JEE (Main) · 2022
JEE (Main) 2022 — 30 July, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
30 Q
- Laws of Motion3
- Kinematics2
- Properties of Bulk Matter2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- Chemical Bonding and Molecular Structure3
- Electrochemistry2
- Organic Chemistry: Some Basic Principles and Techniques2
23 easy · 4 medium · 3 hard
Mathematics
30 Q
- Integral Calculus4
- Matrices and Determinants2
- Binomial Theorem2
1 easy · 25 medium · 4 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Power is energy per unit time: $[P] = [K][T^{-1}]$.
Time in terms of charge and current: $Q = It$, so $[T] = [Q\,I^{-1}]$ and $[T^{-1}] = [I\,Q^{-1}]$.
Hence $[P] = [K\,I\,Q^{-1}]$.
Answer: (B). | ||||
| 2 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Stopping distance $s = \frac{u^2}{2a}$.
Truck A: $s_A = \frac{16^2}{2\times 2} = 64$ m. Truck B: $s_B = \frac{20^2}{2\times 4} = 50$ m.
Together they close $64 + 50 = 114$ m of the 200 m gap (less than 200 m, so they stop before meeting).
Distance left between them: $200 - 114 = 86$ m.
Answer: (C). | ||||
| 3 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion The system starts at rest, so the momenta of bullet and gun are equal and opposite:
$MV = mv \Rightarrow V = \frac{0.01 \times 100}{1} = 1$ m/s.
Answer: (A). | ||||
| 4 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion With extension $x$ the radius of the circle is $(1 + x)$ m and the spring force supplies the centripetal force:
$kx = m\omega^2(1 + x)$
$800x = 0.5 \times 400\,(1 + x) = 200(1 + x)$
$600x = 200 \Rightarrow x = \frac{1}{3}$ m.
Answer: (C). | ||||
| 5 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Kepler's third law: $T^2 \propto r^3$, so $r_X = r_E\left(\frac{T_X}{T_E}\right)^{2/3}$.
With $T_E = 1$ year: $r_X = 1.5\times 10^{11} \times 8^{2/3} = 1.5 \times 10^{11} \times 4 = 6 \times 10^{11}$ m.
Answer: (C). | ||||
| 6 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The wires are massless, so both carry the same tension $Mg$; with the same area both have the same stress $\sigma = \frac{Mg}{A}$.
Strain $= \frac{\sigma}{Y}$, so $\frac{\text{strain}_{steel}}{\text{strain}_{copper}} = \frac{Y_{copper}}{Y_{steel}} = \frac{1.5\times 10^{11}}{2\times 10^{11}} = \frac{3}{4}$.
Answer: (D). | ||||
| 7 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $T_1 = 500$ K, $T_2 = 400$ K, so $\eta = 1 - \frac{400}{500} = 0.2$.
$Q_1 = \frac{W}{\eta} = \frac{1.04\times 10^5}{0.2} = 5.20 \times 10^5$ J.
Answer: (C). | ||||
| 8 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}} \propto \frac{1}{\sqrt{M}}$ at the same temperature.
$v_{N_2} = \sqrt{56}\times\sqrt{\frac{32}{28}} = \sqrt{64} = 8$ m/s.
Answer: (B). | ||||
| 9 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves In a lift accelerating upward the effective gravity is $g_{eff} = g + a = 12.5$ m/s$^2$.
$\frac{T}{T_0} = \sqrt{\frac{g}{g_{eff}}} = \sqrt{\frac{10}{12.5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.
Answer: (B). | ||||
| 10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The semicircle has radius $R = \frac{L}{\pi}$ and linear charge density $\lambda = \frac{q}{L}$.
The field at the centre of a uniformly charged semicircular arc is $E = \frac{2k\lambda}{R} = \frac{\lambda}{2\pi\epsilon_0 R}$ (the components along the diameter cancel).
$E = \frac{q/L}{2\pi\epsilon_0 (L/\pi)} = \frac{q}{2\epsilon_0 L^2}$.
Answer: (A). | ||||
| 11 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Read the circuit: the 9 V battery is on the left branch. The 3 Ω resistor runs from the top of the left branch to the right-hand wire, and the right-hand wire is joined to the bottom wire, i.e. to the other battery terminal. So the 3 Ω resistor is directly across the battery: current $\frac{9}{3} = 3$ A.
The 1.5 Ω resistor goes from the battery's top terminal to a junction J; from J the 6 Ω goes to the right-hand wire and the 2 Ω goes down to the bottom wire, so 6 Ω and 2 Ω are in parallel between J and the battery's other terminal: $\frac{6\times 2}{8} = 1.5$ Ω.
That branch is $1.5 + 1.5 = 3$ Ω, current $\frac{9}{3} = 3$ A.
$I = 3 + 3 = 6$ A.
Answer: (B). | ||||
| 12 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Average power $P = \frac{V_0 I_0}{2}\cos\phi$ with phase difference $\phi = \frac{\pi}{3}$.
$P = \frac{250\times 10}{2}\times\frac{1}{2} = 625$ W.
Answer: (B). | ||||
| 13 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $B_0 = \frac{E_0}{c} = \frac{900}{3\times 10^8} = 3\times 10^{-6}$ T $= 3$ µT.
Answer: (C). | ||||
| 14 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Take the incident light of intensity $I_0$ as unpolarised and the two polaroids as initially parallel ("common axis").
After $P_1$ the intensity is $\frac{I_0}{2}$.
After $P_2$ is turned through 60°, Malus's law gives $I = \frac{I_0}{2}\cos^2 60^\circ = \frac{I_0}{2}\times\frac{1}{4} = \frac{I_0}{8}$.
Answer: (D). | ||||
| 15 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $2\pi r = 8$ m, so $r = \frac{4}{\pi}$ m and area $A = \pi r^2 = \frac{16}{\pi}$ m$^2$.
$m = IA = 0.5\times\frac{16}{\pi} = \frac{8}{\pi}$ A m$^2$.
A current anticlockwise in the y-z plane (seen from the +x side, the usual view) gives a moment along $+\hat{i}$ by the right-hand rule: $\vec{m} = \frac{8}{\pi}\hat{i}$ A m$^2$.
Answer: (B). | ||||
| 16 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter From the figure: the straight wire AB comes in from the left and ends at B, the leftmost point of the circle, level with O; the current then runs over the top and round the right of the circle down to C, the lowest point (three quarters of the circle); from C the straight wire CD runs off to the left, a distance R below O.
AB: O lies on the line of AB, so its field at O is zero.
Arc (3/4 of a circle): $B_1 = \frac{3}{4}\cdot\frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4\pi R}\cdot\frac{3\pi}{2}$, into the page (the current goes clockwise).
