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_000154 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 4x^2 -14x +8=0, let its roots be α and β. What is α²+β²? | {
"A": "8.25",
"B": "16.5",
"C": "11.25",
"D": "7.25"
} | A | 8.25 | By Vieta's relations, α+β = 14/4 and αβ = 8/4. Then α²+β²=(α+β)²−2αβ = (14/4)²−2(8/4) = 8.25. | {
"A": "Correct. The worked calculation gives 8.25.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 8.25, not 16.5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 8.25, not 11.25.",
"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_004226 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 38 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.19. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "170.24 N",
"B": "136.192 N",
"C": "204.288 N",
"D": "168.24 N"
} | A | 170.24 N | Net force=mg(sinθ−μcosθ)=38×10×(0.6−0.19×0.8)=170.24 N. | {
"A": "Correct. The worked calculation gives 170.24 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 170.24 N, not 136.192 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 170.24 N, not 204.288 N.",
"D": "Incorrect. Substituting the st... | [
"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_008737 | 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 0.85 mol L⁻¹ and initially B is absent. At equilibrium, 62% of A has dissociated at constant volume. Find K_c. | {
"A": "1.71968",
"B": "1.43937",
"C": "2.57953",
"D": "3.43937"
} | D | 3.43937 | Let α=0.62. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.85)(0.62)²/(1−0.62)=3.43937. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.43937, not 1.71968.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3.43937, not 1.43937.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.43937, not 2.579... | [
"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_002869 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(0, 9, 8) and v=(9, 5, 3), calculate u·v. | {
"A": "51.75",
"B": "7.9",
"C": "-69",
"D": "69"
} | D | 69 | The dot product is the sum of corresponding component products: (0)(9)+(9)(5)+(8)(3)=69. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 69, not 51.75.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 69, not 7.9.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 69, not -69.",
"D": "Correct. Th... | [
"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_006632 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 1 kg is attached to an ideal spring of force constant 592 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "-1.9178 s",
"B": "1.00822 s",
"C": "0.0822 s",
"D": "2.0822 s"
} | C | 0.0822 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(1/592)=0.0822 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.0822 s, not -1.9178 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.0822 s, not 1.00822 s.",
"C": "Correct. The worked calculation gives 0.0822 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_002759 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(-4, 8, 5) and v=(-7, -6, -2), calculate u·v. | {
"A": "-30",
"B": "-32",
"C": "-22.5",
"D": "-24"
} | A | -30 | The dot product is the sum of corresponding component products: (-4)(-7)+(8)(-6)+(5)(-2)=-30. | {
"A": "Correct. The worked calculation gives -30.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -30, not -32.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -30, not -22.5.",
"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_005128 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +5 μC and +9 μC are separated by 7 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "123.97959 N",
"B": "-82.65306 N",
"C": "82.65306 N",
"D": "81.65306 N"
} | C | 82.65306 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(5×10⁻⁶)(9×10⁻⁶)/(7×10⁻²)²=82.65306 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 82.65306 N, not 123.97959 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 82.65306 N, not -82.65306 N.",
"C": "Correct. The worked calculation gives 82.65306 N.",
"D": "Incorrect. Substitut... | [
"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_006220 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 29 cm in front of a thin convex lens of focal length 6 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "0.19565",
"B": "2.26087",
"C": "1.26087",
"D": "0.26087"
} | D | 0.26087 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(29)(6)/(29−6)=7.56522 cm. Thus |m|=v/u=7.56522/29=0.26087. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.26087, not 0.19565.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.26087, not 2.26087.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.26087, not 1.260... | [
"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_009829 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Fe(OH)_6]^+1, each OH ligand has charge −1. Determine the formal oxidation state of Fe. | {
"A": "3.5",
"B": "10",
"C": "-7",
"D": "7"
} | D | 7 | Let the formal oxidation state of Fe be x. Charge balance gives x+(6)(−1)=+1; hence x=1+6=+7. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7, not 3.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 7, not 10.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7, not -7.",
"D": "Correct. The worke... | [
"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_008951 | 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 0.35 mol L⁻¹ and initially B is absent. At equilibrium, 71% of A has dissociated at constant volume. Find K_c. | {
"A": "4.86717",
"B": "1.24336",
"C": "1.43359",
"D": "2.43359"
} | D | 2.43359 | Let α=0.71. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.35)(0.71)²/(1−0.71)=2.43359. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.43359, not 4.86717.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.43359, not 1.24336.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.43359, not 1.433... | [
"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_008232 | 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 23253 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": "110.16862 g",
"B": "92.80718 g",
"C": "93.80718 g",
"D": "91.80718 g"
} | D | 91.80718 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×12×23253/(2×96500)=91.80718 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 91.80718 g, not 110.16862 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 91.80718 g, not 92.80718 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 91.80... | [
"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_001504 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^8 (6x² -7x -5) dx. | {
"A": "912",
"B": "-760",
"C": "760",
"D": "759"
} | C | 760 | An antiderivative is (6/3)x³+(-7/2)x²+(-5)x. Evaluating between 0 and 8 gives (6/3)(8)³+(-7/2)(8)²+(-5)(8)=760. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 760, not 912.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 760, not -760.",
"C": "Correct. The worked calculation gives 760.",
"D": "Incorrect. Substituting the stated values into the gover... | [
"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_003372 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 46 m s⁻¹ at 40° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "416.77064 m",
"B": "209.38532 m",
"C": "156.28899 m",
"D": "208.38532 m"
} | D | 208.38532 m | Range R=u²sin(2θ)/g. Substituting u=46, θ=40° and g=10 gives R=(46)²sin(80°)/10=208.38532 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 208.38532 m, not 416.77064 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 208.38532 m, not 209.38532 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 20... | [
"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_006813 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 28 mol H₂ and 6 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "24 mol",
