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_009746 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Pd(OH)_5]^-2, each OH ligand has charge −1. Determine the formal oxidation state of Pd. | {
"A": "3",
"B": "4",
"C": "1.3",
"D": "1"
} | A | 3 | Let the formal oxidation state of Pd be x. Charge balance gives x+(5)(−1)=-2; hence x=-2+5=+3. | {
"A": "Correct. The worked calculation gives 3.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3, not 4.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3, not 1.3.",
"D": "Incorrect. Substituting the stated values into the governing rela... | [
"coordination compounds",
"oxidation state",
"charge balance"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_complex_oxidation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001346 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^1 (5x² +6x +3) dx. | {
"A": "7.66667",
"B": "15.33333",
"C": "6.66667",
"D": "9.66667"
} | A | 7.66667 | An antiderivative is (5/3)x³+(6/2)x²+(3)x. Evaluating between 0 and 1 gives (5/3)(1)³+(6/2)(1)²+(3)(1)=7.66667. | {
"A": "Correct. The worked calculation gives 7.66667.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 7.66667, not 15.33333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7.66667, not 6.66667.",
"D": "Incorrect. Substituting the stated v... | [
"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_003480 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 110 m s⁻¹ at 18° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "710.22016 m",
"B": "533.41512 m",
"C": "711.22016 m",
"D": "1066.83023 m"
} | C | 711.22016 m | Range R=u²sin(2θ)/g. Substituting u=110, θ=18° and g=10 gives R=(110)²sin(36°)/10=711.22016 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 711.22016 m, not 710.22016 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 711.22016 m, not 533.41512 m.",
"C": "Correct. The worked calculation gives 711.22016 m.",
"D": "Incorrect. Substi... | [
"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_001072 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-2x³ +5x² +6x+7. Evaluate f′(-2). | {
"A": "-28.5",
"B": "-37",
"C": "-38",
"D": "-36"
} | C | -38 | Differentiate: f′(x)=3(-2)x²+2(5)x+(6). At x=-2, f′(-2)=3(-2)(-2)²+2(5)(-2)+(6)=-38. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -38, not -28.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -38, not -37.",
"C": "Correct. The worked calculation gives -38.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006725 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 23 mol H₂ and 22 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "3.3 mol",
"B": "21 mol",
"C": "23 mol",
"D": "18.4 mol"
} | C | 23 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 23 mol H₂O; O₂ can produce 44 mol H₂O. The limiting condition is H₂; product=23 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 23 mol, not 3.3 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 23 mol, not 21 mol.",
"C": "Correct. The worked calculation gives 23 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_001505 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^2 (-4x² +7x +5) dx. | {
"A": "15.33333",
"B": "16.33333",
"C": "26.66667",
"D": "13.33333"
} | D | 13.33333 | An antiderivative is (-4/3)x³+(7/2)x²+(5)x. Evaluating between 0 and 2 gives (-4/3)(2)³+(7/2)(2)²+(5)(2)=13.33333. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 13.33333, not 15.33333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 13.33333, not 16.33333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 13.33333, not ... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006489 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 20 kg is attached to an ideal spring of force constant 234 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "1.16941 s",
"B": "-1.41529 s",
"C": "0.58471 s",
"D": "3.58471 s"
} | C | 0.58471 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(20/234)=0.58471 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.58471 s, not 1.16941 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.58471 s, not -1.41529 s.",
"C": "Correct. The worked calculation gives 0.58471 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_002249 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 4 red, 10 blue and 5 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "0.35673",
"B": "-2.64327",
"C": "-0.64327",
"D": "3.35673"
} | A | 0.35673 | Favourable pairs = C(4,2)+C(10,2)+C(5,2)=61. Total pairs=C(19,2)=171. Therefore P=61/171=0.35673. | {
"A": "Correct. The worked calculation gives 0.35673.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.35673, not -2.64327.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.35673, not -0.64327.",
"D": "Incorrect. Substituting the stated ... | [
"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_007413 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 54 mL of 0.10 M HCl is mixed with 48 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "2.23045",
"B": "-2.23045",
"C": "4.4609",
"D": "-0.76955"
} | A | 2.23045 | Excess H⁺ moles=0.10×(54−48)/1000=0.000600 mol. Total volume=0.102 L, so [H⁺]=0.005882 M and pH=−log₁₀[H⁺]=2.23045. | {
"A": "Correct. The worked calculation gives 2.23045.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.23045, not -2.23045.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.23045, not 4.4609.",
"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_006745 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 5 mol H₂ and 13 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "5 mol",
"B": "7.5 mol",
"C": "1.5 mol",
"D": "3 mol"
} | A | 5 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 5 mol H₂O; O₂ can produce 26 mol H₂O. The limiting condition is H₂; product=5 mol. | {
"A": "Correct. The worked calculation gives 5 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not 7.5 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not 1.5 mol.",
"D": "Incorrect. Substituting the stated values i... | [
"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_009069 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.34 V, transfers 1 electrons, and has reaction quotient Q=10^-1. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "4.3991 V",
"B": "-0.6009 V",
"C": "0.69955 V",
"D": "1.3991 V"
} | D | 1.3991 V | Apply the Nernst equation: E=1.34−(0.0591/1)(-1)=1.3991 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.3991 V, not 4.3991 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.3991 V, not -0.6009 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.3991 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_007042 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 3 mol of gas at 346 K and 1.25 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "34.07007 L",
"B": "81.76816 L",
"C": "68.14013 L",
"D": "102.2102 L"
} | C | 68.14013 L | PV=nRT, so V=nRT/P=(3)(0.082057)(346)/(1.25)=68.14013 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 68.14013 L, not 34.07007 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 68.14013 L, not 81.76816 L.",
"C": "Correct. The worked calculation gives 68.14013 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_009658 | 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_16H_9N_2X_3. Treat X as the total number of monovalent halogen atoms. | {
"A": "2.2",
"B": "9",
"C": "18",
"D": "12"
} | D | 12 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×16+2+2−9−3)/2=12. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12, not 2.2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12, not 9.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12, not 18.",
