3782 lines
154 KiB
JSON
3782 lines
154 KiB
JSON
[
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Units and Measurements (Set 1)",
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"subtopic": "Dimensions",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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"question_text": "The dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of:",
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"options": {
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"A": "Velocity",
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"B": "Time",
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"C": "Capacitance",
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"D": "Distance"
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},
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"correct_answer": {
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"value": "A"
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},
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"solution_text": "We know that $c = \\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$, where $c$ is the speed of light in vacuum. Therefore, the dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of velocity, which is $[L T^{-1}]$.",
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"language": "English",
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"id": "858a5001-97c3-4d03-873c-b83247223ed7"
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},
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Kinematics (Set 1)",
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"subtopic": "Projectile Motion",
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"difficulty": "Easy",
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"type": "MCQ_Single",
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"question_text": "Two projectiles are thrown with the same initial velocity at angles of $30^\\circ$ and $60^\\circ$ with the horizontal. The ratio of their maximum heights is:",
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"options": {
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"A": "1:3",
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"B": "3:1",
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"C": "1:\\sqrt{3}",
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"D": "1:2"
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},
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"correct_answer": {
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"value": "A"
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},
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"solution_text": "Maximum height $H = \\frac{u^2 \\sin^2 \\theta}{2g}$.\nFor $\\theta_1 = 30^\\circ$, $H_1 = \\frac{u^2 (1/2)^2}{2g} = \\frac{u^2}{8g}$.\nFor $\\theta_2 = 60^\\circ$, $H_2 = \\frac{u^2 (\\sqrt{3}/2)^2}{2g} = \\frac{3u^2}{8g}$.\nRatio $H_1 : H_2 = 1 : 3$.",
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"language": "English",
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"id": "b6e37b17-d6bb-4e79-b3d4-b96a80a060b4"
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},
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Thermodynamics (Set 1)",
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"subtopic": "Carnot Engine",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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"question_text": "A Carnot engine takes heat from a reservoir at $627^\\circ\\text{C}$ and rejects heat to a sink at $27^\\circ\\text{C}$. Its efficiency is:",
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"options": {
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"A": "2/3",
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"B": "1/3",
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"C": "3/4",
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"D": "1/2"
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},
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"correct_answer": {
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"value": "A"
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},
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"solution_text": "Temperature of source, $T_1 = 627 + 273 = 900\\text{ K}$.\nTemperature of sink, $T_2 = 27 + 273 = 300\\text{ K}$.\nEfficiency $\\eta = 1 - \\frac{T_2}{T_1} = 1 - \\frac{300}{900} = 1 - \\frac{1}{3} = \\frac{2}{3}$.",
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"language": "English",
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"id": "c01d4590-27ad-47e9-9f07-217b6dcfa0b9"
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},
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Electrostatics (Set 1)",
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"subtopic": "Capacitors",
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"difficulty": "Hard",
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"type": "MCQ_Single",
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"question_text": "A parallel plate capacitor with air between the plates has a capacitance of $8\\text{ pF}$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?",
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"options": {
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"A": "96 pF",
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"B": "8 pF",
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"C": "48 pF",
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"D": "12 pF"
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},
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"correct_answer": {
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"value": "A"
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},
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"solution_text": "Initial capacitance $C_0 = \\frac{\\varepsilon_0 A}{d} = 8\\text{ pF}$.\nNew distance $d' = d/2$ and new dielectric constant $K = 6$.\nNew capacitance $C' = \\frac{K \\varepsilon_0 A}{d'} = \\frac{6 \\varepsilon_0 A}{d/2} = 12 \\left(\\frac{\\varepsilon_0 A}{d}\\right) = 12 \\times 8\\text{ pF} = 96\\text{ pF}$.",
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"language": "English",
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"id": "4698d5d3-0bbf-42f5-88b6-06cd17d09f75"
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},
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Current Electricity (Set 1)",
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"subtopic": "Meter Bridge",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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"question_text": "In a meter bridge experiment, the null point is obtained at $20\\text{ cm}$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then what will be the new position of the null point from the same end, if one decides to balance a resistance of $4X$ against $Y$?",
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"options": {
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"A": "50 cm",
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"B": "40 cm",
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"C": "80 cm",
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"D": "20 cm"
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},
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"correct_answer": {
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"value": "A"
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},
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"solution_text": "Initially, $\\frac{X}{Y} = \\frac{20}{100-20} = \\frac{20}{80} = \\frac{1}{4}$. So $Y = 4X$.\nFor new resistance $4X$ against $Y$, ratio is $\\frac{4X}{Y} = \\frac{4X}{4X} = 1$.\nLet new null point be $l$, then $\\frac{l}{100-l} = 1 \\implies l = 100 - l \\implies 2l = 100 \\implies l = 50\\text{ cm}$.",
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"language": "English",
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"id": "b845ae53-3ffe-4ade-9c9a-1d7c96cad1ea"
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},
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Magnetism (Set 1)",
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"subtopic": "Biot-Savart Law",
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"difficulty": "Easy",
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"type": "MCQ_Single",
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"question_text": "The magnetic field at the centre of a circular current carrying coil of radius $R$ is $B_0$. If the radius is doubled keeping the current same, the new magnetic field at the centre will be:",
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"options": {
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"A": "$B_0/2$",
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"B": "$B_0/4$",
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"C": "$2B_0$",
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"D": "$B_0$"
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},
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"correct_answer": {
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"value": "A"
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},
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"solution_text": "Magnetic field at the centre of a circular coil is $B = \\frac{\\mu_0 I}{2R}$.\nSince $B \\propto \\frac{1}{R}$, doubling the radius will make the magnetic field half. Thus, new field is $B_0/2$.",
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"language": "English",
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"id": "6c96124e-c093-4070-a60b-937451db625b"
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},
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Alternating Current (Set 1)",
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"subtopic": "LCR Circuit",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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"question_text": "In a series LCR circuit, the voltage across the resistance, capacitance, and inductance is $10\\text{ V}$ each. If the capacitance is short-circuited, the voltage across the inductance will be:",
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"options": {
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"A": "10 V",
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"B": "$10\\sqrt{2}\\text{ V}$",
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"C": "$10/\\sqrt{2}\\text{ V}$",
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"D": "Zero"
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},
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"correct_answer": {
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"value": "C"
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},
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"solution_text": "Initially, $V_R = 10\\text{ V}$, $V_L = 10\\text{ V}$, $V_C = 10\\text{ V}$. Since $V_L = V_C$, the circuit is in resonance. The applied voltage $V = \\sqrt{V_R^2 + (V_L - V_C)^2} = V_R = 10\\text{ V}$.\nIf capacitance is shorted, it becomes an LR circuit. Impedance $Z = \\sqrt{R^2 + X_L^2}$. Since $V_R = V_L$ initially, $R = X_L$, so $Z = R\\sqrt{2}$.\nNew current $I' = \\frac{V}{Z} = \\frac{10}{R\\sqrt{2}}$.\nNew voltage across inductance $V_L' = I' X_L = \\frac{10}{R\\sqrt{2}} \\times R = \\frac{10}{\\sqrt{2}}\\text{ V}$.",
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"language": "English",
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"id": "71dd39a3-1a78-4589-893a-1bbd030006cf"
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},
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Optics (Set 1)",
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"subtopic": "Wave Optics",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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"question_text": "In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:",
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"options": {
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"A": "Half",
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"B": "Four times",
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"C": "One-fourth",
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"D": "Double"
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},
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"correct_answer": {
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"value": "B"
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},
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"solution_text": "Fringe width $\\beta = \\frac{\\lambda D}{d}$.\nNew slit separation $d' = d/2$.\nNew screen distance $D' = 2D$.\nNew fringe width $\\beta' = \\frac{\\lambda (2D)}{(d/2)} = 4 \\frac{\\lambda D}{d} = 4\\beta$.",
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"language": "English",
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"id": "21936288-06c6-4710-9fd5-1d3e085e42b0"
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},
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Modern Physics (Set 1)",
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"subtopic": "Photoelectric Effect",
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"difficulty": "Hard",
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"type": "MCQ_Single",
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"question_text": "When the light of frequency $2\\nu_0$ (where $\\nu_0$ is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5\\nu_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is:",
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"options": {
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"A": "1:2",
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"B": "1:4",
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"C": "4:1",
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"D": "2:1"
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},
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"correct_answer": {
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"value": "A"
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},
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"solution_text": "From Einstein's photoelectric equation: $K_{\\max} = h\\nu - h\\nu_0 = \\frac{1}{2} m v^2$.\nCase 1: $\\frac{1}{2} m v_1^2 = h(2\\nu_0) - h\\nu_0 = h\\nu_0$.\nCase 2: $\\frac{1}{2} m v_2^2 = h(5\\nu_0) - h\\nu_0 = 4h\\nu_0$.\nDividing the two equations: $\\frac{v_1^2}{v_2^2} = \\frac{1}{4} \\implies \\frac{v_1}{v_2} = \\frac{1}{2}$.",
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"language": "English",
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"id": "658541df-4e92-49f1-8861-e1011ad39233"
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},
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{
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"section": "General",
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"subject": "Physics",
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"topic": "Semiconductors (Set 1)",
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"subtopic": "Logic Gates",
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"difficulty": "Easy",
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"type": "MCQ_Single",
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"question_text": "Which logic gate is represented by the boolean expression $Y = \\overline{A \\cdot B}$?",
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"options": {
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"A": "NOR",
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"B": "OR",
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"C": "NAND",
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"D": "AND"
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},
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"correct_answer": {
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"value": "C"
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},
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"solution_text": "The boolean expression $Y = \\overline{A \\cdot B}$ represents the NAND gate. It gives an output of 0 only when both inputs A and B are 1, and 1 otherwise.",
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"language": "English",
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"id": "1c1ade08-e7dd-45b8-94fe-373d320b359e"
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},
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{
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"section": "General",
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"subject": "Chemistry",
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"topic": "Physical Chemistry (Set 1)",
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"subtopic": "Mole Concept",
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"difficulty": "Easy",
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"type": "MCQ_Single",
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"question_text": "The number of water molecules is maximum in:",
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"options": {
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"A": "18 grams of water",
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"B": "18 moles of water",
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"C": "18 molecules of water",
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"D": "1.8 grams of water"
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},
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"correct_answer": {
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"value": "B"
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},
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"solution_text": "18 grams of water = 1 mole = $N_A$ molecules.\n18 moles of water = $18 N_A$ molecules.\n1.8 grams of water = 0.1 mole = $0.1 N_A$ molecules.\nThus, 18 moles of water contains the maximum number of molecules.",
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"language": "English",
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"id": "dd5d5b06-f0a7-4688-9f44-1581d9a2eae6"
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},
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{
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"section": "General",
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"subject": "Chemistry",
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"topic": "Physical Chemistry (Set 1)",
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"subtopic": "Atomic Structure",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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"question_text": "The maximum number of electrons in an atom that can have the quantum numbers $n = 4$ and $m_l = +1$ is:",
