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_009612 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_15H_16X_4. Treat X as the total number of monovalent halogen atoms. | {
"A": "1.6",
"B": "9",
"C": "6",
"D": "5"
} | C | 6 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×15+2+0−16−4)/2=6. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6, not 1.6.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6, not 9.",
"C": "Correct. The worked calculation gives 6.",
"D": "Incorrect. Substituting the stated values into the governing rela... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003386 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 38 m s⁻¹ at 65° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "132.74018 m",
"B": "108.61682 m",
"C": "110.61682 m",
"D": "12.06168 m"
} | C | 110.61682 m | Range R=u²sin(2θ)/g. Substituting u=38, θ=65° and g=10 gives R=(38)²sin(130°)/10=110.61682 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 110.61682 m, not 132.74018 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 110.61682 m, not 108.61682 m.",
"C": "Correct. The worked calculation gives 110.61682 m.",
"D": "Incorrect. Substi... | [
"projectile motion",
"range equation",
"trigonometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_projectile_range | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000639 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 19 and common difference 6. Find the sum of its first 11 terms. | {
"A": "808.5",
"B": "431.2",
"C": "539",
"D": "-539"
} | C | 539 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=19, d=6, n=11: Sₙ=11/2[2(19)+(11−1)(6)]=539. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 539, not 808.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 539, not 431.2.",
"C": "Correct. The worked calculation gives 539.",
"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_009076 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.92 V, transfers 2 electrons, and has reaction quotient Q=10^-3. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "1.00865 V",
"B": "3.00865 V",
"C": "-0.99135 V",
"D": "0.50433 V"
} | A | 1.00865 V | Apply the Nernst equation: E=0.92−(0.0591/2)(-3)=1.00865 V. | {
"A": "Correct. The worked calculation gives 1.00865 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.00865 V, not 3.00865 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.00865 V, not -0.99135 V.",
"D": "Incorrect. Substituting th... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005452 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 40 μF capacitor is charged to a potential difference of 30 V. Calculate the energy stored, expressed in μJ. | {
"A": "18000 μJ",
"B": "1801 μJ",
"C": "13500 μJ",
"D": "18002 μJ"
} | A | 18000 μJ | U=½CV². Since C=40 μF, U in μJ is ½×40×(30)²=18000 μJ. | {
"A": "Correct. The worked calculation gives 18000 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 18000 μJ, not 1801 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 18000 μJ, not 13500 μJ.",
"D": "Incorrect. Substituting the state... | [
"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_001460 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^4 (-4x² -3x +7) dx. | {
"A": "-7.13333",
"B": "-65.06667",
"C": "81.33333",
"D": "-81.33333"
} | D | -81.33333 | An antiderivative is (-4/3)x³+(-3/2)x²+(7)x. Evaluating between 0 and 4 gives (-4/3)(4)³+(-3/2)(4)²+(7)(4)=-81.33333. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -81.33333, not -7.13333.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -81.33333, not -65.06667.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -81.33333, ... | [
"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_007251 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 4 mol of gas at 258 K and 1.98 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "21.38455 L",
"B": "-42.7691 L",
"C": "85.53821 L",
"D": "42.7691 L"
} | D | 42.7691 L | PV=nRT, so V=nRT/P=(4)(0.082057)(258)/(1.98)=42.7691 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 42.7691 L, not 21.38455 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 42.7691 L, not -42.7691 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 42.7691 ... | [
"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_005101 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +8 μC and +2 μC are separated by 8 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "22.5 N",
"B": "3.25 N",
"C": "21.5 N",
"D": "45 N"
} | A | 22.5 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(8×10⁻⁶)(2×10⁻⁶)/(8×10⁻²)²=22.5 N. | {
"A": "Correct. The worked calculation gives 22.5 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 22.5 N, not 3.25 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 22.5 N, not 21.5 N.",
"D": "Incorrect. Substituting the stated values ... | [
"Coulomb's law",
"SI unit conversion",
"inverse-square law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_coulomb | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006083 | physics | ray_optics | convex_lens_magnification | very_hard | provisional_template_level_label | single_correct_mcq | A real object is placed 78 cm in front of a thin convex lens of focal length 17 cm. Using the Cartesian sign convention, what is the magnitude of the linear magnification? | {
"A": "0.27869",
"B": "0.41803",
"C": "0.20902",
"D": "2.27869"
} | A | 0.27869 | Using distances as positive magnitudes, 1/f=1/v+1/u gives v=uf/(u−f)=(78)(17)/(78−17)=21.7377 cm. Thus |m|=v/u=21.7377/78=0.27869. | {
"A": "Correct. The worked calculation gives 0.27869.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.27869, not 0.41803.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.27869, not 0.20902.",
"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_003555 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 17 m s⁻¹ at 70° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "14.86125 m",
"B": "2.85766 m",
"C": "18.57656 m",
"D": "20.57656 m"
} | C | 18.57656 m | Range R=u²sin(2θ)/g. Substituting u=17, θ=70° and g=10 gives R=(17)²sin(140°)/10=18.57656 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 18.57656 m, not 14.86125 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 18.57656 m, not 2.85766 m.",
"C": "Correct. The worked calculation gives 18.57656 m.",
"D": "Incorrect. Substituting... | [
"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_005087 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +8 μC and +8 μC are separated by 12 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "41 N",
"B": "40 N",
"C": "42 N",
"D": "80 N"
} | B | 40 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(8×10⁻⁶)(8×10⁻⁶)/(12×10⁻²)²=40 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 40 N, not 41 N.",
"B": "Correct. The worked calculation gives 40 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 40 N, not 42 N.",
"D": "Incorrect. Substituting the stated values into the g... | [
"Coulomb's law",
"SI unit conversion",
"inverse-square law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_coulomb | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009616 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_18H_30. Treat X as the total number of monovalent halogen atoms. | {
"A": "6",
"B": "4",
"C": "2",
"D": "1.4"
} | B | 4 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×18+2+0−30−0)/2=4. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4, not 6.",
"B": "Correct. The worked calculation gives 4.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4, not 2.",
"D": "Incorrect. Substituting the stated values into the governing relati... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000503 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term -10 and common difference 4. Find the sum of its first 28 terms. | {
"A": "1232",
"B": "124.2",
"C": "924",
"D": "1231"
} | A | 1232 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=-10, d=4, n=28: Sₙ=28/2[2(-10)+(28−1)(4)]=1232. | {
"A": "Correct. The worked calculation gives 1232.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1232, not 124.2.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1232, not 924.",
"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_003891 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 4 kg block initially moves at 6 m s⁻¹. A constant net force does 14 N of work over 9 m in the direction of motion. Find its final kinetic energy. | {
"A": "99 J",
"B": "198 J",
"C": "196 J",
"D": "199 J"
} | B | 198 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(4)(6)²+(14)(9)=198 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 198 J, not 99 J.",