"B": "9 mol",
"C": "10 mol",
"D": "12 mol"
} | D | 12 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 28 mol H₂O; O₂ can produce 12 mol H₂O. The limiting condition is O₂; product=12 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12 mol, not 24 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12 mol, not 9 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12 mol, not 10 mol.",
... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008438 | chemistry | physical_chemistry | boiling_point_elevation | hard | provisional_template_level_label | single_correct_mcq | A non-volatile solute is dissolved in water to produce a solution of molality 1.6 mol kg⁻¹. If its van't Hoff factor is 1 and K_b for water is 0.512 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "1.6384 K",
"B": "-0.1808 K",
"C": "0.8192 K",
"D": "2.8192 K"
} | C | 0.8192 K | ΔT_b=iK_bm. Substitution gives ΔT_b=1×0.512×1.6=0.8192 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.8192 K, not 1.6384 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.8192 K, not -0.1808 K.",
"C": "Correct. The worked calculation gives 0.8192 K.",
"D": "Incorrect. Substituting the sta... | [
"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_001611 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^3 (0x² -6x +5) dx. | {
"A": "-15",
"B": "-12",
"C": "12",
"D": "-9"
} | B | -12 | An antiderivative is (0/3)x³+(-6/2)x²+(5)x. Evaluating between 0 and 3 gives (0/3)(3)³+(-6/2)(3)²+(5)(3)=-12. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -12, not -15.",
"B": "Correct. The worked calculation gives -12.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -12, not 12.",
"D": "Incorrect. Substituting the stated values into the governi... | [
"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_003841 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 7 kg block initially moves at 12 m s⁻¹. A constant net force does 18 N of work over 15 m in the direction of motion. Find its final kinetic energy. | {
"A": "774 J",
"B": "1161 J",
"C": "775 J",
"D": "-774 J"
} | A | 774 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(7)(12)²+(18)(15)=774 J. | {
"A": "Correct. The worked calculation gives 774 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 774 J, not 1161 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 774 J, not 775 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_006111 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 33 cm in front of a thin convex lens of focal length 17 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "0.79688",
"B": "1.0625",
"C": "4.0625",
"D": "1.10625"
} | B | 1.0625 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(33)(17)/(33−17)=35.0625 cm. Thus |m|=v/u=35.0625/33=1.0625. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.0625, not 0.79688.",
"B": "Correct. The worked calculation gives 1.0625.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.0625, not 4.0625.",
"D": "Incorrect. Substituting the stated values... | [
"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_000220 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 5x^2 +17x +3=0, let its roots be α and β. What is α²+β²? | {
"A": "13.36",
"B": "10.36",
"C": "9.36",
"D": "20.72"
} | B | 10.36 | By Vieta's relations, α+β = -17/5 and αβ = 3/5. Then α²+β²=(α+β)²−2αβ = (-17/5)²−2(3/5) = 10.36. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 10.36, not 13.36.",
"B": "Correct. The worked calculation gives 10.36.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 10.36, not 9.36.",
"D": "Incorrect. Substituting the stated values into t... | [
"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_001070 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=3x³ -3x² +8x+7. Evaluate f′(1). | {
"A": "13",
"B": "11",
"C": "14",
"D": "12"
} | B | 11 | Differentiate: f′(x)=3(3)x²+2(-3)x+(8). At x=1, f′(1)=3(3)(1)²+2(-3)(1)+(8)=11. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 11, not 13.",
"B": "Correct. The worked calculation gives 11.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 11, not 14.",
"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_001167 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=2x³ +0x² -12x+7. Evaluate f′(-4). | {
"A": "9.4",
"B": "85",
"C": "82",
"D": "84"
} | D | 84 | Differentiate: f′(x)=3(2)x²+2(0)x+(-12). At x=-4, f′(-4)=3(2)(-4)²+2(0)(-4)+(-12)=84. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 84, not 9.4.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 84, not 85.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 84, not 82.",
"D": "Correct. The wo... | [
"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_003918 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 11 kg block initially moves at 9 m s⁻¹. A constant net force does 6 N of work over 17 m in the direction of motion. Find its final kinetic energy. | {
"A": "273.75 J",
"B": "657 J",
"C": "-547.5 J",
"D": "547.5 J"
} | D | 547.5 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(11)(9)²+(6)(17)=547.5 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 547.5 J, not 273.75 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 547.5 J, not 657 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 547.5 J, not -547.5... | [
"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_000888 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 3 and common ratio 2. What is the sum of its first 9 terms? | {
"A": "1531",
"B": "1535",
"C": "1533",
"D": "1226.4"
} | C | 1533 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=3(2^9−1)/(2−1)=1533. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1533, not 1531.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1533, not 1535.",
"C": "Correct. The worked calculation gives 1533.",
"D": "Incorrect. Substituting the stated values into the g... | [
"geometric progression",
"finite geometric sum"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_gp_sum | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008036 | chemistry | physical_chemistry | faradays_law_copper | very_hard | provisional_template_level_label | single_correct_mcq | An aqueous Cu²⁺ solution is electrolysed using a current of 7 A for 22347 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": "51.46757 g",
"B": "102.93515 g",
"C": "6.14676 g",
"D": "49.46757 g"
} | A | 51.46757 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×7×22347/(2×96500)=51.46757 g. | {
"A": "Correct. The worked calculation gives 51.46757 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 51.46757 g, not 102.93515 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 51.46757 g, not 6.14676 g.",
"D": "Incorrect. Substitutin... | [
"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_005453 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 6 μF capacitor is charged to a potential difference of 6 V. Calculate the energy stored, expressed in μJ. | {
"A": "81 μJ",
"B": "11.8 μJ",
"C": "106 μJ",
"D": "108 μJ"
} | D | 108 μJ | U=½CV². Since C=6 μF, U in μJ is ½×6×(6)²=108 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 108 μJ, not 81 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 108 μJ, not 11.8 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 108 μJ, not 106 μJ.",
... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007255 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 19 mol of gas at 206 K and 2.9 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "110.74865 L",
"B": "12.07487 L",
"C": "166.12298 L",
"D": "55.37433 L"
} | A | 110.74865 L | PV=nRT, so V=nRT/P=(19)(0.082057)(206)/(2.9)=110.74865 L. | {
"A": "Correct. The worked calculation gives 110.74865 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 110.74865 L, not 12.07487 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 110.74865 L, not 166.12298 L.",
"D": "Incorrect. Substit... | [