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"options": {
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"A": "2",
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"B": "4",
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"C": "6",
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"D": "8"
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},
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"correct_answer": {
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"value": "C"
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},
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"solution_text": "For $n = 4$, the possible values of $l$ are 0, 1, 2, 3.\nThe condition $m_l = +1$ is possible for $l = 1$, $l = 2$, and $l = 3$.\nEach subshell ($l=1, 2, 3$) will have one orbital with $m_l = +1$.\nSince each orbital can hold 2 electrons, the total number of electrons is $3 \\times 2 = 6$.",
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"language": "English",
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"id": "0913c5cf-e9b2-4319-a51a-3ffaf1fdf5fd"
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},
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{
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"section": "General",
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"subject": "Chemistry",
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"topic": "Inorganic Chemistry (Set 1)",
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"subtopic": "Chemical Bonding",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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"question_text": "The species having pyramidal shape is:",
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"options": {
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"A": "$\\text{SO}_3$",
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"B": "$\\text{BrF}_3$",
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"C": "$\\text{SiO}_3^{2-}$",
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"D": "$\\text{OSF}_2$"
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},
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"correct_answer": {
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"value": "D"
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},
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"solution_text": "In $\\text{OSF}_2$, the central sulfur atom has 4 electron domains (1 double bond to O, 2 single bonds to F, and 1 lone pair). This corresponds to an $sp^3$ hybridization with a trigonal pyramidal shape. $\\text{SO}_3$ is trigonal planar, $\\text{BrF}_3$ is T-shaped.",
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"language": "English",
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"id": "b0fd946f-02c7-45d2-af59-c2c490019ba6"
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},
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{
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"section": "General",
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"subject": "Chemistry",
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"topic": "Physical Chemistry (Set 1)",
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"subtopic": "Equilibrium",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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"question_text": "For the reversible reaction $\\text{N}_2(\\text{g}) + 3\\text{H}_2(\\text{g}) \\rightleftharpoons 2\\text{NH}_3(\\text{g}) + \\text{Heat}$, the equilibrium shifts in forward direction:",
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"options": {
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"A": "By increasing the concentration of $\\text{NH}_3(\\text{g})$",
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"B": "By decreasing the pressure",
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"C": "By decreasing the concentrations of $\\text{N}_2(\\text{g})$ and $\\text{H}_2(\\text{g})$",
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"D": "By increasing pressure and decreasing temperature"
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},
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"correct_answer": {
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"value": "D"
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},
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"solution_text": "According to Le Chatelier's principle, an exothermic reaction ($\\Delta H < 0$) is favored by low temperatures. A reaction involving a decrease in the number of gaseous moles ($4 \\rightarrow 2$) is favored by high pressures. Therefore, increasing pressure and decreasing temperature will shift the equilibrium in the forward direction.",
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"language": "English",
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"id": "43658efb-4da6-48c1-b6cd-afc56706986f"
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},
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{
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"section": "General",
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"subject": "Chemistry",
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"topic": "Physical Chemistry (Set 1)",
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"subtopic": "Electrochemistry",
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"difficulty": "Hard",
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"type": "MCQ_Single",
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"question_text": "Given $E^\\circ_{\\text{Cr}^{3+}/\\text{Cr}} = -0.74\\text{ V}$; $E^\\circ_{\\text{MnO}_4^-/\\text{Mn}^{2+}} = 1.51\\text{ V}$; $E^\\circ_{\\text{Cr}_2\\text{O}_7^{2-}/\\text{Cr}^{3+}} = 1.33\\text{ V}$; $E^\\circ_{\\text{Cl}/\\text{Cl}^-} = 1.36\\text{ V}$.\nBased on the data given above, strongest oxidizing agent will be:",
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"options": {
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"A": "$\\text{Cl}$",
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"B": "$\\text{Cr}^{3+}$",
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"C": "$\\text{Mn}^{2+}$",
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"D": "$\\text{MnO}_4^-$"
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},
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"correct_answer": {
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"value": "D"
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},
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"solution_text": "Higher the standard reduction potential ($E^\\circ$), stronger is the oxidizing power. The $E^\\circ$ value for $\\text{MnO}_4^-/\\text{Mn}^{2+}$ is the highest ($+1.51\\text{ V}$). Therefore, $\\text{MnO}_4^-$ is the strongest oxidizing agent among the given species.",
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"language": "English",
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"id": "46384cb2-62ea-47cb-88ea-4d8b800f7fd2"
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},
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|
{
|
|
"section": "General",
|
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"subject": "Chemistry",
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"topic": "Organic Chemistry (Set 1)",
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"subtopic": "General Organic Chemistry",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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"question_text": "Which of the following carbocations is most stable?",
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"options": {
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"A": "Allyl carbocation",
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"B": "Primary alkyl carbocation",
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"C": "Tert-butyl carbocation",
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"D": "Triphenylmethyl carbocation"
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},
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"correct_answer": {
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"value": "D"
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},
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"solution_text": "Stability of carbocations is determined by delocalization of charge. The triphenylmethyl carbocation, $(\\text{C}_6\\text{H}_5)_3\\text{C}^+$, has 3 phenyl rings providing extensive resonance stabilization, making it exceptionally stable compared to the others.",
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"language": "English",
|
|
"id": "e2d87153-02ff-4342-88e0-8e496e408532"
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},
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{
|
|
"section": "General",
|
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"subject": "Chemistry",
|
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"topic": "Organic Chemistry (Set 1)",
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"subtopic": "Hydrocarbons",
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"difficulty": "Hard",
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"type": "MCQ_Single",
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"question_text": "Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:",
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"options": {
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"A": "Two isomeric double bonds",
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"B": "A vinyl group",
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"C": "An isopropyl group",
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"D": "An acetylenic triple bond"
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},
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"correct_answer": {
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"value": "B"
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},
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"solution_text": "Ozonolysis cleaves a carbon-carbon double bond and adds oxygen. Formaldehyde ($\\text{HCHO}$) is produced if the original double bond was at the end of a chain, specifically involving a terminal $=\\text{CH}_2$ group, which is characteristic of a vinyl group ($-\\text{CH}=\\text{CH}_2$).",
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"language": "English",
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"id": "b2da608f-c808-4fde-a3cf-45fbd0b4f4ff"
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},
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{
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"section": "General",
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"subject": "Chemistry",
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"topic": "Inorganic Chemistry (Set 1)",
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"subtopic": "Coordination Compounds",
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"difficulty": "Medium",
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"type": "MCQ_Single",
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|
"question_text": "The IUPAC name of $[\\text{Pt}(\\text{NH}_3)_2\\text{Cl}(\\text{NO}_2)]$ is:",
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"options": {
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"A": "Platinum diaminechloronitrite",
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"B": "Chloronitrito-N-ammineplatinum(II)",
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"C": "Diamminechloridonitrito-N-platinum(II)",
|
|
"D": "Diamminechloronitrito-O-platinate(II)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Ligands are named in alphabetical order: ammine, chlorido, nitrito-N (since it coordinates through nitrogen). The oxidation state of Pt is $x + 0 + (-1) + (-1) = 0 \\implies x = +2$. Thus, the correct name is Diamminechloridonitrito-N-platinum(II).",
|
|
"language": "English",
|
|
"id": "e4e0e913-c5a9-4abc-bd86-b2caf406e03f"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 1)",
|
|
"subtopic": "Chemical Kinetics",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The unit of rate constant for a zero order reaction is:",
|
|
"options": {
|
|
"A": "$\\text{s}^{-1}$",
|
|
"B": "$\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"C": "$\\text{L}\\cdot\\text{mol}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"D": "$\\text{L}^2\\cdot\\text{mol}^{-2}\\cdot\\text{s}^{-1}$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "For a zero-order reaction, Rate = $k[A]^0 = k$. The unit of rate is concentration per unit time ($\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$). Thus, the unit of the rate constant $k$ is also $\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$.",
|
|
"language": "English",
|
|
"id": "48f8e5ea-8b89-433f-9aef-4252a70fece4"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 1)",
|
|
"subtopic": "Biomolecules",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human beings?",
|
|
"options": {
|
|
"A": "Thyroxin",
|
|
"B": "Insulin",
|
|
"C": "Adrenaline",
|
|
"D": "Estradiol"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Adrenaline (epinephrine) is the 'fight or flight' hormone released during stress. It promotes the breakdown of glycogen into glucose (glycogenolysis) in the liver to increase blood glucose levels for energy.",
|
|
"language": "English",
|
|
"id": "2d40b28f-5af9-4b15-9fc1-5ada60dd098a"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Cell Structure and Function (Set 1)",
|
|
"subtopic": "Cell Cycle and Cell Division",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Crossing over occurs during which stage of meiosis?",
|
|
"options": {
|
|
"A": "Leptotene",
|
|
"B": "Pachytene",
|
|
"C": "Diplotene",
|
|
"D": "Zygotene"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Crossing over, the exchange of genetic material between non-sister chromatids of homologous chromosomes, occurs during the pachytene stage of prophase I of meiosis.",
|
|
"language": "English",
|
|
"id": "4c3a00cd-776b-4336-a8ce-56dfdf13a834"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 1)",
|
|
"subtopic": "Principles of Inheritance and Variation",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?",
|
|
"options": {
|
|
"A": "TT and Tt",
|
|
"B": "Tt and Tt",
|
|
"C": "TT and TT",
|
|
"D": "Tt and tt"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Since both parents are tall but they produce some dwarf offspring, both parents must carry the recessive allele for dwarfness. Thus, their genotype must be heterozygous (Tt). A cross of Tt x Tt yields a 3:1 ratio of tall to dwarf.",
|
|
"language": "English",
|
|
"id": "6f3fa081-48bc-43f6-a814-005acb92f91b"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 1)",
|
|
"subtopic": "Molecular Basis of Inheritance",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following RNAs should be most abundant in an animal cell?",
|
|
"options": {
|
|
"A": "tRNA",
|
|
"B": "mRNA",
|
|
"C": "miRNA",
|
|
"D": "rRNA"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Ribosomal RNA (rRNA) constitutes about 80% of the total RNA in an actively growing cell. It forms the structural and catalytic core of ribosomes.",
|
|
"language": "English",
|
|
"id": "5466289f-920a-4ab7-9b17-38914bf67664"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 1)",
|
|
"subtopic": "Digestion and Absorption",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following terms describes human dentition?",
|
|
"options": {
|
|
"A": "Thecodont, Diphyodont, Homodont",
|
|
"B": "Thecodont, Diphyodont, Heterodont",
|
|
"C": "Pleurodont, Monophyodont, Homodont",
|
|
"D": "Pleurodont, Diphyodont, Heterodont"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Human dentition is Thecodont (teeth embedded in jaw sockets), Diphyodont (two sets of teeth: deciduous and permanent), and Heterodont (different types of teeth: incisors, canines, premolars, molars).",
|
|
"language": "English",
|
|
"id": "eb80dcf1-5f97-4527-a393-79cf9646a285"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Reproduction (Set 1)",
|
|
"subtopic": "Human Reproduction",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which hormone is responsible for the ovulation in human females?",
|
|
"options": {
|
|
"A": "FSH (Follicle Stimulating Hormone)",
|
|
"B": "Progesterone",
|
|
"C": "LH (Luteinizing Hormone)",
|
|
"D": "Estrogen"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "A surge in LH (Luteinizing Hormone) mid-cycle triggers the rupture of the Graafian follicle and the subsequent release of the ovum, a process known as ovulation.",
|
|
"language": "English",
|
|
"id": "90c4e3a8-0736-47f9-8aca-8bd942cdc2e2"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Plant Physiology (Set 1)",
|
|
"subtopic": "Photosynthesis in Higher Plants",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The first stable product of CO2 fixation in C3 plants is:",
|
|
"options": {
|
|
"A": "Phosphoglyceraldehyde (PGAL)",
|
|
"B": "Malic acid",
|
|
"C": "Oxaloacetic acid (OAA)",
|
|
"D": "3-Phosphoglyceric acid (3-PGA)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In the Calvin cycle of C3 plants, CO2 combines with RuBP to form a 6-carbon unstable intermediate that immediately splits into two molecules of 3-phosphoglyceric acid (3-PGA), the first stable product.",
|
|
"language": "English",
|
|
"id": "a9b47475-4f9b-4758-a5e1-2972379862e8"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Ecology (Set 1)",
|
|
"subtopic": "Ecosystem",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The pyramid of energy is always:",
|
|
"options": {
|
|
"A": "Upright",
|
|
"B": "Inverted",
|
|
"C": "Spindle shaped",
|
|
"D": "Variable"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "The pyramid of energy is always upright because energy is lost as heat at each trophic level (10% law), so energy always decreases as it flows from producers to higher trophic levels.",
|
|
"language": "English",
|
|
"id": "4fd0bba0-8227-4af6-995a-4d536730b45d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Biotechnology (Set 1)",
|
|
"subtopic": "Principles and Processes",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The cutting of DNA at specific locations became possible with the discovery of:",
|
|
"options": {
|
|
"A": "Selectable markers",
|
|
"B": "Ligases",
|
|
"C": "Restriction enzymes",
|
|
"D": "Probes"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Restriction endonucleases (or restriction enzymes) cut DNA at specific recognition sequences, making recombinant DNA technology possible.",
|
|
"language": "English",
|
|
