | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003796 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 10 kg block initially moves at 3 m s⁻¹. A constant net force does 23 N of work over 3 m in the direction of motion. Find its final kinetic energy. | {
"A": "113 J",
"B": "114 J",
"C": "12.4 J",
"D": "85.5 J"
} | B | 114 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(10)(3)²+(23)(3)=114 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 114 J, not 113 J.",
"B": "Correct. The worked calculation gives 114 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 114 J, not 12.4 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_001623 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^4 (6x² +1x +4) dx. | {
"A": "154",
"B": "228",
"C": "152",
"D": "114"
} | C | 152 | An antiderivative is (6/3)x³+(1/2)x²+(4)x. Evaluating between 0 and 4 gives (6/3)(4)³+(1/2)(4)²+(4)(4)=152. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 152, not 154.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 152, not 228.",
"C": "Correct. The worked calculation gives 152.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003454 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 62 m s⁻¹ at 40° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "454.27212 m",
"B": "378.5601 m",
"C": "379.5601 m",
"D": "189.28005 m"
} | B | 378.5601 m | Range R=u²sin(2θ)/g. Substituting u=62, θ=40° and g=10 gives R=(62)²sin(80°)/10=378.5601 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 378.5601 m, not 454.27212 m.",
"B": "Correct. The worked calculation gives 378.5601 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 378.5601 m, not 379.5601 m.",
"D": "Incorrect. Substituti... | [
"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_007428 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 41 mL of 0.10 M HCl is mixed with 58 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "12.23481",
"B": "-12.23481",
"C": "18.35222",
"D": "10.23481"
} | A | 12.23481 | Excess OH⁻ moles=0.10×(58−41)/1000=0.001700 mol. Total volume=0.099 L, so [OH⁻]=0.017172 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.23481. | {
"A": "Correct. The worked calculation gives 12.23481.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.23481, not -12.23481.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12.23481, not 18.35222.",
"D": "Incorrect. Substituting the sta... | [
"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_008804 | 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 1.6 mol L⁻¹ and initially B is absent. At equilibrium, 53% of A has dissociated at constant volume. Find K_c. | {
"A": "5.82502",
"B": "0.82502",
"C": "6.82502",
"D": "3.82502"
} | D | 3.82502 | Let α=0.53. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.6)(0.53)²/(1−0.53)=3.82502. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.82502, not 5.82502.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.82502, not 0.82502.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.82502, not 6.825... | [
"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_007474 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 46 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": "12.46584",
"B": "10.46584",
"C": "14.46584",
"D": "18.69876"
} | A | 12.46584 | Excess OH⁻ moles=0.10×(84−46)/1000=0.003800 mol. Total volume=0.130 L, so [OH⁻]=0.029231 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.46584. | {
"A": "Correct. The worked calculation gives 12.46584.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.46584, not 10.46584.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12.46584, not 14.46584.",
"D": "Incorrect. Substituting the stat... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006254 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 33 cm in front of a thin convex lens of focal length 19 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "-0.64286",
"B": "1.35714",
"C": "1.01786",
"D": "2.03571"
} | B | 1.35714 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(33)(19)/(33−19)=44.78571 cm. Thus |m|=v/u=44.78571/33=1.35714. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.35714, not -0.64286.",
"B": "Correct. The worked calculation gives 1.35714.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.35714, not 1.01786.",
"D": "Incorrect. Substituting the stated v... | [
"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_008272 | 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 8 A for 13043 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": "-34.3308 g",
"B": "35.3308 g",
"C": "34.3308 g",
"D": "36.3308 g"
} | C | 34.3308 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×8×13043/(2×96500)=34.3308 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 34.3308 g, not -34.3308 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 34.3308 g, not 35.3308 g.",
"C": "Correct. The worked calculation gives 34.3308 g.",
"D": "Incorrect. Substituting th... | [
"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_001253 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-5x³ -10x² +0x+7. Evaluate f′(1). | {
"A": "-70",
"B": "-35",
"C": "-34",
"D": "-26.25"
} | B | -35 | Differentiate: f′(x)=3(-5)x²+2(-10)x+(0). At x=1, f′(1)=3(-5)(1)²+2(-10)(1)+(0)=-35. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -35, not -70.",
"B": "Correct. The worked calculation gives -35.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -35, not -34.",
"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_009396 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_3H_3N_2X_1. Treat X as the total number of monovalent halogen atoms. | {
"A": "0",
"B": "5",
"C": "3",
"D": "-3"
} | C | 3 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×3+2+2−3−1)/2=3. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3, not 0.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3, not 5.",
"C": "Correct. The worked calculation gives 3.",