CD: a semi-infinite wire with O on the perpendicular through its end C, at distance R: $B_2 = \frac{\mu_0 I}{4\pi R}$. The current flows to the left below O, so its field at O is also into the page.
Total: $B = \frac{\mu_0 I}{4\pi R}\left(\frac{3}{2}\pi + 1\right)$.
Answer: (C). | ||||
| 17 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $mvr = \frac{nh}{2\pi}$ and $\frac{mv^2}{r} = \frac{ke^2}{r^2}$ give $r = \frac{n^2h^2\epsilon_0}{\pi m e^2}$, so $r \propto n^2$ (0.529 $n^2$ Å for hydrogen).
Answer: (B). | ||||
| 18 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The curve lies in the fourth quadrant: positive voltage with negative (reversed) current, running from the short-circuit current on the I-axis to the open-circuit voltage on the V-axis. The device delivers power, which is the characteristic of a solar cell.
Answer: (D). | ||||
| 19 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter For an electron accelerated through V volts, $\lambda = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}}$ Å.
$\sqrt{15056} \approx 122.7$, so $\lambda = \frac{12.27}{122.7} = 0.1$ Å.
Answer: (A). | ||||
| 20 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter Each side band has amplitude $\frac{\mu A_c}{2}$, where $\mu = \frac{A_m}{A_c}$.
$\frac{\mu \times 5}{2} = 2.5 \Rightarrow \mu = 1$.
$A_m = \mu A_c = 5$ V.
Answer: (D). | ||||
| 21 | Atoms and NucleiRutherford model and alpha scattering | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on rutherford model and alpha scattering, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The estimate is the distance of closest approach in a head-on collision, where all the kinetic energy has become potential energy:
$K = \frac{1}{4\pi\epsilon_0}\frac{(2e)(Ze)}{r_0}$, so $r_0 = \frac{9\times 10^9 \times 2 \times 79 \times (1.6\times 10^{-19})^2}{3.2\times 10^6 \times 1.6\times 10^{-19}}$.
$r_0 = \frac{9\times 10^9 \times 158 \times 1.6\times 10^{-19}}{3.2\times 10^6} = 9\times 158\times 0.5\times 10^{-16} = 711\times 10^{-16}$ m.
Answer: 711. | ||||
| 22 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Seen from outside (air) near-normally, each layer appears shortened by its own refractive index, and the apparent depths add:
$d_{app} = \sum\frac{h_i}{\mu_i} = \frac{10}{1} + \frac{20}{2} + \frac{30}{3} + \frac{40}{4} = 10 + 10 + 10 + 10 = 40$ cm.
Answer: 40. | ||||
| 23 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\omega = 2000$ rad/s.
$X_L = \omega L = 2000\times 0.01 = 20$ Ω; $X_C = \frac{1}{\omega C} = \frac{1}{2000\times 25\times 10^{-6}} = 20$ Ω.
$Z = \sqrt{R^2 + (X_L - X_C)^2} = R = 100$ Ω (the circuit is at resonance).
Answer: 100. | ||||
| 24 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Heat $= \frac{E^2 R}{(R + r)^2}t$. Equal heat in the same time:
$\frac{4}{(4 + r)^2} = \frac{16}{(16 + r)^2} \Rightarrow \frac{16 + r}{4 + r} = 2 \Rightarrow 16 + r = 8 + 2r \Rightarrow r = 8$ Ω.
(In general $r = \sqrt{R_1R_2} = \sqrt{64} = 8$ Ω.)
Answer: 8. | ||||
| 25 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Between $P_1$ and $P_2$ (gap $\frac{d}{3}$) there are three capacitors of areas $\frac{A}{6}$, $\frac{A}{3}$ and $\frac{A}{2}$ side by side, in parallel:
$C_{12} = \frac{\epsilon_0(A/6 + A/3 + A/2)}{d/3} = \frac{3\epsilon_0 A}{d} = 12$ µF.
Between $P_2$ and $P_3$: $C_{23} = \frac{\epsilon_0 A}{d} = 4$ µF.
These two are in series: $C = \frac{12\times 4}{12 + 4} = 3$ µF.
Answer: 3. | ||||
| 26 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At the mean position $v_{max} = A\omega = 20$; at an extreme $a_{max} = A\omega^2 = 25$.
$\omega = \frac{25}{20} = 1.25$ rad/s, so $A = \frac{20}{1.25} = 16$ cm.
Answer: 16. | ||||
| 27 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Volume is conserved: $8\cdot\frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$, so $r = \frac{R}{2} = 0.5$ cm.
Increase in area: $\Delta A = 4\pi(8r^2 - R^2) = 4\pi(8\times 0.25 - 1) = 4\pi$ cm$^2 = 4\pi\times 10^{-4}$ m$^2$.
Gain in surface energy: $T\Delta A = 0.075\times 4\times 3.14\times 10^{-4} = 9.42\times 10^{-5}$ J $= 942\times 10^{-7}$ J.
Answer: 942. | ||||
| 28 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $I = MR^2 = 10\times 0.5^2 = 2.5$ kg m$^2$; $\omega_0 = 360$ rpm $= \frac{360\times 2\pi}{60} = 12\pi$ rad/s; $\theta = 6\times 2\pi = 12\pi$ rad.
Work–energy: $\tau\theta = \frac{1}{2}I\omega_0^2 \Rightarrow \tau = \frac{2.5\times 144\pi^2}{2\times 12\pi} = 15\pi$ N m.
Answer: 15. | ||||
| 29 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | NumericalMedium | |
A medium numerical question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Masses $m$ and $2m$. Momentum: $3m\times 20 = m\times 30 + 2m\,v \Rightarrow v = 15$ m/s.
Initial KE $= \frac{1}{2}(3m)(20)^2 = 600m$. Final KE $= \frac{1}{2}m(30)^2 + \frac{1}{2}(2m)(15)^2 = 450m + 225m = 675m$.
Fractional change $= \frac{675m - 600m}{600m} = \frac{1}{8}$, so $x = 8$.
Answer: 8. | ||||
| 30 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Range $R = \frac{u^2\sin 2\theta}{g}$.
$R_{15} = \frac{u^2\sin 30^\circ}{g} = \frac{u^2}{2g}$, $R_{45} = \frac{u^2}{g}$.
$R_{45} - R_{15} = \frac{u^2}{2g} = 80 \Rightarrow u^2 = 1600 \Rightarrow u = 40$ m/s.
Answer: 40. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Real gases deviate most from ideal behaviour at low temperature (intermolecular attractions matter more) and high pressure (the molecules' own volume matters more).
Of the four, the lowest temperature together with the highest pressure is –100 °C and 8 atm.
Answer: (C). | ||||
| 32 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \propto \frac{1}{\sqrt{K}}$.
With $K \to 9K$, $\lambda \to \frac{\lambda}{\sqrt{9}} = \frac{\lambda}{3}$: one third.