"B": "Correct. The worked calculation gives 198 J.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 198 J, not 196 J.",
"D": "Incorrect. Substituting the stated values into t... | [
"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_000011 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 3x^2 -2x +10=0, let its roots be α and β. What is α²+β²? | {
"A": "-3.11111",
"B": "-9.33333",
"C": "-3.22222",
"D": "-6.22222"
} | D | -6.22222 | By Vieta's relations, α+β = 2/3 and αβ = 10/3. Then α²+β²=(α+β)²−2αβ = (2/3)²−2(10/3) = -6.22222. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives -6.22222, not -3.11111.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -6.22222, not -9.33333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -6.22222, not ... | [
"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_005579 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 56 μF capacitor is charged to a potential difference of 33 V. Calculate the energy stored, expressed in μJ. | {
"A": "30492 μJ",
"B": "36590.4 μJ",
"C": "30491 μJ",
"D": "45738 μJ"
} | A | 30492 μJ | U=½CV². Since C=56 μF, U in μJ is ½×56×(33)²=30492 μJ. | {
"A": "Correct. The worked calculation gives 30492 μJ.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 30492 μJ, not 36590.4 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 30492 μJ, not 30491 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_007731 | 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 87 mmol L⁻¹, what concentration remains after 40 minutes? | {
"A": "21.75 mmol L⁻¹",
"B": "10.875 mmol L⁻¹",
"C": "16.3125 mmol L⁻¹",
"D": "26.1 mmol L⁻¹"
} | A | 21.75 mmol L⁻¹ | 40 minutes corresponds to 2 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=87/2^2=21.75 mmol L⁻¹. | {
"A": "Correct. The worked calculation gives 21.75 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 21.75 mmol L⁻¹, not 10.875 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 21.75 mmol L⁻¹, not 16.3125 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_005845 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 3.3 atm at 224 K. What is its pressure at 625 K? | {
"A": "9.20759 atm",
"B": "-9.20759 atm",
"C": "11.20759 atm",
"D": "10.20759 atm"
} | A | 9.20759 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(3.3)(625)/(224)=9.20759 atm. | {
"A": "Correct. The worked calculation gives 9.20759 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 9.20759 atm, not -9.20759 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 9.20759 atm, not 11.20759 atm.",
"D": "Incorrect. Subs... | [
"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_001686 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 9 elements can be selected from a set of 86 distinct elements? | {
"A": "459856441978",
"B": "459856441980",
"C": "-459856441980",
"D": "459856441982"
} | B | 459856441980 | Order does not matter, so the count is C(86,9)=86!/[9!(77)!]=459856441980. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 459856441980, not 459856441978.",
"B": "Correct. The worked calculation gives 459856441980.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 459856441980, not -459856441980.",
"D": "Incorrect. ... | [
"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_005825 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 3.3 atm at 332 K. What is its pressure at 723 K? | {
"A": "9.18645 atm",
"B": "7.18645 atm",
"C": "14.37289 atm",
"D": "5.18645 atm"
} | B | 7.18645 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(3.3)(723)/(332)=7.18645 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7.18645 atm, not 9.18645 atm.",
"B": "Correct. The worked calculation gives 7.18645 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7.18645 atm, not 14.37289 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_002362 | mathematics | linear_algebra | 2x2_determinant | hard | provisional_template_level_label | single_correct_mcq | Find the determinant of the matrix [[7, 7], [6, 6]]. | {
"A": "-3",
"B": "2",
"C": "1",
"D": "0"
} | D | 0 | For a 2×2 matrix [[a,b],[c,d]], det=ad−bc. Here it is (7)(6)−(7)(6)=0. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0, not -3.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0, not 2.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0, not 1.",
"D": "Correct. The worked c... | [
"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_009553 | chemistry | organic_chemistry | degree_of_unsaturation | very_hard | provisional_template_level_label | single_correct_mcq | Assuming a neutral closed-shell formula, calculate the degree of unsaturation (DBE) for C_15H_21N_3X_4. Treat X as the total number of monovalent halogen atoms. | {
"A": "5",
"B": "3.75",
"C": "7.5",
"D": "2"
} | A | 5 | Use DBE=(2C+2+N−H−X)/2. Substitution gives (2×15+2+3−21−4)/2=5. | {
"A": "Correct. The worked calculation gives 5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5, not 3.75.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5, not 7.5.",
"D": "Incorrect. Substituting the stated values into the governing r... | [
"molecular formula",
"degree of unsaturation",
"organic structure"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_dbe | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009950 | chemistry | inorganic_chemistry | complex_ion_oxidation_state | hard | provisional_template_level_label | single_correct_mcq | For a hypothetical coordination ion [Mn(OH)_6]^+0, each OH ligand has charge −1. Determine the formal oxidation state of Mn. | {
"A": "4",
"B": "8",
"C": "-6",
"D": "6"
} | D | 6 | Let the formal oxidation state of Mn be x. Charge balance gives x+(6)(−1)=+0; hence x=0+6=+6. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 6, not 4.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 6, not 8.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 6, not -6.",
"D": "Correct. The worked c... | [
"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_001932 | mathematics | combinatorics | binomial_coefficient | hard | provisional_template_level_label | single_correct_mcq | How many distinct subsets containing exactly 7 elements can be selected from a set of 68 distinct elements? | {
"A": "969443904",
"B": "727082928",
"C": "969443906",
"D": "969443903"
} | A | 969443904 | Order does not matter, so the count is C(68,7)=68!/[7!(61)!]=969443904. | {
"A": "Correct. The worked calculation gives 969443904.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 969443904, not 727082928.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 969443904, not 969443906.",
"D": "Incorrect. Substituting the... | [
"combinations",
"factorials"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_combinations | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004535 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 13 kg and radius 3 m is acted on by a constant torque of 41 N m about its central axis. Find its angular acceleration. | {
"A": "0.70085 rad s⁻²",
"B": "-0.29915 rad s⁻²",
"C": "-1.29915 rad s⁻²",
"D": "1.40171 rad s⁻²"
} | A | 0.70085 rad s⁻² | For a solid disc, I=½MR²=½(13)(3)²=58.5 kg m². Using τ=Iα, α=41/58.5=0.70085 rad s⁻². | {
"A": "Correct. The worked calculation gives 0.70085 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.70085 rad s⁻², not -0.29915 rad s⁻².",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.70085 rad s⁻², not -1.29915 rad s⁻².",
"... | [
"moment of inertia",
"rotational dynamics",
"torque"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_disc_torque | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_006934 | 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 35 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "-5 mol",
"B": "2 mol",
"C": "3.75 mol",
"D": "5 mol"
} | D | 5 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 5 mol H₂O; O₂ can produce 70 mol H₂O. The limiting condition is H₂; product=5 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not -5 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not 2 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5 mol, not 3.75 mol.",
... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005121 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +4 μC and +6 μC are separated by 11 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "17.85124 N",
"B": "15.85124 N",
"C": "26.77686 N",
"D": "13.38843 N"
} | A | 17.85124 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(4×10⁻⁶)(6×10⁻⁶)/(11×10⁻²)²=17.85124 N. | {
