JEE (Main) · 2021
JEE (Main) 2021 — 16 March, Shift 1
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
- Electromagnetic Induction and Alternating Currents3
- Electromagnetic Waves2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- p-Block Elements3
- Organic Chemistry: Some Basic Principles and Techniques3
- Hydrocarbons3
21 easy · 9 medium · 0 hard
Mathematics
30 Q
- Trigonometry3
- Integral Calculus3
- Matrices and Determinants2
5 easy · 20 medium · 5 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 | 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 $U_e = \frac{1}{2}\varepsilon_0 E^2$ and $U_m = \frac{B^2}{2\mu_0}$. In free space $E = cB$ and $c^2 = \frac{1}{\mu_0\varepsilon_0}$, so $U_e = \frac{1}{2}\varepsilon_0 c^2 B^2 = \frac{B^2}{2\mu_0} = U_m$. The same holds for the averages.
Answer: (A). | ||||
| 2 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Moles: $O_2$: $\frac{16}{32} = 0.5$; $N_2$: $\frac{28}{28} = 1$; $CO_2$: $\frac{44}{44} = 1$. Total $n = 2.5 = \frac{5}{2}$.
By Dalton's law, $P = \frac{nRT}{V} = \frac{5}{2}\frac{RT}{V}$.
Answer: (A). | ||||
| 3 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With a slab of thickness $t$, $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$. Here $t = \frac{3d}{4}$: $d - t + \frac{t}{K} = \frac{d}{4} + \frac{3d}{4K} = \frac{d(K+3)}{4K}$.
So $C' = \frac{4K}{K+3}\cdot\frac{\varepsilon_0 A}{d} = \frac{4K}{K+3}C_0$.
Answer: (A). | ||||
| 4 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane For $0 \le x \le 200$ m the velocity rises linearly from 10 to 50 m/s: $v = 10 + 0.2x$, so $\frac{dv}{dx} = 0.2\ \text{s}^{-1}$.
$a = v\frac{dv}{dx} = 0.2(10 + 0.2x) = 2 + 0.04x$: a straight line from $a = 2$ at $x = 0$ to $a = 18\ \text{m s}^{-2}$ at $x = 200$ m.
For $200 \le x \le 400$ m the velocity is constant, so $a = 0$.
This is the first graph.
Answer: (A). | ||||
| 5 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion The force pulls upward at $\theta$ above the horizontal. Vertically: $N + F\sin\theta = mg$, so $N = mg - F\sin\theta$.
Kinetic friction $\mu_K N$ opposes the motion. Horizontally: $ma = F\cos\theta - \mu_K(mg - F\sin\theta)$.
$a = \frac{F}{m}\cos\theta - \mu_K\left(g - \frac{F}{m}\sin\theta\right)$.
Answer: (D). | ||||
| 6 | Magnetic Effects of Current and MagnetismEarth magnetism | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on earth magnetism, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter With the north pole pointing north, the magnet's field on its equatorial line points south (opposite to $\vec{M}$), so it cancels $B_H$ there: the neutral points lie on the equatorial line.
Half-length $l = 7$ cm, $r = 18$ cm. On the equatorial line $B = \frac{\mu_0}{4\pi}\frac{M}{(r^2 + l^2)^{3/2}}$.
$r^2 + l^2 = 324 + 49 = 373\ \text{cm}^2$, $(373)^{3/2} \approx 7204\ \text{cm}^3 = 7.204\times10^{-3}\ \text{m}^3$.
$M = \frac{B_H (r^2 + l^2)^{3/2}}{10^{-7}} = \frac{0.4\times10^{-4}\times7.204\times10^{-3}}{10^{-7}} \approx 2.88\ \text{J T}^{-1}$.
Answer: (C). | ||||
| 7 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | StatementsEasy | |
A easy statements question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At minimum deviation the ray passes symmetrically: $i = e$ and $r_1 = r_2 = \frac{A}{2}$. For an isosceles prism the ray inside is then parallel to the base, and the incident and emergent rays are symmetric about the prism. So (A), (B) and (C) are true; (D) contradicts $i = e$.