Answer: (A). | ||||
| 33 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics For a pure substance $dG = V\,dP - S\,dT$, so at constant pressure $\left(\frac{\partial G}{\partial T}\right)_P = -S$.
Entropy is positive, so G decreases as temperature rises: the plot slopes downward.
Only plot (A) falls with T; (B) and (C) rise and (D) is flat.
Answer: (A). | ||||
| 34 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | StatementsMedium | |
A medium statements question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter (A) True: with $\frac{1}{n} = 1$, $\frac{x}{m} = KC$, directly proportional to C.
(B) True: with $\frac{1}{n} = 0$, $\frac{x}{m} = KC^0 = K$, independent of C.
(C) and (D) False: n = 0 makes $\frac{1}{n}$ undefined, so the expression gives no plot at all for n = 0. (The flat line of (C) belongs to $\frac{1}{n} = 0$, which is statement (B), not to n = 0.)
So (A) and (B) only.
Answer: (A). | ||||
| 35 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsEasy | |
A easy statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties A. True: down a group more inner shells lie between nucleus and valence electrons, so shielding increases.
B. False: across a period electrons enter the same shell, which shield each other poorly; shielding changes only slightly while the nuclear charge rises.
C. False: ionisation enthalpy decreases down a group as the atomic size grows.
D. True: the outer electron is lost more easily down a group, so metallic character increases.
E. True: electronegativity decreases down a group.
True set: A, D and E.
Answer: (B). | ||||
| 36 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | MatchingEasy | |
A easy matching question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter A. The cyanide (leaching) process extracts gold: A-IV.
B. Froth flotation concentrates (dresses) sulphide ores such as ZnS: B-III.
C. The Hall–Héroult process is the electrolytic extraction of aluminium: C-II.
D. The Mond process refines nickel through volatile $Ni(CO)_4$, a vapour phase refining: D-I.
Answer: (A). | ||||
| 37 | HydrogenHydrogen and its compounds | No matching NCERT chapter | StatementsMedium | |
A medium statements question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Hardness is reported as the mass of $CaCO_3$ that would be equivalent to the hardness-causing salts.
(A) A reason: its molar mass is a round 100 g mol⁻¹ (equivalent mass 50), which makes the conversion simple.
(D) A reason: it is not hygroscopic (moisture insensitive), so it can be weighed accurately as a standard.
(B) Its decomposition temperature has no bearing on expressing hardness, and (C) its insolubility is not why it is chosen as the reference.
So (A) and (D) only.
Answer: (B). | ||||
| 38 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Fajans' rules: covalent character rises with the cation's charge and smallness and with the anion's size.
(A) False: the large $I^-$ is more polarisable, so KI is more covalent than KF.
(B) True: $Sn^{4+}$ has a higher charge and smaller size than $Sn^{2+}$, so it polarises $Cl^-$ more: $SnCl_4 > SnCl_2$.
(C) False: the small $Li^+$ makes LiF more covalent than KF.
(D) False: HCl is a covalent molecule, more covalent than ionic NaCl.
Answer: (B). | ||||
| 39 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry The higher the reduction potential, the stronger the oxidised form is as an oxidising agent.
$E^\circ$: $Hg_2^{2+}$ (+0.79) > $Fe^{3+}$ (+0.77) > $Al^{3+}$ (–1.66) > $Sc^{3+}$ (–2.08).
Answer: (C). | ||||
| 40 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $Co^{3+}$ is $3d^6$.
Low spin (Δ₀ > P): all six pair in $t_{2g}$: $t_{2g}^6 e_g^0$, 6 electrons.
High spin (Δ₀ < P): $t_{2g}^4 e_g^2$, 4 electrons.
Answer: (B). | ||||
| 41 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Photochemical smog forms from nitrogen oxides and hydrocarbons in sunlight; its components include $NO$, $NO_2$, $O_3$, PAN, acrolein and formaldehyde. It is an oxidising smog.
Sulphur dioxide belongs to classical (reducing) smog, not photochemical smog.
Answer: (D). | ||||
| 42 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Glycerol boils at about 563 K and decomposes near its boiling point, so it is distilled under reduced pressure, where it boils at a lower temperature without decomposing. This is how it is recovered from spent-lye in the soap industry.
Answer: (B). | ||||
| 43 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A) Cyclobutadiene is planar, cyclic and fully conjugated with 4 π electrons (4n, n = 1): antiaromatic, highly unstable.
(B) The cyclopentadienyl anion has 6 π electrons in a planar ring: aromatic, stable.
(C) Cycloheptatriene, with one $sp^3$ carbon, is non-aromatic but an ordinary stable compound (if read as the tropylium ion it is aromatic).
(D) Benzene is aromatic.
Answer: (A). | ||||
| 44 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctHard | |
A hard single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The substrate is 3,3-dimethylpent-4-enoic acid, $CH_2{=}CH{-}C(CH_3)_2{-}CH_2{-}COOH$. $NaHCO_3$ turns it into the carboxylate, and $I_2$ forms an iodonium ion on the C=C.
The carboxylate oxygen opens the iodonium ion intramolecularly (iodolactonisation). Attack at the internal carbon (the one bonded to $C(CH_3)_2$) closes a five-membered ring (5-exo), which is favoured, and leaves iodine on the terminal $CH_2$.
The product is the γ-lactone 5-(iodomethyl)-4,4-dimethyldihydrofuran-2(3H)-one, the five-membered lactone with an exocyclic $CH_2I$, which is option (B).
(A) is the six-membered lactone from attack at the terminal carbon; (C) and (D) are open-chain iodo acids, which are not formed.
Answer: (B). | ||||
| 45 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Bromobenzene → benzoic acid needs one carbon added at the ring carbon bearing Br.
Mg in dry ether gives the Grignard reagent $C_6H_5MgBr$; it adds to $CO_2$ to give $C_6H_5COOMgBr$, and hydrolysis with water (acid) gives $C_6H_5COOH$.
(B) and (C) fail because aryl halides do not undergo ordinary nucleophilic substitution by $OH^-$ or $CN^-$; (D) $KMnO_4$ does not touch the C–Br bond.
Answer: (A). | ||||
| 46 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hoffmann bromamide degradation converts a primary amide into a primary amine with one carbon fewer: $RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.
So it makes primary amines only: aniline from benzamide.
(A) and (C) are amides (starting materials, not products); (B) is a secondary amine, which this reaction cannot give.
Answer: (D). | ||||
| 47 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Benzenediazonium chloride with fluoroboric acid, $HBF_4$ (X), precipitates benzenediazonium tetrafluoroborate.
Heating this with aqueous sodium nitrite, $NaNO_2$ (Y), in the presence of copper replaces the diazonium group by $-NO_2$, giving nitrobenzene.