"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_002726 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(8, 6, -3) and v=(3, 5, 9), calculate u·v. | {
"A": "27",
"B": "28",
"C": "-27",
"D": "3.7"
} | A | 27 | The dot product is the sum of corresponding component products: (8)(3)+(6)(5)+(-3)(9)=27. | {
"A": "Correct. The worked calculation gives 27.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 27, not 28.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 27, not -27.",
"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_005949 | 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 4.6 atm at 187 K. What is its pressure at 546 K? | {
"A": "10.07326 atm",
"B": "13.43102 atm",
"C": "11.43102 atm",
"D": "10.43102 atm"
} | B | 13.43102 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(4.6)(546)/(187)=13.43102 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 13.43102 atm, not 10.07326 atm.",
"B": "Correct. The worked calculation gives 13.43102 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 13.43102 atm, not 11.43102 atm.",
"D": "Incorrect. S... | [
"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_000561 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term -4 and common difference 10. Find the sum of its first 28 terms. | {
"A": "3668",
"B": "2934.4",
"C": "367.8",
"D": "2751"
} | A | 3668 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=-4, d=10, n=28: Sₙ=28/2[2(-4)+(28−1)(10)]=3668. | {
"A": "Correct. The worked calculation gives 3668.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3668, not 2934.4.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3668, not 367.8.",
"D": "Incorrect. Substituting the stated values into th... | [
"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_001343 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^8 (5x² +0x +1) dx. | {
"A": "87.13333",
"B": "1292",
"C": "861.33333",
"D": "862.33333"
} | C | 861.33333 | An antiderivative is (5/3)x³+(0/2)x²+(1)x. Evaluating between 0 and 8 gives (5/3)(8)³+(0/2)(8)²+(1)(8)=861.33333. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 861.33333, not 87.13333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 861.33333, not 1292.",
"C": "Correct. The worked calculation gives 861.33333.",
"D": "Incorrect. Substituting the state... | [
"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_000712 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 12 and common ratio -2. What is the sum of its first 9 terms? | {
"A": "1641.6",
"B": "-2052",
"C": "206.2",
"D": "2052"
} | D | 2052 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=12(-2^9−1)/(-2−1)=2052. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2052, not 1641.6.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2052, not -2052.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2052, not 206.2.",
"D": "... | [
"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_001489 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^7 (-4x² +7x +2) dx. | {
"A": "-270.83333",
"B": "-272.83333",
"C": "-271.83333",
"D": "-543.66667"
} | C | -271.83333 | An antiderivative is (-4/3)x³+(7/2)x²+(2)x. Evaluating between 0 and 7 gives (-4/3)(7)³+(7/2)(7)²+(2)(7)=-271.83333. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -271.83333, not -270.83333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -271.83333, not -272.83333.",
"C": "Correct. The worked calculation gives -271.83333.",
"D": "Incorrect. Substitutin... | [
"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_001415 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^3 (-2x² -3x +1) dx. | {
"A": "-28.5",
"B": "-57",
"C": "-34.2",
"D": "-22.8"
} | A | -28.5 | An antiderivative is (-2/3)x³+(-3/2)x²+(1)x. Evaluating between 0 and 3 gives (-2/3)(3)³+(-3/2)(3)²+(1)(3)=-28.5. | {
"A": "Correct. The worked calculation gives -28.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -28.5, not -57.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -28.5, not -34.2.",
"D": "Incorrect. Substituting the stated values into th... | [
"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_004165 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 18 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.11. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "92.16 N",
"B": "46.08 N",
"C": "10.216 N",
"D": "93.16 N"
} | A | 92.16 N | Net force=mg(sinθ−μcosθ)=18×10×(0.6−0.11×0.8)=92.16 N. | {
"A": "Correct. The worked calculation gives 92.16 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 92.16 N, not 46.08 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 92.16 N, not 10.216 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_008777 | 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 1.82 mol L⁻¹ and initially B is absent. At equilibrium, 43% of A has dissociated at constant volume. Find K_c. | {
"A": "2.36153",
"B": "4.36153",
"C": "0.36153",
"D": "-2.36153"
} | A | 2.36153 | Let α=0.43. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.82)(0.43)²/(1−0.43)=2.36153. | {
"A": "Correct. The worked calculation gives 2.36153.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.36153, not 4.36153.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.36153, not 0.36153.",
"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_009230 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.80 V, transfers 3 electrons, and has reaction quotient Q=10^1. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "1.7803 V",
"B": "0.7803 V",
"C": "-0.2197 V",
"D": "1.5606 V"
} | B | 0.7803 V | Apply the Nernst equation: E=0.80−(0.0591/3)(1)=0.7803 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.7803 V, not 1.7803 V.",
"B": "Correct. The worked calculation gives 0.7803 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.7803 V, not -0.2197 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_000143 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 3x^2 +11x +11=0, let its roots be α and β. What is α²+β²? | {