Answer: (B). | ||||
| 8 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The CRO reads the voltage across C. While the square wave is high, C charges through R: $V_C$ rises as $1 - e^{-t/RC}$, steeply at first and then levelling off. While the input is low, C discharges through R: $V_C$ falls as $e^{-t/RC}$, steeply at first and then levelling off. Neither change can be sudden, because the voltage across a capacitor cannot jump. The output is a train of exponential rises and falls, the fourth pattern.
Answer: (D). | ||||
| 9 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion The side wall supplies the centripetal force: $N = m\omega^2 r$ with $\omega = \frac{2\pi}{40} = \frac{\pi}{20}\ \text{rad s}^{-1}$.
$N = 0.2\times\frac{\pi^2}{400}\times0.2 = \frac{0.04\pi^2}{400} = 10^{-4}\pi^2 \approx 9.87\times10^{-4}$ N (with $\pi^2 = 9.859$, $9.859\times10^{-4}$ N).
Answer: (B). | ||||
| 10 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat and work depend on the process followed between two states, not only on the end states; they are not state variables. They are path functions.
Answer: (B). | ||||
| 11 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation Angular momentum about the Sun is conserved, and at the two extreme points the velocity is perpendicular to the radius: $v_{min} r_{max} = v_{max} r_{min}$.
$v_{min} = \frac{6\times10^{4}\times8.0\times10^{10}}{1.6\times10^{12}} = 3.0\times10^{3}$ m/s.
Answer: (B). | ||||
| 12 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The axis through A parallel to diagonal DB is at distance $\frac{l}{\sqrt{2}}$ from DB (half of diagonal AC).
A lies on the axis: 0. D and B lie on DB: each at $\frac{l}{\sqrt{2}}$, giving $2m\cdot\frac{l^2}{2} = ml^2$. C is at the full diagonal $\sqrt{2}l$: $m\cdot2l^2$.
$I = ml^2 + 2ml^2 = 3ml^2$.
Answer: (C). | ||||
| 13 | Communication SystemsAntenna length and wavelength | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on antenna length and wavelength, from Communication Systems. NCERT: No matching NCERT chapter An antenna radiates efficiently when its length is a quarter of the wavelength: $l = \frac{\lambda}{4}$. The height of the tower does not enter.
$\lambda = 4\times25 = 100$ m.
Answer: (D). | ||||
| 14 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $eV_0 = h\nu - \phi$: the stopping potential depends on the frequency of the radiation only; intensity (amplitude) changes the photocurrent, not $V_0$.
Answer: (A). | ||||
| 15 | 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 the lift accelerating upward, $g_{eff} = g + \frac{g}{2} = \frac{3g}{2}$.
$T' = 2\pi\sqrt{\frac{l}{3g/2}} = T\sqrt{\frac{2}{3}}$.
Answer: (D). | ||||
| 16 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Magnitude: $E = cB = 3\times10^{8}\times8.0\times10^{-8} = 24$ V/m.
Direction: $\vec{E}\times\vec{B}$ is along the propagation, $+\hat{y}$. With $\vec{B}$ along $\hat{z}$: $\hat{x}\times\hat{z} = -\hat{y}$, so $\vec{E}$ must be along $-\hat{x}$: $(-\hat{x})\times\hat{z} = \hat{y}$.
$\vec{E} = -24\hat{x}$ V/m.
Answer: (A). | ||||
| 17 | Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Left loop (with $R_1$): as the bar moves left its area falls, so the inward flux falls. By Lenz's law the induced current must add inward flux: clockwise.
Right loop (with $R_2$): its area grows, inward flux grows, so the induced current must make outward flux: anticlockwise.
(Check: the force $q\vec{v}\times\vec{B}$ on charges in the bar drives current down the bar, which is clockwise round the left loop and anticlockwise round the right loop.)
Answer: (B). | ||||
| 18 | Properties of Bulk MatterFluid mechanics | 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 fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $P = P_0 + \rho g h$: $3\times10^{5} = 1\times10^{5} + \rho g h$, so $\rho g h = 2\times10^{5}$ Pa.
At depth $2h$: $P' = 1\times10^{5} + 4\times10^{5} = 5\times10^{5}$ Pa.
Increase $= \frac{2\times10^{5}}{3\times10^{5}}\times100\% = \frac{200}{3}\%$.