"D": "Correct. The wor... | [
"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_009937 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Ni(OH)_2]^+1, each OH ligand has charge −1. Determine the formal oxidation state of Ni. | {
"A": "-3",
"B": "3",
"C": "1",
"D": "6"
} | B | 3 | Let the formal oxidation state of Ni be x. Charge balance gives x+(2)(−1)=+1; hence x=1+2=+3. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3, not -3.",
"B": "Correct. The worked calculation gives 3.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3, not 1.",
"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_004720 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 5 Ω and 13 Ω are connected in series; this combination is connected in parallel with a 16 Ω resistor across a 36 V ideal battery. Find the total current supplied by the battery. | {
"A": "8.5 A",
"B": "3.25 A",
"C": "4.25 A",
"D": "6.25 A"
} | C | 4.25 A | Series branch resistance=5+13=18 Ω. Parallel equivalent R_eq=(18×16)/(18+16)=8.47059 Ω. Total current I=V/R_eq=36/8.47059=4.25 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.25 A, not 8.5 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4.25 A, not 3.25 A.",
"C": "Correct. The worked calculation gives 4.25 A.",
"D": "Incorrect. Substituting the stated values i... | [
"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_006797 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 41 mol H₂ and 2 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "5 mol",
"B": "4 mol",
"C": "2 mol",
"D": "6 mol"
} | B | 4 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 41 mol H₂O; O₂ can produce 4 mol H₂O. The limiting condition is O₂; product=4 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4 mol, not 5 mol.",
"B": "Correct. The worked calculation gives 4 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4 mol, not 2 mol.",
"D": "Incorrect. Substituting the stated values into ... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007832 | 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 69 mmol L⁻¹, what concentration remains after 140 minutes? | {
"A": "0.26953 mmol L⁻¹",
"B": "1.07812 mmol L⁻¹",
"C": "1.53906 mmol L⁻¹",
"D": "0.53906 mmol L⁻¹"
} | D | 0.53906 mmol L⁻¹ | 140 minutes corresponds to 7 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=69/2^7=0.53906 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.53906 mmol L⁻¹, not 0.26953 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.53906 mmol L⁻¹, not 1.07812 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_007953 | 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 55 mmol L⁻¹, what concentration remains after 20 minutes? | {
"A": "-27.5 mmol L⁻¹",
"B": "22 mmol L⁻¹",
"C": "26.5 mmol L⁻¹",
"D": "27.5 mmol L⁻¹"
} | D | 27.5 mmol L⁻¹ | 20 minutes corresponds to 1 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=55/2^1=27.5 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 27.5 mmol L⁻¹, not -27.5 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 27.5 mmol L⁻¹, not 22 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation g... | [
"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_006204 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 25 cm in front of a thin convex lens of focal length 12 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "1.09231",
"B": "-2.07692",
"C": "1.84615",
"D": "0.92308"
} | D | 0.92308 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(25)(12)/(25−12)=23.07692 cm. Thus |m|=v/u=23.07692/25=0.92308. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.92308, not 1.09231.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.92308, not -2.07692.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.92308, not 1.84... | [
"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_007907 | 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 83 mmol L⁻¹, what concentration remains after 160 minutes? | {
"A": "1.03242 mmol L⁻¹",
"B": "0.32422 mmol L⁻¹",
"C": "0.48633 mmol L⁻¹",
"D": "0.64844 mmol L⁻¹"
} | B | 0.32422 mmol L⁻¹ | 160 minutes corresponds to 8 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=83/2^8=0.32422 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.32422 mmol L⁻¹, not 1.03242 mmol L⁻¹.",
"B": "Correct. The worked calculation gives 0.32422 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.32422 mmol L⁻¹, not 0.48633 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_000443 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term -11 and common difference 9. Find the sum of its first 15 terms. | {
"A": "780",
"B": "936",
"C": "390",
"D": "79"
} | A | 780 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=-11, d=9, n=15: Sₙ=15/2[2(-11)+(15−1)(9)]=780. | {
"A": "Correct. The worked calculation gives 780.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 780, not 936.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 780, not 390.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"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_006769 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 50 mol H₂ and 20 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "41 mol",
"B": "32 mol",
"C": "40 mol",
"D": "5 mol"
} | C | 40 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 50 mol H₂O; O₂ can produce 40 mol H₂O. The limiting condition is O₂; product=40 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 40 mol, not 41 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 40 mol, not 32 mol.",
"C": "Correct. The worked calculation gives 40 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_000171 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 1x^2 +16x -4=0, let its roots be α and β. What is α²+β²? | {
"A": "264",
"B": "263",
"C": "198",
"D": "-264"
} | A | 264 | By Vieta's relations, α+β = -16/1 and αβ = -4/1. Then α²+β²=(α+β)²−2αβ = (-16/1)²−2(-4/1) = 264. | {
"A": "Correct. The worked calculation gives 264.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 264, not 263.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 264, not 198.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"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_002121 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 4 red, 4 blue and 12 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "1.41053",
"B": "3.41053",
"C": "0.20526",
"D": "0.41053"
} | D | 0.41053 | Favourable pairs = C(4,2)+C(4,2)+C(12,2)=78. Total pairs=C(20,2)=190. Therefore P=78/190=0.41053. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.41053, not 1.41053.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.41053, not 3.41053.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.41053, not 0.205... | [
"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_006729 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 4 mol H₂ and 17 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "7 mol",
"B": "6 mol",
"C": "4 mol",
"D": "8 mol"
} | C | 4 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 4 mol H₂O; O₂ can produce 34 mol H₂O. The limiting condition is H₂; product=4 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4 mol, not 7 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4 mol, not 6 mol.",
"C": "Correct. The worked calculation gives 4 mol.",
"D": "Incorrect. Substituting the stated values into ... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001108 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=0x³ +1x² -1x+7. Evaluate f′(-3). | {
"A": "0.3",
"B": "-7",
"C": "-14",
"D": "-8"