"A": "6.11111",
"B": "5.11111",
"C": "4.11111",
"D": "9.11111"
} | A | 6.11111 | By Vieta's relations, α+β = -11/3 and αβ = 11/3. Then α²+β²=(α+β)²−2αβ = (-11/3)²−2(11/3) = 6.11111. | {
"A": "Correct. The worked calculation gives 6.11111.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6.11111, not 5.11111.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 6.11111, not 4.11111.",
"D": "Incorrect. Substituting the stated va... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004059 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 42 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": "157.92 N",
"B": "156.92 N",
"C": "78.96 N",
"D": "16.792 N"
} | A | 157.92 N | Net force=mg(sinθ−μcosθ)=42×10×(0.6−0.28×0.8)=157.92 N. | {
"A": "Correct. The worked calculation gives 157.92 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 157.92 N, not 156.92 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 157.92 N, not 78.96 N.",
"D": "Incorrect. Substituting the state... | [
"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_005015 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +6 μC and +7 μC are separated by 2 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "947 N",
"B": "945 N",
"C": "95.5 N",
"D": "946 N"
} | B | 945 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(6×10⁻⁶)(7×10⁻⁶)/(2×10⁻²)²=945 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 945 N, not 947 N.",
"B": "Correct. The worked calculation gives 945 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 945 N, not 95.5 N.",
"D": "Incorrect. Substituting the stated values into... | [
"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_001667 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^7 (-2x² +6x +1) dx. | {
"A": "-56",
"B": "-74.66667",
"C": "-37.33333",
"D": "-76.66667"
} | B | -74.66667 | An antiderivative is (-2/3)x³+(6/2)x²+(1)x. Evaluating between 0 and 7 gives (-2/3)(7)³+(6/2)(7)²+(1)(7)=-74.66667. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -74.66667, not -56.",
"B": "Correct. The worked calculation gives -74.66667.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -74.66667, not -37.33333.",
"D": "Incorrect. Substituting the state... | [
"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_002963 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(-5, -5, 6) and v=(-3, 9, 9), calculate u·v. | {
"A": "23",
"B": "36",
"C": "-24",
"D": "24"
} | D | 24 | The dot product is the sum of corresponding component products: (-5)(-3)+(-5)(9)+(6)(9)=24. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 24, not 23.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 24, not 36.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 24, not -24.",
"D": "Correct. The wo... | [
"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_007852 | 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 35 mmol L⁻¹, what concentration remains after 120 minutes? | {
"A": "1.54688 mmol L⁻¹",
"B": "0.41016 mmol L⁻¹",
"C": "-0.45312 mmol L⁻¹",
"D": "0.54688 mmol L⁻¹"
} | D | 0.54688 mmol L⁻¹ | 120 minutes corresponds to 6 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=35/2^6=0.54688 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.54688 mmol L⁻¹, not 1.54688 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.54688 mmol L⁻¹, not 0.41016 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governi... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005284 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +7 μC and +7 μC are separated by 8 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "82.6875 N",
"B": "69.90625 N",
"C": "55.125 N",
"D": "68.90625 N"
} | D | 68.90625 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(7×10⁻⁶)(7×10⁻⁶)/(8×10⁻²)²=68.90625 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 68.90625 N, not 82.6875 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 68.90625 N, not 69.90625 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 68.9062... | [
"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_008305 | 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 26688 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": "17.56153 g",
"B": "26.3423 g",
"C": "13.17115 g",
"D": "16.56153 g"
} | A | 17.56153 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×2×26688/(2×96500)=17.56153 g. | {
"A": "Correct. The worked calculation gives 17.56153 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 17.56153 g, not 26.3423 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 17.56153 g, not 13.17115 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_004753 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 11 Ω and 3 Ω are connected in series; this combination is connected in parallel with a 3 Ω resistor across a 9 V ideal battery. Find the total current supplied by the battery. | {
"A": "5.64286 A",
"B": "-3.64286 A",
"C": "7.28571 A",
"D": "3.64286 A"
} | D | 3.64286 A | Series branch resistance=11+3=14 Ω. Parallel equivalent R_eq=(14×3)/(14+3)=2.47059 Ω. Total current I=V/R_eq=9/2.47059=3.64286 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.64286 A, not 5.64286 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.64286 A, not -3.64286 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.64286 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_009311 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.38 V, transfers 1 electrons, and has reaction quotient Q=10^-5. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "2.51325 V",
"B": "1.6755 V",
"C": "-0.3245 V",
"D": "3.351 V"
} | B | 1.6755 V | Apply the Nernst equation: E=1.38−(0.0591/1)(-5)=1.6755 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.6755 V, not 2.51325 V.",
"B": "Correct. The worked calculation gives 1.6755 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.6755 V, not -0.3245 V.",