Answer: (C). | ||||
| 19 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity New resistance $R' = \frac{\rho(2l)}{A/2} = \frac{4\rho l}{A}$. With the same V: $I = \frac{V}{R'} = \frac{1}{4}\frac{VA}{\rho l}$.
Answer: (C). | ||||
| 20 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter $n$ VSD $= (n-1)$ MSD, so 1 VSD $= \frac{(n-1)a}{n}$ cm.
Least count $= 1\ \text{MSD} - 1\ \text{VSD} = a - \frac{(n-1)a}{n} = \frac{a}{n}$ cm $= \frac{10a}{n}$ mm.
Answer: (C). | ||||
| 21 | 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 At resonance the impedance is R alone. $V_{rms} = \frac{250}{\sqrt{2}}$.
$P = \frac{V_{rms}^2}{R} = \frac{250^2}{2\times8} = \frac{62500}{16} = 3906.25$ W $\approx 3.9$ kW.
To the nearest integer, $x = 4$.
Answer: 4. | ||||
| 22 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane From the figure, arm AB makes $70^\circ$ with the vertical wall and arm AC makes $180^\circ - 145^\circ = 35^\circ$ with it. So the angle between the arms is $70^\circ - 35^\circ = 35^\circ$, and the vertical force P makes $70^\circ$ (equivalently $110^\circ$) with AB.
Resolving P into two components along AB and AC, the sine rule in the triangle of forces gives the component along AC:
$P_{AC} = \frac{P\sin(\text{angle between }P\text{ and }AB)}{\sin(\text{angle between }AB\text{ and }AC)} = \frac{100\sin110^\circ}{\sin35^\circ} = \frac{100\times0.939}{0.573} \approx 163.9$ N.
To the nearest integer, $x = 164$.
Answer: 164. | ||||
| 23 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Balmer lines end on $n = 2$. First line $3\to2$: $\frac{1}{\lambda_1} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$. Third line $5\to2$: $\frac{1}{\lambda_3} = R\left(\frac{1}{4} - \frac{1}{25}\right) = \frac{21R}{100}$.
$\frac{\lambda_1}{\lambda_3} = \frac{21/100}{5/36} = \frac{756}{500} = 1.512 = 15.12\times10^{-1}$.
To the nearest integer, $x = 15$.
Answer: 15. | ||||
| 24 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The zener holds the load at 15 V. Current through $R_S$: $I_S = \frac{22 - 15}{35} = 0.2$ A. Load current: $I_L = \frac{15}{90} = \frac{1}{6}$ A.
Zener current: $I_Z = 0.2 - \frac{1}{6} = \frac{1}{30}$ A.
$P_Z = 15\times\frac{1}{30} = 0.5$ W $= 5\times10^{-1}$ W.
Answer: 5. | ||||
| 25 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Initial momentum: $10\times10\sqrt{3} = 100\sqrt{3}$ kg m/s along X. Each piece has mass 10 kg.
X: $10\times20\cos\theta = 100\sqrt{3}$, so $\cos\theta = \frac{\sqrt{3}}{2}$.
Y: $10\times10 - 10\times20\sin\theta = 0$, so $\sin\theta = \frac{1}{2}$ (consistent).
$\theta = 30^\circ$.
Answer: 30. | ||||
| 26 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy 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 $I = \frac{1}{2}MR^2 = \frac{1}{2}\times20\times0.04 = 0.4\ \text{kg m}^2$; torque $\tau = FR = 20\times0.2 = 4$ N m.
$\alpha = \frac{\tau}{I} = 10\ \text{rad s}^{-2}$.
$\omega^2 = 2\alpha\theta$: $\theta = \frac{2500}{20} = 125$ rad.
$n = \frac{125}{6.28} \approx 19.9$; to the nearest integer, 20.
Answer: 20. | ||||
| 27 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Reading the circuit: A goes to the NOR gate directly and, through a NOT, to the NAND gate ($\bar{A}$). B passes a NOT ($\bar{B}$) to the NAND gate; from that line a second NOT (drawn upward, crossing the $\bar{A}$ line without joining it) feeds the NOR gate with $\bar{\bar{B}} = B$.
With A = 0, B = 1: NOR output $= \overline{A + B} = \overline{0 + 1} = 0$. NAND output $= \overline{\bar{A}\cdot\bar{B}} = \overline{1\cdot0} = 1$.