"ideal gas equation",
"unit consistency"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_ideal_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008931 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 1.68 mol L⁻¹ and initially B is absent. At equilibrium, 13% of A has dissociated at constant volume. Find K_c. | {
"A": "0.19581",
"B": "-0.86946",
"C": "-0.13054",
"D": "0.13054"
} | D | 0.13054 | Let α=0.13. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.68)(0.13)²/(1−0.13)=0.13054. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.13054, not 0.19581.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.13054, not -0.86946.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.13054, not -0.1... | [
"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_004117 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 28 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.26. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "219.52 N",
"B": "109.76 N",
"C": "11.976 N",
"D": "108.76 N"
} | B | 109.76 N | Net force=mg(sinθ−μcosθ)=28×10×(0.6−0.26×0.8)=109.76 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 109.76 N, not 219.52 N.",
"B": "Correct. The worked calculation gives 109.76 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 109.76 N, not 11.976 N.",
"D": "Incorrect. Substituting the stat... | [
"Newton's laws",
"inclined plane",
"kinetic friction"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_incline_friction | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008768 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 1.38 mol L⁻¹ and initially B is absent. At equilibrium, 17% of A has dissociated at constant volume. Find K_c. | {
"A": "-2.8078",
"B": "-1.8078",
"C": "0.2883",
"D": "0.1922"
} | D | 0.1922 | Let α=0.17. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.38)(0.17)²/(1−0.17)=0.1922. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.1922, not -2.8078.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.1922, not -1.8078.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.1922, not 0.2883."... | [
"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_003631 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 93 m s⁻¹ at 50° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "-851.76023 m",
"B": "425.88011 m",
"C": "638.82017 m",
"D": "851.76023 m"
} | D | 851.76023 m | Range R=u²sin(2θ)/g. Substituting u=93, θ=50° and g=10 gives R=(93)²sin(100°)/10=851.76023 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 851.76023 m, not -851.76023 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 851.76023 m, not 425.88011 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 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_009228 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.89 V, transfers 2 electrons, and has reaction quotient Q=10^-3. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "1.09787 V",
"B": "-2.02135 V",
"C": "0.73399 V",
"D": "0.97865 V"
} | D | 0.97865 V | Apply the Nernst equation: E=0.89−(0.0591/2)(-3)=0.97865 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.97865 V, not 1.09787 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.97865 V, not -2.02135 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.97865 V... | [
"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_005762 | 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 3.8 atm at 372 K. What is its pressure at 441 K? | {
"A": "9.00968 atm",
"B": "4.50484 atm",
"C": "5.50484 atm",
"D": "-4.50484 atm"
} | B | 4.50484 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(3.8)(441)/(372)=4.50484 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.50484 atm, not 9.00968 atm.",
"B": "Correct. The worked calculation gives 4.50484 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4.50484 atm, not 5.50484 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_006911 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 29 mol H₂ and 32 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "29 mol",
"B": "14.5 mol",
"C": "23.2 mol",
"D": "34.8 mol"
} | A | 29 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 29 mol H₂O; O₂ can produce 64 mol H₂O. The limiting condition is H₂; product=29 mol. | {
"A": "Correct. The worked calculation gives 29 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 29 mol, not 14.5 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 29 mol, not 23.2 mol.",
"D": "Incorrect. Substituting the stated val... | [
"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_003570 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 25 m s⁻¹ at 55° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "56.73079 m",
"B": "60.73079 m",
"C": "46.98463 m",
"D": "58.73079 m"
} | D | 58.73079 m | Range R=u²sin(2θ)/g. Substituting u=25, θ=55° and g=10 gives R=(25)²sin(110°)/10=58.73079 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 58.73079 m, not 56.73079 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 58.73079 m, not 60.73079 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 58.730... | [
"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_001031 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-3x³ +5x² -1x+7. Evaluate f′(4). | {
"A": "-52.5",
"B": "-105",
"C": "-106",
"D": "-157.5"
} | B | -105 | Differentiate: f′(x)=3(-3)x²+2(5)x+(-1). At x=4, f′(4)=3(-3)(4)²+2(5)(4)+(-1)=-105. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -105, not -52.5.",
"B": "Correct. The worked calculation gives -105.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -105, not -106.",
"D": "Incorrect. Substituting the stated values into the ... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004955 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 13 Ω and 9 Ω are connected in series; this combination is connected in parallel with a 8 Ω resistor across a 24 V ideal battery. Find the total current supplied by the battery. | {
"A": "6.13636 A",
"B": "4.09091 A",
"C": "2.04545 A",
"D": "6.09091 A"
} | B | 4.09091 A | Series branch resistance=13+9=22 Ω. Parallel equivalent R_eq=(22×8)/(22+8)=5.86667 Ω. Total current I=V/R_eq=24/5.86667=4.09091 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.09091 A, not 6.13636 A.",
"B": "Correct. The worked calculation gives 4.09091 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4.09091 A, not 2.04545 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_006794 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 46 mol H₂ and 30 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "44 mol",
"B": "46 mol",
"C": "45 mol",
"D": "5.6 mol"
} | B | 46 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 46 mol H₂O; O₂ can produce 60 mol H₂O. The limiting condition is H₂; product=46 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 46 mol, not 44 mol.",
"B": "Correct. The worked calculation gives 46 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 46 mol, not 45 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_006453 | 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 719 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "-1.66643 s",
"B": "0.33357 s",
"C": "0.16678 s",
"D": "2.33357 s"
} | B | 0.33357 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(20/719)=0.33357 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.33357 s, not -1.66643 s.",
"B": "Correct. The worked calculation gives 0.33357 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.33357 s, not 0.16678 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_007625 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 16 mL of 0.10 M HCl is mixed with 17 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "11.48149",
"B": "13.48149",
"C": "8.61111",
"D": "22.96297"
} | A | 11.48149 | Excess OH⁻ moles=0.10×(17−16)/1000=0.000100 mol. Total volume=0.033 L, so [OH⁻]=0.003030 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=11.48149. | {