"D": "Incorrect. Substituting the st... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004026 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 48 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": "337.92 N",
"B": "167.96 N",
"C": "169.96 N",
"D": "168.96 N"
} | D | 168.96 N | Net force=mg(sinθ−μcosθ)=48×10×(0.6−0.31×0.8)=168.96 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 168.96 N, not 337.92 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 168.96 N, not 167.96 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 168.96 N, not ... | [
"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_008406 | 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 1.65 mol kg⁻¹. If its van't Hoff factor is 2 and K_b for water is 0.52 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "2.716 K",
"B": "1.716 K",
"C": "1.1716 K",
"D": "-1.716 K"
} | B | 1.716 K | ΔT_b=iK_bm. Substitution gives ΔT_b=2×0.52×1.65=1.716 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.716 K, not 2.716 K.",
"B": "Correct. The worked calculation gives 1.716 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.716 K, not 1.1716 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_004993 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 14 Ω and 7 Ω are connected in series; this combination is connected in parallel with a 14 Ω resistor across a 30 V ideal battery. Find the total current supplied by the battery. | {
"A": "3.57143 A",
"B": "2.57143 A",
"C": "2.67857 A",
"D": "5.57143 A"
} | A | 3.57143 A | Series branch resistance=14+7=21 Ω. Parallel equivalent R_eq=(21×14)/(21+14)=8.4 Ω. Total current I=V/R_eq=30/8.4=3.57143 A. | {
"A": "Correct. The worked calculation gives 3.57143 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.57143 A, not 2.57143 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.57143 A, not 2.67857 A.",
"D": "Incorrect. Substituting the... | [
"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_009004 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.21 V, transfers 3 electrons, and has reaction quotient Q=10^-3. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "-1.2691 V",
"B": "1.90365 V",
"C": "0.63455 V",
"D": "1.2691 V"
} | D | 1.2691 V | Apply the Nernst equation: E=1.21−(0.0591/3)(-3)=1.2691 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.2691 V, not -1.2691 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.2691 V, not 1.90365 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.2691 V, no... | [
"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_000823 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 12 and common ratio 5. What is the sum of its first 10 terms? | {
"A": "2929688.2",
"B": "-29296872",
"C": "29296872",
"D": "29296871"
} | C | 29296872 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=12(5^10−1)/(5−1)=29296872. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 29296872, not 2929688.2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 29296872, not -29296872.",
"C": "Correct. The worked calculation gives 29296872.",
"D": "Incorrect. Substituting the st... | [
"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_000028 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 1x^2 +12x +6=0, let its roots be α and β. What is α²+β²? | {
"A": "14.2",
"B": "-132",
"C": "132",
"D": "99"
} | C | 132 | By Vieta's relations, α+β = -12/1 and αβ = 6/1. Then α²+β²=(α+β)²−2αβ = (-12/1)²−2(6/1) = 132. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 132, not 14.2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 132, not -132.",
"C": "Correct. The worked calculation gives 132.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007321 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 11 mol of gas at 386 K and 1.23 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "424.89515 L",
"B": "566.52687 L",
"C": "284.26343 L",
"D": "283.26343 L"
} | D | 283.26343 L | PV=nRT, so V=nRT/P=(11)(0.082057)(386)/(1.23)=283.26343 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 283.26343 L, not 424.89515 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 283.26343 L, not 566.52687 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 28... | [
"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_008601 | 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 1.96 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": "4.01408 K",
"B": "5.01408 K",
"C": "6.01408 K",
"D": "1.40141 K"
} | A | 4.01408 K | ΔT_b=iK_bm. Substitution gives ΔT_b=4×0.512×1.96=4.01408 K. | {
"A": "Correct. The worked calculation gives 4.01408 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4.01408 K, not 5.01408 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4.01408 K, not 6.01408 K.",
"D": "Incorrect. Substituting the... | [
"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_003787 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 2 kg block initially moves at 7 m s⁻¹. A constant net force does 28 N of work over 8 m in the direction of motion. Find its final kinetic energy. | {
"A": "28.3 J",
"B": "275 J",
"C": "273 J",
"D": "327.6 J"
} | C | 273 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(2)(7)²+(28)(8)=273 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 273 J, not 28.3 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 273 J, not 275 J.",
"C": "Correct. The worked calculation gives 273 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_007892 | 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 27 mmol L⁻¹, what concentration remains after 120 minutes? | {
"A": "0.21094 mmol L⁻¹",
"B": "0.63281 mmol L⁻¹",
"C": "0.42188 mmol L⁻¹",
"D": "3.42188 mmol L⁻¹"
} | C | 0.42188 mmol L⁻¹ | 120 minutes corresponds to 6 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=27/2^6=0.42188 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.42188 mmol L⁻¹, not 0.21094 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.42188 mmol L⁻¹, not 0.63281 mmol L⁻¹.",
"C": "Correct. The worked calculation gives 0.42188 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_001621 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^2 (0x² -1x +6) dx. | {
