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_001471 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^5 (-1x² +5x -2) dx. | {
"A": "10.83333",
"B": "7.83333",
"C": "21.66667",
"D": "9.83333"
} | A | 10.83333 | An antiderivative is (-1/3)x³+(5/2)x²+(-2)x. Evaluating between 0 and 5 gives (-1/3)(5)³+(5/2)(5)²+(-2)(5)=10.83333. | {
"A": "Correct. The worked calculation gives 10.83333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 10.83333, not 7.83333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 10.83333, not 21.66667.",
"D": "Incorrect. Substituting the state... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009050 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.91 V, transfers 1 electrons, and has reaction quotient Q=10^6. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "1.05554 V",
"B": "0.41655 V",
"C": "-0.5554 V",
"D": "0.5554 V"
} | D | 0.5554 V | Apply the Nernst equation: E=0.91−(0.0591/1)(6)=0.5554 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.5554 V, not 1.05554 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.5554 V, not 0.41655 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.5554 V, no... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008295 | 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 13533 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": "53.43081 g",
"B": "28.7154 g",
"C": "25.7154 g",
"D": "26.7154 g"
} | D | 26.7154 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×6×13533/(2×96500)=26.7154 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 26.7154 g, not 53.43081 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 26.7154 g, not 28.7154 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 26.7154 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_001817 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 6 elements can be selected from a set of 30 distinct elements? | {
"A": "593775",
"B": "445331.25",
"C": "890662.5",
"D": "296887.5"
} | A | 593775 | Order does not matter, so the count is C(30,6)=30!/[6!(24)!]=593775. | {
"A": "Correct. The worked calculation gives 593775.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 593775, not 445331.25.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 593775, not 890662.5.",
"D": "Incorrect. Substituting the stated va... | [
"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_001175 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=2x³ -4x² -3x+7. Evaluate f′(1). | {
"A": "-5",
"B": "-3",
"C": "-6",
"D": "-7.5"
} | A | -5 | Differentiate: f′(x)=3(2)x²+2(-4)x+(-3). At x=1, f′(1)=3(2)(1)²+2(-4)(1)+(-3)=-5. | {
"A": "Correct. The worked calculation gives -5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -5, not -3.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -5, not -6.",
"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_000400 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 22 and common difference 7. Find the sum of its first 11 terms. | {
"A": "940.5",
"B": "627",
"C": "629",
"D": "-627"
} | B | 627 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=22, d=7, n=11: Sₙ=11/2[2(22)+(11−1)(7)]=627. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 627, not 940.5.",
"B": "Correct. The worked calculation gives 627.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 627, not 629.",
"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_008150 | chemistry | physical_chemistry | faradays_law_copper | very_hard | provisional_template_level_label | single_correct_mcq | An aqueous Cu²⁺ solution is electrolysed using a current of 8 A for 11552 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": "30.4063 g",
"B": "32.4063 g",
"C": "29.4063 g",
"D": "60.8126 g"
} | A | 30.4063 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×8×11552/(2×96500)=30.4063 g. | {
"A": "Correct. The worked calculation gives 30.4063 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 30.4063 g, not 32.4063 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 30.4063 g, not 29.4063 g.",
"D": "Incorrect. Substituting the... | [
"electrolysis",
"Faraday's laws",
"redox stoichiometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_faraday_copper | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001966 | 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 50 distinct elements? | {
"A": "29400",
"B": "1961",
"C": "19602",
"D": "19600"
} | D | 19600 | Order does not matter, so the count is C(50,3)=50!/[3!(47)!]=19600. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 19600, not 29400.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 19600, not 1961.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 19600, not 19602.",
"D": ... | [
"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_004232 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 17 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.11. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "87.04 N",
"B": "9.704 N",
"C": "69.632 N",
"D": "-87.04 N"
} | A | 87.04 N | Net force=mg(sinθ−μcosθ)=17×10×(0.6−0.11×0.8)=87.04 N. | {
"A": "Correct. The worked calculation gives 87.04 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 87.04 N, not 9.704 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 87.04 N, not 69.632 N.",
"D": "Incorrect. Substituting the stated v... | [
"Newton's laws",
"inclined plane",
"kinetic friction"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_incline_friction | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004770 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 5 Ω and 2 Ω are connected in series; this combination is connected in parallel with a 19 Ω resistor across a 12 V ideal battery. Find the total current supplied by the battery. | {
"A": "1.17293 A",
"B": "1.23459 A",
"C": "2.34586 A",
"D": "4.34586 A"
} | C | 2.34586 A | Series branch resistance=5+2=7 Ω. Parallel equivalent R_eq=(7×19)/(7+19)=5.11538 Ω. Total current I=V/R_eq=12/5.11538=2.34586 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.34586 A, not 1.17293 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.34586 A, not 1.23459 A.",
"C": "Correct. The worked calculation gives 2.34586 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_008997 | 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 1.13 mol L⁻¹ and initially B is absent. At equilibrium, 11% of A has dissociated at constant volume. Find K_c. | {
"A": "0.06145",
"B": "-0.93855",
"C": "3.06145",
"D": "1.06145"
} | A | 0.06145 | Let α=0.11. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(1.13)(0.11)²/(1−0.11)=0.06145. | {
"A": "Correct. The worked calculation gives 0.06145.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.06145, not -0.93855.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.06145, not 3.06145.",
"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_003738 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 14 kg block initially moves at 0 m s⁻¹. A constant net force does 10 N of work over 20 m in the direction of motion. Find its final kinetic energy. | {
"A": "200 J",
"B": "198 J",
"C": "240 J",
"D": "199 J"
} | A | 200 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(14)(0)²+(10)(20)=200 J. | {
"A": "Correct. The worked calculation gives 200 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 200 J, not 198 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 200 J, not 240 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_006335 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 115 cm in front of a thin convex lens of focal length 28 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "0.64368",
"B": "0.24138",
"C": "0.32184",
"D": "2.32184"
} | C | 0.32184 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(115)(28)/(115−28)=37.01149 cm. Thus |m|=v/u=37.01149/115=0.32184. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.32184, not 0.64368.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.32184, not 0.24138.",