Answer: (B). | ||||
| 48 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | StatementsEasy | |
A easy statements question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Statement I: Nylon 6 is made by heating caprolactam with water: correct.
Statement II: Caprolactam is made from cyclohexanone: cyclohexanone + hydroxylamine gives cyclohexanone oxime, which undergoes Beckmann rearrangement to caprolactam: correct.
Answer: (A). | ||||
| 49 | BiomoleculesCarbohydrates | Class 12: Biomolecules | StatementsEasy | |
A easy statements question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Statement I: Glycogen, the storage polysaccharide of animals, is highly branched like amylopectin and is called animal starch: correct.
Statement II: Amylose is the water-soluble component of starch (about 15–20%); amylopectin (80–85%) is the insoluble one. Calling amylose water insoluble is incorrect.
Answer: (C). | ||||
| 50 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter In the brown-ring complex NO is bound as the nitrosonium ion $NO^+$ and water is neutral. The complex ion carries +2 (balanced by $SO_4^{2-}$):
$x + 0 + 1 = +2 \Rightarrow x = +1$.
Answer: (B). | ||||
| 51 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions 4 molal: 4 mol NaOH (160 g) in 1000 g water, so the solution weighs 1160 g.
Volume $= \frac{1160}{1.160} = 1000$ mL = 1 L.
Molarity $= \frac{4}{1} = 4$ M.
Answer: 4. | ||||
| 52 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Bond order $= \frac{N_b - N_a}{2}$:
$O_2$ (16 e⁻): 2. $O_2^+$ (15 e⁻): 2.5. $N_2^{2-}$ (16 e⁻, like $O_2$): 2. $C_2$ (12 e⁻): 2. $B_2$ (10 e⁻): 1.
Bond order two: $O_2$, $N_2^{2-}$, $C_2$, three species.
Answer: 3. | ||||
| 53 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Ethylene glycol $C_2H_6O_2$: M = 62 g mol⁻¹, so 31 g is 0.5 mol, and in 0.5 kg water the molality is 1 mol kg⁻¹.
$\Delta T_f = K_f m = 1.86$ K.
$T_f = 273.15 - 1.86 = 271.29$ K ≈ 271 K (with 273 K, 271.14 K: also 271).
Answer: 271. | ||||
| 54 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Acid: 5 mmol; NaOH: 1 mmol. After reaction: 4 mmol $CH_3COOH$ and 1 mmol $CH_3COO^-$, a buffer.
$pH = pK_a + \log\frac{[salt]}{[acid]} = 4.8 + \log\frac{1}{4} = 4.8 - 2(0.3) = 4.2 = 42\times 10^{-1}$.
Answer: 42. | ||||
| 55 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry For this concentration cell $E^\circ = 0$ and $E = -\frac{0.059}{2}\log\frac{C_1}{C_2}$.
$\Delta G = -nFE = 0$ requires $E = 0$, i.e. $\log\frac{C_1}{C_2} = 0$, so $\frac{C_1}{C_2} = 1$.
Answer: 1. | ||||
| 56 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Doubling the initial concentration doubles the half-life: $t_{1/2} \propto [X]_0$, which is a zero-order reaction.
For zero order, $t_{1/2} = \frac{[X]_0}{2k}$, so $k = \frac{6.0}{2\times 1.0} = 3$ mol L⁻¹ s⁻¹ (check: $\frac{12.0}{2\times 2.0} = 3$).
Answer: 3. | ||||
| 57 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Steric numbers at the central atom:
$XeO_3$: 3 bond pairs + 1 lone pair = 4, $sp^3$.
$BF_4^-$: 4 bond pairs, $sp^3$.
$I_3^-$: 2 bond pairs + 3 lone pairs = 5, $sp^3d$.
$SF_6$: 6, $sp^3d^2$.
$PCl_5$: 5, $sp^3d$.
$sp^3d$: 2; $sp^3$: 2. Ratio = 1.
Answer: 1. | ||||
| 58 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements An ion with unpaired d electrons (partly filled d shell) is both coloured (d–d transitions) and paramagnetic.
$Ti^{3+}$ $3d^1$ ✓; $Co^{2+}$ $3d^7$ ✓; $Ni^{2+}$ $3d^8$ ✓; $Cu^{2+}$ $3d^9$ ✓.
$Cu^+$ $3d^{10}$, $Ti^{4+}$ $3d^0$, $Zn^{2+}$ $3d^{10}$, $Sc^{3+}$ $3d^0$: colourless and diamagnetic.
Count: 4.
Answer: 4. | ||||
| 59 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $C_5H_{10}O_2$ (one degree of unsaturation) compounds that do not react with sodium have no O–H group; the intended set is the esters.
Formates $HCOOC_4H_9$: n-butyl, sec-butyl, isobutyl, tert-butyl: 4.
Acetates $CH_3COOC_3H_7$: n-propyl, isopropyl: 2.
Ethyl propanoate $C_2H_5COOC_2H_5$: 1.
Methyl esters $C_3H_7COOCH_3$: methyl butanoate, methyl 2-methylpropanoate: 2.
Total $4 + 2 + 1 + 2 = 9$.
Answer: 9. | ||||
| 60 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalHard | |
A hard numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Hot acidic $KMnO_4$ cleaves every C=C: $=CH_2$ gives $CO_2$, $=CHR$ gives $RCOOH$, $=CR_2$ gives a ketone. Write R for $-CH_2CH_2COCH_3$ (from the $-CH_2CH_2C(CH_3){=}CH_2$ side chain, which loses $CO_2$).
1. 1-Methyl-4-isopropenylcyclohexane: only the isopropenyl group is cleaved, giving 4-methylcyclohexyl methyl ketone.
2. Cyclobutene with the double bond on the side away from both substituents (ring C=C between two CH): ring opens to two COOH: $HOOC{-}CH(R){-}CH(CH_3){-}COOH$.
3. 1,2-Dimethylcyclopentene with an allyl group on C-3: the ring C=C gives two methyl ketones and the allyl group gives $-CH_2COOH$: $CH_3CO{-}CH_2CH_2{-}CH(CH_2COOH){-}COCH_3$.
4. 1-Methylcyclobutene with the side chain on C-4 (ring C=C carrying the methyl): the methyl-bearing carbon becomes a ketone and the other a COOH: $CH_3CO{-}CH(CH_2CH_2COCH_3){-}CH_2COOH$.
5. Cyclopropene with the side chain on the $sp^3$ carbon: $HOOC{-}CH(R){-}COOH$.
Products 3 and 4 are the same compound: in both a CH carries $-COCH_3$, $-CH_2COOH$ and $-CH_2CH_2COCH_3$. The other three products all differ.
So two hydrocarbons give the same product.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Solve $f(x) = k$: $2x^2 + 3x - 2 = k(x - 5)$, i.e. $2x^2 + (3 - k)x + (5k - 2) = 0$. ($x = 5$ is never a root: it would give $63 = 0$.)