} | B | -7 | Differentiate: f′(x)=3(0)x²+2(1)x+(-1). At x=-3, f′(-3)=3(0)(-3)²+2(1)(-3)+(-1)=-7. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -7, not 0.3.",
"B": "Correct. The worked calculation gives -7.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -7, not -14.",
"D": "Incorrect. Substituting the stated values into the governing... | [
"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_002871 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(7, 7, 1) and v=(-2, 1, 4), calculate u·v. | {
"A": "-3",
"B": "-6",
"C": "3",
"D": "-1"
} | A | -3 | The dot product is the sum of corresponding component products: (7)(-2)+(7)(1)+(1)(4)=-3. | {
"A": "Correct. The worked calculation gives -3.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -3, not -6.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -3, not 3.",
"D": "Incorrect. Substituting the stated values into the governing re... | [
"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_001592 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^5 (-1x² +3x +6) dx. | {
"A": "25.83333",
"B": "3.58333",
"C": "24.83333",
"D": "19.375"
} | A | 25.83333 | An antiderivative is (-1/3)x³+(3/2)x²+(6)x. Evaluating between 0 and 5 gives (-1/3)(5)³+(3/2)(5)²+(6)(5)=25.83333. | {
"A": "Correct. The worked calculation gives 25.83333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 25.83333, not 3.58333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 25.83333, not 24.83333.",
"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_005744 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 3.2 atm at 311 K. What is its pressure at 678 K? | {
"A": "5.97621 atm",
"B": "13.95241 atm",
"C": "1.69762 atm",
"D": "6.97621 atm"
} | D | 6.97621 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(3.2)(678)/(311)=6.97621 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6.97621 atm, not 5.97621 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6.97621 atm, not 13.95241 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 6... | [
"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_001652 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^1 (-2x² -1x +6) dx. | {
"A": "4.83333",
"B": "6.83333",
"C": "7.83333",
"D": "1.83333"
} | A | 4.83333 | An antiderivative is (-2/3)x³+(-1/2)x²+(6)x. Evaluating between 0 and 1 gives (-2/3)(1)³+(-1/2)(1)²+(6)(1)=4.83333. | {
"A": "Correct. The worked calculation gives 4.83333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4.83333, not 6.83333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4.83333, not 7.83333.",
"D": "Incorrect. Substituting the stated va... | [
"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_008097 | chemistry | physical_chemistry | faradays_law_copper | very_hard | provisional_template_level_label | single_correct_mcq | An aqueous Cu²⁺ solution is electrolysed using a current of 3 A for 19932 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": "14.75536 g",
"B": "-19.67381 g",
"C": "39.34763 g",
"D": "19.67381 g"
} | D | 19.67381 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×3×19932/(2×96500)=19.67381 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 19.67381 g, not 14.75536 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 19.67381 g, not -19.67381 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 19.67... | [
"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_007386 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 52 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": "3.01284",
"B": "0.01284",
"C": "-3.01284",
"D": "2.25963"
} | A | 3.01284 | Excess H⁺ moles=0.10×(52−51)/1000=0.000100 mol. Total volume=0.103 L, so [H⁺]=0.000971 M and pH=−log₁₀[H⁺]=3.01284. | {
"A": "Correct. The worked calculation gives 3.01284.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.01284, not 0.01284.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.01284, not -3.01284.",
"D": "Incorrect. Substituting the stated v... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009190 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.31 V, transfers 2 electrons, and has reaction quotient Q=10^2. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "1.2509 V",
"B": "2.2509 V",
"C": "-1.2509 V",
"D": "1.87635 V"
} | A | 1.2509 V | Apply the Nernst equation: E=1.31−(0.0591/2)(2)=1.2509 V. | {
"A": "Correct. The worked calculation gives 1.2509 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.2509 V, not 2.2509 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.2509 V, not -1.2509 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_009679 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Ni(Cl)_6]^+0, each Cl ligand has charge −1. Determine the formal oxidation state of Ni. | {
"A": "-6",
"B": "6",
"C": "9",
"D": "5"
} | B | 6 | Let the formal oxidation state of Ni be x. Charge balance gives x+(6)(−1)=+0; hence x=0+6=+6. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6, not -6.",
"B": "Correct. The worked calculation gives 6.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 6, not 9.",
"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_008650 | 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 1.47 mol kg⁻¹. If its van't Hoff factor is 1 and K_b for water is 0.512 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "1.12896 K",
"B": "0.37632 K",
"C": "0.56448 K",
"D": "0.75264 K"
} | D | 0.75264 K | ΔT_b=iK_bm. Substitution gives ΔT_b=1×0.512×1.47=0.75264 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.75264 K, not 1.12896 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.75264 K, not 0.37632 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.75264 K,... | [
"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_008502 | 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 0.14 mol kg⁻¹. If its van't Hoff factor is 4 and K_b for water is 0.6 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "3.336 K",
"B": "-1.664 K",
"C": "1.336 K",
"D": "0.336 K"
} | D | 0.336 K | ΔT_b=iK_bm. Substitution gives ΔT_b=4×0.6×0.14=0.336 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.336 K, not 3.336 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.336 K, not -1.664 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.336 K, not 1.33... | [
"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_001238 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-1x³ -4x² +8x+7. Evaluate f′(-2). | {
"A": "18",
"B": "12",
"C": "13",
"D": "-12"
} | B | 12 | Differentiate: f′(x)=3(-1)x²+2(-4)x+(8). At x=-2, f′(-2)=3(-1)(-2)²+2(-4)(-2)+(8)=12. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12, not 18.",
"B": "Correct. The worked calculation gives 12.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12, not 13.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003881 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 3 kg block initially moves at 0 m s⁻¹. A constant net force does 10 N of work over 6 m in the direction of motion. Find its final kinetic energy. | {
"A": "-60 J",
"B": "90 J",
"C": "60 J",
"D": "72 J"
} | C | 60 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(3)(0)²+(10)(6)=60 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 60 J, not -60 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 60 J, not 90 J.",
"C": "Correct. The worked calculation gives 60 J.",
"D": "Incorrect. Substituting the stated values into the ... | [
"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_004426 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 5 kg and radius 1 m is acted on by a constant torque of 85 N m about its central axis. Find its angular acceleration. | {
"A": "17 rad s⁻²",
"B": "-34 rad s⁻²",