"C": "Correct. The worked calculation gives 0.32184.",
"D": "Incorrect. Substituting the stated va... | [
"thin lens equation",
"linear magnification",
"real image"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_lens_magnification | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009129 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.60 V, transfers 3 electrons, and has reaction quotient Q=10^3. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "-1.4591 V",
"B": "0.27045 V",
"C": "-0.4591 V",
"D": "0.5409 V"
} | D | 0.5409 V | Apply the Nernst equation: E=0.60−(0.0591/3)(3)=0.5409 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.5409 V, not -1.4591 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.5409 V, not 0.27045 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.5409 V, no... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007352 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 97 mL of 0.10 M HCl is mixed with 16 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "2.28919",
"B": "0.5723",
"C": "-1.85541",
"D": "1.14459"
} | D | 1.14459 | Excess H⁺ moles=0.10×(97−16)/1000=0.008100 mol. Total volume=0.113 L, so [H⁺]=0.071681 M and pH=−log₁₀[H⁺]=1.14459. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.14459, not 2.28919.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.14459, not 0.5723.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.14459, not -1.855... | [
"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_009247 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.99 V, transfers 3 electrons, and has reaction quotient Q=10^-5. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "3.0885 V",
"B": "0.0885 V",
"C": "1.10885 V",
"D": "1.0885 V"
} | D | 1.0885 V | Apply the Nernst equation: E=0.99−(0.0591/3)(-5)=1.0885 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.0885 V, not 3.0885 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.0885 V, not 0.0885 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.0885 V, not ... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003594 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 94 m s⁻¹ at 15° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "220.9 m",
"B": "45.18 m",
"C": "441.8 m",
"D": "439.8 m"
} | C | 441.8 m | Range R=u²sin(2θ)/g. Substituting u=94, θ=15° and g=10 gives R=(94)²sin(30°)/10=441.8 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 441.8 m, not 220.9 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 441.8 m, not 45.18 m.",
"C": "Correct. The worked calculation gives 441.8 m.",
"D": "Incorrect. Substituting the stated va... | [
"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_007168 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 6 mol of gas at 252 K and 1.73 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "53.78765 L",
"B": "57.3735 L",
"C": "71.71687 L",
"D": "69.71687 L"
} | C | 71.71687 L | PV=nRT, so V=nRT/P=(6)(0.082057)(252)/(1.73)=71.71687 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 71.71687 L, not 53.78765 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 71.71687 L, not 57.3735 L.",
"C": "Correct. The worked calculation gives 71.71687 L.",
"D": "Incorrect. Substituting... | [
"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_007948 | 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 28 mmol L⁻¹, what concentration remains after 140 minutes? | {
"A": "1.02188 mmol L⁻¹",
"B": "0.32812 mmol L⁻¹",
"C": "2.21875 mmol L⁻¹",
"D": "0.21875 mmol L⁻¹"
} | D | 0.21875 mmol L⁻¹ | 140 minutes corresponds to 7 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=28/2^7=0.21875 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.21875 mmol L⁻¹, not 1.02188 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.21875 mmol L⁻¹, not 0.32812 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governi... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007729 | 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 50 mmol L⁻¹, what concentration remains after 100 minutes? | {
"A": "1.5625 mmol L⁻¹",
"B": "-1.4375 mmol L⁻¹",
"C": "3.125 mmol L⁻¹",
"D": "2.5625 mmol L⁻¹"
} | A | 1.5625 mmol L⁻¹ | 100 minutes corresponds to 5 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=50/2^5=1.5625 mmol L⁻¹. | {
"A": "Correct. The worked calculation gives 1.5625 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.5625 mmol L⁻¹, not -1.4375 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.5625 mmol L⁻¹, not 3.125 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_009931 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Cr(Cl)_2]^-4, each Cl ligand has charge −1. Determine the formal oxidation state of Cr. | {
"A": "-2",
"B": "-3",
"C": "-4",
"D": "1"
} | A | -2 | Let the formal oxidation state of Cr be x. Charge balance gives x+(2)(−1)=-4; hence x=-4+2=-2. | {
"A": "Correct. The worked calculation gives -2.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -2, not -3.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -2, not -4.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"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_004077 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 43 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.25. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "129 N",
"B": "173 N",
"C": "174 N",
"D": "172 N"
} | D | 172 N | Net force=mg(sinθ−μcosθ)=43×10×(0.6−0.25×0.8)=172 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 172 N, not 129 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 172 N, not 173 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 172 N, not 174 N.",
"D":... | [
"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_007107 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 1 mol of gas at 280 K and 2.69 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "12.81187 L",
"B": "8.54125 L",
"C": "10.54125 L",
"D": "5.54125 L"
} | B | 8.54125 L | PV=nRT, so V=nRT/P=(1)(0.082057)(280)/(2.69)=8.54125 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 8.54125 L, not 12.81187 L.",
"B": "Correct. The worked calculation gives 8.54125 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 8.54125 L, not 10.54125 L.",
"D": "Incorrect. Substituting t... | [
"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_003191 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(11-6i)/(12+12i). What is Re(z), the real part of z? | {
"A": "1.02083",
"B": "-1.79167",
"C": "2.20833",
"D": "0.20833"
} | D | 0.20833 | Multiply numerator and denominator by the conjugate (12-12i). Then Re(z)=(ac+bd)/(c²+d²)=((11)(12)+(-6)(12))/(12²+(12)²)=0.20833. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.20833, not 1.02083.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.20833, not -1.79167.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.20833, not 2.20... | [
"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_005494 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 62 μF capacitor is charged to a potential difference of 12 V. Calculate the energy stored, expressed in μJ. | {
"A": "4463 μJ",
"B": "4464 μJ",
"C": "4465 μJ",
"D": "447.4 μJ"
} | B | 4464 μJ | U=½CV². Since C=62 μF, U in μJ is ½×62×(12)²=4464 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4464 μJ, not 4463 μJ.",
"B": "Correct. The worked calculation gives 4464 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4464 μJ, not 4465 μJ.",
"D": "Incorrect. Substituting the stated va... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007767 | 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 56 mmol L⁻¹, what concentration remains after 60 minutes? | {
"A": "5.25 mmol L⁻¹",
"B": "10 mmol L⁻¹",
"C": "1.7 mmol L⁻¹",
"D": "7 mmol L⁻¹"