"id": "6a84494e-15cb-4c75-86a3-98aeb1c02eba"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 1)",
|
|
"subtopic": "Neural Control and Coordination",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Receptor sites for neurotransmitters are present on:",
|
|
"options": {
|
|
"A": "Pre-synaptic membrane",
|
|
"B": "Tips of axons",
|
|
"C": "Post-synaptic membrane",
|
|
"D": "Membrane of synaptic vesicles"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Neurotransmitters are released from synaptic vesicles in the pre-synaptic neuron and diffuse across the synaptic cleft to bind with specific receptor sites located on the post-synaptic membrane.",
|
|
"language": "English",
|
|
"id": "2427b46d-139b-4526-9eb4-341d192b7124"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Diversity in Living World (Set 1)",
|
|
"subtopic": "Biological Classification",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following are found in extreme saline conditions?",
|
|
"options": {
|
|
"A": "Eubacteria",
|
|
"B": "Cyanobacteria",
|
|
"C": "Mycobacteria",
|
|
"D": "Archaebacteria"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Archaebacteria are known to live in some of the most harsh habitats. Specifically, halophiles live in extreme saline conditions (e.g., Dead Sea).",
|
|
"language": "English",
|
|
"id": "6a691f62-8492-4494-8954-b3089e772ae3"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Units and Measurements (Set 2)",
|
|
"subtopic": "Dimensions",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of:",
|
|
"options": {
|
|
"A": "Velocity",
|
|
"B": "Time",
|
|
"C": "Capacitance",
|
|
"D": "Distance"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "We know that $c = \\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$, where $c$ is the speed of light in vacuum. Therefore, the dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of velocity, which is $[L T^{-1}]$.",
|
|
"language": "English",
|
|
"id": "7b762697-6eec-42af-88d2-89633265013f"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Kinematics (Set 2)",
|
|
"subtopic": "Projectile Motion",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Two projectiles are thrown with the same initial velocity at angles of $30^\\circ$ and $60^\\circ$ with the horizontal. The ratio of their maximum heights is:",
|
|
"options": {
|
|
"A": "1:3",
|
|
"B": "3:1",
|
|
"C": "1:\\sqrt{3}",
|
|
"D": "1:2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Maximum height $H = \\frac{u^2 \\sin^2 \\theta}{2g}$.\nFor $\\theta_1 = 30^\\circ$, $H_1 = \\frac{u^2 (1/2)^2}{2g} = \\frac{u^2}{8g}$.\nFor $\\theta_2 = 60^\\circ$, $H_2 = \\frac{u^2 (\\sqrt{3}/2)^2}{2g} = \\frac{3u^2}{8g}$.\nRatio $H_1 : H_2 = 1 : 3$.",
|
|
"language": "English",
|
|
"id": "360c8296-f0b1-486a-9c9b-5cbf0edeae7e"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Thermodynamics (Set 2)",
|
|
"subtopic": "Carnot Engine",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A Carnot engine takes heat from a reservoir at $627^\\circ\\text{C}$ and rejects heat to a sink at $27^\\circ\\text{C}$. Its efficiency is:",
|
|
"options": {
|
|
"A": "2/3",
|
|
"B": "1/3",
|
|
"C": "3/4",
|
|
"D": "1/2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Temperature of source, $T_1 = 627 + 273 = 900\\text{ K}$.\nTemperature of sink, $T_2 = 27 + 273 = 300\\text{ K}$.\nEfficiency $\\eta = 1 - \\frac{T_2}{T_1} = 1 - \\frac{300}{900} = 1 - \\frac{1}{3} = \\frac{2}{3}$.",
|
|
"language": "English",
|
|
"id": "03d674bb-9da5-436f-be5a-44164bb2b169"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Electrostatics (Set 2)",
|
|
"subtopic": "Capacitors",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A parallel plate capacitor with air between the plates has a capacitance of $8\\text{ pF}$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?",
|
|
"options": {
|
|
"A": "96 pF",
|
|
"B": "8 pF",
|
|
"C": "48 pF",
|
|
"D": "12 pF"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initial capacitance $C_0 = \\frac{\\varepsilon_0 A}{d} = 8\\text{ pF}$.\nNew distance $d' = d/2$ and new dielectric constant $K = 6$.\nNew capacitance $C' = \\frac{K \\varepsilon_0 A}{d'} = \\frac{6 \\varepsilon_0 A}{d/2} = 12 \\left(\\frac{\\varepsilon_0 A}{d}\\right) = 12 \\times 8\\text{ pF} = 96\\text{ pF}$.",
|
|
"language": "English",
|
|
"id": "ca859cf2-0784-4c06-8298-3252a5e9ede0"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Current Electricity (Set 2)",
|
|
"subtopic": "Meter Bridge",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a meter bridge experiment, the null point is obtained at $20\\text{ cm}$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then what will be the new position of the null point from the same end, if one decides to balance a resistance of $4X$ against $Y$?",
|
|
"options": {
|
|
"A": "50 cm",
|
|
"B": "40 cm",
|
|
"C": "80 cm",
|
|
"D": "20 cm"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initially, $\\frac{X}{Y} = \\frac{20}{100-20} = \\frac{20}{80} = \\frac{1}{4}$. So $Y = 4X$.\nFor new resistance $4X$ against $Y$, ratio is $\\frac{4X}{Y} = \\frac{4X}{4X} = 1$.\nLet new null point be $l$, then $\\frac{l}{100-l} = 1 \\implies l = 100 - l \\implies 2l = 100 \\implies l = 50\\text{ cm}$.",
|
|
"language": "English",
|
|
"id": "c3048971-d687-4b55-a166-43df4ab61b2e"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Magnetism (Set 2)",
|
|
"subtopic": "Biot-Savart Law",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The magnetic field at the centre of a circular current carrying coil of radius $R$ is $B_0$. If the radius is doubled keeping the current same, the new magnetic field at the centre will be:",
|
|
"options": {
|
|
"A": "$B_0/2$",
|
|
"B": "$B_0/4$",
|
|
"C": "$2B_0$",
|
|
"D": "$B_0$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Magnetic field at the centre of a circular coil is $B = \\frac{\\mu_0 I}{2R}$.\nSince $B \\propto \\frac{1}{R}$, doubling the radius will make the magnetic field half. Thus, new field is $B_0/2$.",
|
|
"language": "English",
|
|
"id": "ef4bc469-b252-43c9-a8ce-ef0888a44b03"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Alternating Current (Set 2)",
|
|
"subtopic": "LCR Circuit",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a series LCR circuit, the voltage across the resistance, capacitance, and inductance is $10\\text{ V}$ each. If the capacitance is short-circuited, the voltage across the inductance will be:",
|
|
"options": {
|
|
"A": "10 V",
|
|
"B": "$10\\sqrt{2}\\text{ V}$",
|
|
"C": "$10/\\sqrt{2}\\text{ V}$",
|
|
"D": "Zero"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Initially, $V_R = 10\\text{ V}$, $V_L = 10\\text{ V}$, $V_C = 10\\text{ V}$. Since $V_L = V_C$, the circuit is in resonance. The applied voltage $V = \\sqrt{V_R^2 + (V_L - V_C)^2} = V_R = 10\\text{ V}$.\nIf capacitance is shorted, it becomes an LR circuit. Impedance $Z = \\sqrt{R^2 + X_L^2}$. Since $V_R = V_L$ initially, $R = X_L$, so $Z = R\\sqrt{2}$.\nNew current $I' = \\frac{V}{Z} = \\frac{10}{R\\sqrt{2}}$.\nNew voltage across inductance $V_L' = I' X_L = \\frac{10}{R\\sqrt{2}} \\times R = \\frac{10}{\\sqrt{2}}\\text{ V}$.",
|
|
"language": "English",
|
|
"id": "76fe2e5f-d75e-4a67-91e9-1e4e6f214a8b"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Optics (Set 2)",
|
|
"subtopic": "Wave Optics",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:",
|
|
"options": {
|
|
"A": "Half",
|
|
"B": "Four times",
|
|
"C": "One-fourth",
|
|
"D": "Double"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Fringe width $\\beta = \\frac{\\lambda D}{d}$.\nNew slit separation $d' = d/2$.\nNew screen distance $D' = 2D$.\nNew fringe width $\\beta' = \\frac{\\lambda (2D)}{(d/2)} = 4 \\frac{\\lambda D}{d} = 4\\beta$.",
|
|
"language": "English",
|
|
"id": "9bb0969b-efef-4ef1-8790-e0bc31d4cc67"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Modern Physics (Set 2)",
|
|
"subtopic": "Photoelectric Effect",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "When the light of frequency $2\\nu_0$ (where $\\nu_0$ is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5\\nu_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is:",
|
|
"options": {
|
|
"A": "1:2",
|
|
"B": "1:4",
|
|
"C": "4:1",
|
|
"D": "2:1"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "From Einstein's photoelectric equation: $K_{\\max} = h\\nu - h\\nu_0 = \\frac{1}{2} m v^2$.\nCase 1: $\\frac{1}{2} m v_1^2 = h(2\\nu_0) - h\\nu_0 = h\\nu_0$.\nCase 2: $\\frac{1}{2} m v_2^2 = h(5\\nu_0) - h\\nu_0 = 4h\\nu_0$.\nDividing the two equations: $\\frac{v_1^2}{v_2^2} = \\frac{1}{4} \\implies \\frac{v_1}{v_2} = \\frac{1}{2}$.",
|
|
"language": "English",
|
|
"id": "8247e89c-e0a3-429a-ae55-aed2cb229a4b"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Semiconductors (Set 2)",
|
|
"subtopic": "Logic Gates",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which logic gate is represented by the boolean expression $Y = \\overline{A \\cdot B}$?",
|
|
"options": {
|
|
"A": "NOR",
|
|
"B": "OR",
|
|
"C": "NAND",
|
|
"D": "AND"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "The boolean expression $Y = \\overline{A \\cdot B}$ represents the NAND gate. It gives an output of 0 only when both inputs A and B are 1, and 1 otherwise.",
|
|
"language": "English",
|
|
"id": "2dcf2400-ab08-47f6-b5d7-acb58d4eba16"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 2)",
|
|
"subtopic": "Mole Concept",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The number of water molecules is maximum in:",
|
|
"options": {
|
|
"A": "18 grams of water",
|
|
"B": "18 moles of water",
|
|
"C": "18 molecules of water",
|
|
"D": "1.8 grams of water"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "18 grams of water = 1 mole = $N_A$ molecules.\n18 moles of water = $18 N_A$ molecules.\n1.8 grams of water = 0.1 mole = $0.1 N_A$ molecules.\nThus, 18 moles of water contains the maximum number of molecules.",
|
|
"language": "English",
|
|
"id": "8500561b-4357-467a-b112-6f208ed4478f"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 2)",
|
|
"subtopic": "Atomic Structure",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The maximum number of electrons in an atom that can have the quantum numbers $n = 4$ and $m_l = +1$ is:",
|
|
"options": {
|
|
"A": "2",
|
|
"B": "4",
|
|
"C": "6",
|
|
"D": "8"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "For $n = 4$, the possible values of $l$ are 0, 1, 2, 3.\nThe condition $m_l = +1$ is possible for $l = 1$, $l = 2$, and $l = 3$.\nEach subshell ($l=1, 2, 3$) will have one orbital with $m_l = +1$.\nSince each orbital can hold 2 electrons, the total number of electrons is $3 \\times 2 = 6$.",
|
|
"language": "English",
|
|
"id": "486fe0d3-26c6-4e64-8a6f-8bd78d82953e"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 2)",
|
|
"subtopic": "Chemical Bonding",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The species having pyramidal shape is:",
|
|
"options": {
|
|
"A": "$\\text{SO}_3$",
|
|
"B": "$\\text{BrF}_3$",
|
|
"C": "$\\text{SiO}_3^{2-}$",
|
|
"D": "$\\text{OSF}_2$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In $\\text{OSF}_2$, the central sulfur atom has 4 electron domains (1 double bond to O, 2 single bonds to F, and 1 lone pair). This corresponds to an $sp^3$ hybridization with a trigonal pyramidal shape. $\\text{SO}_3$ is trigonal planar, $\\text{BrF}_3$ is T-shaped.",
|
|
"language": "English",
|
|
"id": "de84de18-b5fe-44c1-807c-a7da3e06033d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 2)",
|
|
"subtopic": "Equilibrium",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "For the reversible reaction $\\text{N}_2(\\text{g}) + 3\\text{H}_2(\\text{g}) \\rightleftharpoons 2\\text{NH}_3(\\text{g}) + \\text{Heat}$, the equilibrium shifts in forward direction:",
|
|
"options": {
|
|
"A": "By increasing the concentration of $\\text{NH}_3(\\text{g})$",
|
|
"B": "By decreasing the pressure",
|
|
"C": "By decreasing the concentrations of $\\text{N}_2(\\text{g})$ and $\\text{H}_2(\\text{g})$",
|
|
"D": "By increasing pressure and decreasing temperature"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "According to Le Chatelier's principle, an exothermic reaction ($\\Delta H < 0$) is favored by low temperatures. A reaction involving a decrease in the number of gaseous moles ($4 \\rightarrow 2$) is favored by high pressures. Therefore, increasing pressure and decreasing temperature will shift the equilibrium in the forward direction.",
|
|
"language": "English",
|
|
"id": "8eb9b9c7-d262-4c0e-9366-e76fc9e3311f"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 2)",
|
|
"subtopic": "Electrochemistry",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Given $E^\\circ_{\\text{Cr}^{3+}/\\text{Cr}} = -0.74\\text{ V}$; $E^\\circ_{\\text{MnO}_4^-/\\text{Mn}^{2+}} = 1.51\\text{ V}$; $E^\\circ_{\\text{Cr}_2\\text{O}_7^{2-}/\\text{Cr}^{3+}} = 1.33\\text{ V}$; $E^\\circ_{\\text{Cl}/\\text{Cl}^-} = 1.36\\text{ V}$.\nBased on the data given above, strongest oxidizing agent will be:",
|
|
"options": {
|
|
"A": "$\\text{Cl}$",
|
|
"B": "$\\text{Cr}^{3+}$",
|
|
"C": "$\\text{Mn}^{2+}$",
|
|
"D": "$\\text{MnO}_4^-$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Higher the standard reduction potential ($E^\\circ$), stronger is the oxidizing power. The $E^\\circ$ value for $\\text{MnO}_4^-/\\text{Mn}^{2+}$ is the highest ($+1.51\\text{ V}$). Therefore, $\\text{MnO}_4^-$ is the strongest oxidizing agent among the given species.",
|
|
"language": "English",
|
|
"id": "3c83e93c-1f88-4c55-8e4f-8dbbee38e61e"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 2)",
|
|
"subtopic": "General Organic Chemistry",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following carbocations is most stable?",
|
|
"options": {
|
|
"A": "Allyl carbocation",
|
|
"B": "Primary alkyl carbocation",
|
|
"C": "Tert-butyl carbocation",
|
|
"D": "Triphenylmethyl carbocation"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Stability of carbocations is determined by delocalization of charge. The triphenylmethyl carbocation, $(\\text{C}_6\\text{H}_5)_3\\text{C}^+$, has 3 phenyl rings providing extensive resonance stabilization, making it exceptionally stable compared to the others.",
|
|
"language": "English",
|
|
"id": "7904b9d7-9e66-40cc-8380-b4cf3bd1cb31"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 2)",
|
|
"subtopic": "Hydrocarbons",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:",
|
|
"options": {
|
|
"A": "Two isomeric double bonds",
|
|
"B": "A vinyl group",
|
|
"C": "An isopropyl group",
|
|
"D": "An acetylenic triple bond"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Ozonolysis cleaves a carbon-carbon double bond and adds oxygen. Formaldehyde ($\\text{HCHO}$) is produced if the original double bond was at the end of a chain, specifically involving a terminal $=\\text{CH}_2$ group, which is characteristic of a vinyl group ($-\\text{CH}=\\text{CH}_2$).",
|
|
"language": "English",
|
|
"id": "e97d91f2-e924-48d0-aa02-2541f12b3f85"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 2)",
|
|
"subtopic": "Coordination Compounds",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The IUPAC name of $[\\text{Pt}(\\text{NH}_3)_2\\text{Cl}(\\text{NO}_2)]$ is:",
|
|
"options": {
|
|
"A": "Platinum diaminechloronitrite",
|
|
"B": "Chloronitrito-N-ammineplatinum(II)",
|
|
"C": "Diamminechloridonitrito-N-platinum(II)",
|
|
"D": "Diamminechloronitrito-O-platinate(II)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Ligands are named in alphabetical order: ammine, chlorido, nitrito-N (since it coordinates through nitrogen). The oxidation state of Pt is $x + 0 + (-1) + (-1) = 0 \\implies x = +2$. Thus, the correct name is Diamminechloridonitrito-N-platinum(II).",
|
|
"language": "English",
|
|
"id": "57f1959d-5345-4286-abf1-bb2d2c747a96"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 2)",
|
|
"subtopic": "Chemical Kinetics",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The unit of rate constant for a zero order reaction is:",
|
|
"options": {
|
|
"A": "$\\text{s}^{-1}$",
|
|
"B": "$\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"C": "$\\text{L}\\cdot\\text{mol}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"D": "$\\text{L}^2\\cdot\\text{mol}^{-2}\\cdot\\text{s}^{-1}$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "For a zero-order reaction, Rate = $k[A]^0 = k$. The unit of rate is concentration per unit time ($\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$). Thus, the unit of the rate constant $k$ is also $\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$.",
|
|
"language": "English",
|
|
"id": "90bd6f4a-954b-4e71-930e-e34a41bdc1f4"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 2)",
|
|
"subtopic": "Biomolecules",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human beings?",
|
|
"options": {
|
|
"A": "Thyroxin",
|
|
"B": "Insulin",
|
|
"C": "Adrenaline",
|
|
"D": "Estradiol"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Adrenaline (epinephrine) is the 'fight or flight' hormone released during stress. It promotes the breakdown of glycogen into glucose (glycogenolysis) in the liver to increase blood glucose levels for energy.",
|
|
"language": "English",
|
|
"id": "2e0d89c0-6321-45f4-9508-6ea46c503193"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Cell Structure and Function (Set 2)",
|
|
"subtopic": "Cell Cycle and Cell Division",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Crossing over occurs during which stage of meiosis?",
|
|
"options": {
|
|
"A": "Leptotene",
|
|
"B": "Pachytene",
|
|
"C": "Diplotene",
|
|
"D": "Zygotene"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Crossing over, the exchange of genetic material between non-sister chromatids of homologous chromosomes, occurs during the pachytene stage of prophase I of meiosis.",
|
|
"language": "English",
|
|
"id": "f1cb705f-1484-4481-9e51-9dea8f6267ba"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 2)",
|
|
"subtopic": "Principles of Inheritance and Variation",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?",
|
|
"options": {
|
|
"A": "TT and Tt",
|
|
"B": "Tt and Tt",
|
|
"C": "TT and TT",
|
|
"D": "Tt and tt"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Since both parents are tall but they produce some dwarf offspring, both parents must carry the recessive allele for dwarfness. Thus, their genotype must be heterozygous (Tt). A cross of Tt x Tt yields a 3:1 ratio of tall to dwarf.",
|
|
"language": "English",
|
|
"id": "1f046583-5f89-40bc-9391-3c1ab7cffd62"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 2)",
|
|
"subtopic": "Molecular Basis of Inheritance",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following RNAs should be most abundant in an animal cell?",
|
|
"options": {
|
|
"A": "tRNA",
|
|
"B": "mRNA",
|
|
"C": "miRNA",
|
|
"D": "rRNA"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Ribosomal RNA (rRNA) constitutes about 80% of the total RNA in an actively growing cell. It forms the structural and catalytic core of ribosomes.",