Discriminant: $(3 - k)^2 - 8(5k - 2) = k^2 - 46k + 25$.
This is negative for $23 - \sqrt{504} < k < 23 + \sqrt{504}$ (about $0.55 < k < 45.4$): such k, e.g. $k = 1$, have no preimage, so f is not onto.
For k outside that interval, e.g. $k = 0$ ($2x^2 + 3x - 2 = 0$, $x = \frac{1}{2}, -2$), there are two preimages, so f is not one-one.
Answer: (D). | ||||
| 62 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\frac{1}{z_1} + \frac{1}{z_2} = \frac{1 - 2i}{5} + \frac{2 - i}{5} = \frac{3 - 3i}{5} = \frac{6}{w}$, so $w = \frac{30}{3 - 3i} = \frac{10}{1 - i} = 5 + 5i$.
$\bar{w} = 5 - 5i$, $2 - \bar{w} = -3 + 5i$, $iw = -5 + 5i$.
$z = \frac{-5 + 5i}{-3 + 5i} = \frac{(-5 + 5i)(-3 - 5i)}{34} = \frac{40 + 10i}{34} = \frac{20 + 5i}{17}$.
$(S_1)$: $|z| = \frac{|-5 + 5i|}{|-3 + 5i|} = \frac{5\sqrt{2}}{\sqrt{34}} = \frac{5}{\sqrt{17}}$: correct.
$(S_2)$: $\arg z = \tan^{-1}\frac{1}{4}$, $\arg w = \frac{\pi}{4}$; $\tan^{-1}\frac{1}{4} + \tan^{-1}1 = \tan^{-1}\frac{\frac{1}{4} + 1}{1 - \frac{1}{4}} = \tan^{-1}\frac{5}{3}$: correct.
Answer: (D). | ||||
| 63 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}\lambda & 1 & -1\\ 1 & -1 & -3\\ -1 & 1 & 1\end{vmatrix} = \lambda(-1 + 3) - 1(1 - 3) - 1(1 - 1) = 2\lambda + 2 = 0 \Rightarrow \lambda = -1$.
With $\lambda = -1$: adding the first two equations gives $-4z = 1$, $z = -\frac{1}{4}$; adding the last two gives $-2z = 2 + \mu$, so $\mu = -\frac{3}{2}$. Then the first and third equations both read $-x + y = -\frac{5}{4}$: infinitely many solutions.
Points: $(\lambda + 2\mu, 2\lambda + \mu) = (-4, -\frac{7}{2})$ and $(1, \lambda\mu) = (1, \frac{3}{2})$. Slope $= \frac{3/2 + 7/2}{1 + 4} = 1$.
Line: $y - \frac{3}{2} = x - 1$, i.e. $2x - 2y = -1$.
Answer: (C). | ||||
| 64 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Taking determinants: $1 \cdot |A| \cdot 2 = 1$, so $|A| = \frac{1}{2}$ and $|A^2| = \frac{1}{4}$.
Sum of roots $\frac{3}{4} = -\frac{b}{a}$; product $\frac{1}{8} = \frac{3}{a}$, so $a = 24$ and $b = -18$.
$a + b - ab = 24 - 18 + 432 = 438$.
Answer: (D). | ||||
| 65 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Put $m = n - 1$ (need $m \geq 4$ for ${}^{m}C_4$).
${}^{m}C_4 - {}^{m}C_3 = {}^{m}C_3\left(\frac{m - 3}{4} - 1\right) = \frac{m(m-1)(m-2)(m-7)}{24}$, and ${}^{n-2}P_2 = (m-1)(m-2)$.
Inequality: $(m-1)(m-2)\left[\frac{m(m-7)}{24} - \frac{5}{4}\right] < 0 \iff (m-1)(m-2)(m^2 - 7m - 30) < 0 \iff (m-1)(m-2)(m-10)(m+3) < 0$.
For $m \geq 4$ the factors $(m-1)(m-2)(m+3)$ are positive, so $m < 10$: $m = 4, 5, \ldots, 9$, i.e. $n = 5, 6, \ldots, 10$.
Sum $= 5 + 6 + 7 + 8 + 9 + 10 = 45$.
Answer: (B). | ||||
| 66 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The expression is a geometric series with first term $(1 + 2x)^{2022}$, ratio $\frac{2x}{1 + 2x}$ and 2023 terms:
$S = \frac{(1 + 2x)^{2023} - (2x)^{2023}}{(1 + 2x) - 2x} = (1 + 2x)^{2023} - (2x)^{2023}$.
The coefficient of $x^{1011}$ comes from $(1 + 2x)^{2023}$ only: ${}^{2023}C_{1011}\,2^{1011}$.
Answer: (C). | ||||
| 67 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $5^{4 - 2\alpha} + 5^{2\alpha - 1} = 126$. With $t = 5^{2\alpha}$: $\frac{625}{t} + \frac{t}{5} = 126 \Rightarrow t^2 - 630t + 3125 = 0 \Rightarrow t = 625$ or $5$.
$t = 5$ gives $\alpha = \frac{1}{2}$ (excluded); $t = 625$ gives $\alpha = 2$. The terms are 1, 63, 125, so $d = 62$, and the new A.P. has first term $\alpha^2 = 4$.
$S_{30} - S_{15}$ is the sum of terms 16 to 30: $\frac{15}{2}\left[(4 + 15\cdot 62) + (4 + 29\cdot 62)\right] = 15(4 + 22\cdot 62) = 15 \times 1368 = 20520$.
$30k = 20520 \Rightarrow k = 684$.
Answer: (C). | ||||
| 68 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctHard | |
A hard single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Write the numerator in factorial-friendly form: $(n+1)n(n-1) = n^3 - n$, so $2n^3 + 3n = 2(n+1)n(n-1) + 5n = 2(n+1)n(n-1) + 5(n+1) - 5$.
$\frac{n(2n^2+3)}{(n+1)!} = \frac{2}{(n-2)!} + \frac{5}{n!} - \frac{5}{(n+1)!}$.
Summing from $n = 2$: $\sum\frac{2}{(n-2)!} = 2e$; $\sum_{n\geq 2}\frac{5}{n!} = 5(e - 2)$; $\sum_{n\geq 2}\frac{5}{(n+1)!} = 5\left(e - \frac{5}{2}\right)$.
So $\sum_{n=2}^{\infty}\frac{n(2n^2+3)}{(n+1)!} = 2e + 5(e - 2) - 5\left(e - \frac{5}{2}\right) = 2e + \frac{5}{2}$, and twice this is $4e + 5 \approx 15.87$.
(Check by direct addition: $\frac{22}{6} + \frac{63}{24} + \frac{140}{120} + \frac{265}{720} + \ldots \approx 7.937 = 2e + 2.5$.)