} | D | 7 mmol L⁻¹ | 60 minutes corresponds to 3 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=56/2^3=7 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7 mmol L⁻¹, not 5.25 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 7 mmol L⁻¹, not 10 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7 ... | [
"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_003116 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(-13-3i)/(7+2i). What is Re(z), the real part of z? | {
"A": "-2.83019",
"B": "0.81698",
"C": "-4.83019",
"D": "-1.83019"
} | D | -1.83019 | Multiply numerator and denominator by the conjugate (7-2i). Then Re(z)=(ac+bd)/(c²+d²)=((-13)(7)+(-3)(2))/(7²+(2)²)=-1.83019. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -1.83019, not -2.83019.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -1.83019, not 0.81698.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -1.83019, not -... | [
"complex conjugate",
"complex division",
"real and imaginary parts"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_complex_quotient | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006354 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 14 kg is attached to an ideal spring of force constant 1107 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "1.22492 s",
"B": "2.22492 s",
"C": "0.22492 s",
"D": "-0.77508 s"
} | C | 0.22492 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(14/1107)=0.22492 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.22492 s, not 1.22492 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.22492 s, not 2.22492 s.",
"C": "Correct. The worked calculation gives 0.22492 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_006984 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 27 mol H₂ and 35 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "21.6 mol",
"B": "32.4 mol",
"C": "54 mol",
"D": "27 mol"
} | D | 27 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 27 mol H₂O; O₂ can produce 70 mol H₂O. The limiting condition is H₂; product=27 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 27 mol, not 21.6 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 27 mol, not 32.4 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 27 mol, not 54 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_008300 | 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 3018 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": "5.95781 g",
"B": "23.83125 g",
"C": "-11.91563 g",
"D": "11.91563 g"
} | D | 11.91563 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×12×3018/(2×96500)=11.91563 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 11.91563 g, not 5.95781 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 11.91563 g, not 23.83125 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 11.9156... | [
"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_007557 | 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 45 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "-0.47257",
"B": "1.14557",
"C": "1.52743",
"D": "1.15274"
} | C | 1.52743 | Excess H⁺ moles=0.10×(83−45)/1000=0.003800 mol. Total volume=0.128 L, so [H⁺]=0.029688 M and pH=−log₁₀[H⁺]=1.52743. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.52743, not -0.47257.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.52743, not 1.14557.",
"C": "Correct. The worked calculation gives 1.52743.",
"D": "Incorrect. Substituting the stated v... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001822 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 5 elements can be selected from a set of 67 distinct elements? | {
"A": "9657648",
"B": "9657649",
"C": "4828824",
"D": "7726118.4"
} | A | 9657648 | Order does not matter, so the count is C(67,5)=67!/[5!(62)!]=9657648. | {
"A": "Correct. The worked calculation gives 9657648.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 9657648, not 9657649.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 9657648, not 4828824.",
"D": "Incorrect. Substituting the stated va... | [
"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_005423 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 76 μF capacitor is charged to a potential difference of 34 V. Calculate the energy stored, expressed in μJ. | {
"A": "35142.4 μJ",
"B": "43928 μJ",
"C": "21964 μJ",
"D": "43930 μJ"
} | B | 43928 μJ | U=½CV². Since C=76 μF, U in μJ is ½×76×(34)²=43928 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 43928 μJ, not 35142.4 μJ.",
"B": "Correct. The worked calculation gives 43928 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 43928 μJ, not 21964 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_006919 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 44 mol H₂ and 33 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "22 mol",
"B": "52.8 mol",
"C": "42 mol",
"D": "44 mol"
} | D | 44 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 44 mol H₂O; O₂ can produce 66 mol H₂O. The limiting condition is H₂; product=44 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 44 mol, not 22 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 44 mol, not 52.8 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 44 mol, not 42 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_001574 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^1 (-4x² -4x +6) dx. | {
"A": "3.66667",
"B": "2.66667",
"C": "5.33333",
"D": "2"
} | B | 2.66667 | An antiderivative is (-4/3)x³+(-4/2)x²+(6)x. Evaluating between 0 and 1 gives (-4/3)(1)³+(-4/2)(1)²+(6)(1)=2.66667. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.66667, not 3.66667.",
"B": "Correct. The worked calculation gives 2.66667.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.66667, not 5.33333.",
"D": "Incorrect. Substituting the stated va... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001006 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-5x³ -7x² +4x+7. Evaluate f′(2). | {
"A": "-83",
"B": "-42",
"C": "-7.4",
"D": "-84"
} | D | -84 | Differentiate: f′(x)=3(-5)x²+2(-7)x+(4). At x=2, f′(2)=3(-5)(2)²+2(-7)(2)+(4)=-84. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -84, not -83.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -84, not -42.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -84, not -7.4.",
"D": "Correct. ... | [
"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_007964 | 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 65 mmol L⁻¹, what concentration remains after 140 minutes? | {
"A": "-2.49219 mmol L⁻¹",
"B": "0.50781 mmol L⁻¹",
"C": "-0.49219 mmol L⁻¹",
"D": "0.25391 mmol L⁻¹"
} | B | 0.50781 mmol L⁻¹ | 140 minutes corresponds to 7 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=65/2^7=0.50781 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.50781 mmol L⁻¹, not -2.49219 mmol L⁻¹.",
"B": "Correct. The worked calculation gives 0.50781 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.50781 mmol L⁻¹, not -0.49219 mmol L⁻¹."... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000168 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 5x^2 +5x +11=0, let its roots be α and β. What is α²+β²? | {
"A": "-2.55",
"B": "-6.4",
"C": "-1.4",
"D": "-3.4"
} | D | -3.4 | By Vieta's relations, α+β = -5/5 and αβ = 11/5. Then α²+β²=(α+β)²−2αβ = (-5/5)²−2(11/5) = -3.4. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -3.4, not -2.55.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -3.4, not -6.4.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -3.4, not -1.4.",
"D": "Cor... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006004 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 1.4 atm at 422 K. What is its pressure at 541 K? | {
"A": "3.58957 atm",
"B": "2.69218 atm",
"C": "4.79479 atm",
"D": "1.79479 atm"