"A": "Correct. The worked calculation gives 11.48149.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 11.48149, not 13.48149.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 11.48149, not 8.61111.",
"D": "Incorrect. Substituting the state... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001051 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=6x³ +0x² -3x+7. Evaluate f′(-3). | {
"A": "161",
"B": "119.25",
"C": "159",
"D": "157"
} | C | 159 | Differentiate: f′(x)=3(6)x²+2(0)x+(-3). At x=-3, f′(-3)=3(6)(-3)²+2(0)(-3)+(-3)=159. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 159, not 161.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 159, not 119.25.",
"C": "Correct. The worked calculation gives 159.",
"D": "Incorrect. Substituting the stated values into the gov... | [
"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_004346 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 14 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": "14 rad s⁻²",
"B": "17 rad s⁻²",
"C": "12 rad s⁻²",
"D": "28 rad s⁻²"
} | A | 14 rad s⁻² | For a solid disc, I=½MR²=½(14)(1)²=7 kg m². Using τ=Iα, α=98/7=14 rad s⁻². | {
"A": "Correct. The worked calculation gives 14 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 14 rad s⁻², not 17 rad s⁻².",
"C": "Incorrect. Substituting the stated values into the governing relation gives 14 rad s⁻², not 12 rad s⁻².",
"D": "Incorrect. Substitutin... | [
"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_007816 | chemistry | physical_chemistry | first_order_half_life | hard | provisional_template_level_label | single_correct_mcq | A first-order reaction has a half-life of 20 minutes. If the initial reactant concentration is 61 mmol L⁻¹, what concentration remains after 80 minutes? | {
"A": "5.8125 mmol L⁻¹",
"B": "3.8125 mmol L⁻¹",
"C": "4.8125 mmol L⁻¹",
"D": "5.71875 mmol L⁻¹"
} | B | 3.8125 mmol L⁻¹ | 80 minutes corresponds to 4 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=61/2^4=3.8125 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.8125 mmol L⁻¹, not 5.8125 mmol L⁻¹.",
"B": "Correct. The worked calculation gives 3.8125 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.8125 mmol L⁻¹, not 4.8125 mmol L⁻¹.",
"D"... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003429 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 39 m s⁻¹ at 65° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "116.51536 m",
"B": "-116.51536 m",
"C": "117.51536 m",
"D": "93.21229 m"
} | A | 116.51536 m | Range R=u²sin(2θ)/g. Substituting u=39, θ=65° and g=10 gives R=(39)²sin(130°)/10=116.51536 m. | {
"A": "Correct. The worked calculation gives 116.51536 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 116.51536 m, not -116.51536 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 116.51536 m, not 117.51536 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_003354 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 61 m s⁻¹ at 50° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "366.44696 m",
"B": "549.67045 m",
"C": "368.44696 m",
"D": "439.73636 m"
} | A | 366.44696 m | Range R=u²sin(2θ)/g. Substituting u=61, θ=50° and g=10 gives R=(61)²sin(100°)/10=366.44696 m. | {
"A": "Correct. The worked calculation gives 366.44696 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 366.44696 m, not 549.67045 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 366.44696 m, not 368.44696 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_002426 | mathematics | linear_algebra | 2x2_determinant | hard | provisional_template_level_label | single_correct_mcq | Find the determinant of the matrix [[1, -1], [3, -1]]. | {
"A": "1.2",
"B": "0",
"C": "4",
"D": "2"
} | D | 2 | For a 2×2 matrix [[a,b],[c,d]], det=ad−bc. Here it is (1)(-1)−(-1)(3)=2. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2, not 1.2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2, not 0.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2, not 4.",
"D": "Correct. The worked ... | [
"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_008492 | 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 0.29 mol kg⁻¹. If its van't Hoff factor is 3 and K_b for water is 1.0 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "0.87 K",
"B": "-2.13 K",
"C": "1.087 K",
"D": "0.6525 K"
} | A | 0.87 K | ΔT_b=iK_bm. Substitution gives ΔT_b=3×1.0×0.29=0.87 K. | {
"A": "Correct. The worked calculation gives 0.87 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.87 K, not -2.13 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.87 K, not 1.087 K.",
"D": "Incorrect. Substituting the stated value... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001224 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-4x³ +2x² -3x+7. Evaluate f′(-1). | {
"A": "-19",
"B": "-21",
"C": "-0.9",
"D": "-17"
} | A | -19 | Differentiate: f′(x)=3(-4)x²+2(2)x+(-3). At x=-1, f′(-1)=3(-4)(-1)²+2(2)(-1)+(-3)=-19. | {
"A": "Correct. The worked calculation gives -19.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -19, not -21.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -19, not -0.9.",
"D": "Incorrect. Substituting the stated values into the gover... | [
"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_003500 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 17 m s⁻¹ at 15° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "11.45 m",
"B": "10.8375 m",
"C": "16.45 m",
"D": "14.45 m"
} | D | 14.45 m | Range R=u²sin(2θ)/g. Substituting u=17, θ=15° and g=10 gives R=(17)²sin(30°)/10=14.45 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 14.45 m, not 11.45 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 14.45 m, not 10.8375 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 14.45 m, not 16.... | [
"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_001540 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^1 (2x² -4x -1) dx. | {
"A": "-1.16667",
"B": "-4.33333",
"C": "-2.33333",
"D": "0.66667"
} | C | -2.33333 | An antiderivative is (2/3)x³+(-4/2)x²+(-1)x. Evaluating between 0 and 1 gives (2/3)(1)³+(-4/2)(1)²+(-1)(1)=-2.33333. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -2.33333, not -1.16667.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -2.33333, not -4.33333.",
"C": "Correct. The worked calculation gives -2.33333.",
"D": "Incorrect. Substituting the stat... | [
"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_000512 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 24 and common difference 4. Find the sum of its first 13 terms. | {
"A": "-624",
"B": "499.2",
"C": "624",
"D": "468"
} | C | 624 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=24, d=4, n=13: Sₙ=13/2[2(24)+(13−1)(4)]=624. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 624, not -624.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 624, not 499.2.",
"C": "Correct. The worked calculation gives 624.",
"D": "Incorrect. Substituting the stated values into the gov... | [
"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_000240 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 3x^2 -3x -6=0, let its roots be α and β. What is α²+β²? | {
"A": "2.5",
"B": "7.5",
"C": "5",
"D": "6"
} | C | 5 | By Vieta's relations, α+β = 3/3 and αβ = -6/3. Then α²+β²=(α+β)²−2αβ = (3/3)²−2(-6/3) = 5. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5, not 2.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5, not 7.5.",
"C": "Correct. The worked calculation gives 5.",
"D": "Incorrect. Substituting the stated values into the governing re... | [
"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_000761 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 5 and common ratio 3. What is the sum of its first 10 terms? | {
"A": "177144",
"B": "221430",
"C": "147622",