None of the options equals $4e + 5$; the nearest in value is (C) $2e + 9 \approx 14.44$.
No option is correct as printed. | ||||
| 69 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Left limit: $\frac{\sin((a+3)x)}{x} + \frac{\sin x}{x} \to a + 3 + 1 = a + 4 = 2$, so $a = -2$.
Right side: $\frac{x^{1/3}\left[(1 + 6bx)^{1/3} - 1\right]}{3x^{4/3}} = \frac{(1 + 6bx)^{1/3} - 1}{3x} \to \frac{2b}{3} = 2$, so $b = 3$. Then $ab = -6$.
$f\left(\frac{7}{18}\right) = \frac{(1 + 18\cdot\frac{7}{18})^{1/3} - 1}{3\cdot\frac{7}{18}} = \frac{2 - 1}{7/6} = \frac{6}{7}$.
$(1 - k)\cdot\frac{6}{7} = -6 \Rightarrow 1 - k = -7 \Rightarrow k = 8$.
Answer: (D). | ||||
| 70 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Evaluate the integral: $\int_0^{2x}\sin t\,dt = 1 - \cos 2x$ and $\int_0^{2x} t\sin t\,dt = \sin 2x - 2x\cos 2x$, so
$\int_0^{2x}(x^2 - xt + 1)\sin t\,dt = (x^2 + 1)(1 - \cos 2x) - x(\sin 2x - 2x\cos 2x) = x^2 + 1 + (x^2 - 1)\cos 2x - x\sin 2x$.
Hence $y = \tan x - x^2 - 1 - (x^2 - 1)\cos 2x + x\sin 2x$ and $y' = \sec^2 x - 2x + (2x^2 - 1)\sin 2x$.
$y'' = 2\sec^2 x\tan x - 2 + 4x\sin 2x + 2(2x^2 - 1)\cos 2x$
$y''' = 2\sec^4 x + 4\sec^2 x\tan^2 x + 4\sin 2x + 16x\cos 2x - 4(2x^2 - 1)\sin 2x$.
At $x = \frac{\pi}{3}$: $\tan x = \sqrt{3}$, $\sec^2 x = 4$, $\sin 2x = \frac{\sqrt{3}}{2}$, $\cos 2x = -\frac{1}{2}$:
$y'' = 8\sqrt{3} - 1 + \frac{2\sqrt{3}\pi}{3} - \frac{2\pi^2}{9} \approx 14.29$, $y''' = 80 + 4\sqrt{3} - \frac{8\pi}{3} - \frac{4\sqrt{3}\pi^2}{9} \approx 70.96$.
$2y''' - y'' = 161 - \frac{16\pi}{3} - \frac{2\sqrt{3}\pi}{3} + \frac{2\pi^2}{9} - \frac{8\sqrt{3}\pi^2}{9} \approx 127.6$.
This contains π and matches none of the options; the nearest in value is (C) 126.
No option is correct as printed. | ||||
| 71 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Substitute $x = k\pi + u$: $A_k = e^{-k\pi}(-1)^k\int_0^{\pi}e^{-u}\sin u\,du$, so $|A_k| = e^{-k\pi}A_0$.
$A_0 = \left[-\frac{e^{-u}(\sin u + \cos u)}{2}\right]_0^{\pi} = \frac{e^{-\pi} + 1}{2}$.
$\sum_{k=0}^{20}|A_k| = \frac{1 + e^{-\pi}}{2}\cdot\frac{1 - e^{-21\pi}}{1 - e^{-\pi}} = \frac{1}{2}\left(\frac{e^{\pi} + 1}{e^{\pi} - 1}\right)\left(1 - e^{-21\pi}\right)$.
Answer: (B). | ||||
| 72 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Divide by $n^4$: the sum is $\frac{1}{n}\sum_{k=1}^{n}\frac{1 + (k/n)^2}{1 + (k/n)^2 + (k/n)^4} \to \int_0^1\frac{1 + x^2}{1 + x^2 + x^4}dx$.
Divide numerator and denominator by $x^2$: $\int_0^1\frac{1 + \frac{1}{x^2}}{\left(x - \frac{1}{x}\right)^2 + 3}dx$. With $u = x - \frac{1}{x}$ ($u$ runs from $-\infty$ to 0):
$\int_{-\infty}^{0}\frac{du}{u^2 + 3} = \frac{1}{\sqrt{3}}\left[\tan^{-1}\frac{u}{\sqrt{3}}\right]_{-\infty}^{0} = \frac{1}{\sqrt{3}}\cdot\frac{\pi}{2} = \frac{\pi}{2\sqrt{3}}$.
Answer: (C). | ||||
| 73 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} = \frac{y}{x} - \tan\frac{y}{x}$. Put $y = vx$: $v + x\frac{dv}{dx} = v - \tan v$, so $\cot v\,dv = -\frac{dx}{x}$.
$\ln|\sin v| = -\ln x + c$, i.e. $x\sin\frac{y}{x} = C$. From $y(1) = \frac{\pi}{2}$: $C = 1$.
At $x = \sqrt{2}$: $\sin\frac{y}{\sqrt{2}} = \frac{1}{\sqrt{2}}$, so $\frac{y}{\sqrt{2}} = \frac{\pi}{4}$ or $\frac{3\pi}{4}$, giving $y = \frac{\pi}{2\sqrt{2}}$ or $\frac{3\pi}{2\sqrt{2}}$.
Of these, $\frac{3\pi}{2\sqrt{2}}$ is among the options.
Answer: (D). | ||||
| 74 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations As printed ('= 2') the solution curve is not a closed curve, so it encloses no area; the area $13\pi$ shows that the equation meant is $\frac{dy}{dx} + \frac{x - 2}{y - b} = 0$, which is solved here.
$(y - b)dy + (x - 2)dx = 0 \Rightarrow (x - 2)^2 + (y - b)^2 = r^2$, a circle of area $\pi r^2 = 13\pi$, so $r^2 = 13$.
$y(5) = 0$: $9 + b^2 = 13$, $b = 2$ (b > 0). Circle: centre (2, 2), $r^2 = 13$.
On the y-axis: $4 + (y - 2)^2 = 13$, $y = 5$ or $-1$: $P(0, 5)$, $Q(0, -1)$.
Tangent at P: $(x - 2)(0 - 2) + (y - 2)(5 - 2) = 13$, i.e. $-2x + 3y = 15$. Tangent at Q: $(x - 2)(-2) + (y - 2)(-3) = 13$, i.e. $2x + 3y = -3$.
They meet at $T\left(-\frac{9}{2}, 2\right)$.
Area $= \frac{1}{2}\times PQ\times(\text{distance of T from the y-axis}) = \frac{1}{2}\times 6\times\frac{9}{2} = 13.5$.