"C": "40.8 rad s⁻²",
"D": "34 rad s⁻²"
} | D | 34 rad s⁻² | For a solid disc, I=½MR²=½(5)(1)²=2.5 kg m². Using τ=Iα, α=85/2.5=34 rad s⁻². | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 34 rad s⁻², not 17 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 34 rad s⁻², not -34 rad s⁻².",
"C": "Incorrect. Substituting the stated values into the governing relation gives 34 ra... | [
"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_005341 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 75 μF capacitor is charged to a potential difference of 2 V. Calculate the energy stored, expressed in μJ. | {
"A": "152 μJ",
"B": "150 μJ",
"C": "148 μJ",
"D": "180 μJ"
} | B | 150 μJ | U=½CV². Since C=75 μF, U in μJ is ½×75×(2)²=150 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 150 μJ, not 152 μJ.",
"B": "Correct. The worked calculation gives 150 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 150 μJ, not 148 μJ.",
"D": "Incorrect. Substituting the stated values ... | [
"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_001330 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=1x³ -1x² -3x+7. Evaluate f′(0). | {
"A": "-2.25",
"B": "-3",
"C": "0",
"D": "-6"
} | B | -3 | Differentiate: f′(x)=3(1)x²+2(-1)x+(-3). At x=0, f′(0)=3(1)(0)²+2(-1)(0)+(-3)=-3. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -3, not -2.25.",
"B": "Correct. The worked calculation gives -3.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -3, not 0.",
"D": "Incorrect. Substituting the stated values into the governing... | [
"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_003863 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 10 kg block initially moves at 2 m s⁻¹. A constant net force does 11 N of work over 4 m in the direction of motion. Find its final kinetic energy. | {
"A": "48 J",
"B": "128 J",
"C": "62 J",
"D": "64 J"
} | D | 64 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(10)(2)²+(11)(4)=64 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 64 J, not 48 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 64 J, not 128 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 64 J, not 62 J.",
"D": "Cor... | [
"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_009472 | 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_5N_3X_4. Treat X as the total number of monovalent halogen atoms. | {
"A": "1",
"B": "3",
"C": "0.5",
"D": "2"
} | A | 1 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×3+2+3−5−4)/2=1. | {
"A": "Correct. The worked calculation gives 1.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1, not 3.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1, not 0.5.",
"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_002274 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 12 red, 7 blue and 6 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "0.255",
"B": "0.34",
"C": "1.34",
"D": "0.51"
} | B | 0.34 | Favourable pairs = C(12,2)+C(7,2)+C(6,2)=102. Total pairs=C(25,2)=300. Therefore P=102/300=0.34. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.34, not 0.255.",
"B": "Correct. The worked calculation gives 0.34.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.34, not 1.34.",
"D": "Incorrect. Substituting the stated values into the ... | [
"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_007439 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 35 mL of 0.10 M HCl is mixed with 21 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "2.60206",
"B": "3.60206",
"C": "-1.39794",
"D": "1.60206"
} | D | 1.60206 | Excess H⁺ moles=0.10×(35−21)/1000=0.001400 mol. Total volume=0.056 L, so [H⁺]=0.025000 M and pH=−log₁₀[H⁺]=1.60206. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.60206, not 2.60206.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.60206, not 3.60206.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.60206, not -1.39... | [
"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_003504 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 63 m s⁻¹ at 18° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "174.96897 m",
"B": "233.29197 m",
"C": "-233.29197 m",
"D": "24.3292 m"
} | B | 233.29197 m | Range R=u²sin(2θ)/g. Substituting u=63, θ=18° and g=10 gives R=(63)²sin(36°)/10=233.29197 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 233.29197 m, not 174.96897 m.",
"B": "Correct. The worked calculation gives 233.29197 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 233.29197 m, not -233.29197 m.",
"D": "Incorrect. Subst... | [
"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_008823 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 0.15 mol L⁻¹ and initially B is absent. At equilibrium, 21% of A has dissociated at constant volume. Find K_c. | {
"A": "0.03349",
"B": "0.02512",
"C": "0.06699",
"D": "2.03349"
} | A | 0.03349 | Let α=0.21. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.15)(0.21)²/(1−0.21)=0.03349. | {
"A": "Correct. The worked calculation gives 0.03349.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.03349, not 0.02512.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.03349, not 0.06699.",
"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_005060 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +4 μC and +6 μC are separated by 6 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "60 N",
"B": "120 N",
"C": "-60 N",
"D": "30 N"
} | A | 60 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(4×10⁻⁶)(6×10⁻⁶)/(6×10⁻²)²=60 N. | {
"A": "Correct. The worked calculation gives 60 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 60 N, not 120 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 60 N, not -60 N.",
"D": "Incorrect. Substituting the stated values into the... | [
"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_007457 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 13 mL of 0.10 M HCl is mixed with 82 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "12.86113",
"B": "13.86113",
"C": "14.86113",
"D": "11.86113"
} | A | 12.86113 | Excess OH⁻ moles=0.10×(82−13)/1000=0.006900 mol. Total volume=0.095 L, so [OH⁻]=0.072632 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.86113. | {
"A": "Correct. The worked calculation gives 12.86113.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.86113, not 13.86113.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12.86113, not 14.86113.",
"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_009322 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.51 V, transfers 2 electrons, and has reaction quotient Q=10^1. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "-2.51955 V",
"B": "0.48045 V",
"C": "0.24022 V",
"D": "-0.51955 V"
} | B | 0.48045 V | Apply the Nernst equation: E=0.51−(0.0591/2)(1)=0.48045 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.48045 V, not -2.51955 V.",
"B": "Correct. The worked calculation gives 0.48045 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.48045 V, not 0.24022 V.",
"D": "Incorrect. Substituting th... | [
"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_005749 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 0.5 atm at 269 K. What is its pressure at 385 K? | {
"A": "-1.28439 atm",
"B": "0.71561 atm",
"C": "1.07156 atm",
"D": "0.35781 atm"
} | B | 0.71561 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(0.5)(385)/(269)=0.71561 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.71561 atm, not -1.28439 atm.",
"B": "Correct. The worked calculation gives 0.71561 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.71561 atm, not 1.07156 atm.",
"D": "Incorrect. Subst... | [
"ideal gas law",
"isochoric process",