|
|
"language": "English",
|
|
"id": "3feb4d38-73a0-4d6b-90db-de47a0d690f7"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 2)",
|
|
"subtopic": "Digestion and Absorption",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following terms describes human dentition?",
|
|
"options": {
|
|
"A": "Thecodont, Diphyodont, Homodont",
|
|
"B": "Thecodont, Diphyodont, Heterodont",
|
|
"C": "Pleurodont, Monophyodont, Homodont",
|
|
"D": "Pleurodont, Diphyodont, Heterodont"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Human dentition is Thecodont (teeth embedded in jaw sockets), Diphyodont (two sets of teeth: deciduous and permanent), and Heterodont (different types of teeth: incisors, canines, premolars, molars).",
|
|
"language": "English",
|
|
"id": "84049894-bf1b-4b34-bf54-188d81ddb524"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Reproduction (Set 2)",
|
|
"subtopic": "Human Reproduction",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which hormone is responsible for the ovulation in human females?",
|
|
"options": {
|
|
"A": "FSH (Follicle Stimulating Hormone)",
|
|
"B": "Progesterone",
|
|
"C": "LH (Luteinizing Hormone)",
|
|
"D": "Estrogen"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "A surge in LH (Luteinizing Hormone) mid-cycle triggers the rupture of the Graafian follicle and the subsequent release of the ovum, a process known as ovulation.",
|
|
"language": "English",
|
|
"id": "decff3ac-b594-4fe4-b876-408926384c47"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Plant Physiology (Set 2)",
|
|
"subtopic": "Photosynthesis in Higher Plants",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The first stable product of CO2 fixation in C3 plants is:",
|
|
"options": {
|
|
"A": "Phosphoglyceraldehyde (PGAL)",
|
|
"B": "Malic acid",
|
|
"C": "Oxaloacetic acid (OAA)",
|
|
"D": "3-Phosphoglyceric acid (3-PGA)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In the Calvin cycle of C3 plants, CO2 combines with RuBP to form a 6-carbon unstable intermediate that immediately splits into two molecules of 3-phosphoglyceric acid (3-PGA), the first stable product.",
|
|
"language": "English",
|
|
"id": "ce615ee2-8ae5-41fe-9cb8-df2d0974fa70"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Ecology (Set 2)",
|
|
"subtopic": "Ecosystem",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The pyramid of energy is always:",
|
|
"options": {
|
|
"A": "Upright",
|
|
"B": "Inverted",
|
|
"C": "Spindle shaped",
|
|
"D": "Variable"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "The pyramid of energy is always upright because energy is lost as heat at each trophic level (10% law), so energy always decreases as it flows from producers to higher trophic levels.",
|
|
"language": "English",
|
|
"id": "b8931a19-6f8b-4bc0-8ef3-19c7bd151780"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Biotechnology (Set 2)",
|
|
"subtopic": "Principles and Processes",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The cutting of DNA at specific locations became possible with the discovery of:",
|
|
"options": {
|
|
"A": "Selectable markers",
|
|
"B": "Ligases",
|
|
"C": "Restriction enzymes",
|
|
"D": "Probes"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Restriction endonucleases (or restriction enzymes) cut DNA at specific recognition sequences, making recombinant DNA technology possible.",
|
|
"language": "English",
|
|
"id": "a26ad786-743a-4669-82cb-adcc472ae82a"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 2)",
|
|
"subtopic": "Neural Control and Coordination",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Receptor sites for neurotransmitters are present on:",
|
|
"options": {
|
|
"A": "Pre-synaptic membrane",
|
|
"B": "Tips of axons",
|
|
"C": "Post-synaptic membrane",
|
|
"D": "Membrane of synaptic vesicles"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Neurotransmitters are released from synaptic vesicles in the pre-synaptic neuron and diffuse across the synaptic cleft to bind with specific receptor sites located on the post-synaptic membrane.",
|
|
"language": "English",
|
|
"id": "cf1d3d36-d823-4045-8574-4670011b4da1"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Diversity in Living World (Set 2)",
|
|
"subtopic": "Biological Classification",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following are found in extreme saline conditions?",
|
|
"options": {
|
|
"A": "Eubacteria",
|
|
"B": "Cyanobacteria",
|
|
"C": "Mycobacteria",
|
|
"D": "Archaebacteria"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Archaebacteria are known to live in some of the most harsh habitats. Specifically, halophiles live in extreme saline conditions (e.g., Dead Sea).",
|
|
"language": "English",
|
|
"id": "acb7d418-b090-4fa9-9dbe-d11d80dc769d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Units and Measurements (Set 3)",
|
|
"subtopic": "Dimensions",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of:",
|
|
"options": {
|
|
"A": "Velocity",
|
|
"B": "Time",
|
|
"C": "Capacitance",
|
|
"D": "Distance"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "We know that $c = \\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$, where $c$ is the speed of light in vacuum. Therefore, the dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of velocity, which is $[L T^{-1}]$.",
|
|
"language": "English",
|
|
"id": "4f6acacf-edec-4889-afa0-fccbb163fd14"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Kinematics (Set 3)",
|
|
"subtopic": "Projectile Motion",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Two projectiles are thrown with the same initial velocity at angles of $30^\\circ$ and $60^\\circ$ with the horizontal. The ratio of their maximum heights is:",
|
|
"options": {
|
|
"A": "1:3",
|
|
"B": "3:1",
|
|
"C": "1:\\sqrt{3}",
|
|
"D": "1:2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Maximum height $H = \\frac{u^2 \\sin^2 \\theta}{2g}$.\nFor $\\theta_1 = 30^\\circ$, $H_1 = \\frac{u^2 (1/2)^2}{2g} = \\frac{u^2}{8g}$.\nFor $\\theta_2 = 60^\\circ$, $H_2 = \\frac{u^2 (\\sqrt{3}/2)^2}{2g} = \\frac{3u^2}{8g}$.\nRatio $H_1 : H_2 = 1 : 3$.",
|
|
"language": "English",
|
|
"id": "b99d4c8f-591b-4f11-9109-852f1f14be6a"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Thermodynamics (Set 3)",
|
|
"subtopic": "Carnot Engine",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A Carnot engine takes heat from a reservoir at $627^\\circ\\text{C}$ and rejects heat to a sink at $27^\\circ\\text{C}$. Its efficiency is:",
|
|
"options": {
|
|
"A": "2/3",
|
|
"B": "1/3",
|
|
"C": "3/4",
|
|
"D": "1/2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Temperature of source, $T_1 = 627 + 273 = 900\\text{ K}$.\nTemperature of sink, $T_2 = 27 + 273 = 300\\text{ K}$.\nEfficiency $\\eta = 1 - \\frac{T_2}{T_1} = 1 - \\frac{300}{900} = 1 - \\frac{1}{3} = \\frac{2}{3}$.",
|
|
"language": "English",
|
|
"id": "88ae576d-44d1-42bc-aeac-295d0c841a97"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Electrostatics (Set 3)",
|
|
"subtopic": "Capacitors",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A parallel plate capacitor with air between the plates has a capacitance of $8\\text{ pF}$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?",
|
|
"options": {
|
|
"A": "96 pF",
|
|
"B": "8 pF",
|
|
"C": "48 pF",
|
|
"D": "12 pF"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initial capacitance $C_0 = \\frac{\\varepsilon_0 A}{d} = 8\\text{ pF}$.\nNew distance $d' = d/2$ and new dielectric constant $K = 6$.\nNew capacitance $C' = \\frac{K \\varepsilon_0 A}{d'} = \\frac{6 \\varepsilon_0 A}{d/2} = 12 \\left(\\frac{\\varepsilon_0 A}{d}\\right) = 12 \\times 8\\text{ pF} = 96\\text{ pF}$.",
|
|
"language": "English",
|
|
"id": "823af790-369b-4ad6-ba99-6cfc54295d63"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Current Electricity (Set 3)",
|
|
"subtopic": "Meter Bridge",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a meter bridge experiment, the null point is obtained at $20\\text{ cm}$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then what will be the new position of the null point from the same end, if one decides to balance a resistance of $4X$ against $Y$?",
|
|
"options": {
|
|
"A": "50 cm",
|
|
"B": "40 cm",
|
|
"C": "80 cm",
|
|
"D": "20 cm"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initially, $\\frac{X}{Y} = \\frac{20}{100-20} = \\frac{20}{80} = \\frac{1}{4}$. So $Y = 4X$.\nFor new resistance $4X$ against $Y$, ratio is $\\frac{4X}{Y} = \\frac{4X}{4X} = 1$.\nLet new null point be $l$, then $\\frac{l}{100-l} = 1 \\implies l = 100 - l \\implies 2l = 100 \\implies l = 50\\text{ cm}$.",
|
|
"language": "English",
|
|
"id": "c067d0b8-b2f4-4d86-add5-b4f8a711235c"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Magnetism (Set 3)",
|
|
"subtopic": "Biot-Savart Law",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The magnetic field at the centre of a circular current carrying coil of radius $R$ is $B_0$. If the radius is doubled keeping the current same, the new magnetic field at the centre will be:",
|
|
"options": {
|
|
"A": "$B_0/2$",
|
|
"B": "$B_0/4$",
|
|
"C": "$2B_0$",
|
|
"D": "$B_0$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Magnetic field at the centre of a circular coil is $B = \\frac{\\mu_0 I}{2R}$.\nSince $B \\propto \\frac{1}{R}$, doubling the radius will make the magnetic field half. Thus, new field is $B_0/2$.",
|
|
"language": "English",
|
|
"id": "b4101faa-9c9d-4c51-815e-306614afbdb0"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Alternating Current (Set 3)",
|
|
"subtopic": "LCR Circuit",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a series LCR circuit, the voltage across the resistance, capacitance, and inductance is $10\\text{ V}$ each. If the capacitance is short-circuited, the voltage across the inductance will be:",
|
|
"options": {
|
|
"A": "10 V",
|
|
"B": "$10\\sqrt{2}\\text{ V}$",
|
|
"C": "$10/\\sqrt{2}\\text{ V}$",
|
|
"D": "Zero"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Initially, $V_R = 10\\text{ V}$, $V_L = 10\\text{ V}$, $V_C = 10\\text{ V}$. Since $V_L = V_C$, the circuit is in resonance. The applied voltage $V = \\sqrt{V_R^2 + (V_L - V_C)^2} = V_R = 10\\text{ V}$.\nIf capacitance is shorted, it becomes an LR circuit. Impedance $Z = \\sqrt{R^2 + X_L^2}$. Since $V_R = V_L$ initially, $R = X_L$, so $Z = R\\sqrt{2}$.\nNew current $I' = \\frac{V}{Z} = \\frac{10}{R\\sqrt{2}}$.\nNew voltage across inductance $V_L' = I' X_L = \\frac{10}{R\\sqrt{2}} \\times R = \\frac{10}{\\sqrt{2}}\\text{ V}$.",
|
|
"language": "English",
|
|
"id": "7869286a-9de6-4a20-8589-602bf552ed8d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Optics (Set 3)",
|
|
"subtopic": "Wave Optics",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:",
|
|
"options": {
|
|
"A": "Half",
|
|
"B": "Four times",
|
|
"C": "One-fourth",
|
|
"D": "Double"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Fringe width $\\beta = \\frac{\\lambda D}{d}$.\nNew slit separation $d' = d/2$.\nNew screen distance $D' = 2D$.\nNew fringe width $\\beta' = \\frac{\\lambda (2D)}{(d/2)} = 4 \\frac{\\lambda D}{d} = 4\\beta$.",
|
|
"language": "English",
|
|
"id": "b53b002e-9b01-41e6-b43c-105894b69fd9"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Modern Physics (Set 3)",
|
|
"subtopic": "Photoelectric Effect",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "When the light of frequency $2\\nu_0$ (where $\\nu_0$ is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5\\nu_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is:",
|
|
"options": {
|
|
"A": "1:2",
|
|
"B": "1:4",
|
|
"C": "4:1",
|
|
"D": "2:1"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "From Einstein's photoelectric equation: $K_{\\max} = h\\nu - h\\nu_0 = \\frac{1}{2} m v^2$.\nCase 1: $\\frac{1}{2} m v_1^2 = h(2\\nu_0) - h\\nu_0 = h\\nu_0$.\nCase 2: $\\frac{1}{2} m v_2^2 = h(5\\nu_0) - h\\nu_0 = 4h\\nu_0$.\nDividing the two equations: $\\frac{v_1^2}{v_2^2} = \\frac{1}{4} \\implies \\frac{v_1}{v_2} = \\frac{1}{2}$.",
|
|
"language": "English",
|
|
"id": "bc8b21fa-04f2-4fd6-901e-303b2f1a8c60"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Semiconductors (Set 3)",
|
|
"subtopic": "Logic Gates",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which logic gate is represented by the boolean expression $Y = \\overline{A \\cdot B}$?",
|
|
"options": {
|
|
"A": "NOR",
|
|
"B": "OR",
|
|
"C": "NAND",
|
|
"D": "AND"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "The boolean expression $Y = \\overline{A \\cdot B}$ represents the NAND gate. It gives an output of 0 only when both inputs A and B are 1, and 1 otherwise.",
|
|
"language": "English",
|
|
"id": "90148afe-d488-4e11-b3b1-38fd0733f024"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 3)",
|
|
"subtopic": "Mole Concept",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The number of water molecules is maximum in:",
|
|
"options": {
|
|
"A": "18 grams of water",
|
|
"B": "18 moles of water",
|
|
"C": "18 molecules of water",
|
|
"D": "1.8 grams of water"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "18 grams of water = 1 mole = $N_A$ molecules.\n18 moles of water = $18 N_A$ molecules.\n1.8 grams of water = 0.1 mole = $0.1 N_A$ molecules.\nThus, 18 moles of water contains the maximum number of molecules.",
|
|
"language": "English",
|
|
"id": "8e30b99f-bc31-4edc-bfe7-e1ff8d1d8062"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 3)",
|
|
"subtopic": "Atomic Structure",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The maximum number of electrons in an atom that can have the quantum numbers $n = 4$ and $m_l = +1$ is:",
|
|
"options": {
|
|
"A": "2",
|
|
"B": "4",
|
|
"C": "6",
|
|
"D": "8"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "For $n = 4$, the possible values of $l$ are 0, 1, 2, 3.\nThe condition $m_l = +1$ is possible for $l = 1$, $l = 2$, and $l = 3$.\nEach subshell ($l=1, 2, 3$) will have one orbital with $m_l = +1$.\nSince each orbital can hold 2 electrons, the total number of electrons is $3 \\times 2 = 6$.",
|
|
"language": "English",
|
|
"id": "b956e495-8e61-4cd2-9b90-d36a3a235946"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 3)",
|
|
"subtopic": "Chemical Bonding",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The species having pyramidal shape is:",
|
|
"options": {
|
|
"A": "$\\text{SO}_3$",
|
|
"B": "$\\text{BrF}_3$",
|
|
"C": "$\\text{SiO}_3^{2-}$",
|
|
"D": "$\\text{OSF}_2$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In $\\text{OSF}_2$, the central sulfur atom has 4 electron domains (1 double bond to O, 2 single bonds to F, and 1 lone pair). This corresponds to an $sp^3$ hybridization with a trigonal pyramidal shape. $\\text{SO}_3$ is trigonal planar, $\\text{BrF}_3$ is T-shaped.",
|
|
"language": "English",
|
|
"id": "d629e2eb-96f3-4c6b-9a8d-097965bd330c"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 3)",
|
|
"subtopic": "Equilibrium",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "For the reversible reaction $\\text{N}_2(\\text{g}) + 3\\text{H}_2(\\text{g}) \\rightleftharpoons 2\\text{NH}_3(\\text{g}) + \\text{Heat}$, the equilibrium shifts in forward direction:",
|
|
"options": {
|
|
"A": "By increasing the concentration of $\\text{NH}_3(\\text{g})$",
|
|
"B": "By decreasing the pressure",
|
|
"C": "By decreasing the concentrations of $\\text{N}_2(\\text{g})$ and $\\text{H}_2(\\text{g})$",
|
|
"D": "By increasing pressure and decreasing temperature"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "According to Le Chatelier's principle, an exothermic reaction ($\\Delta H < 0$) is favored by low temperatures. A reaction involving a decrease in the number of gaseous moles ($4 \\rightarrow 2$) is favored by high pressures. Therefore, increasing pressure and decreasing temperature will shift the equilibrium in the forward direction.",
|
|
"language": "English",
|
|
"id": "0b78c170-738d-4d7b-b02d-cac8c9dc2191"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 3)",
|
|
"subtopic": "Electrochemistry",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Given $E^\\circ_{\\text{Cr}^{3+}/\\text{Cr}} = -0.74\\text{ V}$; $E^\\circ_{\\text{MnO}_4^-/\\text{Mn}^{2+}} = 1.51\\text{ V}$; $E^\\circ_{\\text{Cr}_2\\text{O}_7^{2-}/\\text{Cr}^{3+}} = 1.33\\text{ V}$; $E^\\circ_{\\text{Cl}/\\text{Cl}^-} = 1.36\\text{ V}$.\nBased on the data given above, strongest oxidizing agent will be:",
|
|
"options": {
|
|
"A": "$\\text{Cl}$",
|
|
"B": "$\\text{Cr}^{3+}$",
|
|
"C": "$\\text{Mn}^{2+}$",
|
|
"D": "$\\text{MnO}_4^-$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Higher the standard reduction potential ($E^\\circ$), stronger is the oxidizing power. The $E^\\circ$ value for $\\text{MnO}_4^-/\\text{Mn}^{2+}$ is the highest ($+1.51\\text{ V}$). Therefore, $\\text{MnO}_4^-$ is the strongest oxidizing agent among the given species.",
|
|
"language": "English",
|
|
"id": "4e54fdf6-c2c0-4cf0-9181-112b46da2bd7"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 3)",
|
|
"subtopic": "General Organic Chemistry",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following carbocations is most stable?",
|
|
"options": {
|
|
"A": "Allyl carbocation",
|
|
"B": "Primary alkyl carbocation",
|
|
"C": "Tert-butyl carbocation",
|
|
"D": "Triphenylmethyl carbocation"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Stability of carbocations is determined by delocalization of charge. The triphenylmethyl carbocation, $(\\text{C}_6\\text{H}_5)_3\\text{C}^+$, has 3 phenyl rings providing extensive resonance stabilization, making it exceptionally stable compared to the others.",
|
|
"language": "English",
|
|
"id": "61cf97ef-68f5-4aae-9916-a6f28bf2777f"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 3)",
|
|
"subtopic": "Hydrocarbons",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:",
|
|
"options": {
|
|
"A": "Two isomeric double bonds",
|
|
"B": "A vinyl group",
|
|
"C": "An isopropyl group",
|
|
"D": "An acetylenic triple bond"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Ozonolysis cleaves a carbon-carbon double bond and adds oxygen. Formaldehyde ($\\text{HCHO}$) is produced if the original double bond was at the end of a chain, specifically involving a terminal $=\\text{CH}_2$ group, which is characteristic of a vinyl group ($-\\text{CH}=\\text{CH}_2$).",
|
|
"language": "English",
|
|
"id": "fe32e132-b359-4db0-a078-84942d8738cc"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 3)",
|
|
"subtopic": "Coordination Compounds",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The IUPAC name of $[\\text{Pt}(\\text{NH}_3)_2\\text{Cl}(\\text{NO}_2)]$ is:",
|
|
"options": {
|
|
"A": "Platinum diaminechloronitrite",
|
|
"B": "Chloronitrito-N-ammineplatinum(II)",
|
|
"C": "Diamminechloridonitrito-N-platinum(II)",
|
|
"D": "Diamminechloronitrito-O-platinate(II)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Ligands are named in alphabetical order: ammine, chlorido, nitrito-N (since it coordinates through nitrogen). The oxidation state of Pt is $x + 0 + (-1) + (-1) = 0 \\implies x = +2$. Thus, the correct name is Diamminechloridonitrito-N-platinum(II).",