"A": "2",
"B": "13",
"C": "10",
"D": "15"
} | C | 10 | An antiderivative is (0/3)x³+(-1/2)x²+(6)x. Evaluating between 0 and 2 gives (0/3)(2)³+(-1/2)(2)²+(6)(2)=10. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 10, not 2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 10, not 13.",
"C": "Correct. The worked calculation gives 10.",
"D": "Incorrect. Substituting the stated values into the governing re... | [
"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_006412 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 25 kg is attached to an ideal spring of force constant 144 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "1.08333 s",
"B": "0.625 s",
"C": "0.83333 s",
"D": "-2.16667 s"
} | C | 0.83333 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(25/144)=0.83333 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.83333 s, not 1.08333 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.83333 s, not 0.625 s.",
"C": "Correct. The worked calculation gives 0.83333 s.",
"D": "Incorrect. Substituting the s... | [
"simple harmonic motion",
"mass–spring oscillator",
"time period"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_shm_period | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005907 | 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 1.7 atm at 294 K. What is its pressure at 585 K? | {
"A": "3.38265 atm",
"B": "5.07398 atm",
"C": "0.38265 atm",
"D": "2.53699 atm"
} | A | 3.38265 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(1.7)(585)/(294)=3.38265 atm. | {
"A": "Correct. The worked calculation gives 3.38265 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.38265 atm, not 5.07398 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.38265 atm, not 0.38265 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_009597 | 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_37N_1. Treat X as the total number of monovalent halogen atoms. | {
"A": "0",
"B": "3",
"C": "-2",
"D": "1"
} | A | 0 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×17+2+1−37−0)/2=0. | {
"A": "Correct. The worked calculation gives 0.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0, not 3.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0, not -2.",
"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_006732 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 46 mol H₂ and 4 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "12 mol",
"B": "8 mol",
"C": "1.8 mol",
"D": "4 mol"
} | B | 8 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 46 mol H₂O; O₂ can produce 8 mol H₂O. The limiting condition is O₂; product=8 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 8 mol, not 12 mol.",
"B": "Correct. The worked calculation gives 8 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 8 mol, not 1.8 mol.",
"D": "Incorrect. Substituting the stated values in... | [
"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_005040 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +4 μC and +9 μC are separated by 6 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "72 N",
"B": "89 N",
"C": "90 N",
"D": "91 N"
} | C | 90 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(4×10⁻⁶)(9×10⁻⁶)/(6×10⁻²)²=90 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 90 N, not 72 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 90 N, not 89 N.",
"C": "Correct. The worked calculation gives 90 N.",
"D": "Incorrect. Substituting the stated values into the g... | [
"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_001603 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^6 (-1x² -3x +1) dx. | {
"A": "-144",
"B": "-90",
"C": "-240",
"D": "-120"
} | D | -120 | An antiderivative is (-1/3)x³+(-3/2)x²+(1)x. Evaluating between 0 and 6 gives (-1/3)(6)³+(-3/2)(6)²+(1)(6)=-120. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -120, not -144.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -120, not -90.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -120, not -240.",
"D": "Corre... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009277 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.02 V, transfers 1 electrons, and has reaction quotient Q=10^2. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "1.9018 V",
"B": "1.8036 V",
"C": "0.67635 V",
"D": "0.9018 V"
} | D | 0.9018 V | Apply the Nernst equation: E=1.02−(0.0591/1)(2)=0.9018 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.9018 V, not 1.9018 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.9018 V, not 1.8036 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.9018 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_008468 | 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 0.74 mol kg⁻¹. If its van't Hoff factor is 1 and K_b for water is 0.52 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "0.5772 K",
"B": "0.2886 K",
"C": "1.3848 K",
"D": "0.3848 K"
} | D | 0.3848 K | ΔT_b=iK_bm. Substitution gives ΔT_b=1×0.52×0.74=0.3848 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.3848 K, not 0.5772 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.3848 K, not 0.2886 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.3848 K, not ... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007699 | 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 40 mmol L⁻¹, what concentration remains after 60 minutes? | {
"A": "5 mmol L⁻¹",
"B": "8 mmol L⁻¹",
"C": "2.5 mmol L⁻¹",
"D": "6 mmol L⁻¹"
} | A | 5 mmol L⁻¹ | 60 minutes corresponds to 3 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=40/2^3=5 mmol L⁻¹. | {
"A": "Correct. The worked calculation gives 5 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5 mmol L⁻¹, not 8 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5 mmol L⁻¹, not 2.5 mmol L⁻¹.",
"D": "Incorrect. Substitut... | [