"A": "Correct. The worked calculation gives 17.85124 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 17.85124 N, not 15.85124 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 17.85124 N, not 26.77686 N.",
"D": "Incorrect. Substitutin... | [
"Coulomb's law",
"SI unit conversion",
"inverse-square law"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_coulomb | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003713 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 11 kg block initially moves at 6 m s⁻¹. A constant net force does 5 N of work over 4 m in the direction of motion. Find its final kinetic energy. | {
"A": "219 J",
"B": "174.4 J",
"C": "218 J",
"D": "-218 J"
} | C | 218 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(11)(6)²+(5)(4)=218 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 218 J, not 219 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 218 J, not 174.4 J.",
"C": "Correct. The worked calculation gives 218 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_007097 | 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 579 K and 0.9 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "1266.96008 L",
"B": "1055.80007 L",
"C": "2111.60013 L",
"D": "1056.80007 L"
} | B | 1055.80007 L | PV=nRT, so V=nRT/P=(20)(0.082057)(579)/(0.9)=1055.80007 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1055.80007 L, not 1266.96008 L.",
"B": "Correct. The worked calculation gives 1055.80007 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1055.80007 L, not 2111.60013 L.",
"D": "Incorrect. S... | [
"ideal gas equation",
"unit consistency"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_ideal_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008482 | 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 0.36 mol kg⁻¹. If its van't Hoff factor is 1 and K_b for water is 0.6 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "-0.784 K",
"B": "0.324 K",
"C": "3.216 K",
"D": "0.216 K"
} | D | 0.216 K | ΔT_b=iK_bm. Substitution gives ΔT_b=1×0.6×0.36=0.216 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.216 K, not -0.784 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.216 K, not 0.324 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.216 K, not 3.21... | [
"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_005099 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +4 μC and +9 μC are separated by 9 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "20 N",
"B": "39 N",
"C": "-40 N",
"D": "40 N"
} | D | 40 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(4×10⁻⁶)(9×10⁻⁶)/(9×10⁻²)²=40 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 40 N, not 20 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 40 N, not 39 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 40 N, not -40 N.",
"D": "Cor... | [
"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_003189 | mathematics | complex_numbers | real_part_of_complex_quotient | very_hard | provisional_template_level_label | single_correct_mcq | Let z=(-9+10i)/(4+7i). What is Re(z), the real part of z? | {
"A": "0.52308",
"B": "1.05231",
"C": "0.39231",
"D": "2.52308"
} | A | 0.52308 | Multiply numerator and denominator by the conjugate (4-7i). Then Re(z)=(ac+bd)/(c²+d²)=((-9)(4)+(10)(7))/(4²+(7)²)=0.52308. | {
"A": "Correct. The worked calculation gives 0.52308.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.52308, not 1.05231.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.52308, not 0.39231.",
"D": "Incorrect. Substituting the stated va... | [
"complex conjugate",
"complex division",
"real and imaginary parts"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_complex_quotient | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003668 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 97 m s⁻¹ at 60° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "814.8433 m",
"B": "813.8433 m",
"C": "815.8433 m",
"D": "1629.6866 m"
} | A | 814.8433 m | Range R=u²sin(2θ)/g. Substituting u=97, θ=60° and g=10 gives R=(97)²sin(120°)/10=814.8433 m. | {
"A": "Correct. The worked calculation gives 814.8433 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 814.8433 m, not 813.8433 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 814.8433 m, not 815.8433 m.",
"D": "Incorrect. Substitutin... | [
"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_009199 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=1.44 V, transfers 2 electrons, and has reaction quotient Q=10^5. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "0.29225 V",
"B": "-1.29225 V",
"C": "2.29225 V",
"D": "1.29225 V"
} | D | 1.29225 V | Apply the Nernst equation: E=1.44−(0.0591/2)(5)=1.29225 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.29225 V, not 0.29225 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.29225 V, not -1.29225 V.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.29225 V... | [
"electrochemical cells",
"Nernst equation",
"logarithms"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_nernst | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008625 | chemistry | physical_chemistry | boiling_point_elevation | hard | provisional_template_level_label | single_correct_mcq | A non-volatile solute is dissolved in water to produce a solution of molality 1.01 mol kg⁻¹. If its van't Hoff factor is 1 and K_b for water is 0.52 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "1.0504 K",
"B": "0.3939 K",
"C": "0.2626 K",
"D": "0.5252 K"
} | D | 0.5252 K | ΔT_b=iK_bm. Substitution gives ΔT_b=1×0.52×1.01=0.5252 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.5252 K, not 1.0504 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.5252 K, not 0.3939 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.5252 K, not ... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004741 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 5 Ω and 7 Ω are connected in series; this combination is connected in parallel with a 12 Ω resistor across a 24 V ideal battery. Find the total current supplied by the battery. | {
"A": "5 A",
"B": "6 A",
"C": "4 A",
"D": "7 A"
} | C | 4 A | Series branch resistance=5+7=12 Ω. Parallel equivalent R_eq=(12×12)/(12+12)=6 Ω. Total current I=V/R_eq=24/6=4 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4 A, not 5 A.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4 A, not 6 A.",
"C": "Correct. The worked calculation gives 4 A.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"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_004286 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 6 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.19. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "26.88 N",
"B": "20.16 N",
"C": "53.76 N",
"D": "32.256 N"
} | A | 26.88 N | Net force=mg(sinθ−μcosθ)=6×10×(0.6−0.19×0.8)=26.88 N. | {
"A": "Correct. The worked calculation gives 26.88 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 26.88 N, not 20.16 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 26.88 N, not 53.76 N.",
"D": "Incorrect. Substituting the stated va... | [
"Newton's laws",
"inclined plane",
"kinetic friction"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_incline_friction | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005157 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +6 μC and +9 μC are separated by 8 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "91.125 N",
"B": "60.75 N",
"C": "75.9375 N",
"D": "76.9375 N"
} | C | 75.9375 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(6×10⁻⁶)(9×10⁻⁶)/(8×10⁻²)²=75.9375 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 75.9375 N, not 91.125 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 75.9375 N, not 60.75 N.",
"C": "Correct. The worked calculation gives 75.9375 N.",
"D": "Incorrect. Substituting the st... | [
"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_007787 | 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 13 mmol L⁻¹, what concentration remains after 160 minutes? | {
"A": "0.07617 mmol L⁻¹",
"B": "-1.94922 mmol L⁻¹",
"C": "-2.94922 mmol L⁻¹",
"D": "0.05078 mmol L⁻¹"
} | D | 0.05078 mmol L⁻¹ | 160 minutes corresponds to 8 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=13/2^8=0.05078 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.05078 mmol L⁻¹, not 0.07617 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.05078 mmol L⁻¹, not -1.94922 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the govern... | [