Answer: (A). | ||||
| 75 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse: $c^2 = 25 - 9 = 16$, foci $(\pm 4, 0)$. Hyperbola: $a^2 + 12 = 16$, so $a^2 = 4$; take $a = 2$, giving the point (2, 3).
The second hyperbola, $\frac{y^2}{B^2} - \frac{x^2}{A^2} = 1$ with $A^2 + B^2 = 10$, passes through (2, 3):
$\frac{9}{B^2} - \frac{4}{10 - B^2} = 1 \Rightarrow 9(10 - B^2) - 4B^2 = B^2(10 - B^2) \Rightarrow B^4 - 23B^2 + 90 = 0 \Rightarrow B^2 = 5$ (18 is too large).
$e = \frac{\sqrt{10}}{B} = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}$.
Answer: (B). | ||||
| 76 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent of slope m to $y^2 = 4Ax$ (here $4A = k$): $y = mx + \frac{A}{m} = \frac{x}{2} + \frac{k}{2}$.
For it to touch $xy = -1$: $x\left(\frac{x}{2} + \frac{k}{2}\right) = -1 \Rightarrow x^2 + kx + 2 = 0$ must have equal roots: $k^2 = 8$, $k = \pm 2\sqrt{2} \approx \pm 2.83$.
$2\sqrt{2}$ lies in (2, 4] and in (1, 3); $-2\sqrt{2}$ lies in (−3, −1). Neither value lies in [−4, −3).
Answer: (C). | ||||
| 77 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Both constants are positive. The normals give $(2)(3) + (-1)(-2) + (-2)(6) = -4 < 0$, so the bisector of the angle containing the origin is the acute-angle bisector:
$\frac{2x - y - 2z + 3}{3} = +\frac{3x - 2y + 6z + 8}{7}$
$14x - 7y - 14z + 21 = 9x - 6y + 18z + 24 \Rightarrow 5x - y - 32z - 3 = 0$, i.e. $-5x + y + 32z + 3 = 0$.
(Check: the angle between $(5, -1, -32)$ and $(2, -1, -2)$ is about 39.5°, half the acute angle 79° between the planes.)
$a + b + c = -5 + 1 + 32 = 28$.
Answer: (B). | ||||
| 78 | StatisticsMean, median and mode | Class 11: Statistics | Single correctMedium | |
A medium single correct question on mean, median and mode, from Statistics. NCERT: Class 11: Statistics $3 + 6 + 2 + x + y = 20 \Rightarrow x + y = 9$.
Median: $\frac{N}{2} = 10$; cumulative frequencies 3, 9, 11, … so the median class is 20 – 30: median $= 20 + \frac{10 - 9}{2}\times 10 = 25$.
Mean $= \frac{5(3) + 15(6) + 25(2) + 35x + 45y}{20} = \frac{155 + 35x + 45y}{20} = 25 \Rightarrow 35x + 45y = 345 \Rightarrow 7x + 9y = 69$.
With $x + y = 9$: $63 + 2y = 69$, $y = 3$, $x = 6$.
$xy^2 = 6 \times 9 = 54$.
Answer: (A). | ||||
| 79 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let the foot of the tower be O and its height h. At A (due north of O) the elevation is 45°, so $OA = h$.
B is 50 m west of A, so $OA \perp AB$ and $OB^2 = h^2 + 50^2$.
At B the elevation is 30°: $OB = h\cot 30^\circ = h\sqrt{3}$.
$3h^2 = h^2 + 2500 \Rightarrow h^2 = 1250 \Rightarrow h = 25\sqrt{2}$ m.
Answer: (A). | ||||
| 80 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter (B) $\sim(p \Rightarrow q) \equiv p \wedge \sim q$, so (B) is $p \wedge \sim q \wedge \sim p$, false for every truth value: a contradiction.
(A) $(\sim p \vee q) \equiv (p \Rightarrow q)$, so (A) is $(p \Rightarrow q) \Rightarrow \sim(p \Rightarrow q) \equiv \sim(p \Rightarrow q)$, true when p is T and q is F.
(C) is true when p and q are both T. (D) is true when both are F.
Answer: (B). | ||||
| 81 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Since $|x^2 + 3x + 2| \geq 0$, we need $|x + 5| \leq 2$, i.e. $-7 \leq x \leq -3$. There $x^2 + 3x + 2 = (x + 1)(x + 2) > 0$.
For $-5 \leq x \leq -3$: $x^2 + 3x + 2 + x + 5 = 2 \Rightarrow x^2 + 4x + 5 = 0$, discriminant $-4 < 0$: no root.
For $-7 \leq x < -5$: $x^2 + 3x + 2 - x - 5 = 2 \Rightarrow x^2 + 2x - 5 = 0$, $x = -1 \pm\sqrt{6}$ ($\approx 1.45$ or $-3.45$), neither in $[-7, -5)$.
No solutions.
Answer: 0. | ||||
| 82 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $x^2 + 2 + \frac{1}{x^2} = \left(\frac{1 + x^2}{x}\right)^2$, so the expression is $\frac{x^{10}}{(1 + x^2)^{10}}(1 + x^2)^{40} = x^{10}(1 + x^2)^{30}$.
The coefficient of $x^{30}$ is the coefficient of $x^{20}$ in $(1 + x^2)^{30}$: ${}^{30}C_{10} = 30045015$.
By Legendre's formula the prime exponents in $\frac{30!}{10!\,20!}$ are: 2: $26 - 8 - 18 = 0$; 3: $14 - 4 - 8 = 2$; 5: $7 - 2 - 4 = 1$; 7: $4 - 1 - 2 = 1$; 11: 1; 13: 1; 17: 0; 19: 0; 23: 1; 29: 1.
${}^{30}C_{10} = 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 23\cdot 29$. The primes in decreasing order: 29, 23, 13, 11, 7, 5, 3.
$p_1 = 23$, $p_2 = 11$: $p_1 + p_2 = 34$.
Answer: 34. | ||||
| 83 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f(x) = 2x^3 - 3x^2 - 12x + b$, $f'(x) = 6(x - 2)(x + 1)$: local maximum at $x = -1$, local minimum at $x = 2$.
Three distinct real roots need $f(-1) > 0$ and $f(2) < 0$: $f(-1) = 7 + b > 0$ and $f(2) = b - 20 < 0$.
$-7 < b < 20$: the integers $-6, -5, \ldots, 19$, which are 26.
Answer: 26. | ||||
| 84 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let the shorter parallel side be a. With base angles 60° and height x, the longer side is $a + \frac{2x}{\sqrt{3}}$ and each slant side is $\frac{x}{\sin 60^\circ} = \frac{2x}{\sqrt{3}}$.