"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_005332 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +2 μC and +7 μC are separated by 5 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "60.48 N",
"B": "50.4 N",
"C": "-50.4 N",
"D": "75.6 N"
} | B | 50.4 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(2×10⁻⁶)(7×10⁻⁶)/(5×10⁻²)²=50.4 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 50.4 N, not 60.48 N.",
"B": "Correct. The worked calculation gives 50.4 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 50.4 N, not -50.4 N.",
"D": "Incorrect. Substituting the stated value... | [
"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_006039 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 97 cm in front of a thin convex lens of focal length 23 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "-1.68919",
"B": "0.23311",
"C": "0.15541",
"D": "0.31081"
} | D | 0.31081 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(97)(23)/(97−23)=30.14865 cm. Thus |m|=v/u=30.14865/97=0.31081. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.31081, not -1.68919.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.31081, not 0.23311.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.31081, not 0.15... | [
"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_000117 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 2x^2 +17x +11=0, let its roots be α and β. What is α²+β²? | {
"A": "62.25",
"B": "63.25",
"C": "122.5",
"D": "61.25"
} | D | 61.25 | By Vieta's relations, α+β = -17/2 and αβ = 11/2. Then α²+β²=(α+β)²−2αβ = (-17/2)²−2(11/2) = 61.25. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 61.25, not 62.25.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 61.25, not 63.25.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 61.25, not 122.5.",
"D":... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006605 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 7 kg is attached to an ideal spring of force constant 363 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "3.27773 s",
"B": "0.27773 s",
"C": "-1.72227 s",
"D": "1.27773 s"
} | B | 0.27773 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(7/363)=0.27773 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.27773 s, not 3.27773 s.",
"B": "Correct. The worked calculation gives 0.27773 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.27773 s, not -1.72227 s.",
"D": "Incorrect. Substituting th... | [
"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_001722 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 5 elements can be selected from a set of 46 distinct elements? | {
"A": "137076.4",
"B": "685377",
"C": "1370756",
"D": "1370754"
} | D | 1370754 | Order does not matter, so the count is C(46,5)=46!/[5!(41)!]=1370754. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1370754, not 137076.4.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1370754, not 685377.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1370754, not 13707... | [
"combinations",
"factorials"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_combinations | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004221 | 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 0.45. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "8.92 N",
"B": "77.2 N",
"C": "79.2 N",
"D": "118.8 N"
} | C | 79.2 N | Net force=mg(sinθ−μcosθ)=33×10×(0.6−0.45×0.8)=79.2 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 79.2 N, not 8.92 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 79.2 N, not 77.2 N.",
"C": "Correct. The worked calculation gives 79.2 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_006883 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 21 mol H₂ and 18 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "20 mol",
"B": "-21 mol",
"C": "15.75 mol",
"D": "21 mol"
} | D | 21 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 21 mol H₂O; O₂ can produce 36 mol H₂O. The limiting condition is H₂; product=21 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 21 mol, not 20 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 21 mol, not -21 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 21 mol, not 15.75 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_001078 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-4x³ +10x² +8x+7. Evaluate f′(-2). | {
"A": "-80",
"B": "-40",
"C": "-64",
"D": "-160"
} | A | -80 | Differentiate: f′(x)=3(-4)x²+2(10)x+(8). At x=-2, f′(-2)=3(-4)(-2)²+2(10)(-2)+(8)=-80. | {
"A": "Correct. The worked calculation gives -80.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -80, not -40.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -80, not -64.",
"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_009367 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_3H_2N_2X_4. Treat X as the total number of monovalent halogen atoms. | {
"A": "1",
"B": "4",
"C": "2",
"D": "1.5"
} | C | 2 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×3+2+2−2−4)/2=2. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2, not 1.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2, not 4.",
"C": "Correct. The worked calculation gives 2.",
"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_009080 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.63 V, transfers 1 electrons, and has reaction quotient Q=10^-4. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "3.8664 V",
"B": "-0.8664 V",
"C": "2.8664 V",
"D": "0.8664 V"
} | D | 0.8664 V | Apply the Nernst equation: E=0.63−(0.0591/1)(-4)=0.8664 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.8664 V, not 3.8664 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.8664 V, not -0.8664 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.8664 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_008795 | 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 0.34 mol L⁻¹ and initially B is absent. At equilibrium, 44% of A has dissociated at constant volume. Find K_c. | {
"A": "0.47017",
"B": "1.04702",
"C": "0.70526",
"D": "3.47017"