"D": "147620"
} | D | 147620 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=5(3^10−1)/(3−1)=147620. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 147620, not 177144.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 147620, not 221430.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 147620, not 147622.",
... | [
"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_009453 | 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_5H_1N_2X_1. Treat X as the total number of monovalent halogen atoms. | {
"A": "12",
"B": "6",
"C": "3",
"D": "4"
} | B | 6 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×5+2+2−1−1)/2=6. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6, not 12.",
"B": "Correct. The worked calculation gives 6.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 6, not 3.",
"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_006566 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 7 kg is attached to an ideal spring of force constant 176 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "0.39886 s",
"B": "1.03989 s",
"C": "0.79772 s",
"D": "0.19943 s"
} | A | 0.39886 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(7/176)=0.39886 s. | {
"A": "Correct. The worked calculation gives 0.39886 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.39886 s, not 1.03989 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.39886 s, not 0.79772 s.",
"D": "Incorrect. Substituting the... | [
"simple harmonic motion",
"mass–spring oscillator",
"time period"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_shm_period | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007995 | 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 12 mmol L⁻¹, what concentration remains after 60 minutes? | {
"A": "1.5 mmol L⁻¹",
"B": "-1.5 mmol L⁻¹",
"C": "-0.5 mmol L⁻¹",
"D": "1.15 mmol L⁻¹"
} | A | 1.5 mmol L⁻¹ | 60 minutes corresponds to 3 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=12/2^3=1.5 mmol L⁻¹. | {
"A": "Correct. The worked calculation gives 1.5 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.5 mmol L⁻¹, not -1.5 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.5 mmol L⁻¹, not -0.5 mmol L⁻¹.",
"D": "Incorrect.... | [
"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_003163 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(10-12i)/(11+6i). What is Re(z), the real part of z? | {
"A": "-0.75796",
"B": "0.24204",
"C": "2.24204",
"D": "-2.75796"
} | B | 0.24204 | Multiply numerator and denominator by the conjugate (11-6i). Then Re(z)=(ac+bd)/(c²+d²)=((10)(11)+(-12)(6))/(11²+(6)²)=0.24204. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.24204, not -0.75796.",
"B": "Correct. The worked calculation gives 0.24204.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.24204, not 2.24204.",
"D": "Incorrect. Substituting the stated v... | [
"complex conjugate",
"complex division",
"real and imaginary parts"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_complex_quotient | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005400 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 57 μF capacitor is charged to a potential difference of 27 V. Calculate the energy stored, expressed in μJ. | {
"A": "2078.65 μJ",
"B": "31164.75 μJ",
"C": "20776.5 μJ",
"D": "41553 μJ"
} | C | 20776.5 μJ | U=½CV². Since C=57 μF, U in μJ is ½×57×(27)²=20776.5 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 20776.5 μJ, not 2078.65 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 20776.5 μJ, not 31164.75 μJ.",
"C": "Correct. The worked calculation gives 20776.5 μJ.",
"D": "Incorrect. Substituti... | [
"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_001971 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 3 elements can be selected from a set of 33 distinct elements? | {
"A": "5455",
"B": "10912",
"C": "6547.2",
"D": "5456"
} | D | 5456 | Order does not matter, so the count is C(33,3)=33!/[3!(30)!]=5456. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5456, not 5455.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5456, not 10912.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5456, not 6547.2.",
"D": "C... | [
"combinations",
"factorials"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_combinations | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006831 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 11 mol H₂ and 14 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "10 mol",
"B": "12 mol",
"C": "9 mol",
"D": "11 mol"
} | D | 11 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 11 mol H₂O; O₂ can produce 28 mol H₂O. The limiting condition is H₂; product=11 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 11 mol, not 10 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 11 mol, not 12 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 11 mol, not 9 mol.",
... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004842 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 6 Ω and 5 Ω are connected in series; this combination is connected in parallel with a 5 Ω resistor across a 30 V ideal battery. Find the total current supplied by the battery. | {
"A": "10.72727 A",
"B": "8.72727 A",
"C": "17.45455 A",
"D": "11.72727 A"
} | B | 8.72727 A | Series branch resistance=6+5=11 Ω. Parallel equivalent R_eq=(11×5)/(11+5)=3.4375 Ω. Total current I=V/R_eq=30/3.4375=8.72727 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 8.72727 A, not 10.72727 A.",
"B": "Correct. The worked calculation gives 8.72727 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 8.72727 A, not 17.45455 A.",
"D": "Incorrect. Substituting t... | [
"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_008545 | 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 1.56 mol kg⁻¹. If its van't Hoff factor is 1 and K_b for water is 0.6 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "1.404 K",
"B": "1.872 K",
"C": "0.936 K",
"D": "2.936 K"
} | C | 0.936 K | ΔT_b=iK_bm. Substitution gives ΔT_b=1×0.6×1.56=0.936 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.936 K, not 1.404 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.936 K, not 1.872 K.",
"C": "Correct. The worked calculation gives 0.936 K.",
"D": "Incorrect. Substituting the stated va... | [
"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_003101 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(-7-14i)/(5+10i). What is Re(z), the real part of z? | {
"A": "-1.4",
"B": "1.6",
"C": "-2.1",
"D": "-4.4"
} | A | -1.4 | Multiply numerator and denominator by the conjugate (5-10i). Then Re(z)=(ac+bd)/(c²+d²)=((-7)(5)+(-14)(10))/(5²+(10)²)=-1.4. | {
"A": "Correct. The worked calculation gives -1.4.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -1.4, not 1.6.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -1.4, not -2.1.",
"D": "Incorrect. Substituting the stated values into the go... | [
"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_001591 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^8 (-2x² +7x +1) dx. | {
"A": "-109.33333",
"B": "-164",
"C": "-110.33333",
"D": "-111.33333"
} | A | -109.33333 | An antiderivative is (-2/3)x³+(7/2)x²+(1)x. Evaluating between 0 and 8 gives (-2/3)(8)³+(7/2)(8)²+(1)(8)=-109.33333. | {
"A": "Correct. The worked calculation gives -109.33333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -109.33333, not -164.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -109.33333, not -110.33333.",
"D": "Incorrect. Substituting the ... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009952 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Pt(OH)_5]^+0, each OH ligand has charge −1. Determine the formal oxidation state of Pt. | {
"A": "7.5",
"B": "5",
"C": "4",
"D": "2"
} | B | 5 | Let the formal oxidation state of Pt be x. Charge balance gives x+(5)(−1)=+0; hence x=0+5=+5. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5, not 7.5.",