|
|
"language": "English",
|
|
"id": "d7a0f59f-0e20-450f-96af-e46e066bff59"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 3)",
|
|
"subtopic": "Chemical Kinetics",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The unit of rate constant for a zero order reaction is:",
|
|
"options": {
|
|
"A": "$\\text{s}^{-1}$",
|
|
"B": "$\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"C": "$\\text{L}\\cdot\\text{mol}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"D": "$\\text{L}^2\\cdot\\text{mol}^{-2}\\cdot\\text{s}^{-1}$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "For a zero-order reaction, Rate = $k[A]^0 = k$. The unit of rate is concentration per unit time ($\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$). Thus, the unit of the rate constant $k$ is also $\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$.",
|
|
"language": "English",
|
|
"id": "07db4654-f99d-4d08-ba02-fd29d7848f61"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 3)",
|
|
"subtopic": "Biomolecules",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human beings?",
|
|
"options": {
|
|
"A": "Thyroxin",
|
|
"B": "Insulin",
|
|
"C": "Adrenaline",
|
|
"D": "Estradiol"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Adrenaline (epinephrine) is the 'fight or flight' hormone released during stress. It promotes the breakdown of glycogen into glucose (glycogenolysis) in the liver to increase blood glucose levels for energy.",
|
|
"language": "English",
|
|
"id": "aaf5a7d6-8337-4efd-920b-400820bcfa07"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Cell Structure and Function (Set 3)",
|
|
"subtopic": "Cell Cycle and Cell Division",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Crossing over occurs during which stage of meiosis?",
|
|
"options": {
|
|
"A": "Leptotene",
|
|
"B": "Pachytene",
|
|
"C": "Diplotene",
|
|
"D": "Zygotene"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Crossing over, the exchange of genetic material between non-sister chromatids of homologous chromosomes, occurs during the pachytene stage of prophase I of meiosis.",
|
|
"language": "English",
|
|
"id": "b3899a79-4864-4482-a1c1-7d049124a643"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 3)",
|
|
"subtopic": "Principles of Inheritance and Variation",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?",
|
|
"options": {
|
|
"A": "TT and Tt",
|
|
"B": "Tt and Tt",
|
|
"C": "TT and TT",
|
|
"D": "Tt and tt"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Since both parents are tall but they produce some dwarf offspring, both parents must carry the recessive allele for dwarfness. Thus, their genotype must be heterozygous (Tt). A cross of Tt x Tt yields a 3:1 ratio of tall to dwarf.",
|
|
"language": "English",
|
|
"id": "94f205a1-1e11-4253-8ee6-77b3b61e392c"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 3)",
|
|
"subtopic": "Molecular Basis of Inheritance",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following RNAs should be most abundant in an animal cell?",
|
|
"options": {
|
|
"A": "tRNA",
|
|
"B": "mRNA",
|
|
"C": "miRNA",
|
|
"D": "rRNA"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Ribosomal RNA (rRNA) constitutes about 80% of the total RNA in an actively growing cell. It forms the structural and catalytic core of ribosomes.",
|
|
"language": "English",
|
|
"id": "d5114d8a-9d77-4d3f-b2bb-36d82ad3bd26"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 3)",
|
|
"subtopic": "Digestion and Absorption",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following terms describes human dentition?",
|
|
"options": {
|
|
"A": "Thecodont, Diphyodont, Homodont",
|
|
"B": "Thecodont, Diphyodont, Heterodont",
|
|
"C": "Pleurodont, Monophyodont, Homodont",
|
|
"D": "Pleurodont, Diphyodont, Heterodont"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Human dentition is Thecodont (teeth embedded in jaw sockets), Diphyodont (two sets of teeth: deciduous and permanent), and Heterodont (different types of teeth: incisors, canines, premolars, molars).",
|
|
"language": "English",
|
|
"id": "50c3586b-339f-4430-8f73-63e1b6780f14"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Reproduction (Set 3)",
|
|
"subtopic": "Human Reproduction",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which hormone is responsible for the ovulation in human females?",
|
|
"options": {
|
|
"A": "FSH (Follicle Stimulating Hormone)",
|
|
"B": "Progesterone",
|
|
"C": "LH (Luteinizing Hormone)",
|
|
"D": "Estrogen"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "A surge in LH (Luteinizing Hormone) mid-cycle triggers the rupture of the Graafian follicle and the subsequent release of the ovum, a process known as ovulation.",
|
|
"language": "English",
|
|
"id": "d1a663d5-10dd-4885-9655-71060103c3bd"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Plant Physiology (Set 3)",
|
|
"subtopic": "Photosynthesis in Higher Plants",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The first stable product of CO2 fixation in C3 plants is:",
|
|
"options": {
|
|
"A": "Phosphoglyceraldehyde (PGAL)",
|
|
"B": "Malic acid",
|
|
"C": "Oxaloacetic acid (OAA)",
|
|
"D": "3-Phosphoglyceric acid (3-PGA)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In the Calvin cycle of C3 plants, CO2 combines with RuBP to form a 6-carbon unstable intermediate that immediately splits into two molecules of 3-phosphoglyceric acid (3-PGA), the first stable product.",
|
|
"language": "English",
|
|
"id": "1abfefc9-f273-4cd4-add4-df7ad5ec1c3e"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Ecology (Set 3)",
|
|
"subtopic": "Ecosystem",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The pyramid of energy is always:",
|
|
"options": {
|
|
"A": "Upright",
|
|
"B": "Inverted",
|
|
"C": "Spindle shaped",
|
|
"D": "Variable"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "The pyramid of energy is always upright because energy is lost as heat at each trophic level (10% law), so energy always decreases as it flows from producers to higher trophic levels.",
|
|
"language": "English",
|
|
"id": "2b9d9743-53fc-46d6-a9f6-55d4fcab0d4b"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Biotechnology (Set 3)",
|
|
"subtopic": "Principles and Processes",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The cutting of DNA at specific locations became possible with the discovery of:",
|
|
"options": {
|
|
"A": "Selectable markers",
|
|
"B": "Ligases",
|
|
"C": "Restriction enzymes",
|
|
"D": "Probes"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Restriction endonucleases (or restriction enzymes) cut DNA at specific recognition sequences, making recombinant DNA technology possible.",
|
|
"language": "English",
|
|
"id": "03ea2344-2db9-4eb9-a231-9a9fb00d2d0c"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 3)",
|
|
"subtopic": "Neural Control and Coordination",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Receptor sites for neurotransmitters are present on:",
|
|
"options": {
|
|
"A": "Pre-synaptic membrane",
|
|
"B": "Tips of axons",
|
|
"C": "Post-synaptic membrane",
|
|
"D": "Membrane of synaptic vesicles"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Neurotransmitters are released from synaptic vesicles in the pre-synaptic neuron and diffuse across the synaptic cleft to bind with specific receptor sites located on the post-synaptic membrane.",
|
|
"language": "English",
|
|
"id": "ade2064c-c01e-4257-89cf-57f3db564e05"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Diversity in Living World (Set 3)",
|
|
"subtopic": "Biological Classification",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following are found in extreme saline conditions?",
|
|
"options": {
|
|
"A": "Eubacteria",
|
|
"B": "Cyanobacteria",
|
|
"C": "Mycobacteria",
|
|
"D": "Archaebacteria"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Archaebacteria are known to live in some of the most harsh habitats. Specifically, halophiles live in extreme saline conditions (e.g., Dead Sea).",
|
|
"language": "English",
|
|
"id": "71dd9778-d209-4c67-aa11-e8da44d9cc0d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Units and Measurements (Set 4)",
|
|
"subtopic": "Dimensions",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of:",
|
|
"options": {
|
|
"A": "Velocity",
|
|
"B": "Time",
|
|
"C": "Capacitance",
|
|
"D": "Distance"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "We know that $c = \\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$, where $c$ is the speed of light in vacuum. Therefore, the dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of velocity, which is $[L T^{-1}]$.",
|
|
"language": "English",
|
|
"id": "125dc892-386f-4774-93de-1577084cea04"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Kinematics (Set 4)",
|
|
"subtopic": "Projectile Motion",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Two projectiles are thrown with the same initial velocity at angles of $30^\\circ$ and $60^\\circ$ with the horizontal. The ratio of their maximum heights is:",
|
|
"options": {
|
|
"A": "1:3",
|
|
"B": "3:1",
|
|
"C": "1:\\sqrt{3}",
|
|
"D": "1:2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Maximum height $H = \\frac{u^2 \\sin^2 \\theta}{2g}$.\nFor $\\theta_1 = 30^\\circ$, $H_1 = \\frac{u^2 (1/2)^2}{2g} = \\frac{u^2}{8g}$.\nFor $\\theta_2 = 60^\\circ$, $H_2 = \\frac{u^2 (\\sqrt{3}/2)^2}{2g} = \\frac{3u^2}{8g}$.\nRatio $H_1 : H_2 = 1 : 3$.",
|
|
"language": "English",
|
|
"id": "00f4623c-1416-4d3a-b58a-a5222d810e54"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Thermodynamics (Set 4)",
|
|
"subtopic": "Carnot Engine",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A Carnot engine takes heat from a reservoir at $627^\\circ\\text{C}$ and rejects heat to a sink at $27^\\circ\\text{C}$. Its efficiency is:",
|
|
"options": {
|
|
"A": "2/3",
|
|
"B": "1/3",
|
|
"C": "3/4",
|
|
"D": "1/2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Temperature of source, $T_1 = 627 + 273 = 900\\text{ K}$.\nTemperature of sink, $T_2 = 27 + 273 = 300\\text{ K}$.\nEfficiency $\\eta = 1 - \\frac{T_2}{T_1} = 1 - \\frac{300}{900} = 1 - \\frac{1}{3} = \\frac{2}{3}$.",
|
|
"language": "English",
|
|
"id": "69d7036d-c152-41f0-9edb-a008d62a5b0f"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Electrostatics (Set 4)",
|
|
"subtopic": "Capacitors",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A parallel plate capacitor with air between the plates has a capacitance of $8\\text{ pF}$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?",
|
|
"options": {
|
|
"A": "96 pF",
|
|
"B": "8 pF",
|
|
"C": "48 pF",
|
|
"D": "12 pF"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initial capacitance $C_0 = \\frac{\\varepsilon_0 A}{d} = 8\\text{ pF}$.\nNew distance $d' = d/2$ and new dielectric constant $K = 6$.\nNew capacitance $C' = \\frac{K \\varepsilon_0 A}{d'} = \\frac{6 \\varepsilon_0 A}{d/2} = 12 \\left(\\frac{\\varepsilon_0 A}{d}\\right) = 12 \\times 8\\text{ pF} = 96\\text{ pF}$.",
|
|
"language": "English",
|
|
"id": "8461826b-c62a-4494-9bb0-681d1bbfe967"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Current Electricity (Set 4)",
|
|
"subtopic": "Meter Bridge",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a meter bridge experiment, the null point is obtained at $20\\text{ cm}$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then what will be the new position of the null point from the same end, if one decides to balance a resistance of $4X$ against $Y$?",
|
|
"options": {
|
|
"A": "50 cm",
|
|
"B": "40 cm",
|
|
"C": "80 cm",
|
|
"D": "20 cm"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initially, $\\frac{X}{Y} = \\frac{20}{100-20} = \\frac{20}{80} = \\frac{1}{4}$. So $Y = 4X$.\nFor new resistance $4X$ against $Y$, ratio is $\\frac{4X}{Y} = \\frac{4X}{4X} = 1$.\nLet new null point be $l$, then $\\frac{l}{100-l} = 1 \\implies l = 100 - l \\implies 2l = 100 \\implies l = 50\\text{ cm}$.",
|
|
"language": "English",
|
|
"id": "e16f6881-5dd0-42c2-b91d-59bfa3a8b307"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Magnetism (Set 4)",
|
|
"subtopic": "Biot-Savart Law",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The magnetic field at the centre of a circular current carrying coil of radius $R$ is $B_0$. If the radius is doubled keeping the current same, the new magnetic field at the centre will be:",
|
|
"options": {
|
|
"A": "$B_0/2$",
|
|
"B": "$B_0/4$",
|
|
"C": "$2B_0$",
|
|
"D": "$B_0$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Magnetic field at the centre of a circular coil is $B = \\frac{\\mu_0 I}{2R}$.\nSince $B \\propto \\frac{1}{R}$, doubling the radius will make the magnetic field half. Thus, new field is $B_0/2$.",
|
|
"language": "English",
|
|
"id": "f4a6d234-002b-41ad-a4f2-69fb2fd1cd7d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Alternating Current (Set 4)",
|
|
"subtopic": "LCR Circuit",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a series LCR circuit, the voltage across the resistance, capacitance, and inductance is $10\\text{ V}$ each. If the capacitance is short-circuited, the voltage across the inductance will be:",
|
|
"options": {
|
|
"A": "10 V",
|
|
"B": "$10\\sqrt{2}\\text{ V}$",
|
|
"C": "$10/\\sqrt{2}\\text{ V}$",
|
|
"D": "Zero"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Initially, $V_R = 10\\text{ V}$, $V_L = 10\\text{ V}$, $V_C = 10\\text{ V}$. Since $V_L = V_C$, the circuit is in resonance. The applied voltage $V = \\sqrt{V_R^2 + (V_L - V_C)^2} = V_R = 10\\text{ V}$.\nIf capacitance is shorted, it becomes an LR circuit. Impedance $Z = \\sqrt{R^2 + X_L^2}$. Since $V_R = V_L$ initially, $R = X_L$, so $Z = R\\sqrt{2}$.\nNew current $I' = \\frac{V}{Z} = \\frac{10}{R\\sqrt{2}}$.\nNew voltage across inductance $V_L' = I' X_L = \\frac{10}{R\\sqrt{2}} \\times R = \\frac{10}{\\sqrt{2}}\\text{ V}$.",
|
|
"language": "English",
|
|
"id": "f6f9fedc-a555-41f5-b24d-09ff0a4e3af9"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Optics (Set 4)",
|
|
"subtopic": "Wave Optics",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:",
|
|
"options": {
|
|
"A": "Half",
|
|
"B": "Four times",
|
|
"C": "One-fourth",
|
|
"D": "Double"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Fringe width $\\beta = \\frac{\\lambda D}{d}$.\nNew slit separation $d' = d/2$.\nNew screen distance $D' = 2D$.\nNew fringe width $\\beta' = \\frac{\\lambda (2D)}{(d/2)} = 4 \\frac{\\lambda D}{d} = 4\\beta$.",
|
|
"language": "English",
|
|
"id": "47133d9d-a7f2-4b1a-8913-0920e21006a4"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Modern Physics (Set 4)",
|
|
"subtopic": "Photoelectric Effect",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "When the light of frequency $2\\nu_0$ (where $\\nu_0$ is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5\\nu_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is:",
|
|
"options": {
|
|
"A": "1:2",
|
|
"B": "1:4",
|
|
"C": "4:1",
|
|
"D": "2:1"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "From Einstein's photoelectric equation: $K_{\\max} = h\\nu - h\\nu_0 = \\frac{1}{2} m v^2$.\nCase 1: $\\frac{1}{2} m v_1^2 = h(2\\nu_0) - h\\nu_0 = h\\nu_0$.\nCase 2: $\\frac{1}{2} m v_2^2 = h(5\\nu_0) - h\\nu_0 = 4h\\nu_0$.\nDividing the two equations: $\\frac{v_1^2}{v_2^2} = \\frac{1}{4} \\implies \\frac{v_1}{v_2} = \\frac{1}{2}$.",
|
|
"language": "English",
|
|
"id": "c26de02a-bede-4dd8-8e08-2d0e8a9e1170"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Semiconductors (Set 4)",
|
|
"subtopic": "Logic Gates",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which logic gate is represented by the boolean expression $Y = \\overline{A \\cdot B}$?",
|
|
"options": {
|
|
"A": "NOR",
|
|
"B": "OR",
|
|
"C": "NAND",
|
|
"D": "AND"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "The boolean expression $Y = \\overline{A \\cdot B}$ represents the NAND gate. It gives an output of 0 only when both inputs A and B are 1, and 1 otherwise.",
|
|
"language": "English",
|
|
"id": "a0a61994-1783-4988-bba8-697192131e2a"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 4)",
|
|
"subtopic": "Mole Concept",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The number of water molecules is maximum in:",
|
|
"options": {
|
|
"A": "18 grams of water",
|
|
"B": "18 moles of water",
|
|
"C": "18 molecules of water",
|
|
"D": "1.8 grams of water"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "18 grams of water = 1 mole = $N_A$ molecules.\n18 moles of water = $18 N_A$ molecules.\n1.8 grams of water = 0.1 mole = $0.1 N_A$ molecules.\nThus, 18 moles of water contains the maximum number of molecules.",
|
|
"language": "English",
|
|
"id": "a485be71-df65-433d-b4b1-0e28e7425c9f"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 4)",
|
|
"subtopic": "Atomic Structure",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The maximum number of electrons in an atom that can have the quantum numbers $n = 4$ and $m_l = +1$ is:",
|
|
"options": {
|
|
"A": "2",
|
|
"B": "4",
|
|
"C": "6",
|
|
"D": "8"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "For $n = 4$, the possible values of $l$ are 0, 1, 2, 3.\nThe condition $m_l = +1$ is possible for $l = 1$, $l = 2$, and $l = 3$.\nEach subshell ($l=1, 2, 3$) will have one orbital with $m_l = +1$.\nSince each orbital can hold 2 electrons, the total number of electrons is $3 \\times 2 = 6$.",
|
|
"language": "English",
|
|
"id": "a45008df-c27b-4b28-8aa7-381aac5a0f05"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 4)",
|
|
"subtopic": "Chemical Bonding",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The species having pyramidal shape is:",
|
|
"options": {
|
|
"A": "$\\text{SO}_3$",
|
|
"B": "$\\text{BrF}_3$",
|
|
"C": "$\\text{SiO}_3^{2-}$",
|
|
"D": "$\\text{OSF}_2$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In $\\text{OSF}_2$, the central sulfur atom has 4 electron domains (1 double bond to O, 2 single bonds to F, and 1 lone pair). This corresponds to an $sp^3$ hybridization with a trigonal pyramidal shape. $\\text{SO}_3$ is trigonal planar, $\\text{BrF}_3$ is T-shaped.",
|
|
"language": "English",
|
|
"id": "12111a4a-c754-4c2f-8a09-e5e62832fa6a"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 4)",
|
|
"subtopic": "Equilibrium",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "For the reversible reaction $\\text{N}_2(\\text{g}) + 3\\text{H}_2(\\text{g}) \\rightleftharpoons 2\\text{NH}_3(\\text{g}) + \\text{Heat}$, the equilibrium shifts in forward direction:",
|
|
"options": {
|
|
"A": "By increasing the concentration of $\\text{NH}_3(\\text{g})$",
|
|
"B": "By decreasing the pressure",
|
|
"C": "By decreasing the concentrations of $\\text{N}_2(\\text{g})$ and $\\text{H}_2(\\text{g})$",
|
|