} | A | 0.47017 | Let α=0.44. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.34)(0.44)²/(1−0.44)=0.47017. | {
"A": "Correct. The worked calculation gives 0.47017.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.47017, not 1.04702.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.47017, not 0.70526.",
"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_003229 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(11+12i)/(4+5i). What is Re(z), the real part of z? | {
"A": "1.53659",
"B": "3.80488",
"C": "2.53659",
"D": "1.25366"
} | C | 2.53659 | Multiply numerator and denominator by the conjugate (4-5i). Then Re(z)=(ac+bd)/(c²+d²)=((11)(4)+(12)(5))/(4²+(5)²)=2.53659. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.53659, not 1.53659.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.53659, not 3.80488.",
"C": "Correct. The worked calculation gives 2.53659.",
"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_003674 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 3 kg block initially moves at 3 m s⁻¹. A constant net force does 17 N of work over 8 m in the direction of motion. Find its final kinetic energy. | {
"A": "148.5 J",
"B": "147.5 J",
"C": "149.5 J",
"D": "299 J"
} | C | 149.5 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(3)(3)²+(17)(8)=149.5 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 149.5 J, not 148.5 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 149.5 J, not 147.5 J.",
"C": "Correct. The worked calculation gives 149.5 J.",
"D": "Incorrect. Substituting the stated va... | [
"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_007551 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 43 mL of 0.10 M HCl is mixed with 69 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "12.36576",
"B": "14.36576",
"C": "2.23658",
"D": "24.73151"
} | A | 12.36576 | Excess OH⁻ moles=0.10×(69−43)/1000=0.002600 mol. Total volume=0.112 L, so [OH⁻]=0.023214 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.36576. | {
"A": "Correct. The worked calculation gives 12.36576.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.36576, not 14.36576.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12.36576, not 2.23658.",
"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_004681 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 15 Ω and 5 Ω are connected in series; this combination is connected in parallel with a 13 Ω resistor across a 12 V ideal battery. Find the total current supplied by the battery. | {
"A": "1.52308 A",
"B": "4.52308 A",
"C": "0.52308 A",
"D": "2.52308 A"
} | A | 1.52308 A | Series branch resistance=15+5=20 Ω. Parallel equivalent R_eq=(20×13)/(20+13)=7.87879 Ω. Total current I=V/R_eq=12/7.87879=1.52308 A. | {
"A": "Correct. The worked calculation gives 1.52308 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.52308 A, not 4.52308 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.52308 A, not 0.52308 A.",
"D": "Incorrect. Substituting the... | [
"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_008940 | 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 0.95 mol L⁻¹ and initially B is absent. At equilibrium, 43% of A has dissociated at constant volume. Find K_c. | {
"A": "0.23267",
"B": "2.46533",
"C": "-1.76733",
"D": "1.23267"
} | D | 1.23267 | Let α=0.43. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.95)(0.43)²/(1−0.43)=1.23267. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.23267, not 0.23267.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.23267, not 2.46533.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.23267, not -1.76... | [
"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_007201 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 19 mol of gas at 331 K and 1.78 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "579.83873 L",
"B": "288.91937 L",
"C": "231.93549 L",
"D": "289.91937 L"
} | D | 289.91937 L | PV=nRT, so V=nRT/P=(19)(0.082057)(331)/(1.78)=289.91937 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 289.91937 L, not 579.83873 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 289.91937 L, not 288.91937 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 28... | [
"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_001679 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 4 elements can be selected from a set of 41 distinct elements? | {
"A": "81016",
"B": "101268",
"C": "101270",
"D": "151905"
} | C | 101270 | Order does not matter, so the count is C(41,4)=41!/[4!(37)!]=101270. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 101270, not 81016.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 101270, not 101268.",
"C": "Correct. The worked calculation gives 101270.",
"D": "Incorrect. Substituting the stated values i... | [
"combinations",
"factorials"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_combinations | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003470 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 89 m s⁻¹ at 15° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "475.26 m",
"B": "-396.05 m",
"C": "397.05 m",
"D": "396.05 m"
} | D | 396.05 m | Range R=u²sin(2θ)/g. Substituting u=89, θ=15° and g=10 gives R=(89)²sin(30°)/10=396.05 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 396.05 m, not 475.26 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 396.05 m, not -396.05 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 396.05 m, not... | [
"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_004449 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 4 kg and radius 4 m is acted on by a constant torque of 97 N m about its central axis. Find its angular acceleration. | {
"A": "2.27344 rad s⁻²",
"B": "-3.03125 rad s⁻²",
"C": "3.03125 rad s⁻²",
"D": "0.03125 rad s⁻²"
} | C | 3.03125 rad s⁻² | For a solid disc, I=½MR²=½(4)(4)²=32 kg m². Using τ=Iα, α=97/32=3.03125 rad s⁻². | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.03125 rad s⁻², not 2.27344 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.03125 rad s⁻², not -3.03125 rad s⁻².",