"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_008707 | chemistry | physical_chemistry | equilibrium_degree_dissociation | very_hard | provisional_template_level_label | single_correct_mcq | For A(g) ⇌ 2B(g), the initial concentration of A is 0.85 mol L⁻¹ and initially B is absent. At equilibrium, 48% of A has dissociated at constant volume. Find K_c. | {
"A": "3.01292",
"B": "1.15065",
"C": "2.25969",
"D": "1.50646"
} | D | 1.50646 | Let α=0.48. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.85)(0.48)²/(1−0.48)=1.50646. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.50646, not 3.01292.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.50646, not 1.15065.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.50646, not 2.259... | [
"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_007286 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 16 mol of gas at 263 K and 1.78 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "193.98644 L",
"B": "232.78372 L",
"C": "195.98644 L",
"D": "20.39864 L"
} | A | 193.98644 L | PV=nRT, so V=nRT/P=(16)(0.082057)(263)/(1.78)=193.98644 L. | {
"A": "Correct. The worked calculation gives 193.98644 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 193.98644 L, not 232.78372 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 193.98644 L, not 195.98644 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_006857 | 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 12 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "23 mol",
"B": "3.4 mol",
"C": "24 mol",
"D": "28.8 mol"
} | C | 24 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 44 mol H₂O; O₂ can produce 24 mol H₂O. The limiting condition is O₂; product=24 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 24 mol, not 23 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 24 mol, not 3.4 mol.",
"C": "Correct. The worked calculation gives 24 mol.",
"D": "Incorrect. Substituting the stated values... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005833 | 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 218 K. What is its pressure at 344 K? | {
"A": "2.20917 atm",
"B": "4.20917 atm",
"C": "1.65688 atm",
"D": "1.22092 atm"
} | A | 2.20917 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(1.4)(344)/(218)=2.20917 atm. | {
"A": "Correct. The worked calculation gives 2.20917 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.20917 atm, not 4.20917 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.20917 atm, not 1.65688 atm.",
"D": "Incorrect. Substi... | [
"ideal gas law",
"isochoric process",
"absolute temperature"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_isochoric_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005865 | 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 2.3 atm at 353 K. What is its pressure at 655 K? | {
"A": "6.26771 atm",
"B": "-4.26771 atm",
"C": "2.13385 atm",
"D": "4.26771 atm"
} | D | 4.26771 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(2.3)(655)/(353)=4.26771 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4.26771 atm, not 6.26771 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4.26771 atm, not -4.26771 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 4... | [
"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_007867 | 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 63 mmol L⁻¹, what concentration remains after 160 minutes? | {
"A": "0.24609 mmol L⁻¹",
"B": "-0.24609 mmol L⁻¹",
"C": "-0.75391 mmol L⁻¹",
"D": "1.02461 mmol L⁻¹"
} | A | 0.24609 mmol L⁻¹ | 160 minutes corresponds to 8 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=63/2^8=0.24609 mmol L⁻¹. | {
"A": "Correct. The worked calculation gives 0.24609 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.24609 mmol L⁻¹, not -0.24609 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.24609 mmol L⁻¹, not -0.75391 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_008049 | chemistry | physical_chemistry | faradays_law_copper | very_hard | provisional_template_level_label | single_correct_mcq | An aqueous Cu²⁺ solution is electrolysed using a current of 3 A for 19289 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": "28.55871 g",
"B": "19.03914 g",
"C": "22.84697 g",
"D": "-19.03914 g"
} | B | 19.03914 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×3×19289/(2×96500)=19.03914 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 19.03914 g, not 28.55871 g.",
"B": "Correct. The worked calculation gives 19.03914 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 19.03914 g, not 22.84697 g.",
"D": "Incorrect. Substitutin... | [
"electrolysis",
"Faraday's laws",
"redox stoichiometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_faraday_copper | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000420 | mathematics | sequences_and_series | arithmetic_progression | hard | provisional_template_level_label | single_correct_mcq | An arithmetic progression has first term 28 and common difference 13. Find the sum of its first 28 terms. | {
"A": "5698",
"B": "4273.5",
"C": "5700",
"D": "4558.4"
} | A | 5698 | Use Sₙ = n/2[2a+(n−1)d]. Substituting a=28, d=13, n=28: Sₙ=28/2[2(28)+(28−1)(13)]=5698. | {
"A": "Correct. The worked calculation gives 5698.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5698, not 4273.5.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5698, not 5700.",
"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_004873 | physics | electricity | series_parallel_circuit | very_hard | provisional_template_level_label | single_correct_mcq | Resistors 15 Ω and 6 Ω are connected in series; this combination is connected in parallel with a 12 Ω resistor across a 24 V ideal battery. Find the total current supplied by the battery. | {
"A": "1.57143 A",
"B": "3.14286 A",
"C": "1.14286 A",
"D": "-3.14286 A"
} | B | 3.14286 A | Series branch resistance=15+6=21 Ω. Parallel equivalent R_eq=(21×12)/(21+12)=7.63636 Ω. Total current I=V/R_eq=24/7.63636=3.14286 A. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3.14286 A, not 1.57143 A.",
"B": "Correct. The worked calculation gives 3.14286 A.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 3.14286 A, not 1.14286 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_006821 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 47 mol H₂ and 14 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "14 mol",
"B": "42 mol",
"C": "28 mol",
"D": "3.8 mol"
} | C | 28 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 47 mol H₂O; O₂ can produce 28 mol H₂O. The limiting condition is O₂; product=28 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 28 mol, not 14 mol.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 28 mol, not 42 mol.",
"C": "Correct. The worked calculation gives 28 mol.",
"D": "Incorrect. Substituting the stated values ... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003411 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 75 m s⁻¹ at 40° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "551.95436 m",
"B": "443.16349 m",
"C": "553.95436 m",
"D": "1107.90872 m"
} | C | 553.95436 m | Range R=u²sin(2θ)/g. Substituting u=75, θ=40° and g=10 gives R=(75)²sin(80°)/10=553.95436 m. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 553.95436 m, not 551.95436 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 553.95436 m, not 443.16349 m.",
"C": "Correct. The worked calculation gives 553.95436 m.",
"D": "Incorrect. Substi... | [
"projectile motion",
"range equation",
"trigonometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_projectile_range | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007631 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 11 mL of 0.10 M HCl is mixed with 32 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "14.68875",
"B": "19.03313",
"C": "12.68875",
"D": "13.68875"
} | C | 12.68875 | Excess OH⁻ moles=0.10×(32−11)/1000=0.002100 mol. Total volume=0.043 L, so [OH⁻]=0.048837 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.68875. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12.68875, not 14.68875.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.68875, not 19.03313.",
"C": "Correct. The worked calculation gives 12.68875.",
"D": "Incorrect. Substituting the stat... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001259 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=1x³ +9x² +3x+7. Evaluate f′(0). | {
"A": "1.5",
"B": "1.3",
"C": "3",
"D": "1"
} | C | 3 | Differentiate: f′(x)=3(1)x²+2(9)x+(3). At x=0, f′(0)=3(1)(0)²+2(9)(0)+(3)=3. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 3, not 1.5.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 3, not 1.3.",
"C": "Correct. The worked calculation gives 3.",