"D": "By increasing pressure and decreasing temperature"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "According to Le Chatelier's principle, an exothermic reaction ($\\Delta H < 0$) is favored by low temperatures. A reaction involving a decrease in the number of gaseous moles ($4 \\rightarrow 2$) is favored by high pressures. Therefore, increasing pressure and decreasing temperature will shift the equilibrium in the forward direction.",
|
|
"language": "English",
|
|
"id": "11a28113-cb35-4715-8fea-7025b97d98f6"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 4)",
|
|
"subtopic": "Electrochemistry",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Given $E^\\circ_{\\text{Cr}^{3+}/\\text{Cr}} = -0.74\\text{ V}$; $E^\\circ_{\\text{MnO}_4^-/\\text{Mn}^{2+}} = 1.51\\text{ V}$; $E^\\circ_{\\text{Cr}_2\\text{O}_7^{2-}/\\text{Cr}^{3+}} = 1.33\\text{ V}$; $E^\\circ_{\\text{Cl}/\\text{Cl}^-} = 1.36\\text{ V}$.\nBased on the data given above, strongest oxidizing agent will be:",
|
|
"options": {
|
|
"A": "$\\text{Cl}$",
|
|
"B": "$\\text{Cr}^{3+}$",
|
|
"C": "$\\text{Mn}^{2+}$",
|
|
"D": "$\\text{MnO}_4^-$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Higher the standard reduction potential ($E^\\circ$), stronger is the oxidizing power. The $E^\\circ$ value for $\\text{MnO}_4^-/\\text{Mn}^{2+}$ is the highest ($+1.51\\text{ V}$). Therefore, $\\text{MnO}_4^-$ is the strongest oxidizing agent among the given species.",
|
|
"language": "English",
|
|
"id": "bfa2a8aa-fa96-431a-896c-9b969280068e"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 4)",
|
|
"subtopic": "General Organic Chemistry",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following carbocations is most stable?",
|
|
"options": {
|
|
"A": "Allyl carbocation",
|
|
"B": "Primary alkyl carbocation",
|
|
"C": "Tert-butyl carbocation",
|
|
"D": "Triphenylmethyl carbocation"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Stability of carbocations is determined by delocalization of charge. The triphenylmethyl carbocation, $(\\text{C}_6\\text{H}_5)_3\\text{C}^+$, has 3 phenyl rings providing extensive resonance stabilization, making it exceptionally stable compared to the others.",
|
|
"language": "English",
|
|
"id": "c12d1985-e70b-4ba8-a8c9-967e001dbf46"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 4)",
|
|
"subtopic": "Hydrocarbons",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:",
|
|
"options": {
|
|
"A": "Two isomeric double bonds",
|
|
"B": "A vinyl group",
|
|
"C": "An isopropyl group",
|
|
"D": "An acetylenic triple bond"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Ozonolysis cleaves a carbon-carbon double bond and adds oxygen. Formaldehyde ($\\text{HCHO}$) is produced if the original double bond was at the end of a chain, specifically involving a terminal $=\\text{CH}_2$ group, which is characteristic of a vinyl group ($-\\text{CH}=\\text{CH}_2$).",
|
|
"language": "English",
|
|
"id": "b6759d7f-dff9-4a3a-86ee-d054b701aff8"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 4)",
|
|
"subtopic": "Coordination Compounds",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The IUPAC name of $[\\text{Pt}(\\text{NH}_3)_2\\text{Cl}(\\text{NO}_2)]$ is:",
|
|
"options": {
|
|
"A": "Platinum diaminechloronitrite",
|
|
"B": "Chloronitrito-N-ammineplatinum(II)",
|
|
"C": "Diamminechloridonitrito-N-platinum(II)",
|
|
"D": "Diamminechloronitrito-O-platinate(II)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Ligands are named in alphabetical order: ammine, chlorido, nitrito-N (since it coordinates through nitrogen). The oxidation state of Pt is $x + 0 + (-1) + (-1) = 0 \\implies x = +2$. Thus, the correct name is Diamminechloridonitrito-N-platinum(II).",
|
|
"language": "English",
|
|
"id": "7d335b70-4352-4ee5-b17b-281edbbabd30"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 4)",
|
|
"subtopic": "Chemical Kinetics",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The unit of rate constant for a zero order reaction is:",
|
|
"options": {
|
|
"A": "$\\text{s}^{-1}$",
|
|
"B": "$\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"C": "$\\text{L}\\cdot\\text{mol}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"D": "$\\text{L}^2\\cdot\\text{mol}^{-2}\\cdot\\text{s}^{-1}$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "For a zero-order reaction, Rate = $k[A]^0 = k$. The unit of rate is concentration per unit time ($\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$). Thus, the unit of the rate constant $k$ is also $\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$.",
|
|
"language": "English",
|
|
"id": "5ca190aa-4050-423c-93d8-a739a54d9dc6"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 4)",
|
|
"subtopic": "Biomolecules",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human beings?",
|
|
"options": {
|
|
"A": "Thyroxin",
|
|
"B": "Insulin",
|
|
"C": "Adrenaline",
|
|
"D": "Estradiol"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Adrenaline (epinephrine) is the 'fight or flight' hormone released during stress. It promotes the breakdown of glycogen into glucose (glycogenolysis) in the liver to increase blood glucose levels for energy.",
|
|
"language": "English",
|
|
"id": "0a0fc9fd-6a7e-4b8f-90c8-aeaf0bbc6619"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Cell Structure and Function (Set 4)",
|
|
"subtopic": "Cell Cycle and Cell Division",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Crossing over occurs during which stage of meiosis?",
|
|
"options": {
|
|
"A": "Leptotene",
|
|
"B": "Pachytene",
|
|
"C": "Diplotene",
|
|
"D": "Zygotene"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Crossing over, the exchange of genetic material between non-sister chromatids of homologous chromosomes, occurs during the pachytene stage of prophase I of meiosis.",
|
|
"language": "English",
|
|
"id": "746b509c-3d31-41e2-82a0-dbc633bfd99e"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 4)",
|
|
"subtopic": "Principles of Inheritance and Variation",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?",
|
|
"options": {
|
|
"A": "TT and Tt",
|
|
"B": "Tt and Tt",
|
|
"C": "TT and TT",
|
|
"D": "Tt and tt"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Since both parents are tall but they produce some dwarf offspring, both parents must carry the recessive allele for dwarfness. Thus, their genotype must be heterozygous (Tt). A cross of Tt x Tt yields a 3:1 ratio of tall to dwarf.",
|
|
"language": "English",
|
|
"id": "68389613-4924-4439-893d-cbcabfe50a4d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 4)",
|
|
"subtopic": "Molecular Basis of Inheritance",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following RNAs should be most abundant in an animal cell?",
|
|
"options": {
|
|
"A": "tRNA",
|
|
"B": "mRNA",
|
|
"C": "miRNA",
|
|
"D": "rRNA"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Ribosomal RNA (rRNA) constitutes about 80% of the total RNA in an actively growing cell. It forms the structural and catalytic core of ribosomes.",
|
|
"language": "English",
|
|
"id": "153c10b8-304e-457e-b753-4be0dfda153d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 4)",
|
|
"subtopic": "Digestion and Absorption",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following terms describes human dentition?",
|
|
"options": {
|
|
"A": "Thecodont, Diphyodont, Homodont",
|
|
"B": "Thecodont, Diphyodont, Heterodont",
|
|
"C": "Pleurodont, Monophyodont, Homodont",
|
|
"D": "Pleurodont, Diphyodont, Heterodont"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Human dentition is Thecodont (teeth embedded in jaw sockets), Diphyodont (two sets of teeth: deciduous and permanent), and Heterodont (different types of teeth: incisors, canines, premolars, molars).",
|
|
"language": "English",
|
|
"id": "e5a4e179-10c6-4a03-8483-9a7320a211ec"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Reproduction (Set 4)",
|
|
"subtopic": "Human Reproduction",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which hormone is responsible for the ovulation in human females?",
|
|
"options": {
|
|
"A": "FSH (Follicle Stimulating Hormone)",
|
|
"B": "Progesterone",
|
|
"C": "LH (Luteinizing Hormone)",
|
|
"D": "Estrogen"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "A surge in LH (Luteinizing Hormone) mid-cycle triggers the rupture of the Graafian follicle and the subsequent release of the ovum, a process known as ovulation.",
|
|
"language": "English",
|
|
"id": "039a4deb-fbd5-4b17-bbab-a48cc325a2fd"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Plant Physiology (Set 4)",
|
|
"subtopic": "Photosynthesis in Higher Plants",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The first stable product of CO2 fixation in C3 plants is:",
|
|
"options": {
|
|
"A": "Phosphoglyceraldehyde (PGAL)",
|
|
"B": "Malic acid",
|
|
"C": "Oxaloacetic acid (OAA)",
|
|
"D": "3-Phosphoglyceric acid (3-PGA)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In the Calvin cycle of C3 plants, CO2 combines with RuBP to form a 6-carbon unstable intermediate that immediately splits into two molecules of 3-phosphoglyceric acid (3-PGA), the first stable product.",
|
|
"language": "English",
|
|
"id": "b3dcd78a-0342-4036-9102-5b517706649b"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Ecology (Set 4)",
|
|
"subtopic": "Ecosystem",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The pyramid of energy is always:",
|
|
"options": {
|
|
"A": "Upright",
|
|
"B": "Inverted",
|
|
"C": "Spindle shaped",
|
|
"D": "Variable"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "The pyramid of energy is always upright because energy is lost as heat at each trophic level (10% law), so energy always decreases as it flows from producers to higher trophic levels.",
|
|
"language": "English",
|
|
"id": "ee00348e-d154-4bee-9fb6-1dd3875b05e0"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Biotechnology (Set 4)",
|
|
"subtopic": "Principles and Processes",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The cutting of DNA at specific locations became possible with the discovery of:",
|
|
"options": {
|
|
"A": "Selectable markers",
|
|
"B": "Ligases",
|
|
"C": "Restriction enzymes",
|
|
"D": "Probes"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Restriction endonucleases (or restriction enzymes) cut DNA at specific recognition sequences, making recombinant DNA technology possible.",
|
|
"language": "English",
|
|
"id": "c90a3d1d-a327-4090-a4a4-2478622c8c74"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 4)",
|
|
"subtopic": "Neural Control and Coordination",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Receptor sites for neurotransmitters are present on:",
|
|
"options": {
|
|
"A": "Pre-synaptic membrane",
|
|
"B": "Tips of axons",
|
|
"C": "Post-synaptic membrane",
|
|
"D": "Membrane of synaptic vesicles"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Neurotransmitters are released from synaptic vesicles in the pre-synaptic neuron and diffuse across the synaptic cleft to bind with specific receptor sites located on the post-synaptic membrane.",
|
|
"language": "English",
|
|
"id": "400b1d45-f375-47b6-aeb1-39ba6c53b56b"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Diversity in Living World (Set 4)",
|
|
"subtopic": "Biological Classification",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following are found in extreme saline conditions?",
|
|
"options": {
|
|
"A": "Eubacteria",
|
|
"B": "Cyanobacteria",
|
|
"C": "Mycobacteria",
|
|
"D": "Archaebacteria"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Archaebacteria are known to live in some of the most harsh habitats. Specifically, halophiles live in extreme saline conditions (e.g., Dead Sea).",
|
|
"language": "English",
|
|
"id": "cd4b967a-d552-49ed-ba86-bccdac81e439"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Units and Measurements (Set 5)",
|
|
"subtopic": "Dimensions",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of:",
|
|
"options": {
|
|
"A": "Velocity",
|
|
"B": "Time",
|
|
"C": "Capacitance",
|
|
"D": "Distance"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "We know that $c = \\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$, where $c$ is the speed of light in vacuum. Therefore, the dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of velocity, which is $[L T^{-1}]$.",
|
|
"language": "English",
|
|
"id": "b2f04447-2710-46bc-a2af-c7e918701550"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Kinematics (Set 5)",
|
|
"subtopic": "Projectile Motion",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Two projectiles are thrown with the same initial velocity at angles of $30^\\circ$ and $60^\\circ$ with the horizontal. The ratio of their maximum heights is:",
|
|
"options": {
|
|
"A": "1:3",
|
|
"B": "3:1",
|
|
"C": "1:\\sqrt{3}",
|
|
"D": "1:2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Maximum height $H = \\frac{u^2 \\sin^2 \\theta}{2g}$.\nFor $\\theta_1 = 30^\\circ$, $H_1 = \\frac{u^2 (1/2)^2}{2g} = \\frac{u^2}{8g}$.\nFor $\\theta_2 = 60^\\circ$, $H_2 = \\frac{u^2 (\\sqrt{3}/2)^2}{2g} = \\frac{3u^2}{8g}$.\nRatio $H_1 : H_2 = 1 : 3$.",
|
|
"language": "English",
|
|
"id": "2b66c506-0b95-413f-925c-0d5e83c4d9d8"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Thermodynamics (Set 5)",
|
|
"subtopic": "Carnot Engine",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A Carnot engine takes heat from a reservoir at $627^\\circ\\text{C}$ and rejects heat to a sink at $27^\\circ\\text{C}$. Its efficiency is:",
|
|
"options": {
|
|
"A": "2/3",
|
|
"B": "1/3",
|
|
"C": "3/4",
|
|
"D": "1/2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Temperature of source, $T_1 = 627 + 273 = 900\\text{ K}$.\nTemperature of sink, $T_2 = 27 + 273 = 300\\text{ K}$.\nEfficiency $\\eta = 1 - \\frac{T_2}{T_1} = 1 - \\frac{300}{900} = 1 - \\frac{1}{3} = \\frac{2}{3}$.",
|
|
"language": "English",
|
|
"id": "348fa2cc-1306-46da-9a1e-3723beed6dd7"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Electrostatics (Set 5)",
|
|
"subtopic": "Capacitors",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A parallel plate capacitor with air between the plates has a capacitance of $8\\text{ pF}$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?",
|
|
"options": {
|
|
"A": "96 pF",
|
|
"B": "8 pF",
|
|
"C": "48 pF",
|
|
"D": "12 pF"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initial capacitance $C_0 = \\frac{\\varepsilon_0 A}{d} = 8\\text{ pF}$.\nNew distance $d' = d/2$ and new dielectric constant $K = 6$.\nNew capacitance $C' = \\frac{K \\varepsilon_0 A}{d'} = \\frac{6 \\varepsilon_0 A}{d/2} = 12 \\left(\\frac{\\varepsilon_0 A}{d}\\right) = 12 \\times 8\\text{ pF} = 96\\text{ pF}$.",
|
|
"language": "English",
|
|
"id": "133d4adf-5cba-47bc-a919-8983d8bf6c99"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Current Electricity (Set 5)",
|
|
"subtopic": "Meter Bridge",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a meter bridge experiment, the null point is obtained at $20\\text{ cm}$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then what will be the new position of the null point from the same end, if one decides to balance a resistance of $4X$ against $Y$?",
|
|
"options": {
|
|
"A": "50 cm",
|
|
"B": "40 cm",
|
|
"C": "80 cm",
|
|
"D": "20 cm"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initially, $\\frac{X}{Y} = \\frac{20}{100-20} = \\frac{20}{80} = \\frac{1}{4}$. So $Y = 4X$.\nFor new resistance $4X$ against $Y$, ratio is $\\frac{4X}{Y} = \\frac{4X}{4X} = 1$.\nLet new null point be $l$, then $\\frac{l}{100-l} = 1 \\implies l = 100 - l \\implies 2l = 100 \\implies l = 50\\text{ cm}$.",
|
|
"language": "English",
|
|
"id": "fc788d01-fddc-4615-82d4-e0761143edf6"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Magnetism (Set 5)",
|
|
"subtopic": "Biot-Savart Law",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The magnetic field at the centre of a circular current carrying coil of radius $R$ is $B_0$. If the radius is doubled keeping the current same, the new magnetic field at the centre will be:",
|
|
"options": {
|
|
"A": "$B_0/2$",
|
|
"B": "$B_0/4$",
|
|
"C": "$2B_0$",
|
|
"D": "$B_0$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Magnetic field at the centre of a circular coil is $B = \\frac{\\mu_0 I}{2R}$.\nSince $B \\propto \\frac{1}{R}$, doubling the radius will make the magnetic field half. Thus, new field is $B_0/2$.",
|
|
"language": "English",
|
|
"id": "9be2d264-5f15-4205-acf6-971f7df348e3"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Alternating Current (Set 5)",
|
|
"subtopic": "LCR Circuit",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a series LCR circuit, the voltage across the resistance, capacitance, and inductance is $10\\text{ V}$ each. If the capacitance is short-circuited, the voltage across the inductance will be:",
|
|
"options": {
|
|
"A": "10 V",
|
|
"B": "$10\\sqrt{2}\\text{ V}$",
|
|
"C": "$10/\\sqrt{2}\\text{ V}$",
|
|
"D": "Zero"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Initially, $V_R = 10\\text{ V}$, $V_L = 10\\text{ V}$, $V_C = 10\\text{ V}$. Since $V_L = V_C$, the circuit is in resonance. The applied voltage $V = \\sqrt{V_R^2 + (V_L - V_C)^2} = V_R = 10\\text{ V}$.\nIf capacitance is shorted, it becomes an LR circuit. Impedance $Z = \\sqrt{R^2 + X_L^2}$. Since $V_R = V_L$ initially, $R = X_L$, so $Z = R\\sqrt{2}$.\nNew current $I' = \\frac{V}{Z} = \\frac{10}{R\\sqrt{2}}$.\nNew voltage across inductance $V_L' = I' X_L = \\frac{10}{R\\sqrt{2}} \\times R = \\frac{10}{\\sqrt{2}}\\text{ V}$.",
|
|
"language": "English",
|
|
"id": "6ed6639d-484a-4cd7-b1bb-c2b70dfc2f3a"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Optics (Set 5)",
|
|
"subtopic": "Wave Optics",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:",
|
|
"options": {
|
|
"A": "Half",
|
|
"B": "Four times",
|
|
"C": "One-fourth",
|
|
"D": "Double"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Fringe width $\\beta = \\frac{\\lambda D}{d}$.\nNew slit separation $d' = d/2$.\nNew screen distance $D' = 2D$.\nNew fringe width $\\beta' = \\frac{\\lambda (2D)}{(d/2)} = 4 \\frac{\\lambda D}{d} = 4\\beta$.",
|
|
"language": "English",
|
|
"id": "1005e9e4-0e0d-4ed7-a9d2-e7b57f011f0d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Modern Physics (Set 5)",
|
|
"subtopic": "Photoelectric Effect",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "When the light of frequency $2\\nu_0$ (where $\\nu_0$ is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5\\nu_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is:",
|
|
"options": {
|
|
"A": "1:2",
|
|
"B": "1:4",
|
|
"C": "4:1",
|
|
"D": "2:1"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "From Einstein's photoelectric equation: $K_{\\max} = h\\nu - h\\nu_0 = \\frac{1}{2} m v^2$.\nCase 1: $\\frac{1}{2} m v_1^2 = h(2\\nu_0) - h\\nu_0 = h\\nu_0$.\nCase 2: $\\frac{1}{2} m v_2^2 = h(5\\nu_0) - h\\nu_0 = 4h\\nu_0$.\nDividing the two equations: $\\frac{v_1^2}{v_2^2} = \\frac{1}{4} \\implies \\frac{v_1}{v_2} = \\frac{1}{2}$.",
|
|
"language": "English",
|
|