Area: $\left(a + \frac{x}{\sqrt{3}}\right)x = 1.5 \Rightarrow a = \frac{1.5}{x} - \frac{x}{\sqrt{3}}$.
The duct is open front and back, so the sheet area is the perimeter times 1 m:
$S = 2a + \frac{2x}{\sqrt{3}} + \frac{4x}{\sqrt{3}} = \frac{3}{x} + \frac{4x}{\sqrt{3}}$.
$\frac{dS}{dx} = -\frac{3}{x^2} + \frac{4}{\sqrt{3}} = 0 \Rightarrow x_0^2 = \frac{3\sqrt{3}}{4}$ (a minimum, since $S'' > 0$).
$x_0^4 = \frac{27}{16}$, so $16x_0^4 = 27$.
Answer: 27. | ||||
| 85 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Read the region as the part above the line $y = \frac{x}{2}$ of the two regions: (i) for $-1 \leq x \leq 0$, between $y = x^{1/3}$ (below) and $y = x^2$ (above), closed by $x = -1$; (ii) for $0 \leq x \leq 1$, between $y = x^2$ (below) and $y = x^{1/3}$ (above), which meet at $x = 0$ and $x = 1$.
(i) On $[-1, 0]$ the line $y = \frac{x}{2}$ lies above $y = x^{1/3}$ (they meet only at $|x| = 2\sqrt{2}$ and 0) and below $y = x^2$, so the part above the line is $\int_{-1}^{0}\left(x^2 - \frac{x}{2}\right)dx = \frac{1}{3} + \frac{1}{4} = \frac{7}{12}$.
(ii) On $[0, 1]$ the lower boundary of the part above the line is $\max\left(x^2, \frac{x}{2}\right)$: $\frac{x}{2}$ on $[0, \frac{1}{2}]$ and $x^2$ on $[\frac{1}{2}, 1]$:
$\int_0^1 x^{1/3}dx - \int_0^{1/2}\frac{x}{2}dx - \int_{1/2}^{1}x^2dx = \frac{3}{4} - \frac{1}{16} - \frac{7}{24} = \frac{19}{48}$.
Total $= \frac{28}{48} + \frac{19}{48} = \frac{47}{48} = \frac{n}{n+1}$, so $n = 47$.
Answer: 47. | ||||
| 86 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines AD: $\frac{x}{2} + \frac{y}{a + 2} = 1 \Rightarrow a + 2 = \frac{2y}{2 - x}$. BC: $-\frac{x}{2} + \frac{y}{a} = 1 \Rightarrow a = \frac{2y}{2 + x}$.
Subtract: $2 = \frac{2y}{2 - x} - \frac{2y}{2 + x} = \frac{4xy}{4 - x^2}$, so the locus is $x^2 + 2xy - 4 = 0$.
At $x = 4$: $16 + 8\gamma - 4 = 0$, $\gamma = -\frac{3}{2}$.
Slope: $2x + 2y + 2xy' = 0 \Rightarrow y' = -\frac{x + y}{x} = -\frac{5/2}{4} = -\frac{5}{8}$.
Tangent: $y + \frac{3}{2} = -\frac{5}{8}(x - 4) \Rightarrow 5x + 8y = 8 \Rightarrow \frac{x}{8/5} + \frac{y}{1} = 1$: $\alpha = \frac{8}{5}$, $\beta = 1$.
$\alpha(\beta - \gamma) = \frac{8}{5}\cdot\frac{5}{2} = 4$.
Answer: 4. | ||||
| 87 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $L_1$ passes through $(0, \frac{1}{4}, 0)$ and $(0, 0, \frac{1}{C})$: direction $\propto (0, -C, 4)$.
$L_2$ passes through $(\frac{1}{3}, 0, 0)$ and $(0, 0, -\frac{1}{C})$: direction $\propto (C, 0, 3)$.
$\vec{d_1}\times\vec{d_2} = (-3C, 4C, C^2)$, of length $C\sqrt{25 + C^2}$.
Join $(0, 0, \frac{1}{C})$ to $(0, 0, -\frac{1}{C})$: $(0, 0, -\frac{2}{C})$; its dot product with the cross product has size $2C$.
Shortest distance $= \frac{2C}{C\sqrt{25 + C^2}} = \frac{2}{\sqrt{25 + C^2}} = \frac{1}{5} \Rightarrow 25 + C^2 = 100 \Rightarrow C^2 = 75$.
Answer: 75. | ||||
| 88 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Acute angles with all axes: $9 - a^2 > 0$ and $a - 1 > 0$, so with $a \in \mathbb{N}$, $a = 2$ and $P = (2, 2, 2)$.
In parallelogram OABC, $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = (b + 2, 6, -2)$.
Projection: $\frac{2(b + 2) + 12 - 4}{\sqrt{(b + 2)^2 + 40}} = 2 \Rightarrow (b + 6)^2 = (b + 2)^2 + 40 \Rightarrow 8b = 8 \Rightarrow b = 1$.
$\overrightarrow{OB} = (3, 6, -2)$, $|\overrightarrow{OB}|^2 = 49$; $\overrightarrow{AC} = (-1, -2, 8)$, $|\overrightarrow{AC}|^2 = 69$.
Sum $= 118$.
Answer: 118. | ||||
| 89 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $E(X) = \sum\frac{i}{2^i} = 2$ and $E(X^2) = \sum\frac{i^2}{2^i} = 6$ (from $\sum i x^i = \frac{x}{(1-x)^2}$ and $\sum i^2x^i = \frac{x(1+x)}{(1-x)^3}$ at $x = \frac{1}{2}$).
$\text{Var}(X) = 6 - 2^2 = 2$.
Answer: 2. | ||||
| 90 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Work interval by interval:
$[0, \frac{\pi}{2})$: the sum is $\theta + \theta + \theta = 3\theta = \frac{4\pi}{5}$, $\theta = \frac{4\pi}{15}$ ✓.
$(\frac{\pi}{2}, \pi]$: $(\pi - \theta) + \theta + (\theta - \pi) = \theta = \frac{4\pi}{5}$ ✓.
$(\pi, \frac{3\pi}{2})$: $(\pi - \theta) + (2\pi - \theta) + (\theta - \pi) = 2\pi - \theta = \frac{4\pi}{5}$, $\theta = \frac{6\pi}{5}$ ✓.
$(\frac{3\pi}{2}, 2\pi]$: $(\theta - 2\pi) + (2\pi - \theta) + (\theta - 2\pi) = \theta - 2\pi = \frac{4\pi}{5}$ gives $\theta = \frac{14\pi}{5} > 2\pi$ ✗.
$\sum\theta = \frac{4\pi}{15} + \frac{12\pi}{15} + \frac{18\pi}{15} = \frac{34\pi}{15}$, and $\frac{30}{\pi}\cdot\frac{34\pi}{15} = 68$.
Answer: 68. | ||||
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