"C": "Correct. The worked calculation gives 3.03125 rad s⁻².",
"D... | [
"moment of inertia",
"rotational dynamics",
"torque"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_disc_torque | 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_007392 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 44 mL of 0.10 M HCl is mixed with 38 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "2.13566",
"B": "5.13566",
"C": "3.20349",
"D": "-0.86434"
} | A | 2.13566 | Excess H⁺ moles=0.10×(44−38)/1000=0.000600 mol. Total volume=0.082 L, so [H⁺]=0.007317 M and pH=−log₁₀[H⁺]=2.13566. | {
"A": "Correct. The worked calculation gives 2.13566.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.13566, not 5.13566.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.13566, not 3.20349.",
"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_007484 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 73 mL of 0.10 M HCl is mixed with 51 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "-0.249",
"B": "3.502",
"C": "-1.249",
"D": "1.751"
} | D | 1.751 | Excess H⁺ moles=0.10×(73−51)/1000=0.002200 mol. Total volume=0.124 L, so [H⁺]=0.017742 M and pH=−log₁₀[H⁺]=1.751. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.751, not -0.249.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.751, not 3.502.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.751, not -1.249.",
"D... | [
"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_004180 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 18 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.18. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "80.08 N",
"B": "61.56 N",
"C": "9.208 N",
"D": "82.08 N"
} | D | 82.08 N | Net force=mg(sinθ−μcosθ)=18×10×(0.6−0.18×0.8)=82.08 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 82.08 N, not 80.08 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 82.08 N, not 61.56 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 82.08 N, not 9.208... | [
"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_009505 | 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_29N_1. Treat X as the total number of monovalent halogen atoms. | {
"A": "6",
"B": "-4",
"C": "3",
"D": "4"
} | D | 4 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×17+2+1−29−0)/2=4. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4, not 6.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4, not -4.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4, not 3.",
"D": "Correct. The worked c... | [
"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_009816 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Pt(OH)_4]^-4, each OH ligand has charge −1. Determine the formal oxidation state of Pt. | {
"A": "1",
"B": "0",
"C": "-3",
"D": "-1"
} | B | 0 | Let the formal oxidation state of Pt be x. Charge balance gives x+(4)(−1)=-4; hence x=-4+4=+0. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0, not 1.",
"B": "Correct. The worked calculation gives 0.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0, not -3.",
"D": "Incorrect. Substituting the stated values into the governing relat... | [
"coordination compounds",
"oxidation state",
"charge balance"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_complex_oxidation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007071 | 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 575 K and 0.51 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "1850.3049 L",
"B": "1387.72868 L",
"C": "925.15245 L",
"D": "1848.3049 L"
} | A | 1850.3049 L | PV=nRT, so V=nRT/P=(20)(0.082057)(575)/(0.51)=1850.3049 L. | {
"A": "Correct. The worked calculation gives 1850.3049 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1850.3049 L, not 1387.72868 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1850.3049 L, not 925.15245 L.",
"D": "Incorrect. Subst... | [
"ideal gas equation",
"unit consistency"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_ideal_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000643 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 5 and common difference 2. Find the sum of its first 8 terms. | {
"A": "97",
"B": "72",
"C": "96",
"D": "144"
} | C | 96 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=5, d=2, n=8: Sₙ=8/2[2(5)+(8−1)(2)]=96. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 96, not 97.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 96, not 72.",
"C": "Correct. The worked calculation gives 96.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"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_009101 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.65 V, transfers 1 electrons, and has reaction quotient Q=10^2. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "0.39885 V",
"B": "-1.4682 V",
"C": "0.5318 V",
"D": "-0.5318 V"
} | C | 0.5318 V | Apply the Nernst equation: E=0.65−(0.0591/1)(2)=0.5318 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.5318 V, not 0.39885 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.5318 V, not -1.4682 V.",
"C": "Correct. The worked calculation gives 0.5318 V.",
"D": "Incorrect. Substituting the st... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005058 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +3 μC and +6 μC are separated by 9 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "20 N",
"B": "-20 N",
"C": "3 N",
"D": "40 N"
} | A | 20 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(3×10⁻⁶)(6×10⁻⁶)/(9×10⁻²)²=20 N. | {
"A": "Correct. The worked calculation gives 20 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 20 N, not -20 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 20 N, not 3 N.",
"D": "Incorrect. Substituting the stated values into the g... | [
"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 |
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