"id": "59778b7a-fe97-417b-ac19-739e04688ccf"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Semiconductors (Set 5)",
|
|
"subtopic": "Logic Gates",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which logic gate is represented by the boolean expression $Y = \\overline{A \\cdot B}$?",
|
|
"options": {
|
|
"A": "NOR",
|
|
"B": "OR",
|
|
"C": "NAND",
|
|
"D": "AND"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "The boolean expression $Y = \\overline{A \\cdot B}$ represents the NAND gate. It gives an output of 0 only when both inputs A and B are 1, and 1 otherwise.",
|
|
"language": "English",
|
|
"id": "adac2d1f-cc43-4634-a49b-8bb47e0139e7"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 5)",
|
|
"subtopic": "Mole Concept",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The number of water molecules is maximum in:",
|
|
"options": {
|
|
"A": "18 grams of water",
|
|
"B": "18 moles of water",
|
|
"C": "18 molecules of water",
|
|
"D": "1.8 grams of water"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "18 grams of water = 1 mole = $N_A$ molecules.\n18 moles of water = $18 N_A$ molecules.\n1.8 grams of water = 0.1 mole = $0.1 N_A$ molecules.\nThus, 18 moles of water contains the maximum number of molecules.",
|
|
"language": "English",
|
|
"id": "ff604fd9-3cd8-4622-b8e5-953e99100e6a"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 5)",
|
|
"subtopic": "Atomic Structure",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The maximum number of electrons in an atom that can have the quantum numbers $n = 4$ and $m_l = +1$ is:",
|
|
"options": {
|
|
"A": "2",
|
|
"B": "4",
|
|
"C": "6",
|
|
"D": "8"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "For $n = 4$, the possible values of $l$ are 0, 1, 2, 3.\nThe condition $m_l = +1$ is possible for $l = 1$, $l = 2$, and $l = 3$.\nEach subshell ($l=1, 2, 3$) will have one orbital with $m_l = +1$.\nSince each orbital can hold 2 electrons, the total number of electrons is $3 \\times 2 = 6$.",
|
|
"language": "English",
|
|
"id": "5d975bc3-87b3-4715-90f6-6f33cf559566"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 5)",
|
|
"subtopic": "Chemical Bonding",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The species having pyramidal shape is:",
|
|
"options": {
|
|
"A": "$\\text{SO}_3$",
|
|
"B": "$\\text{BrF}_3$",
|
|
"C": "$\\text{SiO}_3^{2-}$",
|
|
"D": "$\\text{OSF}_2$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In $\\text{OSF}_2$, the central sulfur atom has 4 electron domains (1 double bond to O, 2 single bonds to F, and 1 lone pair). This corresponds to an $sp^3$ hybridization with a trigonal pyramidal shape. $\\text{SO}_3$ is trigonal planar, $\\text{BrF}_3$ is T-shaped.",
|
|
"language": "English",
|
|
"id": "64a9af48-1868-42db-a1d4-b2186f81bb11"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 5)",
|
|
"subtopic": "Equilibrium",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "For the reversible reaction $\\text{N}_2(\\text{g}) + 3\\text{H}_2(\\text{g}) \\rightleftharpoons 2\\text{NH}_3(\\text{g}) + \\text{Heat}$, the equilibrium shifts in forward direction:",
|
|
"options": {
|
|
"A": "By increasing the concentration of $\\text{NH}_3(\\text{g})$",
|
|
"B": "By decreasing the pressure",
|
|
"C": "By decreasing the concentrations of $\\text{N}_2(\\text{g})$ and $\\text{H}_2(\\text{g})$",
|
|
"D": "By increasing pressure and decreasing temperature"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "According to Le Chatelier's principle, an exothermic reaction ($\\Delta H < 0$) is favored by low temperatures. A reaction involving a decrease in the number of gaseous moles ($4 \\rightarrow 2$) is favored by high pressures. Therefore, increasing pressure and decreasing temperature will shift the equilibrium in the forward direction.",
|
|
"language": "English",
|
|
"id": "6d191cad-6a77-4dcd-8f19-ba3134024517"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 5)",
|
|
"subtopic": "Electrochemistry",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Given $E^\\circ_{\\text{Cr}^{3+}/\\text{Cr}} = -0.74\\text{ V}$; $E^\\circ_{\\text{MnO}_4^-/\\text{Mn}^{2+}} = 1.51\\text{ V}$; $E^\\circ_{\\text{Cr}_2\\text{O}_7^{2-}/\\text{Cr}^{3+}} = 1.33\\text{ V}$; $E^\\circ_{\\text{Cl}/\\text{Cl}^-} = 1.36\\text{ V}$.\nBased on the data given above, strongest oxidizing agent will be:",
|
|
"options": {
|
|
"A": "$\\text{Cl}$",
|
|
"B": "$\\text{Cr}^{3+}$",
|
|
"C": "$\\text{Mn}^{2+}$",
|
|
"D": "$\\text{MnO}_4^-$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Higher the standard reduction potential ($E^\\circ$), stronger is the oxidizing power. The $E^\\circ$ value for $\\text{MnO}_4^-/\\text{Mn}^{2+}$ is the highest ($+1.51\\text{ V}$). Therefore, $\\text{MnO}_4^-$ is the strongest oxidizing agent among the given species.",
|
|
"language": "English",
|
|
"id": "fc75431f-f3ab-4f54-9296-7e53de4c4531"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 5)",
|
|
"subtopic": "General Organic Chemistry",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following carbocations is most stable?",
|
|
"options": {
|
|
"A": "Allyl carbocation",
|
|
"B": "Primary alkyl carbocation",
|
|
"C": "Tert-butyl carbocation",
|
|
"D": "Triphenylmethyl carbocation"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Stability of carbocations is determined by delocalization of charge. The triphenylmethyl carbocation, $(\\text{C}_6\\text{H}_5)_3\\text{C}^+$, has 3 phenyl rings providing extensive resonance stabilization, making it exceptionally stable compared to the others.",
|
|
"language": "English",
|
|
"id": "315893aa-46ad-4b6e-9d65-fda1187eba80"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 5)",
|
|
"subtopic": "Hydrocarbons",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:",
|
|
"options": {
|
|
"A": "Two isomeric double bonds",
|
|
"B": "A vinyl group",
|
|
"C": "An isopropyl group",
|
|
"D": "An acetylenic triple bond"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Ozonolysis cleaves a carbon-carbon double bond and adds oxygen. Formaldehyde ($\\text{HCHO}$) is produced if the original double bond was at the end of a chain, specifically involving a terminal $=\\text{CH}_2$ group, which is characteristic of a vinyl group ($-\\text{CH}=\\text{CH}_2$).",
|
|
"language": "English",
|
|
"id": "5592a3c2-e4e3-493b-8ec7-d36f366a6dad"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 5)",
|
|
"subtopic": "Coordination Compounds",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The IUPAC name of $[\\text{Pt}(\\text{NH}_3)_2\\text{Cl}(\\text{NO}_2)]$ is:",
|
|
"options": {
|
|
"A": "Platinum diaminechloronitrite",
|
|
"B": "Chloronitrito-N-ammineplatinum(II)",
|
|
"C": "Diamminechloridonitrito-N-platinum(II)",
|
|
"D": "Diamminechloronitrito-O-platinate(II)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Ligands are named in alphabetical order: ammine, chlorido, nitrito-N (since it coordinates through nitrogen). The oxidation state of Pt is $x + 0 + (-1) + (-1) = 0 \\implies x = +2$. Thus, the correct name is Diamminechloridonitrito-N-platinum(II).",
|
|
"language": "English",
|
|
"id": "dde0d306-3a56-44fa-a01d-8eec040f4ba7"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 5)",
|
|
"subtopic": "Chemical Kinetics",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The unit of rate constant for a zero order reaction is:",
|
|
"options": {
|
|
"A": "$\\text{s}^{-1}$",
|
|
"B": "$\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"C": "$\\text{L}\\cdot\\text{mol}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"D": "$\\text{L}^2\\cdot\\text{mol}^{-2}\\cdot\\text{s}^{-1}$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "For a zero-order reaction, Rate = $k[A]^0 = k$. The unit of rate is concentration per unit time ($\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$). Thus, the unit of the rate constant $k$ is also $\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$.",
|
|
"language": "English",
|
|
"id": "242645d1-3f78-4a8d-b6e9-4fe1dfec026d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 5)",
|
|
"subtopic": "Biomolecules",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human beings?",
|
|
"options": {
|
|
"A": "Thyroxin",
|
|
"B": "Insulin",
|
|
"C": "Adrenaline",
|
|
"D": "Estradiol"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Adrenaline (epinephrine) is the 'fight or flight' hormone released during stress. It promotes the breakdown of glycogen into glucose (glycogenolysis) in the liver to increase blood glucose levels for energy.",
|
|
"language": "English",
|
|
"id": "2d74deba-b576-413c-bdc9-b38ebc4133ff"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Cell Structure and Function (Set 5)",
|
|
"subtopic": "Cell Cycle and Cell Division",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Crossing over occurs during which stage of meiosis?",
|
|
"options": {
|
|
"A": "Leptotene",
|
|
"B": "Pachytene",
|
|
"C": "Diplotene",
|
|
"D": "Zygotene"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Crossing over, the exchange of genetic material between non-sister chromatids of homologous chromosomes, occurs during the pachytene stage of prophase I of meiosis.",
|
|
"language": "English",
|
|
"id": "d35a85de-ae37-43c4-aefb-08a8634e97d9"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 5)",
|
|
"subtopic": "Principles of Inheritance and Variation",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?",
|
|
"options": {
|
|
"A": "TT and Tt",
|
|
"B": "Tt and Tt",
|
|
"C": "TT and TT",
|
|
"D": "Tt and tt"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Since both parents are tall but they produce some dwarf offspring, both parents must carry the recessive allele for dwarfness. Thus, their genotype must be heterozygous (Tt). A cross of Tt x Tt yields a 3:1 ratio of tall to dwarf.",
|
|
"language": "English",
|
|
"id": "74e69024-34e1-436b-91c1-05795ddc8991"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 5)",
|
|
"subtopic": "Molecular Basis of Inheritance",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following RNAs should be most abundant in an animal cell?",
|
|
"options": {
|
|
"A": "tRNA",
|
|
"B": "mRNA",
|
|
"C": "miRNA",
|
|
"D": "rRNA"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Ribosomal RNA (rRNA) constitutes about 80% of the total RNA in an actively growing cell. It forms the structural and catalytic core of ribosomes.",
|
|
"language": "English",
|
|
"id": "428e0b29-ea6a-4373-98c5-3496a5efd50a"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 5)",
|
|
"subtopic": "Digestion and Absorption",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following terms describes human dentition?",
|
|
"options": {
|
|
"A": "Thecodont, Diphyodont, Homodont",
|
|
"B": "Thecodont, Diphyodont, Heterodont",
|
|
"C": "Pleurodont, Monophyodont, Homodont",
|
|
"D": "Pleurodont, Diphyodont, Heterodont"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Human dentition is Thecodont (teeth embedded in jaw sockets), Diphyodont (two sets of teeth: deciduous and permanent), and Heterodont (different types of teeth: incisors, canines, premolars, molars).",
|
|
"language": "English",
|
|
"id": "7902ca11-3492-42bb-ad58-61bda135d377"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Reproduction (Set 5)",
|
|
"subtopic": "Human Reproduction",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which hormone is responsible for the ovulation in human females?",
|
|
"options": {
|
|
"A": "FSH (Follicle Stimulating Hormone)",
|
|
"B": "Progesterone",
|
|
"C": "LH (Luteinizing Hormone)",
|
|
"D": "Estrogen"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "A surge in LH (Luteinizing Hormone) mid-cycle triggers the rupture of the Graafian follicle and the subsequent release of the ovum, a process known as ovulation.",
|
|
"language": "English",
|
|
"id": "dfe577bd-71d7-4e0f-9321-defb739fcaa5"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Plant Physiology (Set 5)",
|
|
"subtopic": "Photosynthesis in Higher Plants",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The first stable product of CO2 fixation in C3 plants is:",
|
|
"options": {
|
|
"A": "Phosphoglyceraldehyde (PGAL)",
|
|
"B": "Malic acid",
|
|
"C": "Oxaloacetic acid (OAA)",
|
|
"D": "3-Phosphoglyceric acid (3-PGA)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In the Calvin cycle of C3 plants, CO2 combines with RuBP to form a 6-carbon unstable intermediate that immediately splits into two molecules of 3-phosphoglyceric acid (3-PGA), the first stable product.",
|
|
"language": "English",
|
|
"id": "656e0b14-518f-4145-a244-20fc9ab3b7f0"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Ecology (Set 5)",
|
|
"subtopic": "Ecosystem",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The pyramid of energy is always:",
|
|
"options": {
|
|
"A": "Upright",
|
|
"B": "Inverted",
|
|
"C": "Spindle shaped",
|
|
"D": "Variable"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "The pyramid of energy is always upright because energy is lost as heat at each trophic level (10% law), so energy always decreases as it flows from producers to higher trophic levels.",
|
|
"language": "English",
|
|
"id": "2e290702-ad9e-4f58-9ae4-c8a2fc10df09"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Biotechnology (Set 5)",
|
|
"subtopic": "Principles and Processes",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The cutting of DNA at specific locations became possible with the discovery of:",
|
|
"options": {
|
|
"A": "Selectable markers",
|
|
"B": "Ligases",
|
|
"C": "Restriction enzymes",
|
|
"D": "Probes"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Restriction endonucleases (or restriction enzymes) cut DNA at specific recognition sequences, making recombinant DNA technology possible.",
|
|
"language": "English",
|
|
"id": "b471b34c-bb0e-4c19-a08d-0bfd44dd970b"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 5)",
|
|
"subtopic": "Neural Control and Coordination",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Receptor sites for neurotransmitters are present on:",
|
|
"options": {
|
|
"A": "Pre-synaptic membrane",
|
|
"B": "Tips of axons",
|
|
"C": "Post-synaptic membrane",
|
|
"D": "Membrane of synaptic vesicles"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Neurotransmitters are released from synaptic vesicles in the pre-synaptic neuron and diffuse across the synaptic cleft to bind with specific receptor sites located on the post-synaptic membrane.",
|
|
"language": "English",
|
|
"id": "d132f31d-b66c-4873-885a-8d76aad4eb59"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Diversity in Living World (Set 5)",
|
|
"subtopic": "Biological Classification",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following are found in extreme saline conditions?",
|
|
"options": {
|
|
"A": "Eubacteria",
|
|
"B": "Cyanobacteria",
|
|
"C": "Mycobacteria",
|
|
"D": "Archaebacteria"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Archaebacteria are known to live in some of the most harsh habitats. Specifically, halophiles live in extreme saline conditions (e.g., Dead Sea).",
|
|
"language": "English",
|
|
"id": "cd148387-e310-4c96-9378-fad03647b30d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Units and Measurements (Set 6)",
|
|
"subtopic": "Dimensions",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of:",
|
|
"options": {
|
|
"A": "Velocity",
|
|
"B": "Time",
|
|
"C": "Capacitance",
|
|
"D": "Distance"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "We know that $c = \\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$, where $c$ is the speed of light in vacuum. Therefore, the dimension of $\\frac{1}{\\sqrt{\\mu_0 \\varepsilon_0}}$ is that of velocity, which is $[L T^{-1}]$.",
|
|
"language": "English",
|
|
"id": "ff3df799-7390-43f9-82a8-da1891d371bb"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Kinematics (Set 6)",
|
|
"subtopic": "Projectile Motion",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Two projectiles are thrown with the same initial velocity at angles of $30^\\circ$ and $60^\\circ$ with the horizontal. The ratio of their maximum heights is:",
|
|
"options": {
|
|
"A": "1:3",
|
|
"B": "3:1",
|
|
"C": "1:\\sqrt{3}",
|
|
"D": "1:2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Maximum height $H = \\frac{u^2 \\sin^2 \\theta}{2g}$.\nFor $\\theta_1 = 30^\\circ$, $H_1 = \\frac{u^2 (1/2)^2}{2g} = \\frac{u^2}{8g}$.\nFor $\\theta_2 = 60^\\circ$, $H_2 = \\frac{u^2 (\\sqrt{3}/2)^2}{2g} = \\frac{3u^2}{8g}$.\nRatio $H_1 : H_2 = 1 : 3$.",
|
|
"language": "English",
|
|
"id": "2a49c38b-9c45-4ed3-93c7-1b98c13c7418"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Thermodynamics (Set 6)",
|
|
"subtopic": "Carnot Engine",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A Carnot engine takes heat from a reservoir at $627^\\circ\\text{C}$ and rejects heat to a sink at $27^\\circ\\text{C}$. Its efficiency is:",
|
|
"options": {
|
|
"A": "2/3",
|
|
"B": "1/3",
|
|
"C": "3/4",
|
|
"D": "1/2"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Temperature of source, $T_1 = 627 + 273 = 900\\text{ K}$.\nTemperature of sink, $T_2 = 27 + 273 = 300\\text{ K}$.\nEfficiency $\\eta = 1 - \\frac{T_2}{T_1} = 1 - \\frac{300}{900} = 1 - \\frac{1}{3} = \\frac{2}{3}$.",
|
|
"language": "English",
|
|
"id": "d096dcda-f450-44d2-93a4-ba122e8f59f3"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Electrostatics (Set 6)",
|
|
"subtopic": "Capacitors",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A parallel plate capacitor with air between the plates has a capacitance of $8\\text{ pF}$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?",
|
|
"options": {
|
|
"A": "96 pF",
|
|
"B": "8 pF",
|
|
"C": "48 pF",
|
|
"D": "12 pF"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initial capacitance $C_0 = \\frac{\\varepsilon_0 A}{d} = 8\\text{ pF}$.\nNew distance $d' = d/2$ and new dielectric constant $K = 6$.\nNew capacitance $C' = \\frac{K \\varepsilon_0 A}{d'} = \\frac{6 \\varepsilon_0 A}{d/2} = 12 \\left(\\frac{\\varepsilon_0 A}{d}\\right) = 12 \\times 8\\text{ pF} = 96\\text{ pF}$.",
|
|
"language": "English",
|
|
"id": "a26c476c-25ae-4da3-ae3f-072830afda60"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Current Electricity (Set 6)",
|
|
"subtopic": "Meter Bridge",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a meter bridge experiment, the null point is obtained at $20\\text{ cm}$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then what will be the new position of the null point from the same end, if one decides to balance a resistance of $4X$ against $Y$?",
|
|
"options": {
|
|
"A": "50 cm",
|
|
"B": "40 cm",
|
|
"C": "80 cm",
|
|
"D": "20 cm"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Initially, $\\frac{X}{Y} = \\frac{20}{100-20} = \\frac{20}{80} = \\frac{1}{4}$. So $Y = 4X$.\nFor new resistance $4X$ against $Y$, ratio is $\\frac{4X}{Y} = \\frac{4X}{4X} = 1$.\nLet new null point be $l$, then $\\frac{l}{100-l} = 1 \\implies l = 100 - l \\implies 2l = 100 \\implies l = 50\\text{ cm}$.",