} | D | 1.79479 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(1.4)(541)/(422)=1.79479 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.79479 atm, not 3.58957 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.79479 atm, not 2.69218 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.... | [
"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_000298 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 4x^2 +9x -5=0, let its roots be α and β. What is α²+β²? | {
"A": "5.5625",
"B": "7.5625",
"C": "15.125",
"D": "4.5625"
} | B | 7.5625 | By Vieta's relations, α+β = -9/4 and αβ = -5/4. Then α²+β²=(α+β)²−2αβ = (-9/4)²−2(-5/4) = 7.5625. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7.5625, not 5.5625.",
"B": "Correct. The worked calculation gives 7.5625.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7.5625, not 15.125.",
"D": "Incorrect. Substituting the stated values ... | [
"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_006455 | 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 683 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "0.68449 s",
"B": "0.34224 s",
"C": "-0.65776 s",
"D": "2.34224 s"
} | B | 0.34224 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(20/683)=0.34224 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.34224 s, not 0.68449 s.",
"B": "Correct. The worked calculation gives 0.34224 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.34224 s, not -0.65776 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_000811 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 7 and common ratio 4. What is the sum of its first 7 terms? | {
"A": "-38227",
"B": "38227",
"C": "19113.5",
"D": "38225"
} | B | 38227 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=7(4^7−1)/(4−1)=38227. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 38227, not -38227.",
"B": "Correct. The worked calculation gives 38227.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 38227, not 19113.5.",
"D": "Incorrect. Substituting the stated values in... | [
"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_009726 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Fe(Cl)_3]^-2, each Cl ligand has charge −1. Determine the formal oxidation state of Fe. | {
"A": "1",
"B": "0",
"C": "3",
"D": "-1"
} | A | 1 | Let the formal oxidation state of Fe be x. Charge balance gives x+(3)(−1)=-2; hence x=-2+3=+1. | {
"A": "Correct. The worked calculation gives 1.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1, not 0.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1, not 3.",
"D": "Incorrect. Substituting the stated values into the governing relati... | [
"coordination compounds",
"oxidation state",
"charge balance"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_complex_oxidation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004082 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 25 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.23. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "106 N",
"B": "124.8 N",
"C": "208 N",
"D": "104 N"
} | D | 104 N | Net force=mg(sinθ−μcosθ)=25×10×(0.6−0.23×0.8)=104 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 104 N, not 106 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 104 N, not 124.8 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 104 N, not 208 N.",
"D... | [
"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_006815 | 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 23 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "34.5 mol",
"B": "-46 mol",
"C": "46 mol",
"D": "45 mol"
} | C | 46 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 46 mol H₂O; O₂ can produce 46 mol H₂O. The limiting condition is neither; stoichiometric; product=46 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 46 mol, not 34.5 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 46 mol, not -46 mol.",
"C": "Correct. The worked calculation gives 46 mol.",
"D": "Incorrect. Substituting the stated valu... | [
"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_002910 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(8, 2, 7) and v=(-5, -3, 2), calculate u·v. | {
"A": "-31",
"B": "-30",
"C": "-32",
"D": "-16"
} | C | -32 | The dot product is the sum of corresponding component products: (8)(-5)+(2)(-3)+(7)(2)=-32. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -32, not -31.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -32, not -30.",
"C": "Correct. The worked calculation gives -32.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"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_006493 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 17 kg is attached to an ideal spring of force constant 366 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "-2.56896 s",
"B": "-0.43104 s",
"C": "0.21552 s",
"D": "0.43104 s"
} | D | 0.43104 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(17/366)=0.43104 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.43104 s, not -2.56896 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.43104 s, not -0.43104 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.43104 ... | [
"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_001123 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=-5x³ -8x² -7x+7. Evaluate f′(0). | {
"A": "-6",
"B": "-7",
"C": "-4",
"D": "-10"
} | B | -7 | Differentiate: f′(x)=3(-5)x²+2(-8)x+(-7). At x=0, f′(0)=3(-5)(0)²+2(-8)(0)+(-7)=-7. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -7, not -6.",
"B": "Correct. The worked calculation gives -7.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -7, not -4.",
"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_002740 | mathematics | vectors | dot_product | hard | provisional_template_level_label | single_correct_mcq | Given vectors u=(7, 8, 4) and v=(0, 1, 6), calculate u·v. | {
"A": "32",
"B": "48",
"C": "34",
"D": "38.4"
} | A | 32 | The dot product is the sum of corresponding component products: (7)(0)+(8)(1)+(4)(6)=32. | {
"A": "Correct. The worked calculation gives 32.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 32, not 48.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 32, not 34.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"three-dimensional vectors",
"scalar product"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_vector_dot | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006190 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 23 cm in front of a thin convex lens of focal length 5 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "1.02778",
"B": "1.27778",
"C": "-0.72222",
"D": "0.27778"
} | D | 0.27778 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(23)(5)/(23−5)=6.38889 cm. Thus |m|=v/u=6.38889/23=0.27778. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.27778, not 1.02778.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.27778, not 1.27778.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.27778, not -0.72... | [
"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_007220 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 20 mol of gas at 273 K and 2.53 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "265.63116 L",
"B": "177.08744 L",
"C": "-177.08744 L",
"D": "175.08744 L"
} | B | 177.08744 L | PV=nRT, so V=nRT/P=(20)(0.082057)(273)/(2.53)=177.08744 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 177.08744 L, not 265.63116 L.",
"B": "Correct. The worked calculation gives 177.08744 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 177.08744 L, not -177.08744 L.",
"D": "Incorrect. Subst... | [
"ideal gas equation",
"unit consistency"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_ideal_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002399 | mathematics | linear_algebra | 2x2_determinant | hard | provisional_template_level_label | single_correct_mcq | Find the determinant of the matrix [[2, -6], [9, -1]]. | {
"A": "41.6",
"B": "52",
"C": "78",
"D": "51"