"B": "Correct. The worked calculation gives 5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5, not 4.",
"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_007857 | chemistry | physical_chemistry | first_order_half_life | hard | provisional_template_level_label | single_correct_mcq | A first-order reaction has a half-life of 20 minutes. If the initial reactant concentration is 27 mmol L⁻¹, what concentration remains after 20 minutes? | {
"A": "2.35 mmol L⁻¹",
"B": "15.5 mmol L⁻¹",
"C": "13.5 mmol L⁻¹",
"D": "11.5 mmol L⁻¹"
} | C | 13.5 mmol L⁻¹ | 20 minutes corresponds to 1 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=27/2^1=13.5 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 13.5 mmol L⁻¹, not 2.35 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 13.5 mmol L⁻¹, not 15.5 mmol L⁻¹.",
"C": "Correct. The worked calculation gives 13.5 mmol L⁻¹.",
"D": "Incorre... | [
"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_002141 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 11 red, 9 blue and 10 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "-0.31264",
"B": "0.31264",
"C": "-1.68736",
"D": "0.15632"
} | B | 0.31264 | Favourable pairs = C(11,2)+C(9,2)+C(10,2)=136. Total pairs=C(30,2)=435. Therefore P=136/435=0.31264. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.31264, not -0.31264.",
"B": "Correct. The worked calculation gives 0.31264.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.31264, not -1.68736.",
"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_003044 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(7-1i)/(12+1i). What is Re(z), the real part of z? | {
"A": "1.57241",
"B": "-2.42759",
"C": "1.14483",
"D": "0.57241"
} | D | 0.57241 | Multiply numerator and denominator by the conjugate (12-1i). Then Re(z)=(ac+bd)/(c²+d²)=((7)(12)+(-1)(1))/(12²+(1)²)=0.57241. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.57241, not 1.57241.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.57241, not -2.42759.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.57241, not 1.14... | [
"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_005667 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 74 μF capacitor is charged to a potential difference of 17 V. Calculate the energy stored, expressed in μJ. | {
"A": "5346.5 μJ",
"B": "8554.4 μJ",
"C": "10693 μJ",
"D": "10692 μJ"
} | C | 10693 μJ | U=½CV². Since C=74 μF, U in μJ is ½×74×(17)²=10693 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 10693 μJ, not 5346.5 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 10693 μJ, not 8554.4 μJ.",
"C": "Correct. The worked calculation gives 10693 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_004320 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 6 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.39. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "20.736 N",
"B": "8.64 N",
"C": "-17.28 N",
"D": "17.28 N"
} | D | 17.28 N | Net force=mg(sinθ−μcosθ)=6×10×(0.6−0.39×0.8)=17.28 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 17.28 N, not 20.736 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 17.28 N, not 8.64 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 17.28 N, not -17.2... | [
"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_000867 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 12 and common ratio 2. What is the sum of its first 5 terms? | {
"A": "297.6",
"B": "-372",
"C": "372",
"D": "373"
} | C | 372 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=12(2^5−1)/(2−1)=372. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 372, not 297.6.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 372, not -372.",
"C": "Correct. The worked calculation gives 372.",
"D": "Incorrect. Substituting the stated values into the gov... | [
"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_008238 | chemistry | physical_chemistry | faradays_law_copper | very_hard | provisional_template_level_label | single_correct_mcq | An aqueous Cu²⁺ solution is electrolysed using a current of 13 A for 6510 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": "-27.84459 g",
"B": "22.27567 g",
"C": "41.76688 g",
"D": "27.84459 g"
} | D | 27.84459 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×13×6510/(2×96500)=27.84459 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 27.84459 g, not -27.84459 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 27.84459 g, not 22.27567 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 27.84... | [
"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_008529 | 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 1.9 mol kg⁻¹. If its van't Hoff factor is 2 and K_b for water is 0.512 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "2.9184 K",
"B": "1.19456 K",
"C": "0.9728 K",
"D": "1.9456 K"
} | D | 1.9456 K | ΔT_b=iK_bm. Substitution gives ΔT_b=2×0.512×1.9=1.9456 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.9456 K, not 2.9184 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.9456 K, not 1.19456 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.9456 K, not... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004718 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 5 Ω and 7 Ω are connected in series; this combination is connected in parallel with a 17 Ω resistor across a 6 V ideal battery. Find the total current supplied by the battery. | {
"A": "-2.14706 A",
"B": "1.08529 A",
"C": "-1.14706 A",
"D": "0.85294 A"
} | D | 0.85294 A | Series branch resistance=5+7=12 Ω. Parallel equivalent R_eq=(12×17)/(12+17)=7.03448 Ω. Total current I=V/R_eq=6/7.03448=0.85294 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.85294 A, not -2.14706 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.85294 A, not 1.08529 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.85294 A... | [
"series resistors",
"parallel resistors",
"Ohm's law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_series_parallel | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005509 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 51 μF capacitor is charged to a potential difference of 25 V. Calculate the energy stored, expressed in μJ. | {
"A": "12750 μJ",
"B": "7968.75 μJ",
"C": "15937.5 μJ",
"D": "31875 μJ"
} | C | 15937.5 μJ | U=½CV². Since C=51 μF, U in μJ is ½×51×(25)²=15937.5 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 15937.5 μJ, not 12750 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 15937.5 μJ, not 7968.75 μJ.",
"C": "Correct. The worked calculation gives 15937.5 μJ.",
"D": "Incorrect. Substituting ... | [
"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_006950 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 48 mol H₂ and 13 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "25 mol",
"B": "19.5 mol",
"C": "26 mol",
"D": "24 mol"
} | C | 26 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 48 mol H₂O; O₂ can produce 26 mol H₂O. The limiting condition is O₂; product=26 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 26 mol, not 25 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 26 mol, not 19.5 mol.",
"C": "Correct. The worked calculation gives 26 mol.",
"D": "Incorrect. Substituting the stated value... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002022 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 7 red, 11 blue and 11 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "1.32266",
"B": "0.32266",
"C": "0.64532",
"D": "-0.67734"
} | B | 0.32266 | Favourable pairs = C(7,2)+C(11,2)+C(11,2)=131. Total pairs=C(29,2)=406. Therefore P=131/406=0.32266. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.32266, not 1.32266.",
"B": "Correct. The worked calculation gives 0.32266.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.32266, not 0.64532.",