|
|
"language": "English",
|
|
"id": "027fff15-6c63-4b2b-b472-b4a853e6f707"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Magnetism (Set 6)",
|
|
"subtopic": "Biot-Savart Law",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The magnetic field at the centre of a circular current carrying coil of radius $R$ is $B_0$. If the radius is doubled keeping the current same, the new magnetic field at the centre will be:",
|
|
"options": {
|
|
"A": "$B_0/2$",
|
|
"B": "$B_0/4$",
|
|
"C": "$2B_0$",
|
|
"D": "$B_0$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "Magnetic field at the centre of a circular coil is $B = \\frac{\\mu_0 I}{2R}$.\nSince $B \\propto \\frac{1}{R}$, doubling the radius will make the magnetic field half. Thus, new field is $B_0/2$.",
|
|
"language": "English",
|
|
"id": "657cc163-4800-4c4c-8c84-d0bd96eb0535"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Alternating Current (Set 6)",
|
|
"subtopic": "LCR Circuit",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In a series LCR circuit, the voltage across the resistance, capacitance, and inductance is $10\\text{ V}$ each. If the capacitance is short-circuited, the voltage across the inductance will be:",
|
|
"options": {
|
|
"A": "10 V",
|
|
"B": "$10\\sqrt{2}\\text{ V}$",
|
|
"C": "$10/\\sqrt{2}\\text{ V}$",
|
|
"D": "Zero"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Initially, $V_R = 10\\text{ V}$, $V_L = 10\\text{ V}$, $V_C = 10\\text{ V}$. Since $V_L = V_C$, the circuit is in resonance. The applied voltage $V = \\sqrt{V_R^2 + (V_L - V_C)^2} = V_R = 10\\text{ V}$.\nIf capacitance is shorted, it becomes an LR circuit. Impedance $Z = \\sqrt{R^2 + X_L^2}$. Since $V_R = V_L$ initially, $R = X_L$, so $Z = R\\sqrt{2}$.\nNew current $I' = \\frac{V}{Z} = \\frac{10}{R\\sqrt{2}}$.\nNew voltage across inductance $V_L' = I' X_L = \\frac{10}{R\\sqrt{2}} \\times R = \\frac{10}{\\sqrt{2}}\\text{ V}$.",
|
|
"language": "English",
|
|
"id": "95c1c552-22e9-4859-946e-e5ea233c427d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Optics (Set 6)",
|
|
"subtopic": "Wave Optics",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:",
|
|
"options": {
|
|
"A": "Half",
|
|
"B": "Four times",
|
|
"C": "One-fourth",
|
|
"D": "Double"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Fringe width $\\beta = \\frac{\\lambda D}{d}$.\nNew slit separation $d' = d/2$.\nNew screen distance $D' = 2D$.\nNew fringe width $\\beta' = \\frac{\\lambda (2D)}{(d/2)} = 4 \\frac{\\lambda D}{d} = 4\\beta$.",
|
|
"language": "English",
|
|
"id": "eb6c484a-b031-428f-be49-8ae3880c0ee5"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Modern Physics (Set 6)",
|
|
"subtopic": "Photoelectric Effect",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "When the light of frequency $2\\nu_0$ (where $\\nu_0$ is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5\\nu_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is:",
|
|
"options": {
|
|
"A": "1:2",
|
|
"B": "1:4",
|
|
"C": "4:1",
|
|
"D": "2:1"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "From Einstein's photoelectric equation: $K_{\\max} = h\\nu - h\\nu_0 = \\frac{1}{2} m v^2$.\nCase 1: $\\frac{1}{2} m v_1^2 = h(2\\nu_0) - h\\nu_0 = h\\nu_0$.\nCase 2: $\\frac{1}{2} m v_2^2 = h(5\\nu_0) - h\\nu_0 = 4h\\nu_0$.\nDividing the two equations: $\\frac{v_1^2}{v_2^2} = \\frac{1}{4} \\implies \\frac{v_1}{v_2} = \\frac{1}{2}$.",
|
|
"language": "English",
|
|
"id": "b6164e29-a4e2-4782-94a4-28763cf3878b"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Physics",
|
|
"topic": "Semiconductors (Set 6)",
|
|
"subtopic": "Logic Gates",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which logic gate is represented by the boolean expression $Y = \\overline{A \\cdot B}$?",
|
|
"options": {
|
|
"A": "NOR",
|
|
"B": "OR",
|
|
"C": "NAND",
|
|
"D": "AND"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "The boolean expression $Y = \\overline{A \\cdot B}$ represents the NAND gate. It gives an output of 0 only when both inputs A and B are 1, and 1 otherwise.",
|
|
"language": "English",
|
|
"id": "490441d3-ebee-4c9a-8398-07d009223c22"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 6)",
|
|
"subtopic": "Mole Concept",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The number of water molecules is maximum in:",
|
|
"options": {
|
|
"A": "18 grams of water",
|
|
"B": "18 moles of water",
|
|
"C": "18 molecules of water",
|
|
"D": "1.8 grams of water"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "18 grams of water = 1 mole = $N_A$ molecules.\n18 moles of water = $18 N_A$ molecules.\n1.8 grams of water = 0.1 mole = $0.1 N_A$ molecules.\nThus, 18 moles of water contains the maximum number of molecules.",
|
|
"language": "English",
|
|
"id": "de008807-5fc1-461f-9139-d6a1771c5525"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 6)",
|
|
"subtopic": "Atomic Structure",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The maximum number of electrons in an atom that can have the quantum numbers $n = 4$ and $m_l = +1$ is:",
|
|
"options": {
|
|
"A": "2",
|
|
"B": "4",
|
|
"C": "6",
|
|
"D": "8"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "For $n = 4$, the possible values of $l$ are 0, 1, 2, 3.\nThe condition $m_l = +1$ is possible for $l = 1$, $l = 2$, and $l = 3$.\nEach subshell ($l=1, 2, 3$) will have one orbital with $m_l = +1$.\nSince each orbital can hold 2 electrons, the total number of electrons is $3 \\times 2 = 6$.",
|
|
"language": "English",
|
|
"id": "f9e4a4ec-b2a0-49f3-bbc1-cc050ed33432"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 6)",
|
|
"subtopic": "Chemical Bonding",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The species having pyramidal shape is:",
|
|
"options": {
|
|
"A": "$\\text{SO}_3$",
|
|
"B": "$\\text{BrF}_3$",
|
|
"C": "$\\text{SiO}_3^{2-}$",
|
|
"D": "$\\text{OSF}_2$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In $\\text{OSF}_2$, the central sulfur atom has 4 electron domains (1 double bond to O, 2 single bonds to F, and 1 lone pair). This corresponds to an $sp^3$ hybridization with a trigonal pyramidal shape. $\\text{SO}_3$ is trigonal planar, $\\text{BrF}_3$ is T-shaped.",
|
|
"language": "English",
|
|
"id": "2d265d5d-17f2-4aed-8c06-c2c99dce4515"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 6)",
|
|
"subtopic": "Equilibrium",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "For the reversible reaction $\\text{N}_2(\\text{g}) + 3\\text{H}_2(\\text{g}) \\rightleftharpoons 2\\text{NH}_3(\\text{g}) + \\text{Heat}$, the equilibrium shifts in forward direction:",
|
|
"options": {
|
|
"A": "By increasing the concentration of $\\text{NH}_3(\\text{g})$",
|
|
"B": "By decreasing the pressure",
|
|
"C": "By decreasing the concentrations of $\\text{N}_2(\\text{g})$ and $\\text{H}_2(\\text{g})$",
|
|
"D": "By increasing pressure and decreasing temperature"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "According to Le Chatelier's principle, an exothermic reaction ($\\Delta H < 0$) is favored by low temperatures. A reaction involving a decrease in the number of gaseous moles ($4 \\rightarrow 2$) is favored by high pressures. Therefore, increasing pressure and decreasing temperature will shift the equilibrium in the forward direction.",
|
|
"language": "English",
|
|
"id": "c1663444-4775-48ad-a006-a801c5bb4366"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 6)",
|
|
"subtopic": "Electrochemistry",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Given $E^\\circ_{\\text{Cr}^{3+}/\\text{Cr}} = -0.74\\text{ V}$; $E^\\circ_{\\text{MnO}_4^-/\\text{Mn}^{2+}} = 1.51\\text{ V}$; $E^\\circ_{\\text{Cr}_2\\text{O}_7^{2-}/\\text{Cr}^{3+}} = 1.33\\text{ V}$; $E^\\circ_{\\text{Cl}/\\text{Cl}^-} = 1.36\\text{ V}$.\nBased on the data given above, strongest oxidizing agent will be:",
|
|
"options": {
|
|
"A": "$\\text{Cl}$",
|
|
"B": "$\\text{Cr}^{3+}$",
|
|
"C": "$\\text{Mn}^{2+}$",
|
|
"D": "$\\text{MnO}_4^-$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Higher the standard reduction potential ($E^\\circ$), stronger is the oxidizing power. The $E^\\circ$ value for $\\text{MnO}_4^-/\\text{Mn}^{2+}$ is the highest ($+1.51\\text{ V}$). Therefore, $\\text{MnO}_4^-$ is the strongest oxidizing agent among the given species.",
|
|
"language": "English",
|
|
"id": "9ac3c12a-3bb9-4233-bcc3-7eb8f5f0e060"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 6)",
|
|
"subtopic": "General Organic Chemistry",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following carbocations is most stable?",
|
|
"options": {
|
|
"A": "Allyl carbocation",
|
|
"B": "Primary alkyl carbocation",
|
|
"C": "Tert-butyl carbocation",
|
|
"D": "Triphenylmethyl carbocation"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Stability of carbocations is determined by delocalization of charge. The triphenylmethyl carbocation, $(\\text{C}_6\\text{H}_5)_3\\text{C}^+$, has 3 phenyl rings providing extensive resonance stabilization, making it exceptionally stable compared to the others.",
|
|
"language": "English",
|
|
"id": "533ebd7e-f374-471f-8163-7b4f858db478"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 6)",
|
|
"subtopic": "Hydrocarbons",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:",
|
|
"options": {
|
|
"A": "Two isomeric double bonds",
|
|
"B": "A vinyl group",
|
|
"C": "An isopropyl group",
|
|
"D": "An acetylenic triple bond"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Ozonolysis cleaves a carbon-carbon double bond and adds oxygen. Formaldehyde ($\\text{HCHO}$) is produced if the original double bond was at the end of a chain, specifically involving a terminal $=\\text{CH}_2$ group, which is characteristic of a vinyl group ($-\\text{CH}=\\text{CH}_2$).",
|
|
"language": "English",
|
|
"id": "5ca357c9-3d85-466f-82a0-73352cee31b2"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Inorganic Chemistry (Set 6)",
|
|
"subtopic": "Coordination Compounds",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The IUPAC name of $[\\text{Pt}(\\text{NH}_3)_2\\text{Cl}(\\text{NO}_2)]$ is:",
|
|
"options": {
|
|
"A": "Platinum diaminechloronitrite",
|
|
"B": "Chloronitrito-N-ammineplatinum(II)",
|
|
"C": "Diamminechloridonitrito-N-platinum(II)",
|
|
"D": "Diamminechloronitrito-O-platinate(II)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Ligands are named in alphabetical order: ammine, chlorido, nitrito-N (since it coordinates through nitrogen). The oxidation state of Pt is $x + 0 + (-1) + (-1) = 0 \\implies x = +2$. Thus, the correct name is Diamminechloridonitrito-N-platinum(II).",
|
|
"language": "English",
|
|
"id": "3fcb25d5-2b6e-420b-86cb-1cc004e59d14"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Physical Chemistry (Set 6)",
|
|
"subtopic": "Chemical Kinetics",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The unit of rate constant for a zero order reaction is:",
|
|
"options": {
|
|
"A": "$\\text{s}^{-1}$",
|
|
"B": "$\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"C": "$\\text{L}\\cdot\\text{mol}^{-1}\\cdot\\text{s}^{-1}$",
|
|
"D": "$\\text{L}^2\\cdot\\text{mol}^{-2}\\cdot\\text{s}^{-1}$"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "For a zero-order reaction, Rate = $k[A]^0 = k$. The unit of rate is concentration per unit time ($\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$). Thus, the unit of the rate constant $k$ is also $\\text{mol}\\cdot\\text{L}^{-1}\\cdot\\text{s}^{-1}$.",
|
|
"language": "English",
|
|
"id": "2174d7a8-3b23-4ca0-91f8-975aa8e34c04"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Chemistry",
|
|
"topic": "Organic Chemistry (Set 6)",
|
|
"subtopic": "Biomolecules",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human beings?",
|
|
"options": {
|
|
"A": "Thyroxin",
|
|
"B": "Insulin",
|
|
"C": "Adrenaline",
|
|
"D": "Estradiol"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Adrenaline (epinephrine) is the 'fight or flight' hormone released during stress. It promotes the breakdown of glycogen into glucose (glycogenolysis) in the liver to increase blood glucose levels for energy.",
|
|
"language": "English",
|
|
"id": "88c7ec81-dc43-4eda-b6d4-60d14d49fa85"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Cell Structure and Function (Set 6)",
|
|
"subtopic": "Cell Cycle and Cell Division",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Crossing over occurs during which stage of meiosis?",
|
|
"options": {
|
|
"A": "Leptotene",
|
|
"B": "Pachytene",
|
|
"C": "Diplotene",
|
|
"D": "Zygotene"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Crossing over, the exchange of genetic material between non-sister chromatids of homologous chromosomes, occurs during the pachytene stage of prophase I of meiosis.",
|
|
"language": "English",
|
|
"id": "2a0e2fd6-5c37-43bc-92d6-766ef87cf765"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 6)",
|
|
"subtopic": "Principles of Inheritance and Variation",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?",
|
|
"options": {
|
|
"A": "TT and Tt",
|
|
"B": "Tt and Tt",
|
|
"C": "TT and TT",
|
|
"D": "Tt and tt"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Since both parents are tall but they produce some dwarf offspring, both parents must carry the recessive allele for dwarfness. Thus, their genotype must be heterozygous (Tt). A cross of Tt x Tt yields a 3:1 ratio of tall to dwarf.",
|
|
"language": "English",
|
|
"id": "2d687e42-b22d-4b2b-9551-3d9afdab6e0f"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Genetics and Evolution (Set 6)",
|
|
"subtopic": "Molecular Basis of Inheritance",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following RNAs should be most abundant in an animal cell?",
|
|
"options": {
|
|
"A": "tRNA",
|
|
"B": "mRNA",
|
|
"C": "miRNA",
|
|
"D": "rRNA"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Ribosomal RNA (rRNA) constitutes about 80% of the total RNA in an actively growing cell. It forms the structural and catalytic core of ribosomes.",
|
|
"language": "English",
|
|
"id": "21bf4e9b-fb36-404e-bf18-b362ac089093"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 6)",
|
|
"subtopic": "Digestion and Absorption",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following terms describes human dentition?",
|
|
"options": {
|
|
"A": "Thecodont, Diphyodont, Homodont",
|
|
"B": "Thecodont, Diphyodont, Heterodont",
|
|
"C": "Pleurodont, Monophyodont, Homodont",
|
|
"D": "Pleurodont, Diphyodont, Heterodont"
|
|
},
|
|
"correct_answer": {
|
|
"value": "B"
|
|
},
|
|
"solution_text": "Human dentition is Thecodont (teeth embedded in jaw sockets), Diphyodont (two sets of teeth: deciduous and permanent), and Heterodont (different types of teeth: incisors, canines, premolars, molars).",
|
|
"language": "English",
|
|
"id": "69b32d02-bce7-46ce-89d8-835699c1f509"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Reproduction (Set 6)",
|
|
"subtopic": "Human Reproduction",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which hormone is responsible for the ovulation in human females?",
|
|
"options": {
|
|
"A": "FSH (Follicle Stimulating Hormone)",
|
|
"B": "Progesterone",
|
|
"C": "LH (Luteinizing Hormone)",
|
|
"D": "Estrogen"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "A surge in LH (Luteinizing Hormone) mid-cycle triggers the rupture of the Graafian follicle and the subsequent release of the ovum, a process known as ovulation.",
|
|
"language": "English",
|
|
"id": "92e6ed7d-649b-4aec-b393-939dbeb78eee"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Plant Physiology (Set 6)",
|
|
"subtopic": "Photosynthesis in Higher Plants",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The first stable product of CO2 fixation in C3 plants is:",
|
|
"options": {
|
|
"A": "Phosphoglyceraldehyde (PGAL)",
|
|
"B": "Malic acid",
|
|
"C": "Oxaloacetic acid (OAA)",
|
|
"D": "3-Phosphoglyceric acid (3-PGA)"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "In the Calvin cycle of C3 plants, CO2 combines with RuBP to form a 6-carbon unstable intermediate that immediately splits into two molecules of 3-phosphoglyceric acid (3-PGA), the first stable product.",
|
|
"language": "English",
|
|
"id": "0956708f-eff5-4517-9d40-90ca12f90b93"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Ecology (Set 6)",
|
|
"subtopic": "Ecosystem",
|
|
"difficulty": "Easy",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The pyramid of energy is always:",
|
|
"options": {
|
|
"A": "Upright",
|
|
"B": "Inverted",
|
|
"C": "Spindle shaped",
|
|
"D": "Variable"
|
|
},
|
|
"correct_answer": {
|
|
"value": "A"
|
|
},
|
|
"solution_text": "The pyramid of energy is always upright because energy is lost as heat at each trophic level (10% law), so energy always decreases as it flows from producers to higher trophic levels.",
|
|
"language": "English",
|
|
"id": "b64efac0-7a70-487d-8c6d-7ee3c218b704"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Biotechnology (Set 6)",
|
|
"subtopic": "Principles and Processes",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "The cutting of DNA at specific locations became possible with the discovery of:",
|
|
"options": {
|
|
"A": "Selectable markers",
|
|
"B": "Ligases",
|
|
"C": "Restriction enzymes",
|
|
"D": "Probes"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Restriction endonucleases (or restriction enzymes) cut DNA at specific recognition sequences, making recombinant DNA technology possible.",
|
|
"language": "English",
|
|
"id": "bee1707b-4169-418c-8e09-c3e37feb355d"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Human Physiology (Set 6)",
|
|
"subtopic": "Neural Control and Coordination",
|
|
"difficulty": "Hard",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Receptor sites for neurotransmitters are present on:",
|
|
"options": {
|
|
"A": "Pre-synaptic membrane",
|
|
"B": "Tips of axons",
|
|
"C": "Post-synaptic membrane",
|
|
"D": "Membrane of synaptic vesicles"
|
|
},
|
|
"correct_answer": {
|
|
"value": "C"
|
|
},
|
|
"solution_text": "Neurotransmitters are released from synaptic vesicles in the pre-synaptic neuron and diffuse across the synaptic cleft to bind with specific receptor sites located on the post-synaptic membrane.",
|
|
"language": "English",
|
|
"id": "a53f69e9-44f8-4b63-a0b9-4db7c5a175df"
|
|
},
|
|
{
|
|
"section": "General",
|
|
"subject": "Biology",
|
|
"topic": "Diversity in Living World (Set 6)",
|
|
"subtopic": "Biological Classification",
|
|
"difficulty": "Medium",
|
|
"type": "MCQ_Single",
|
|
"question_text": "Which of the following are found in extreme saline conditions?",
|
|
"options": {
|
|
"A": "Eubacteria",
|
|
"B": "Cyanobacteria",
|
|
"C": "Mycobacteria",
|
|
"D": "Archaebacteria"
|
|
},
|
|
"correct_answer": {
|
|
"value": "D"
|
|
},
|
|
"solution_text": "Archaebacteria are known to live in some of the most harsh habitats. Specifically, halophiles live in extreme saline conditions (e.g., Dead Sea).",
|
|
"language": "English",
|
|
"id": "5978c4fe-94ec-4e92-a6a0-c5a1af6d960a"
|
|
}
|
|
] |