"absolute temperature"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_isochoric_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009634 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_13H_1N_3X_4. Treat X as the total number of monovalent halogen atoms. | {
"A": "16",
"B": "6.5",
"C": "12",
"D": "13"
} | D | 13 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×13+2+3−1−4)/2=13. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 13, not 16.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 13, not 6.5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 13, not 12.",
"D": "Correct. The wo... | [
"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_003728 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 6 kg block initially moves at 6 m s⁻¹. A constant net force does 8 N of work over 12 m in the direction of motion. Find its final kinetic energy. | {
"A": "102 J",
"B": "408 J",
"C": "204 J",
"D": "244.8 J"
} | C | 204 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(6)(6)²+(8)(12)=204 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 204 J, not 102 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 204 J, not 408 J.",
"C": "Correct. The worked calculation gives 204 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_005074 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +4 μC and +4 μC are separated by 5 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "69.12 N",
"B": "115.2 N",
"C": "57.6 N",
"D": "58.6 N"
} | C | 57.6 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(4×10⁻⁶)(4×10⁻⁶)/(5×10⁻²)²=57.6 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 57.6 N, not 69.12 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 57.6 N, not 115.2 N.",
"C": "Correct. The worked calculation gives 57.6 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_009303 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.13 V, transfers 3 electrons, and has reaction quotient Q=10^-2. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "0.1694 V",
"B": "0.87705 V",
"C": "1.1694 V",
"D": "3.1694 V"
} | C | 1.1694 V | Apply the Nernst equation: E=1.13−(0.0591/3)(-2)=1.1694 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.1694 V, not 0.1694 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.1694 V, not 0.87705 V.",
"C": "Correct. The worked calculation gives 1.1694 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_004641 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 7 kg and radius 2 m is acted on by a constant torque of 13 N m about its central axis. Find its angular acceleration. | {
"A": "0.69643 rad s⁻²",
"B": "0.46429 rad s⁻²",
"C": "0.92857 rad s⁻²",
"D": "3.92857 rad s⁻²"
} | C | 0.92857 rad s⁻² | For a solid disc, I=½MR²=½(7)(2)²=14 kg m². Using τ=Iα, α=13/14=0.92857 rad s⁻². | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.92857 rad s⁻², not 0.69643 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.92857 rad s⁻², not 0.46429 rad s⁻².",
"C": "Correct. The worked calculation gives 0.92857 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_000259 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 2x^2 -13x +3=0, let its roots be α and β. What is α²+β²? | {
"A": "19.625",
"B": "78.5",
"C": "39.25",
"D": "29.4375"
} | C | 39.25 | By Vieta's relations, α+β = 13/2 and αβ = 3/2. Then α²+β²=(α+β)²−2αβ = (13/2)²−2(3/2) = 39.25. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 39.25, not 19.625.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 39.25, not 78.5.",
"C": "Correct. The worked calculation gives 39.25.",
"D": "Incorrect. Substituting the stated values into ... | [
"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_004088 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 27 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.43. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "7.912 N",
"B": "69.12 N",
"C": "103.68 N",
"D": "67.12 N"
} | B | 69.12 N | Net force=mg(sinθ−μcosθ)=27×10×(0.6−0.43×0.8)=69.12 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 69.12 N, not 7.912 N.",
"B": "Correct. The worked calculation gives 69.12 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 69.12 N, not 103.68 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_008271 | 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 12 A for 22932 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": "181.07963 g",
"B": "67.90486 g",
"C": "45.26991 g",
"D": "90.53981 g"
} | D | 90.53981 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×12×22932/(2×96500)=90.53981 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 90.53981 g, not 181.07963 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 90.53981 g, not 67.90486 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 90.53... | [
"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_000081 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 2x^2 -9x +12=0, let its roots be α and β. What is α²+β²? | {
"A": "8.25",
"B": "12.375",
"C": "6.25",
"D": "5.25"
} | A | 8.25 | By Vieta's relations, α+β = 9/2 and αβ = 12/2. Then α²+β²=(α+β)²−2αβ = (9/2)²−2(12/2) = 8.25. | {
"A": "Correct. The worked calculation gives 8.25.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 8.25, not 12.375.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 8.25, not 6.25.",
"D": "Incorrect. Substituting the stated values into the... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001184 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=3x³ +2x² -10x+7. Evaluate f′(-3). | {
"A": "60",
"B": "61",
"C": "59",
"D": "47.2"
} | C | 59 | Differentiate: f′(x)=3(3)x²+2(2)x+(-10). At x=-3, f′(-3)=3(3)(-3)²+2(2)(-3)+(-10)=59. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 59, not 60.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 59, not 61.",
"C": "Correct. The worked calculation gives 59.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004851 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 4 Ω and 10 Ω are connected in series; this combination is connected in parallel with a 5 Ω resistor across a 12 V ideal battery. Find the total current supplied by the battery. | {
"A": "3.25714 A",
"B": "1.62857 A",
"C": "6.25714 A",
"D": "0.25714 A"
} | A | 3.25714 A | Series branch resistance=4+10=14 Ω. Parallel equivalent R_eq=(14×5)/(14+5)=3.68421 Ω. Total current I=V/R_eq=12/3.68421=3.25714 A. | {
"A": "Correct. The worked calculation gives 3.25714 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.25714 A, not 1.62857 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.25714 A, not 6.25714 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_006037 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 30 cm in front of a thin convex lens of focal length 8 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "2.36364",
"B": "0.36364",
"C": "0.72727",
"D": "1.36364"
} | B | 0.36364 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(30)(8)/(30−8)=10.90909 cm. Thus |m|=v/u=10.90909/30=0.36364. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.36364, not 2.36364.",
"B": "Correct. The worked calculation gives 0.36364.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.36364, not 0.72727.",
"D": "Incorrect. Substituting the stated va... | [
"thin lens equation",
"linear magnification",
"real image"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_lens_magnification | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004157 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 50 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.25. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "199 N",
"B": "160 N",
"C": "200 N",