} | B | 52 | For a 2×2 matrix [[a,b],[c,d]], det=ad−bc. Here it is (2)(-1)−(-6)(9)=52. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 52, not 41.6.",
"B": "Correct. The worked calculation gives 52.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 52, not 78.",
"D": "Incorrect. Substituting the stated values into the governing... | [
"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_000733 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 7 and common ratio 4. What is the sum of its first 12 terms? | {
"A": "19573417.5",
"B": "46976202",
"C": "39146835",
"D": "3914684.5"
} | C | 39146835 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=7(4^12−1)/(4−1)=39146835. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 39146835, not 19573417.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 39146835, not 46976202.",
"C": "Correct. The worked calculation gives 39146835.",
"D": "Incorrect. Substituting the st... | [
"geometric progression",
"finite geometric sum"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_gp_sum | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005313 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +7 μC and +3 μC are separated by 11 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "14.61983 N",
"B": "18.7438 N",
"C": "15.61983 N",
"D": "16.61983 N"
} | C | 15.61983 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(7×10⁻⁶)(3×10⁻⁶)/(11×10⁻²)²=15.61983 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 15.61983 N, not 14.61983 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 15.61983 N, not 18.7438 N.",
"C": "Correct. The worked calculation gives 15.61983 N.",
"D": "Incorrect. Substituting... | [
"Coulomb's law",
"SI unit conversion",
"inverse-square law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_coulomb | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000388 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 18 and common difference 5. Find the sum of its first 29 terms. | {
"A": "1276",
"B": "2553",
"C": "1914",
"D": "2552"
} | D | 2552 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=18, d=5, n=29: Sₙ=29/2[2(18)+(29−1)(5)]=2552. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2552, not 1276.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2552, not 2553.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2552, not 1914.",
"D": "Corr... | [
"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_003162 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(-2+12i)/(3+3i). What is Re(z), the real part of z? | {
"A": "-1.66667",
"B": "-0.33333",
"C": "0.66667",
"D": "1.66667"
} | D | 1.66667 | Multiply numerator and denominator by the conjugate (3-3i). Then Re(z)=(ac+bd)/(c²+d²)=((-2)(3)+(12)(3))/(3²+(3)²)=1.66667. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.66667, not -1.66667.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.66667, not -0.33333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.66667, not 0.6... | [
"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_001542 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^7 (-2x² +1x +7) dx. | {
"A": "-77.58333",
"B": "-154.16667",
"C": "-155.16667",
"D": "-156.16667"
} | C | -155.16667 | An antiderivative is (-2/3)x³+(1/2)x²+(7)x. Evaluating between 0 and 7 gives (-2/3)(7)³+(1/2)(7)²+(7)(7)=-155.16667. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -155.16667, not -77.58333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -155.16667, not -154.16667.",
"C": "Correct. The worked calculation gives -155.16667.",
"D": "Incorrect. Substituting... | [
"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_008358 | 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 0.38 mol kg⁻¹. If its van't Hoff factor is 3 and K_b for water is 0.512 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "-2.41632 K",
"B": "-1.41632 K",
"C": "0.58368 K",
"D": "2.58368 K"
} | C | 0.58368 K | ΔT_b=iK_bm. Substitution gives ΔT_b=3×0.512×0.38=0.58368 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.58368 K, not -2.41632 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.58368 K, not -1.41632 K.",
"C": "Correct. The worked calculation gives 0.58368 K.",
"D": "Incorrect. Substituting t... | [
"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_007060 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 6 mol of gas at 475 K and 1.2 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "195.88538 L",
"B": "292.32806 L",
"C": "194.88538 L",
"D": "193.88538 L"
} | C | 194.88538 L | PV=nRT, so V=nRT/P=(6)(0.082057)(475)/(1.2)=194.88538 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 194.88538 L, not 195.88538 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 194.88538 L, not 292.32806 L.",
"C": "Correct. The worked calculation gives 194.88538 L.",
"D": "Incorrect. Substi... | [
"ideal gas equation",
"unit consistency"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_ideal_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002314 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 4 red, 9 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": "0.35145",
"B": "1.03514",
"C": "-0.35145",
"D": "1.35145"
} | A | 0.35145 | Favourable pairs = C(4,2)+C(9,2)+C(11,2)=97. Total pairs=C(24,2)=276. Therefore P=97/276=0.35145. | {
"A": "Correct. The worked calculation gives 0.35145.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.35145, not 1.03514.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.35145, not -0.35145.",
"D": "Incorrect. Substituting the stated v... | [
"classical probability",
"combinations",
"without replacement"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_probability_colours | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005401 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 38 μF capacitor is charged to a potential difference of 23 V. Calculate the energy stored, expressed in μJ. | {
"A": "20102 μJ",
"B": "10052 μJ",
"C": "10051 μJ",
"D": "8040.8 μJ"
} | C | 10051 μJ | U=½CV². Since C=38 μF, U in μJ is ½×38×(23)²=10051 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 10051 μJ, not 20102 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 10051 μJ, not 10052 μJ.",
"C": "Correct. The worked calculation gives 10051 μJ.",
"D": "Incorrect. Substituting the stat... | [
"capacitance",
"energy stored in a capacitor"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_capacitor_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006733 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 10 mol H₂ and 8 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "12 mol",
"B": "-10 mol",
"C": "2 mol",
"D": "10 mol"
} | D | 10 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 10 mol H₂O; O₂ can produce 16 mol H₂O. The limiting condition is H₂; product=10 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 10 mol, not 12 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 10 mol, not -10 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 10 mol, not 2 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_003014 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(12+5i)/(7+10i). What is Re(z), the real part of z? | {
"A": "0.89933",
"B": "2.89933",
"C": "-1.10067",
"D": "3.89933"
} | A | 0.89933 | Multiply numerator and denominator by the conjugate (7-10i). Then Re(z)=(ac+bd)/(c²+d²)=((12)(7)+(5)(10))/(7²+(10)²)=0.89933. | {
"A": "Correct. The worked calculation gives 0.89933.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.89933, not 2.89933.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.89933, not -1.10067.",
"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_007742 | 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 26 mmol L⁻¹, what concentration remains after 80 minutes? | {
"A": "4.625 mmol L⁻¹",
"B": "0.625 mmol L⁻¹",
"C": "3.25 mmol L⁻¹",