"D": "Incorrect. Substituting the stated values into the governing re... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_001076 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=3x³ -8x² +10x+7. Evaluate f′(3). | {
"A": "44",
"B": "43",
"C": "34.4",
"D": "21.5"
} | B | 43 | Differentiate: f′(x)=3(3)x²+2(-8)x+(10). At x=3, f′(3)=3(3)(3)²+2(-8)(3)+(10)=43. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 43, not 44.",
"B": "Correct. The worked calculation gives 43.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 43, not 34.4.",
"D": "Incorrect. Substituting the stated values into the governing... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_009014 | chemistry | physical_chemistry | nernst_equation_298K | hard | provisional_template_level_label | single_correct_mcq | At 298 K, a galvanic cell has E°cell=0.74 V, transfers 1 electrons, and has reaction quotient Q=10^-5. Using E=E°−(0.0591/n)log₁₀Q, calculate Ecell. | {
"A": "2.0355 V",
"B": "1.10355 V",
"C": "1.0355 V",
"D": "2.071 V"
} | C | 1.0355 V | Apply the Nernst equation: E=0.74−(0.0591/1)(-5)=1.0355 V. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.0355 V, not 2.0355 V.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.0355 V, not 1.10355 V.",
"C": "Correct. The worked calculation gives 1.0355 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_005410 | physics | electrostatics | capacitor_energy | hard | provisional_template_level_label | single_correct_mcq | A 17 μF capacitor is charged to a potential difference of 30 V. Calculate the energy stored, expressed in μJ. | {
"A": "11475 μJ",
"B": "7650 μJ",
"C": "3825 μJ",
"D": "6120 μJ"
} | B | 7650 μJ | U=½CV². Since C=17 μF, U in μJ is ½×17×(30)²=7650 μJ. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7650 μJ, not 11475 μJ.",
"B": "Correct. The worked calculation gives 7650 μJ.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7650 μJ, not 3825 μJ.",
"D": "Incorrect. Substituting the stated v... | [
"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_006808 | chemistry | stoichiometry | limiting_reagent_water | very_hard | provisional_template_level_label | single_correct_mcq | Hydrogen reacts with oxygen according to 2H₂ + O₂ → 2H₂O. If 16 mol H₂ and 22 mol O₂ are mixed and the reaction goes to completion, how many moles of H₂O form? | {
"A": "18 mol",
"B": "16 mol",
"C": "24 mol",
"D": "-16 mol"
} | B | 16 mol | From 2 mol H₂ and 1 mol O₂, 2 mol H₂O form. H₂ can produce 16 mol H₂O; O₂ can produce 44 mol H₂O. The limiting condition is H₂; product=16 mol. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 16 mol, not 18 mol.",
"B": "Correct. The worked calculation gives 16 mol.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 16 mol, not 24 mol.",
"D": "Incorrect. Substituting the stated values ... | [
"balanced chemical equations",
"limiting reagent",
"mole ratios"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_limiting_reagent | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_003469 | physics | mechanics | projectile_range | hard | provisional_template_level_label | single_correct_mcq | A projectile is launched from level ground at 43 m s⁻¹ at 70° to the horizontal. Take g=10 m s⁻². Find its horizontal range. | {
"A": "118.85143 m",
"B": "89.13857 m",
"C": "142.62171 m",
"D": "95.08114 m"
} | A | 118.85143 m | Range R=u²sin(2θ)/g. Substituting u=43, θ=70° and g=10 gives R=(43)²sin(140°)/10=118.85143 m. | {
"A": "Correct. The worked calculation gives 118.85143 m.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 118.85143 m, not 89.13857 m.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 118.85143 m, not 142.62171 m.",
"D": "Incorrect. Substit... | [
"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_003963 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 13 kg block initially moves at 12 m s⁻¹. A constant net force does 10 N of work over 2 m in the direction of motion. Find its final kinetic energy. | {
"A": "717 J",
"B": "1147.2 J",
"C": "956 J",
"D": "764.8 J"
} | C | 956 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(13)(12)²+(10)(2)=956 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 956 J, not 717 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 956 J, not 1147.2 J.",
"C": "Correct. The worked calculation gives 956 J.",
"D": "Incorrect. Substituting the stated values in... | [
"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_007384 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 24 mL of 0.10 M HCl is mixed with 62 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "6.32264",
"B": "9.48396",
"C": "12.64529",
"D": "2.26453"
} | C | 12.64529 | Excess OH⁻ moles=0.10×(62−24)/1000=0.003800 mol. Total volume=0.086 L, so [OH⁻]=0.044186 M, pOH=−log₁₀[OH⁻], and pH=14−pOH=12.64529. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 12.64529, not 6.32264.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.64529, not 9.48396.",
"C": "Correct. The worked calculation gives 12.64529.",
"D": "Incorrect. Substituting the stated... | [
"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_002546 | mathematics | linear_algebra | 2x2_determinant | hard | provisional_template_level_label | single_correct_mcq | Find the determinant of the matrix [[4, -1], [8, -7]]. | {
"A": "-20",
"B": "-40",
"C": "-24",
"D": "20"
} | A | -20 | For a 2×2 matrix [[a,b],[c,d]], det=ad−bc. Here it is (4)(-7)−(-1)(8)=-20. | {
"A": "Correct. The worked calculation gives -20.",
"B": "Incorrect. Substituting the stated values into the governing relation gives -20, not -40.",
"C": "Incorrect. Substituting the stated values into the governing relation gives -20, not -24.",
"D": "Incorrect. Substituting the stated values into the govern... | [
"determinants",
"matrix operations"
] | 2 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_determinant_2x2 | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005719 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 0.5 atm at 408 K. What is its pressure at 531 K? | {
"A": "1.65074 atm",
"B": "-0.34926 atm",
"C": "2.65074 atm",
"D": "0.65074 atm"
} | D | 0.65074 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(0.5)(531)/(408)=0.65074 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.65074 atm, not 1.65074 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.65074 atm, not -0.34926 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0... | [
"ideal gas law",
"isochoric process",
"absolute temperature"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_isochoric_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004129 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 10 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.30. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "38 N",
"B": "18 N",
"C": "37 N",
"D": "36 N"
} | D | 36 N | Net force=mg(sinθ−μcosθ)=10×10×(0.6−0.30×0.8)=36 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 36 N, not 38 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 36 N, not 18 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 36 N, not 37 N.",
"D": "Corr... | [
"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_001637 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^2 (6x² +1x -2) dx. | {
"A": "14",
"B": "12",
"C": "-14",
"D": "21"
} | A | 14 | An antiderivative is (6/3)x³+(1/2)x²+(-2)x. Evaluating between 0 and 2 gives (6/3)(2)³+(1/2)(2)²+(-2)(2)=14. | {
"A": "Correct. The worked calculation gives 14.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 14, not 12.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 14, not -14.",
"D": "Incorrect. Substituting the stated values into the governing ... | [
"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_001060 | mathematics | calculus | polynomial_differentiation | hard | provisional_template_level_label | single_correct_mcq | Let f(x)=7x³ +3x² +12x+7. Evaluate f′(5). | {
"A": "567",
"B": "566",
"C": "680.4",
"D": "569"
} | A | 567 | Differentiate: f′(x)=3(7)x²+2(3)x+(12). At x=5, f′(5)=3(7)(5)²+2(3)(5)+(12)=567. | {
"A": "Correct. The worked calculation gives 567.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 567, not 566.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 567, not 680.4.",
"D": "Incorrect. Substituting the stated values into the gove... | [
"power rule",
"polynomial derivative"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_derivative_cubic | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008662 | chemistry | physical_chemistry | boiling_point_elevation | hard | provisional_template_level_label | single_correct_mcq | A non-volatile solute is dissolved in water to produce a solution of molality 1.51 mol kg⁻¹. If its van't Hoff factor is 3 and K_b for water is 0.52 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "5.3556 K",