"D": "300 N"
} | C | 200 N | Net force=mg(sinθ−μcosθ)=50×10×(0.6−0.25×0.8)=200 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 200 N, not 199 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 200 N, not 160 N.",
"C": "Correct. The worked calculation gives 200 N.",
"D": "Incorrect. Substituting the stated values into ... | [
"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_005684 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 1.9 atm at 362 K. What is its pressure at 770 K? | {
"A": "2.04144 atm",
"B": "1.40414 atm",
"C": "4.04144 atm",
"D": "1.04144 atm"
} | C | 4.04144 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(1.9)(770)/(362)=4.04144 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.04144 atm, not 2.04144 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4.04144 atm, not 1.40414 atm.",
"C": "Correct. The worked calculation gives 4.04144 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_008039 | chemistry | physical_chemistry | faradays_law_copper | very_hard | provisional_template_level_label | single_correct_mcq | An aqueous Cu²⁺ solution is electrolysed using a current of 9 A for 2196 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": "9.50266 g",
"B": "9.75399 g",
"C": "1.65027 g",
"D": "6.50266 g"
} | D | 6.50266 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×9×2196/(2×96500)=6.50266 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6.50266 g, not 9.50266 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6.50266 g, not 9.75399 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 6.50266 g,... | [
"electrolysis",
"Faraday's laws",
"redox stoichiometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_faraday_copper | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009564 | 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_6H_14N_2X_2. Treat X as the total number of monovalent halogen atoms. | {
"A": "-3",
"B": "0",
"C": "2",
"D": "-2"
} | B | 0 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×6+2+2−14−2)/2=0. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0, not -3.",
"B": "Correct. The worked calculation gives 0.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0, not 2.",
"D": "Incorrect. Substituting the stated values into the governing relat... | [
"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_005486 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 25 μF capacitor is charged to a potential difference of 28 V. Calculate the energy stored, expressed in μJ. | {
"A": "19600 μJ",
"B": "14700 μJ",
"C": "9800 μJ",
"D": "7350 μJ"
} | C | 9800 μJ | U=½CV². Since C=25 μF, U in μJ is ½×25×(28)²=9800 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 9800 μJ, not 19600 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 9800 μJ, not 14700 μJ.",
"C": "Correct. The worked calculation gives 9800 μJ.",
"D": "Incorrect. Substituting the stated ... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000305 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 3x^2 -11x -9=0, let its roots be α and β. What is α²+β²? | {
"A": "19.44444",
"B": "14.58333",
"C": "38.88889",
"D": "23.33333"
} | A | 19.44444 | By Vieta's relations, α+β = 11/3 and αβ = -9/3. Then α²+β²=(α+β)²−2αβ = (11/3)²−2(-9/3) = 19.44444. | {
"A": "Correct. The worked calculation gives 19.44444.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 19.44444, not 14.58333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 19.44444, not 38.88889.",
"D": "Incorrect. Substituting the stat... | [
"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_003074 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(12-11i)/(3+7i). What is Re(z), the real part of z? | {
"A": "-1.41379",
"B": "0.2931",
"C": "-0.7069",
"D": "0.7069"
} | C | -0.7069 | Multiply numerator and denominator by the conjugate (3-7i). Then Re(z)=(ac+bd)/(c²+d²)=((12)(3)+(-11)(7))/(3²+(7)²)=-0.7069. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -0.7069, not -1.41379.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -0.7069, not 0.2931.",
"C": "Correct. The worked calculation gives -0.7069.",
"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_009663 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_9H_8N_3X_3. Treat X as the total number of monovalent halogen atoms. | {
"A": "6",
"B": "7",
"C": "5",
"D": "1.6"
} | A | 6 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×9+2+3−8−3)/2=6. | {
"A": "Correct. The worked calculation gives 6.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6, not 7.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 6, not 5.",
"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_008604 | 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 0.86 mol kg⁻¹. If its van't Hoff factor is 1 and K_b for water is 1.0 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "1.086 K",
"B": "3.86 K",
"C": "-0.86 K",
"D": "0.86 K"
} | D | 0.86 K | ΔT_b=iK_bm. Substitution gives ΔT_b=1×1.0×0.86=0.86 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.86 K, not 1.086 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.86 K, not 3.86 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.86 K, not -0.86 K."... | [
"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_001272 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=4x³ +0x² +0x+7. Evaluate f′(1). | {
"A": "9",
"B": "10",
"C": "12",
"D": "15"
} | C | 12 | Differentiate: f′(x)=3(4)x²+2(0)x+(0). At x=1, f′(1)=3(4)(1)²+2(0)(1)+(0)=12. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12, not 9.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12, not 10.",
"C": "Correct. The worked calculation gives 12.",
"D": "Incorrect. Substituting the stated values into the governing re... | [
"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_007976 | 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 88 mmol L⁻¹, what concentration remains after 40 minutes? | {
"A": "33 mmol L⁻¹",
"B": "22 mmol L⁻¹",
"C": "23 mmol L⁻¹",
"D": "21 mmol L⁻¹"
} | B | 22 mmol L⁻¹ | 40 minutes corresponds to 2 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=88/2^2=22 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 22 mmol L⁻¹, not 33 mmol L⁻¹.",
"B": "Correct. The worked calculation gives 22 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 22 mmol L⁻¹, not 23 mmol L⁻¹.",
"D": "Incorrect. Substi... | [
"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_002599 | mathematics | linear_algebra | 2x2_determinant | hard | provisional_template_level_label | single_correct_mcq | Find the determinant of the matrix [[2, 5], [4, 1]]. | {
"A": "-18",
"B": "-16",
"C": "-27",
"D": "-14.4"
} | A | -18 | For a 2×2 matrix [[a,b],[c,d]], det=ad−bc. Here it is (2)(1)−(5)(4)=-18. | {
"A": "Correct. The worked calculation gives -18.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -18, not -16.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -18, not -27.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"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_000854 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 1 and common ratio 4. What is the sum of its first 10 terms? | {
"A": "419430",
"B": "349525",
"C": "349526",
"D": "262143.75"
} | B | 349525 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=1(4^10−1)/(4−1)=349525. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 349525, not 419430.",
"B": "Correct. The worked calculation gives 349525.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 349525, not 349526.",
"D": "Incorrect. Substituting the stated values ... | [
"geometric progression",