"D": "1.625 mmol L⁻¹"
} | D | 1.625 mmol L⁻¹ | 80 minutes corresponds to 4 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=26/2^4=1.625 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.625 mmol L⁻¹, not 4.625 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.625 mmol L⁻¹, not 0.625 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relat... | [
"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_003724 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 2 kg block initially moves at 0 m s⁻¹. A constant net force does 13 N of work over 16 m in the direction of motion. Find its final kinetic energy. | {
"A": "208 J",
"B": "249.6 J",
"C": "156 J",
"D": "416 J"
} | A | 208 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(2)(0)²+(13)(16)=208 J. | {
"A": "Correct. The worked calculation gives 208 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 208 J, not 249.6 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 208 J, not 156 J.",
"D": "Incorrect. Substituting the stated values int... | [
"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_006384 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 16 kg is attached to an ideal spring of force constant 897 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "0.20033 s",
"B": "0.26711 s",
"C": "0.40067 s",
"D": "-2.73289 s"
} | B | 0.26711 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(16/897)=0.26711 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.26711 s, not 0.20033 s.",
"B": "Correct. The worked calculation gives 0.26711 s.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.26711 s, not 0.40067 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_002477 | mathematics | linear_algebra | 2x2_determinant | hard | provisional_template_level_label | single_correct_mcq | Find the determinant of the matrix [[1, -6], [6, 7]]. | {
"A": "-43",
"B": "43",
"C": "45",
"D": "21.5"
} | B | 43 | For a 2×2 matrix [[a,b],[c,d]], det=ad−bc. Here it is (1)(7)−(-6)(6)=43. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 43, not -43.",
"B": "Correct. The worked calculation gives 43.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 43, not 45.",
"D": "Incorrect. Substituting the stated values into the governing ... | [
"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_004582 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 2 kg and radius 2 m is acted on by a constant torque of 71 N m about its central axis. Find its angular acceleration. | {
"A": "18.75 rad s⁻²",
"B": "17.75 rad s⁻²",
"C": "26.625 rad s⁻²",
"D": "15.75 rad s⁻²"
} | B | 17.75 rad s⁻² | For a solid disc, I=½MR²=½(2)(2)²=4 kg m². Using τ=Iα, α=71/4=17.75 rad s⁻². | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 17.75 rad s⁻², not 18.75 rad s⁻².",
"B": "Correct. The worked calculation gives 17.75 rad s⁻².",
"C": "Incorrect. Substituting the stated values into the governing relation gives 17.75 rad s⁻², not 26.625 rad s⁻².",
"D": "Incorr... | [
"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_008829 | 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 0.99 mol L⁻¹ and initially B is absent. At equilibrium, 42% of A has dissociated at constant volume. Find K_c. | {
"A": "-1.79561",
"B": "1.80658",
"C": "2.20439",
"D": "1.20439"
} | D | 1.20439 | Let α=0.42. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.99)(0.42)²/(1−0.42)=1.20439. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.20439, not -1.79561.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.20439, not 1.80658.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.20439, not 2.20... | [
"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_007822 | 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 31 mmol L⁻¹, what concentration remains after 160 minutes? | {
"A": "-2.87891 mmol L⁻¹",
"B": "0.12109 mmol L⁻¹",
"C": "0.18164 mmol L⁻¹",
"D": "0.06055 mmol L⁻¹"
} | B | 0.12109 mmol L⁻¹ | 160 minutes corresponds to 8 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=31/2^8=0.12109 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.12109 mmol L⁻¹, not -2.87891 mmol L⁻¹.",
"B": "Correct. The worked calculation gives 0.12109 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.12109 mmol L⁻¹, not 0.18164 mmol L⁻¹.",... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002174 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 12 red, 6 blue and 6 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "0.17391",
"B": "1.34783",
"C": "0.26087",
"D": "0.34783"
} | D | 0.34783 | Favourable pairs = C(12,2)+C(6,2)+C(6,2)=96. Total pairs=C(24,2)=276. Therefore P=96/276=0.34783. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.34783, not 0.17391.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.34783, not 1.34783.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.34783, not 0.260... | [
"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_006735 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 5 mol H₂ and 32 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "6 mol",
"B": "5 mol",
"C": "1.5 mol",
"D": "-5 mol"
} | B | 5 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 5 mol H₂O; O₂ can produce 64 mol H₂O. The limiting condition is H₂; product=5 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not 6 mol.",
"B": "Correct. The worked calculation gives 5 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not 1.5 mol.",
"D": "Incorrect. Substituting the stated values int... | [
"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_003675 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 8 kg block initially moves at 1 m s⁻¹. A constant net force does 5 N of work over 17 m in the direction of motion. Find its final kinetic energy. | {
"A": "133.5 J",
"B": "88 J",
"C": "178 J",
"D": "89 J"
} | D | 89 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(8)(1)²+(5)(17)=89 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 89 J, not 133.5 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 89 J, not 88 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 89 J, not 178 J.",
"D": "... | [
"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_006325 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 22 cm in front of a thin convex lens of focal length 16 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "-0.33333",
"B": "1.26667",
"C": "2.66667",
"D": "0.66667"
} | C | 2.66667 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(22)(16)/(22−16)=58.66667 cm. Thus |m|=v/u=58.66667/22=2.66667. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.66667, not -0.33333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.66667, not 1.26667.",
"C": "Correct. The worked calculation gives 2.66667.",
"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_006499 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 11 kg is attached to an ideal spring of force constant 1053 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "1.02044 s",
"B": "1.20441 s",
"C": "0.20441 s",
"D": "-1.79559 s"
} | C | 0.20441 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(11/1053)=0.20441 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.20441 s, not 1.02044 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.20441 s, not 1.20441 s.",
"C": "Correct. The worked calculation gives 0.20441 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_001380 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^6 (3x² -2x +1) dx. | {
"A": "188",
"B": "19.6",
"C": "186",
"D": "372"
} | C | 186 | An antiderivative is (3/3)x³+(-2/2)x²+(1)x. Evaluating between 0 and 6 gives (3/3)(6)³+(-2/2)(6)²+(1)(6)=186. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 186, not 188.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 186, not 19.6.",
"C": "Correct. The worked calculation gives 186.",