"D": "Incorrect. Substituting the stated va... | [
"classical probability",
"combinations",
"without replacement"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_probability_colours | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006101 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 79 cm in front of a thin convex lens of focal length 18 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "-0.29508",
"B": "0.22131",
"C": "0.29508",
"D": "0.44262"
} | C | 0.29508 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(79)(18)/(79−18)=23.31148 cm. Thus |m|=v/u=23.31148/79=0.29508. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.29508, not -0.29508.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.29508, not 0.22131.",
"C": "Correct. The worked calculation gives 0.29508.",
"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_009576 | 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_2N_1X_1. Treat X as the total number of monovalent halogen atoms. | {
"A": "15",
"B": "16",
"C": "17",
"D": "2.6"
} | B | 16 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×16+2+1−2−1)/2=16. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 16, not 15.",
"B": "Correct. The worked calculation gives 16.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 16, not 17.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004044 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 44 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.12. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "266.112 N",
"B": "443.52 N",
"C": "223.76 N",
"D": "221.76 N"
} | D | 221.76 N | Net force=mg(sinθ−μcosθ)=44×10×(0.6−0.12×0.8)=221.76 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 221.76 N, not 266.112 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 221.76 N, not 443.52 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 221.76 N, not... | [
"Newton's laws",
"inclined plane",
"kinetic friction"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_incline_friction | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008087 | 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 3441 s. Assuming 100% current efficiency, calculate the mass of copper deposited. Use M(Cu)=63.5 g mol⁻¹ and F=96500 C mol⁻¹. | {
"A": "7.79285 g",
"B": "8.79285 g",
"C": "5.79285 g",
"D": "6.79285 g"
} | D | 6.79285 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×6×3441/(2×96500)=6.79285 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6.79285 g, not 7.79285 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6.79285 g, not 8.79285 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 6.79285 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_007494 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 65 mL of 0.10 M HCl is mixed with 80 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "12.01472",
"B": "24.02945",
"C": "9.01104",
"D": "14.01472"
} | A | 12.01472 | Excess OH⁻ moles=0.10×(80−65)/1000=0.001500 mol. Total volume=0.145 L, so [OH⁻]=0.010345 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.01472. | {
"A": "Correct. The worked calculation gives 12.01472.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.01472, not 24.02945.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12.01472, not 9.01104.",
"D": "Incorrect. Substituting the state... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005072 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +7 μC and +9 μC are separated by 11 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "46.8595 N",
"B": "23.42975 N",
"C": "56.2314 N",
"D": "47.8595 N"
} | A | 46.8595 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(7×10⁻⁶)(9×10⁻⁶)/(11×10⁻²)²=46.8595 N. | {
"A": "Correct. The worked calculation gives 46.8595 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 46.8595 N, not 23.42975 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 46.8595 N, not 56.2314 N.",
"D": "Incorrect. Substituting th... | [
"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_008930 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 1.01 mol L⁻¹ and initially B is absent. At equilibrium, 10% of A has dissociated at constant volume. Find K_c. | {
"A": "-0.95511",
"B": "0.02244",
"C": "0.04489",
"D": "-1.95511"
} | C | 0.04489 | Let α=0.1. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.01)(0.1)²/(1−0.1)=0.04489. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.04489, not -0.95511.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.04489, not 0.02244.",
"C": "Correct. The worked calculation gives 0.04489.",
"D": "Incorrect. Substituting the stated v... | [
"chemical equilibrium",
"equilibrium constant",
"degree of dissociation"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_equilibrium_alpha | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008876 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 1.34 mol L⁻¹ and initially B is absent. At equilibrium, 36% of A has dissociated at constant volume. Find K_c. | {
"A": "-1.0854",
"B": "1.10854",
"C": "0.81405",
"D": "1.0854"
} | D | 1.0854 | Let α=0.36. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.34)(0.36)²/(1−0.36)=1.0854. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.0854, not -1.0854.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.0854, not 1.10854.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.0854, not 0.81405.... | [
"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_002074 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 9 red, 3 blue and 9 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "2.35714",
"B": "0.35714",
"C": "1.35714",
"D": "0.71429"
} | B | 0.35714 | Favourable pairs = C(9,2)+C(3,2)+C(9,2)=75. Total pairs=C(21,2)=210. Therefore P=75/210=0.35714. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.35714, not 2.35714.",
"B": "Correct. The worked calculation gives 0.35714.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.35714, not 1.35714.",
"D": "Incorrect. Substituting the stated va... | [
"classical probability",
"combinations",
"without replacement"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_probability_colours | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007305 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 11 mol of gas at 189 K and 1.31 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "65.11317 L",
"B": "132.22634 L",
"C": "130.22634 L",
"D": "195.33951 L"
} | C | 130.22634 L | PV=nRT, so V=nRT/P=(11)(0.082057)(189)/(1.31)=130.22634 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 130.22634 L, not 65.11317 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 130.22634 L, not 132.22634 L.",
"C": "Correct. The worked calculation gives 130.22634 L.",
"D": "Incorrect. Substit... | [
"ideal gas equation",
"unit consistency"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_ideal_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006043 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 69 cm in front of a thin convex lens of focal length 18 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "-1.64706",
"B": "0.35294",
"C": "0.52941",
"D": "1.03529"
} | B | 0.35294 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(69)(18)/(69−18)=24.35294 cm. Thus |m|=v/u=24.35294/69=0.35294. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.35294, not -1.64706.",
"B": "Correct. The worked calculation gives 0.35294.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.35294, not 0.52941.",
"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_008994 | 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 0.23 mol L⁻¹ and initially B is absent. At equilibrium, 58% of A has dissociated at constant volume. Find K_c. | {
"A": "0.36844",
"B": "0.73688",
"C": "-1.26312",
"D": "1.07369"