"finite geometric sum"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_gp_sum | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004717 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 6 Ω and 3 Ω are connected in series; this combination is connected in parallel with a 8 Ω resistor across a 12 V ideal battery. Find the total current supplied by the battery. | {
"A": "1.28333 A",
"B": "3.83333 A",
"C": "2.83333 A",
"D": "-0.16667 A"
} | C | 2.83333 A | Series branch resistance=6+3=9 Ω. Parallel equivalent R_eq=(9×8)/(9+8)=4.23529 Ω. Total current I=V/R_eq=12/4.23529=2.83333 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.83333 A, not 1.28333 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.83333 A, not 3.83333 A.",
"C": "Correct. The worked calculation gives 2.83333 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_004952 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 8 Ω and 3 Ω are connected in series; this combination is connected in parallel with a 7 Ω resistor across a 12 V ideal battery. Find the total current supplied by the battery. | {
"A": "3.80519 A",
"B": "2.80519 A",
"C": "1.4026 A",
"D": "1.28052 A"
} | B | 2.80519 A | Series branch resistance=8+3=11 Ω. Parallel equivalent R_eq=(11×7)/(11+7)=4.27778 Ω. Total current I=V/R_eq=12/4.27778=2.80519 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.80519 A, not 3.80519 A.",
"B": "Correct. The worked calculation gives 2.80519 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.80519 A, not 1.4026 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_009085 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.85 V, transfers 1 electrons, and has reaction quotient Q=10^4. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "-0.6136 V",
"B": "1.6136 V",
"C": "1.06136 V",
"D": "0.6136 V"
} | D | 0.6136 V | Apply the Nernst equation: E=0.85−(0.0591/1)(4)=0.6136 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.6136 V, not -0.6136 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.6136 V, not 1.6136 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.6136 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_005380 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 56 μF capacitor is charged to a potential difference of 29 V. Calculate the energy stored, expressed in μJ. | {
"A": "35322 μJ",
"B": "23548 μJ",
"C": "28257.6 μJ",
"D": "-23548 μJ"
} | B | 23548 μJ | U=½CV². Since C=56 μF, U in μJ is ½×56×(29)²=23548 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 23548 μJ, not 35322 μJ.",
"B": "Correct. The worked calculation gives 23548 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 23548 μJ, not 28257.6 μJ.",
"D": "Incorrect. Substituting the st... | [
"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_006050 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 27 cm in front of a thin convex lens of focal length 8 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "0.42105",
"B": "0.21053",
"C": "-0.57895",
"D": "0.31579"
} | A | 0.42105 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(27)(8)/(27−8)=11.36842 cm. Thus |m|=v/u=11.36842/27=0.42105. | {
"A": "Correct. The worked calculation gives 0.42105.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.42105, not 0.21053.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.42105, not -0.57895.",
"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_000547 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 5 and common difference 11. Find the sum of its first 22 terms. | {
"A": "266.1",
"B": "2652",
"C": "2651",
"D": "1988.25"
} | C | 2651 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=5, d=11, n=22: Sₙ=22/2[2(5)+(22−1)(11)]=2651. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2651, not 266.1.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2651, not 2652.",
"C": "Correct. The worked calculation gives 2651.",
"D": "Incorrect. Substituting the stated values into the ... | [
"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_009965 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Fe(Cl)_3]^-4, each Cl ligand has charge −1. Determine the formal oxidation state of Fe. | {
"A": "-1.5",
"B": "-1",
"C": "1",
"D": "-2"
} | B | -1 | Let the formal oxidation state of Fe be x. Charge balance gives x+(3)(−1)=-4; hence x=-4+3=-1. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -1, not -1.5.",
"B": "Correct. The worked calculation gives -1.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -1, not 1.",
"D": "Incorrect. Substituting the stated values into the governing ... | [
"coordination compounds",
"oxidation state",
"charge balance"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_complex_oxidation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004514 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 7 kg and radius 1 m is acted on by a constant torque of 98 N m about its central axis. Find its angular acceleration. | {
"A": "3.8 rad s⁻²",
"B": "30 rad s⁻²",
"C": "28 rad s⁻²",
"D": "21 rad s⁻²"
} | C | 28 rad s⁻² | For a solid disc, I=½MR²=½(7)(1)²=3.5 kg m². Using τ=Iα, α=98/3.5=28 rad s⁻². | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 28 rad s⁻², not 3.8 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 28 rad s⁻², not 30 rad s⁻².",
"C": "Correct. The worked calculation gives 28 rad s⁻².",
"D": "Incorrect. Substituti... | [
"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_003461 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 75 m s⁻¹ at 30° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "365.35447 m",
"B": "-487.13929 m",
"C": "730.70893 m",
"D": "487.13929 m"
} | D | 487.13929 m | Range R=u²sin(2θ)/g. Substituting u=75, θ=30° and g=10 gives R=(75)²sin(60°)/10=487.13929 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 487.13929 m, not 365.35447 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 487.13929 m, not -487.13929 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4... | [
"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_003695 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 11 kg block initially moves at 5 m s⁻¹. A constant net force does 9 N of work over 15 m in the direction of motion. Find its final kinetic energy. | {
"A": "408.75 J",
"B": "270.5 J",
"C": "204.375 J",
"D": "272.5 J"
} | D | 272.5 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(11)(5)²+(9)(15)=272.5 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 272.5 J, not 408.75 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 272.5 J, not 270.5 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 272.5 J, not 204.... | [
"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_008417 | 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 0.78 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": "-0.80192 K",
"B": "-1.80192 K",
"C": "-1.19808 K",
"D": "1.19808 K"
} | D | 1.19808 K | ΔT_b=iK_bm. Substitution gives ΔT_b=3×0.512×0.78=1.19808 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.19808 K, not -0.80192 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.19808 K, not -1.80192 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.19808 ... | [
"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_006360 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 21 kg is attached to an ideal spring of force constant 929 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "-1.6993 s",
"B": "0.3007 s",
"C": "0.45105 s",
"D": "3.3007 s"
} | B | 0.3007 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(21/929)=0.3007 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.3007 s, not -1.6993 s.",
"B": "Correct. The worked calculation gives 0.3007 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.3007 s, not 0.45105 s.",