"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_008597 | 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 0.12 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": "-0.64 K",
"B": "-0.36 K",
"C": "0.36 K",
"D": "0.27 K"
} | C | 0.36 K | ΔT_b=iK_bm. Substitution gives ΔT_b=3×1.0×0.12=0.36 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.36 K, not -0.64 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.36 K, not -0.36 K.",
"C": "Correct. The worked calculation gives 0.36 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_002292 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 11 red, 8 blue and 5 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "-0.66304",
"B": "0.33696",
"C": "-2.66304",
"D": "2.33696"
} | B | 0.33696 | Favourable pairs = C(11,2)+C(8,2)+C(5,2)=93. Total pairs=C(24,2)=276. Therefore P=93/276=0.33696. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.33696, not -0.66304.",
"B": "Correct. The worked calculation gives 0.33696.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.33696, not -2.66304.",
"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_009323 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.12 V, transfers 1 electrons, and has reaction quotient Q=10^-4. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "-0.6436 V",
"B": "1.3564 V",
"C": "2.0346 V",
"D": "3.3564 V"
} | B | 1.3564 V | Apply the Nernst equation: E=1.12−(0.0591/1)(-4)=1.3564 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.3564 V, not -0.6436 V.",
"B": "Correct. The worked calculation gives 1.3564 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.3564 V, not 2.0346 V.",
"D": "Incorrect. Substituting the sta... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002094 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 4 red, 5 blue and 6 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "3.29524",
"B": "-0.70476",
"C": "0.22143",
"D": "0.29524"
} | D | 0.29524 | Favourable pairs = C(4,2)+C(5,2)+C(6,2)=31. Total pairs=C(15,2)=105. Therefore P=31/105=0.29524. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.29524, not 3.29524.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.29524, not -0.70476.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.29524, not 0.22... | [
"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_007725 | 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 80 mmol L⁻¹, what concentration remains after 100 minutes? | {
"A": "1.25 mmol L⁻¹",
"B": "2.5 mmol L⁻¹",
"C": "3.75 mmol L⁻¹",
"D": "3.5 mmol L⁻¹"
} | B | 2.5 mmol L⁻¹ | 100 minutes corresponds to 5 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=80/2^5=2.5 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.5 mmol L⁻¹, not 1.25 mmol L⁻¹.",
"B": "Correct. The worked calculation gives 2.5 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.5 mmol L⁻¹, not 3.75 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_007709 | 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 84 mmol L⁻¹, what concentration remains after 140 minutes? | {
"A": "0.65625 mmol L⁻¹",
"B": "-0.65625 mmol L⁻¹",
"C": "1.65625 mmol L⁻¹",
"D": "-1.34375 mmol L⁻¹"
} | A | 0.65625 mmol L⁻¹ | 140 minutes corresponds to 7 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=84/2^7=0.65625 mmol L⁻¹. | {
"A": "Correct. The worked calculation gives 0.65625 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.65625 mmol L⁻¹, not -0.65625 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.65625 mmol L⁻¹, not 1.65625 mmol L⁻¹.",... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005846 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 4.5 atm at 314 K. What is its pressure at 515 K? | {
"A": "3.69029 atm",
"B": "11.07086 atm",
"C": "7.38057 atm",
"D": "8.38057 atm"
} | C | 7.38057 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(4.5)(515)/(314)=7.38057 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7.38057 atm, not 3.69029 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 7.38057 atm, not 11.07086 atm.",
"C": "Correct. The worked calculation gives 7.38057 atm.",
"D": "Incorrect. Subst... | [
"ideal gas law",
"isochoric process",
"absolute temperature"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_isochoric_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000283 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 4x^2 -12x +3=0, let its roots be α and β. What is α²+β²? | {
"A": "6.5",
"B": "7.5",
"C": "1.75",
"D": "4.5"
} | B | 7.5 | By Vieta's relations, α+β = 12/4 and αβ = 3/4. Then α²+β²=(α+β)²−2αβ = (12/4)²−2(3/4) = 7.5. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7.5, not 6.5.",
"B": "Correct. The worked calculation gives 7.5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7.5, not 1.75.",
"D": "Incorrect. Substituting the stated values into the gover... | [
"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_008309 | 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 29367 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": "-125.60859 g",
"B": "188.41289 g",
"C": "124.60859 g",
"D": "125.60859 g"
} | D | 125.60859 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×13×29367/(2×96500)=125.60859 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 125.60859 g, not -125.60859 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 125.60859 g, not 188.41289 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1... | [
"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_001865 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 8 elements can be selected from a set of 28 distinct elements? | {
"A": "310811.5",
"B": "3108105",
"C": "3108103",
"D": "3729726"
} | B | 3108105 | Order does not matter, so the count is C(28,8)=28!/[8!(20)!]=3108105. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3108105, not 310811.5.",
"B": "Correct. The worked calculation gives 3108105.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3108105, not 3108103.",
"D": "Incorrect. Substituting the stated v... | [
"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_002528 | mathematics | linear_algebra | 2x2_determinant | hard | provisional_template_level_label | single_correct_mcq | Find the determinant of the matrix [[8, 7], [3, -6]]. | {
"A": "-138",
"B": "-82.8",
"C": "69",
"D": "-69"
} | D | -69 | For a 2×2 matrix [[a,b],[c,d]], det=ad−bc. Here it is (8)(-6)−(7)(3)=-69. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -69, not -138.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -69, not -82.8.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -69, not 69.",
"D": "Correct.... | [
"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_004144 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 25 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.18. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "112 N",
"B": "91.2 N",
"C": "114 N",
"D": "12.4 N"
} | C | 114 N | Net force=mg(sinθ−μcosθ)=25×10×(0.6−0.18×0.8)=114 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 114 N, not 112 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 114 N, not 91.2 N.",
"C": "Correct. The worked calculation gives 114 N.",
"D": "Incorrect. Substituting the stated values into... | [
"Newton's laws",
"inclined plane",
"kinetic friction"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_incline_friction | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000779 | mathematics | sequences_and_series | geometric_progression | hard | provisional_template_level_label | single_correct_mcq | A geometric progression has first term 7 and common ratio 2. What is the sum of its first 7 terms? | {
"A": "888",
"B": "89.9",
"C": "666.75",
"D": "889"
} | D | 889 | For r≠1, Sₙ=a(rⁿ−1)/(r−1). Thus Sₙ=7(2^7−1)/(2−1)=889. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 889, not 888.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 889, not 89.9.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 889, not 666.75.",