"B": "-2.3556 K",
"C": "2.3556 K",
"D": "1.3556 K"
} | C | 2.3556 K | ΔT_b=iK_bm. Substitution gives ΔT_b=3×0.52×1.51=2.3556 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.3556 K, not 5.3556 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.3556 K, not -2.3556 K.",
"C": "Correct. The worked calculation gives 2.3556 K.",
"D": "Incorrect. Substituting the sta... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007427 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 69 mL of 0.10 M HCl is mixed with 46 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "0.69897",
"B": "3.69897",
"C": "1.69897",
"D": "4.69897"
} | C | 1.69897 | Excess H⁺ moles=0.10×(69−46)/1000=0.002300 mol. Total volume=0.115 L, so [H⁺]=0.020000 M and pH=−log₁₀[H⁺]=1.69897. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.69897, not 0.69897.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.69897, not 3.69897.",
"C": "Correct. The worked calculation gives 1.69897.",
"D": "Incorrect. Substituting the stated va... | [
"strong acid–strong base neutralization",
"molarity",
"pH and pOH"
] | 5 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_acid_base_pH | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008824 | 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 0.69 mol L⁻¹ and initially B is absent. At equilibrium, 21% of A has dissociated at constant volume. Find K_c. | {
"A": "0.23111",
"B": "0.15407",
"C": "-0.84593",
"D": "2.15407"
} | B | 0.15407 | Let α=0.21. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.69)(0.21)²/(1−0.21)=0.15407. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.15407, not 0.23111.",
"B": "Correct. The worked calculation gives 0.15407.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.15407, not -0.84593.",
"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_008153 | 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 20590 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": "32.51726 g",
"B": "39.64658 g",
"C": "40.64658 g",
"D": "81.29316 g"
} | C | 40.64658 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×6×20590/(2×96500)=40.64658 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 40.64658 g, not 32.51726 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 40.64658 g, not 39.64658 g.",
"C": "Correct. The worked calculation gives 40.64658 g.",
"D": "Incorrect. Substitutin... | [
"electrolysis",
"Faraday's laws",
"redox stoichiometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_faraday_copper | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_005894 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 0.7 atm at 261 K. What is its pressure at 541 K? | {
"A": "2.17644 atm",
"B": "1.1451 atm",
"C": "0.45096 atm",
"D": "1.45096 atm"
} | D | 1.45096 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(0.7)(541)/(261)=1.45096 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.45096 atm, not 2.17644 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.45096 atm, not 1.1451 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.4... | [
"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_002063 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 11 red, 4 blue and 10 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "-1.64667",
"B": "0.35333",
"C": "1.35333",
"D": "-0.64667"
} | B | 0.35333 | Favourable pairs = C(11,2)+C(4,2)+C(10,2)=106. Total pairs=C(25,2)=300. Therefore P=106/300=0.35333. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.35333, not -1.64667.",
"B": "Correct. The worked calculation gives 0.35333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.35333, not 1.35333.",
"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_003707 | physics | mechanics | work_energy_theorem | very_hard | provisional_template_level_label | single_correct_mcq | A 3 kg block initially moves at 0 m s⁻¹. A constant net force does 11 N of work over 7 m in the direction of motion. Find its final kinetic energy. | {
"A": "154 J",
"B": "57.75 J",
"C": "77 J",
"D": "76 J"
} | C | 77 J | By the work–energy theorem, K_f=K_i+W=½mu²+Fs=½(3)(0)²+(11)(7)=77 J. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 77 J, not 154 J.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 77 J, not 57.75 J.",
"C": "Correct. The worked calculation gives 77 J.",
"D": "Incorrect. Substituting the stated values into t... | [
"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_007956 | 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 20 minutes? | {
"A": "41 mmol L⁻¹",
"B": "31.5 mmol L⁻¹",
"C": "40 mmol L⁻¹",
"D": "42 mmol L⁻¹"
} | D | 42 mmol L⁻¹ | 20 minutes corresponds to 1 half-lives. The remaining concentration is [A]₀(1/2)ⁿ=84/2^1=42 mmol L⁻¹. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 42 mmol L⁻¹, not 41 mmol L⁻¹.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 42 mmol L⁻¹, not 31.5 mmol L⁻¹.",
"C": "Incorrect. Substituting the stated values into the governing relation gives ... | [
"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_008032 | 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 5 A for 28734 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": "47.26966 g",
"B": "35.45225 g",
"C": "49.26966 g",
"D": "46.26966 g"
} | A | 47.26966 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×5×28734/(2×96500)=47.26966 g. | {
"A": "Correct. The worked calculation gives 47.26966 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 47.26966 g, not 35.45225 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 47.26966 g, not 49.26966 g.",
"D": "Incorrect. Substitutin... | [
"electrolysis",
"Faraday's laws",
"redox stoichiometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_faraday_copper | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004266 | physics | mechanics | inclined_plane_with_friction | very_hard | provisional_template_level_label | single_correct_mcq | A 3 kg block slides down a 37° rough incline. The coefficient of kinetic friction is 0.30. Taking g=10 m s⁻², find the net force down the plane. Use sin37°=0.6 and cos37°=0.8. | {
"A": "10.8 N",
"B": "8.8 N",
"C": "12.8 N",
"D": "16.2 N"
} | A | 10.8 N | Net force=mg(sinθ−μcosθ)=3×10×(0.6−0.30×0.8)=10.8 N. | {
"A": "Correct. The worked calculation gives 10.8 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 10.8 N, not 8.8 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 10.8 N, not 12.8 N.",
"D": "Incorrect. Substituting the stated values i... | [
"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_005752 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 3.3 atm at 388 K. What is its pressure at 598 K? | {
"A": "6.08608 atm",
"B": "3.08608 atm",
"C": "8.08608 atm",
"D": "5.08608 atm"
} | D | 5.08608 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(3.3)(598)/(388)=5.08608 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 5.08608 atm, not 6.08608 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 5.08608 atm, not 3.08608 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 5.... | [
"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_005297 | physics | electrostatics | coulombs_law | hard | provisional_template_level_label | single_correct_mcq | Two point charges of +5 μC and +6 μC are separated by 10 cm in vacuum. Using k=9×10⁹ N m² C⁻², find the magnitude of the electrostatic force. | {
"A": "25 N",
"B": "28 N",
"C": "54 N",
"D": "27 N"
} | D | 27 N | Convert charges to coulombs and distance to metres. F=k|q₁q₂|/r²=9×10⁹(5×10⁻⁶)(6×10⁻⁶)/(10×10⁻²)²=27 N. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 27 N, not 25 N.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 27 N, not 28 N.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 27 N, not 54 N.",
"D": "Corr... | [
"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_005851 | 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 2.9 atm at 287 K. What is its pressure at 730 K? | {
"A": "14.75261 atm",
"B": "5.37631 atm",
"C": "7.37631 atm",
"D": "8.37631 atm"
} | C | 7.37631 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(2.9)(730)/(287)=7.37631 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7.37631 atm, not 14.75261 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 7.37631 atm, not 5.37631 atm.",
"C": "Correct. The worked calculation gives 7.37631 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_008136 | 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 5 A for 18566 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": "30.54251 g",