"D": "Incorrect. Substituting the st... | [
"simple harmonic motion",
"mass–spring oscillator",
"time period"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_shm_period | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000579 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term -10 and common difference 4. Find the sum of its first 11 terms. | {
"A": "112",
"B": "108",
"C": "82.5",
"D": "110"
} | D | 110 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=-10, d=4, n=11: Sₙ=11/2[2(-10)+(11−1)(4)]=110. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 110, not 112.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 110, not 108.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 110, not 82.5.",
"D": "Correct. ... | [
"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_005029 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +1 μC and +5 μC are separated by 4 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "30.125 N",
"B": "28.125 N",
"C": "56.25 N",
"D": "26.125 N"
} | B | 28.125 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(1×10⁻⁶)(5×10⁻⁶)/(4×10⁻²)²=28.125 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 28.125 N, not 30.125 N.",
"B": "Correct. The worked calculation gives 28.125 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 28.125 N, not 56.25 N.",
"D": "Incorrect. Substituting the state... | [
"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_002766 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(-4, 4, -7) and v=(9, 9, 1), calculate u·v. | {
"A": "-4",
"B": "-3.5",
"C": "7",
"D": "-7"
} | D | -7 | The dot product is the sum of corresponding component products: (-4)(9)+(4)(9)+(-7)(1)=-7. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -7, not -4.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -7, not -3.5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -7, not 7.",
"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_003384 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 67 m s⁻¹ at 45° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "359.12 m",
"B": "449.9 m",
"C": "45.89 m",
"D": "448.9 m"
} | D | 448.9 m | Range R=u²sin(2θ)/g. Substituting u=67, θ=45° and g=10 gives R=(67)²sin(90°)/10=448.9 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 448.9 m, not 359.12 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 448.9 m, not 449.9 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 448.9 m, not 45.8... | [
"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_008206 | 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 11055 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": "43.6472 g",
"B": "29.09813 g",
"C": "34.91776 g",
"D": "3.90981 g"
} | B | 29.09813 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×8×11055/(2×96500)=29.09813 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 29.09813 g, not 43.6472 g.",
"B": "Correct. The worked calculation gives 29.09813 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 29.09813 g, not 34.91776 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_008883 | 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 0.6 mol L⁻¹ and initially B is absent. At equilibrium, 33% of A has dissociated at constant volume. Find K_c. | {
"A": "0.39009",
"B": "3.39009",
"C": "0.78018",
"D": "2.39009"
} | A | 0.39009 | Let α=0.33. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.6)(0.33)²/(1−0.33)=0.39009. | {
"A": "Correct. The worked calculation gives 0.39009.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.39009, not 3.39009.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.39009, not 0.78018.",
"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_004588 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 6 kg and radius 1 m is acted on by a constant torque of 100 N m about its central axis. Find its angular acceleration. | {
"A": "34.33333 rad s⁻²",
"B": "33.33333 rad s⁻²",
"C": "26.66667 rad s⁻²",
"D": "50 rad s⁻²"
} | B | 33.33333 rad s⁻² | For a solid disc, I=½MR²=½(6)(1)²=3 kg m². Using τ=Iα, α=100/3=33.33333 rad s⁻². | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 33.33333 rad s⁻², not 34.33333 rad s⁻².",
"B": "Correct. The worked calculation gives 33.33333 rad s⁻².",
"C": "Incorrect. Substituting the stated values into the governing relation gives 33.33333 rad s⁻², not 26.66667 rad s⁻².",
... | [
"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_003328 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(-15-8i)/(12+2i). What is Re(z), the real part of z? | {
"A": "-0.99324",
"B": "-1.32432",
"C": "-3.32432",
"D": "-2.64865"
} | B | -1.32432 | Multiply numerator and denominator by the conjugate (12-2i). Then Re(z)=(ac+bd)/(c²+d²)=((-15)(12)+(-8)(2))/(12²+(2)²)=-1.32432. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -1.32432, not -0.99324.",
"B": "Correct. The worked calculation gives -1.32432.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -1.32432, not -3.32432.",
"D": "Incorrect. Substituting the stat... | [
"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_008151 | 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 6 A for 22662 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": "89.4738 g",
"B": "33.55268 g",
"C": "42.7369 g",
"D": "44.7369 g"
} | D | 44.7369 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×6×22662/(2×96500)=44.7369 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 44.7369 g, not 89.4738 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 44.7369 g, not 33.55268 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 44.7369 g... | [
"electrolysis",
"Faraday's laws",
"redox stoichiometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_faraday_copper | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006773 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 3 mol H₂ and 26 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "3 mol",
"B": "6 mol",
"C": "2 mol",
"D": "-3 mol"
} | A | 3 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 3 mol H₂O; O₂ can produce 52 mol H₂O. The limiting condition is H₂; product=3 mol. | {
"A": "Correct. The worked calculation gives 3 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3 mol, not 6 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3 mol, not 2 mol.",
"D": "Incorrect. Substituting the stated values into ... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002950 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(-3, -2, -4) and v=(3, 3, 2), calculate u·v. | {
"A": "-22",
"B": "-34.5",
"C": "-23",
"D": "-17.25"
} | C | -23 | The dot product is the sum of corresponding component products: (-3)(3)+(-2)(3)+(-4)(2)=-23. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -23, not -22.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -23, not -34.5.",
"C": "Correct. The worked calculation gives -23.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"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_002145 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 8 red, 4 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": "0.24123",
"B": "3.32164",
"C": "-0.32164",
"D": "0.32164"
} | D | 0.32164 | Favourable pairs = C(8,2)+C(4,2)+C(7,2)=55. Total pairs=C(19,2)=171. Therefore P=55/171=0.32164. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.32164, not 0.24123.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.32164, not 3.32164.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.32164, not -0.32... | [
"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 |
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