"D": "Correc... | [
"geometric progression",
"finite geometric sum"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_gp_sum | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004663 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 8 kg and radius 6 m is acted on by a constant torque of 56 N m about its central axis. Find its angular acceleration. | {
"A": "0.29167 rad s⁻²",
"B": "-2.61111 rad s⁻²",
"C": "3.38889 rad s⁻²",
"D": "0.38889 rad s⁻²"
} | D | 0.38889 rad s⁻² | For a solid disc, I=½MR²=½(8)(6)²=144 kg m². Using τ=Iα, α=56/144=0.38889 rad s⁻². | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.38889 rad s⁻², not 0.29167 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.38889 rad s⁻², not -2.61111 rad s⁻².",
"C": "Incorrect. Substituting the stated values into the governing ... | [
"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_007661 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 89 mL of 0.10 M HCl is mixed with 53 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "3.19197",
"B": "1.59599",
"C": "-1.40401",
"D": "2.59599"
} | B | 1.59599 | Excess H⁺ moles=0.10×(89−53)/1000=0.003600 mol. Total volume=0.142 L, so [H⁺]=0.025352 M and pH=−log₁₀[H⁺]=1.59599. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.59599, not 3.19197.",
"B": "Correct. The worked calculation gives 1.59599.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.59599, not -1.40401.",
"D": "Incorrect. Substituting the stated v... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000445 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term -3 and common difference 6. Find the sum of its first 27 terms. | {
"A": "4050",
"B": "2025",
"C": "-2025",
"D": "1518.75"
} | B | 2025 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=-3, d=6, n=27: Sₙ=27/2[2(-3)+(27−1)(6)]=2025. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2025, not 4050.",
"B": "Correct. The worked calculation gives 2025.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2025, not -2025.",
"D": "Incorrect. Substituting the stated values into the ... | [
"arithmetic progression",
"finite series"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_ap_sum | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000427 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 25 and common difference 12. Find the sum of its first 13 terms. | {
"A": "1260",
"B": "1263",
"C": "1261",
"D": "630.5"
} | C | 1261 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=25, d=12, n=13: Sₙ=13/2[2(25)+(13−1)(12)]=1261. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1261, not 1260.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1261, not 1263.",
"C": "Correct. The worked calculation gives 1261.",
"D": "Incorrect. Substituting the stated values into the g... | [
"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_007858 | 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 96 mmol L⁻¹, what concentration remains after 80 minutes? | {
"A": "1.6 mmol L⁻¹",
"B": "6 mmol L⁻¹",
"C": "9 mmol L⁻¹",
"D": "4 mmol L⁻¹"
} | B | 6 mmol L⁻¹ | 80 minutes corresponds to 4 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=96/2^4=6 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6 mmol L⁻¹, not 1.6 mmol L⁻¹.",
"B": "Correct. The worked calculation gives 6 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 6 mmol L⁻¹, not 9 mmol L⁻¹.",
"D": "Incorrect. Substitut... | [
"first-order kinetics",
"half-life",
"exponential decay"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_first_order_halflife | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001401 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^8 (0x² +0x -2) dx. | {
"A": "-12",
"B": "-16",
"C": "-32",
"D": "-0.6"
} | B | -16 | An antiderivative is (0/3)x³+(0/2)x²+(-2)x. Evaluating between 0 and 8 gives (0/3)(8)³+(0/2)(8)²+(-2)(8)=-16. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -16, not -12.",
"B": "Correct. The worked calculation gives -16.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -16, not -32.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"definite integration",
"power rule",
"limits of integration"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_integral_poly | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004282 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 49 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.44. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "121.52 N",
"B": "119.52 N",
"C": "145.824 N",
"D": "-121.52 N"
} | A | 121.52 N | Net force=mg(sinθ−μcosθ)=49×10×(0.6−0.44×0.8)=121.52 N. | {
"A": "Correct. The worked calculation gives 121.52 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 121.52 N, not 119.52 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 121.52 N, not 145.824 N.",
"D": "Incorrect. Substituting the sta... | [
"Newton's laws",
"inclined plane",
"kinetic friction"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_incline_friction | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003976 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 11 kg block initially moves at 5 m s⁻¹. A constant net force does 24 N of work over 10 m in the direction of motion. Find its final kinetic energy. | {
"A": "38.75 J",
"B": "377.5 J",
"C": "379.5 J",
"D": "566.25 J"
} | B | 377.5 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(11)(5)²+(24)(10)=377.5 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 377.5 J, not 38.75 J.",
"B": "Correct. The worked calculation gives 377.5 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 377.5 J, not 379.5 J.",
"D": "Incorrect. Substituting the stated va... | [
"work–energy theorem",
"kinetic energy",
"work by constant force"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_work_energy | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000428 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term -19 and common difference 7. Find the sum of its first 17 terms. | {
"A": "629",
"B": "754.8",
"C": "314.5",
"D": "-629"
} | A | 629 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=-19, d=7, n=17: Sₙ=17/2[2(-19)+(17−1)(7)]=629. | {
"A": "Correct. The worked calculation gives 629.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 629, not 754.8.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 629, not 314.5.",
"D": "Incorrect. Substituting the stated values into the go... | [
"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_005230 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +1 μC and +9 μC are separated by 10 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "8.1 N",
"B": "10.1 N",
"C": "6.1 N",
"D": "12.15 N"
} | A | 8.1 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(1×10⁻⁶)(9×10⁻⁶)/(10×10⁻²)²=8.1 N. | {
"A": "Correct. The worked calculation gives 8.1 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 8.1 N, not 10.1 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 8.1 N, not 6.1 N.",
"D": "Incorrect. Substituting the stated values into... | [
"Coulomb's law",
"SI unit conversion",
"inverse-square law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_coulomb | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007117 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 17 mol of gas at 234 K and 2.68 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "91.34965 L",
"B": "121.79953 L",
"C": "13.17995 L",
"D": "120.79953 L"
} | B | 121.79953 L | PV=nRT, so V=nRT/P=(17)(0.082057)(234)/(2.68)=121.79953 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 121.79953 L, not 91.34965 L.",
"B": "Correct. The worked calculation gives 121.79953 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 121.79953 L, not 13.17995 L.",
"D": "Incorrect. Substitu... | [
"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 |
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