"B": "-30.54251 g",
"C": "4.05425 g",
"D": "22.90688 g"
} | A | 30.54251 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×5×18566/(2×96500)=30.54251 g. | {
"A": "Correct. The worked calculation gives 30.54251 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 30.54251 g, not -30.54251 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 30.54251 g, not 4.05425 g.",
"D": "Incorrect. Substitutin... | [
"electrolysis",
"Faraday's laws",
"redox stoichiometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_faraday_copper | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002148 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 6 red, 7 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": "0.24728",
"B": "0.32971",
"C": "-1.67029",
"D": "-0.32971"
} | B | 0.32971 | Favourable pairs = C(6,2)+C(7,2)+C(11,2)=91. Total pairs=C(24,2)=276. Therefore P=91/276=0.32971. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.32971, not 0.24728.",
"B": "Correct. The worked calculation gives 0.32971.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.32971, not -1.67029.",
"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_008378 | 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 0.93 mol kg⁻¹. If its van't Hoff factor is 1 and K_b for water is 0.6 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "3.558 K",
"B": "-0.558 K",
"C": "0.4185 K",
"D": "0.558 K"
} | D | 0.558 K | ΔT_b=iK_bm. Substitution gives ΔT_b=1×0.6×0.93=0.558 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.558 K, not 3.558 K.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.558 K, not -0.558 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.558 K, not 0.41... | [
"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_006463 | physics | oscillations | mass_spring_period_over_pi | hard | provisional_template_level_label | single_correct_mcq | A mass of 4 kg is attached to an ideal spring of force constant 89 N m⁻¹ and oscillates horizontally without damping. What is the time period divided by π? | {
"A": "-1.576 s",
"B": "0.636 s",
"C": "0.424 s",
"D": "1.0424 s"
} | C | 0.424 s | For a mass–spring oscillator, T=2π√(m/k). Therefore T/π=2√(4/89)=0.424 s. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.424 s, not -1.576 s.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.424 s, not 0.636 s.",
"C": "Correct. The worked calculation gives 0.424 s.",
"D": "Incorrect. Substituting the stated v... | [
"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_008589 | chemistry | physical_chemistry | boiling_point_elevation | hard | provisional_template_level_label | single_correct_mcq | A non-volatile solute is dissolved in water to produce a solution of molality 1.93 mol kg⁻¹. If its van't Hoff factor is 4 and K_b for water is 1.0 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "4.72 K",
"B": "7.72 K",
"C": "9.72 K",
"D": "8.72 K"
} | B | 7.72 K | ΔT_b=iK_bm. Substitution gives ΔT_b=4×1.0×1.93=7.72 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7.72 K, not 4.72 K.",
"B": "Correct. The worked calculation gives 7.72 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7.72 K, not 9.72 K.",
"D": "Incorrect. Substituting the stated values ... | [
"colligative properties",
"van't Hoff factor",
"molality"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_boiling_elevation | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007477 | chemistry | physical_chemistry | strong_acid_base_mixture | very_hard | provisional_template_level_label | single_correct_mcq | 74 mL of 0.10 M HCl is mixed with 17 mL of 0.10 M NaOH at 25°C. Assuming complete dissociation and additive volumes, find the final pH. | {
"A": "-0.79683",
"B": "-1.79683",
"C": "1.20317",
"D": "1.12032"
} | C | 1.20317 | Excess H⁺ moles=0.10×(74−17)/1000=0.005700 mol. Total volume=0.091 L, so [H⁺]=0.062637 M and pH=−log₁₀[H⁺]=1.20317. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.20317, not -0.79683.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.20317, not -1.79683.",
"C": "Correct. The worked calculation gives 1.20317.",
"D": "Incorrect. Substituting the stated ... | [
"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_001397 | mathematics | calculus | definite_integral_polynomial | very_hard | provisional_template_level_label | single_correct_mcq | Evaluate the definite integral ∫₀^2 (4x² +0x -4) dx. | {
"A": "3.66667",
"B": "-0.33333",
"C": "1.26667",
"D": "2.66667"
} | D | 2.66667 | An antiderivative is (4/3)x³+(0/2)x²+(-4)x. Evaluating between 0 and 2 gives (4/3)(2)³+(0/2)(2)²+(-4)(2)=2.66667. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.66667, not 3.66667.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.66667, not -0.33333.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.66667, not 1.26... | [
"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_005971 | 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 4.1 atm at 344 K. What is its pressure at 639 K? | {
"A": "3.80799 atm",
"B": "7.61599 atm",
"C": "10.61599 atm",
"D": "9.61599 atm"
} | B | 7.61599 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(4.1)(639)/(344)=7.61599 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 7.61599 atm, not 3.80799 atm.",
"B": "Correct. The worked calculation gives 7.61599 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 7.61599 atm, not 10.61599 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_000218 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 4x^2 +17x +7=0, let its roots be α and β. What is α²+β²? | {
"A": "29.125",
"B": "-14.5625",
"C": "14.5625",
"D": "21.84375"
} | C | 14.5625 | By Vieta's relations, α+β = -17/4 and αβ = 7/4. Then α²+β²=(α+β)²−2αβ = (-17/4)²−2(7/4) = 14.5625. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 14.5625, not 29.125.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 14.5625, not -14.5625.",
"C": "Correct. The worked calculation gives 14.5625.",
"D": "Incorrect. Substituting the stated va... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_000170 | mathematics | algebra | quadratic_roots | hard | provisional_template_level_label | single_correct_mcq | For the quadratic equation 2x^2 -13x -2=0, let its roots be α and β. What is α²+β²? | {
"A": "-44.25",
"B": "43.25",
"C": "46.25",
"D": "44.25"
} | D | 44.25 | By Vieta's relations, α+β = 13/2 and αβ = -2/2. Then α²+β²=(α+β)²−2αβ = (13/2)²−2(-2/2) = 44.25. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 44.25, not -44.25.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 44.25, not 43.25.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 44.25, not 46.25.",
"D"... | [
"Vieta's formulas",
"symmetric polynomial expressions"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | math_quadratic_vieta | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004431 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 5 kg and radius 3 m is acted on by a constant torque of 27 N m about its central axis. Find its angular acceleration. | {
"A": "1.2 rad s⁻²",
"B": "3.2 rad s⁻²",
"C": "-0.8 rad s⁻²",
"D": "-1.8 rad s⁻²"
} | A | 1.2 rad s⁻² | For a solid disc, I=½MR²=½(5)(3)²=22.5 kg m². Using τ=Iα, α=27/22.5=1.2 rad s⁻². | {
"A": "Correct. The worked calculation gives 1.2 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 1.2 rad s⁻², not 3.2 rad s⁻².",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.2 rad s⁻², not -0.8 rad s⁻².",
"D": "Incorrect. Subst... | [
"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_005805 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 1.5 atm at 415 K. What is its pressure at 733 K? | {
"A": "0.6494 atm",
"B": "1.6494 atm",
"C": "5.6494 atm",
"D": "2.6494 atm"
} | D | 2.6494 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(1.5)(733)/(415)=2.6494 atm. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.6494 atm, not 0.6494 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2.6494 atm, not 1.6494 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.6494... | [
"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_008203 | chemistry | physical_chemistry | faradays_law_copper | very_hard | provisional_template_level_label | single_correct_mcq | An aqueous Cu²⁺ solution is electrolysed using a current of 9 A for 4693 s. Assuming 100% current efficiency, calculate the mass of copper deposited. Use M(Cu)=63.5 g mol⁻¹ and F=96500 C mol⁻¹. | {
"A": "13.89663 g",
"B": "11.89663 g",
"C": "10.42247 g",
"D": "20.84494 g"
} | A | 13.89663 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×9×4693/(2×96500)=13.89663 g. | {