} | B | 0.73688 | Let α=0.58. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.23)(0.58)²/(1−0.58)=0.73688. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.73688, not 0.36844.",
"B": "Correct. The worked calculation gives 0.73688.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.73688, not -1.26312.",
"D": "Incorrect. Substituting the stated v... | [
"chemical equilibrium",
"equilibrium constant",
"degree of dissociation"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_equilibrium_alpha | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007586 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 83 mL of 0.10 M HCl is mixed with 66 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "0.94274",
"B": "4.94274",
"C": "1.45705",
"D": "1.94274"
} | D | 1.94274 | Excess H⁺ moles=0.10×(83−66)/1000=0.001700 mol. Total volume=0.149 L, so [H⁺]=0.011409 M and pH=−log₁₀[H⁺]=1.94274. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.94274, not 0.94274.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.94274, not 4.94274.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.94274, not 1.457... | [
"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_005555 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 33 μF capacitor is charged to a potential difference of 7 V. Calculate the energy stored, expressed in μJ. | {
"A": "808.5 μJ",
"B": "404.25 μJ",
"C": "-808.5 μJ",
"D": "1212.75 μJ"
} | A | 808.5 μJ | U=½CV². Since C=33 μF, U in μJ is ½×33×(7)²=808.5 μJ. | {
"A": "Correct. The worked calculation gives 808.5 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 808.5 μJ, not 404.25 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 808.5 μJ, not -808.5 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_007764 | 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 82 mmol L⁻¹, what concentration remains after 80 minutes? | {
"A": "1.5125 mmol L⁻¹",
"B": "10.25 mmol L⁻¹",
"C": "7.125 mmol L⁻¹",
"D": "5.125 mmol L⁻¹"
} | D | 5.125 mmol L⁻¹ | 80 minutes corresponds to 4 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=82/2^4=5.125 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5.125 mmol L⁻¹, not 1.5125 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5.125 mmol L⁻¹, not 10.25 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing rela... | [
"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_009486 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_10H_7N_1. Treat X as the total number of monovalent halogen atoms. | {
"A": "6",
"B": "7",
"C": "16",
"D": "8"
} | D | 8 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×10+2+1−7−0)/2=8. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 8, not 6.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 8, not 7.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 8, not 16.",
"D": "Correct. The worked c... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000593 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 7 and common difference 9. Find the sum of its first 11 terms. | {
"A": "572",
"B": "58.2",
"C": "1144",
"D": "574"
} | A | 572 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=7, d=9, n=11: Sₙ=11/2[2(7)+(11−1)(9)]=572. | {
"A": "Correct. The worked calculation gives 572.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 572, not 58.2.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 572, not 1144.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"arithmetic progression",
"finite series"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_ap_sum | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007115 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 4 mol of gas at 291 K and 1.55 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "-61.62216 L",
"B": "62.62216 L",
"C": "59.62216 L",
"D": "61.62216 L"
} | D | 61.62216 L | PV=nRT, so V=nRT/P=(4)(0.082057)(291)/(1.55)=61.62216 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 61.62216 L, not -61.62216 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 61.62216 L, not 62.62216 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 61.62... | [
"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_005666 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 12 μF capacitor is charged to a potential difference of 22 V. Calculate the energy stored, expressed in μJ. | {
"A": "2323.2 μJ",
"B": "-2904 μJ",
"C": "291.4 μJ",
"D": "2904 μJ"
} | D | 2904 μJ | U=½CV². Since C=12 μF, U in μJ is ½×12×(22)²=2904 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2904 μJ, not 2323.2 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2904 μJ, not -2904 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2904 μJ, not 29... | [
"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_003281 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(4+0i)/(4+2i). What is Re(z), the real part of z? | {
"A": "0.8",
"B": "-2.2",
"C": "1.08",
"D": "3.8"
} | A | 0.8 | Multiply numerator and denominator by the conjugate (4-2i). Then Re(z)=(ac+bd)/(c²+d²)=((4)(4)+(0)(2))/(4²+(2)²)=0.8. | {
"A": "Correct. The worked calculation gives 0.8.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.8, not -2.2.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.8, not 1.08.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"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_009271 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.67 V, transfers 1 electrons, and has reaction quotient Q=10^0. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "0.335 V",
"B": "0.67 V",
"C": "3.67 V",
"D": "2.67 V"
} | B | 0.67 V | Apply the Nernst equation: E=0.67−(0.0591/1)(0)=0.67 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.67 V, not 0.335 V.",
"B": "Correct. The worked calculation gives 0.67 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.67 V, not 3.67 V.",
"D": "Incorrect. Substituting the stated values... | [
"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_008592 | 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 0.22 mol kg⁻¹. If its van't Hoff factor is 3 and K_b for water is 0.6 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "0.594 K",
"B": "0.396 K",
"C": "0.792 K",
"D": "1.0396 K"
} | B | 0.396 K | ΔT_b=iK_bm. Substitution gives ΔT_b=3×0.6×0.22=0.396 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.396 K, not 0.594 K.",
"B": "Correct. The worked calculation gives 0.396 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.396 K, not 0.792 K.",
"D": "Incorrect. Substituting the stated va... | [
"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_007035 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 15 mol of gas at 422 K and 0.25 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "2493.21989 L",
"B": "2077.68324 L",
"C": "2075.68324 L",
"D": "1558.26243 L"
} | B | 2077.68324 L | PV=nRT, so V=nRT/P=(15)(0.082057)(422)/(0.25)=2077.68324 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2077.68324 L, not 2493.21989 L.",
"B": "Correct. The worked calculation gives 2077.68324 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2077.68324 L, not 2075.68324 L.",
"D": "Incorrect. S... | [
"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_001835 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 2 elements can be selected from a set of 49 distinct elements? | {
"A": "2352",
"B": "1411.2",
"C": "1176",
"D": "588"
} | C | 1176 | Order does not matter, so the count is C(49,2)=49!/[2!(47)!]=1176. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1176, not 2352.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1176, not 1411.2.",
"C": "Correct. The worked calculation gives 1176.",
"D": "Incorrect. Substituting the stated values into the... | [
"combinations",
"factorials"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_combinations | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
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