"A": "Correct. The worked calculation gives 13.89663 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 13.89663 g, not 11.89663 g.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 13.89663 g, not 10.42247 g.",
"D": "Incorrect. Substitutin... | [
"electrolysis",
"Faraday's laws",
"redox stoichiometry"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_faraday_copper | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_002029 | mathematics | probability | same_colour_without_replacement | very_hard | provisional_template_level_label | single_correct_mcq | A bag contains 12 red, 3 blue and 8 green balls. Two balls are drawn uniformly at random without replacement. What is the probability that both have the same colour? | {
"A": "2.3834",
"B": "1.3834",
"C": "0.28755",
"D": "0.3834"
} | D | 0.3834 | Favourable pairs = C(12,2)+C(3,2)+C(8,2)=97. Total pairs=C(23,2)=253. Therefore P=97/253=0.3834. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.3834, not 2.3834.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.3834, not 1.3834.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.3834, not 0.28755.",... | [
"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_007170 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 19 mol of gas at 578 K and 0.22 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "4915.36349 L",
"B": "2048.06812 L",
"C": "4096.13625 L",
"D": "4094.13625 L"
} | C | 4096.13625 L | PV=nRT, so V=nRT/P=(19)(0.082057)(578)/(0.22)=4096.13625 L. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 4096.13625 L, not 4915.36349 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 4096.13625 L, not 2048.06812 L.",
"C": "Correct. The worked calculation gives 4096.13625 L.",
"D": "Incorrect. S... | [
"ideal gas equation",
"unit consistency"
] | 3 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_ideal_gas | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_004359 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 3 kg and radius 6 m is acted on by a constant torque of 19 N m about its central axis. Find its angular acceleration. | {
"A": "0.35185 rad s⁻²",
"B": "3.35185 rad s⁻²",
"C": "0.52778 rad s⁻²",
"D": "-2.64815 rad s⁻²"
} | A | 0.35185 rad s⁻² | For a solid disc, I=½MR²=½(3)(6)²=54 kg m². Using τ=Iα, α=19/54=0.35185 rad s⁻². | {
"A": "Correct. The worked calculation gives 0.35185 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.35185 rad s⁻², not 3.35185 rad s⁻².",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.35185 rad s⁻², not 0.52778 rad s⁻².",
"D"... | [
"moment of inertia",
"rotational dynamics",
"torque"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_disc_torque | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008280 | 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 14 A for 22263 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": "101.54822 g",
"B": "51.27411 g",
"C": "102.54822 g",
"D": "103.54822 g"
} | C | 102.54822 g | Cu²⁺+2e⁻→Cu, so z=2. By Faraday's law m=MIt/(zF)=63.5×14×22263/(2×96500)=102.54822 g. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 102.54822 g, not 101.54822 g.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 102.54822 g, not 51.27411 g.",
"C": "Correct. The worked calculation gives 102.54822 g.",
"D": "Incorrect. Substit... | [
"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_005700 | physics | thermodynamics | ideal_gas_isochoric_pressure | hard | provisional_template_level_label | single_correct_mcq | A fixed amount of ideal gas is kept in a rigid container. Its pressure is 4.2 atm at 184 K. What is its pressure at 549 K? | {
"A": "12.53152 atm",
"B": "13.53152 atm",
"C": "-12.53152 atm",
"D": "18.79728 atm"
} | A | 12.53152 atm | At constant volume, P/T is constant. P₂=P₁T₂/T₁=(4.2)(549)/(184)=12.53152 atm. | {
"A": "Correct. The worked calculation gives 12.53152 atm.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 12.53152 atm, not 13.53152 atm.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 12.53152 atm, not -12.53152 atm.",
"D": "Incorrect. ... | [
"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_008982 | 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 0.54 mol L⁻¹ and initially B is absent. At equilibrium, 66% of A has dissociated at constant volume. Find K_c. | {
"A": "0.76734",
"B": "2.76734",
"C": "5.53468",
"D": "2.07551"
} | B | 2.76734 | Let α=0.66. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.54)(0.66)²/(1−0.66)=2.76734. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 2.76734, not 0.76734.",
"B": "Correct. The worked calculation gives 2.76734.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2.76734, not 5.53468.",
"D": "Incorrect. Substituting the stated va... | [
"chemical equilibrium",
"equilibrium constant",
"degree of dissociation"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | chem_equilibrium_alpha | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_007288 | chemistry | physical_chemistry | ideal_gas_volume | hard | provisional_template_level_label | single_correct_mcq | Assuming ideal behaviour, what volume is occupied by 8 mol of gas at 456 K and 0.1 atm? Use R=0.082057 L atm mol⁻¹ K⁻¹. | {
"A": "2993.43936 L",
"B": "300.34394 L",
"C": "4490.15904 L",
"D": "1496.71968 L"
} | A | 2993.43936 L | PV=nRT, so V=nRT/P=(8)(0.082057)(456)/(0.1)=2993.43936 L. | {
"A": "Correct. The worked calculation gives 2993.43936 L.",
"B": "Incorrect. Substituting the stated values into the governing relation gives 2993.43936 L, not 300.34394 L.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 2993.43936 L, not 4490.15904 L.",
"D": "Incorrect. Su... | [
"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_008999 | 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 0.17 mol L⁻¹ and initially B is absent. At equilibrium, 50% of A has dissociated at constant volume. Find K_c. | {
"A": "0.68",
"B": "0.34",
"C": "2.34",
"D": "-1.66"
} | B | 0.34 | Let α=0.5. At equilibrium [A]=c₀(1−α) and [B]=2c₀α. Thus K_c=[B]²/[A]=4c₀α²/(1−α)=4(0.17)(0.5)²/(1−0.5)=0.34. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.34, not 0.68.",
"B": "Correct. The worked calculation gives 0.34.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 0.34, not 2.34.",
"D": "Incorrect. Substituting the stated values into the g... | [
"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_004470 | physics | rotational_mechanics | disc_angular_acceleration | hard | provisional_template_level_label | single_correct_mcq | A uniform solid disc of mass 9 kg and radius 4 m is acted on by a constant torque of 7 N m about its central axis. Find its angular acceleration. | {
"A": "0.14583 rad s⁻²",
"B": "-2.90278 rad s⁻²",
"C": "0.09722 rad s⁻²",
"D": "-0.90278 rad s⁻²"
} | C | 0.09722 rad s⁻² | For a solid disc, I=½MR²=½(9)(4)²=72 kg m². Using τ=Iα, α=7/72=0.09722 rad s⁻². | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 0.09722 rad s⁻², not 0.14583 rad s⁻².",
"B": "Incorrect. Substituting the stated values into the governing relation gives 0.09722 rad s⁻², not -2.90278 rad s⁻².",
"C": "Correct. The worked calculation gives 0.09722 rad s⁻².",
"D... | [
"moment of inertia",
"rotational dynamics",
"torque"
] | 4 | numeric | en | synthetic_parametric | AxiloLabz deterministic STEM question generator | physics_disc_torque | template_formula_generated_structurally_validated_not_expert_reviewed | Answer is computed by the explicit equation in the generator template; this is not an independent expert proof. | Difficulty is approximate and not independently calibrated by JEE examiners; no ultra-hard claim is made in v0.1. | original_synthetic_question | correct_answer_option_matches_generated_numeric_value |
openstem_008661 | chemistry | physical_chemistry | boiling_point_elevation | hard | provisional_template_level_label | single_correct_mcq | A non-volatile solute is dissolved in water to produce a solution of molality 1.18 mol kg⁻¹. If its van't Hoff factor is 2 and K_b for water is 0.52 K kg mol⁻¹, find the boiling-point elevation. | {
"A": "3.2272 K",
"B": "1.2272 K",
"C": "4.2272 K",
"D": "-1.7728 K"
} | B | 1.2272 K | ΔT_b=iK_bm. Substitution gives ΔT_b=2×0.52×1.18=1.2272 K. | {
"A": "Incorrect. Substituting the stated values into the governing relation gives 1.2272 K, not 3.2272 K.",
"B": "Correct. The worked calculation gives 1.2272 K.",
"C": "Incorrect. Substituting the stated values into the governing relation gives 1.2272 K, not 4.2272 K.",
"D": "Incorrect. Substituting the stat... | [
"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 |
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