$Y = 0\cdot1 = 0$.
Answer: 0. | ||||
| 28 | 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 Three 3 k$\Omega$ resistors in parallel: 1 k$\Omega$. Total resistance with the 5 k$\Omega$ resistor and the 1 k$\Omega$ internal resistance: $5 + 1 + 1 = 7$ k$\Omega$.
$I = \frac{21}{7\ \text{k}\Omega} = 3$ mA.
Answer: 3. | ||||
| 29 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\beta = \frac{\lambda D}{d}$, so $\lambda = \frac{\beta d}{D} = \frac{6\times10^{-3}\times1\times10^{-3}}{10} = 6\times10^{-7}$ m $= 600$ nm.
Answer: 600. | ||||
| 30 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalEasy | |
A easy numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\frac{\Delta R}{R}\times100 = \left(\frac{\Delta V}{V} + \frac{\Delta I}{I}\right)\times100 = \left(\frac{2}{50} + \frac{0.2}{20}\right)\times100 = 4 + 1 = 5\%$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Assertion and reasonEasy | |
A easy assertion and reason question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Water has four electron pairs on O (two bond pairs, two lone pairs), tetrahedrally arranged. The lone pairs repel more strongly than bond pairs (lp$-$lp > lp$-$bp > bp$-$bp), squeezing the H$-$O$-$H angle from $109.5^\circ$ to $104.5^\circ$. Both statements are true and R explains A.
Answer: (A). | ||||
| 32 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Haber's process makes ammonia (iii); Ostwald's process oxidises ammonia to nitric acid (i); the Contact process makes sulphuric acid (iv); the Hall$-$Heroult process is the electrolytic extraction of aluminium (ii).
Answer: (B). | ||||
| 33 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Down group 15 the E$-$H bond weakens, so the reducing power of the hydrides rises: $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$. Bismuth is also the one true metal of the group.
Answer: (A). | ||||
| 34 | Isolation of MetalsRoasting and calcination | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on roasting and calcination, from Isolation of Metals. NCERT: No matching NCERT chapter Roasting heats a sulphide ore in excess air, driving off sulphur as $SO_2$ (for example $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$).
Answer: (B). | ||||
| 35 | HydrogenHydrogen and its compounds | No matching NCERT chapter | StatementsEasy | |
A easy statements question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Statement I: in basic medium $H_2O_2$ oxidises (e.g. $2Fe^{2+} \rightarrow 2Fe^{3+}$) and reduces (e.g. $2MnO_4^- + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2H_2O + 2OH^-$). True.
Statement II: the basic idea of the hydrogen economy is to store and transmit energy as dihydrogen instead of electricity. True.
Answer: (A). | ||||
| 36 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | StatementsMedium | |
A medium statements question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Statement I: the hydrated chloride of calcium is dehydrated on heating, but hydrated magnesium chloride is hydrolysed on heating (giving MgO / basic chloride and HCl) rather than simply dehydrated. False.
Statement II: BeO is amphoteric, but the oxides of the other group 2 elements (MgO, CaO, SrO, BaO) are basic, not acidic. False.
Answer: (B). | ||||
| 37 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Hypophosphorous acid $H_3PO_2$: P is +1 (v). Orthophosphoric acid $H_3PO_4$: +5 (i). Hypophosphoric acid $H_4P_2O_6$: +4 (ii). Orthophosphorous acid $H_3PO_3$: +3 (iii).
Answer: (D). | ||||
| 38 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Assertion and reasonEasy | |
A easy assertion and reason question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Bk (Z = 97) comes after Np (Z = 93) in the actinoid series; the $M^{3+}$ radii fall steadily across the series, so $Bk^{3+}$ is smaller than $Np^{3+}$. A is true.
The cause is the actinoid contraction (poor shielding by 5f electrons), not the lanthanoid contraction. R is false.
Answer: (C). | ||||
| 39 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | StatementsEasy | |
A easy statements question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Statement I: $E^\circ(Ce^{4+}/Ce^{3+}) = +1.74$ V, the standard value. Correct.
Statement II: the large positive potential shows that $Ce^{4+}$ is readily reduced to $Ce^{3+}$; $Ce^{4+}$ is a strong oxidant and +3 is the more stable state. Incorrect.
Answer: (C). | ||||
| 40 | 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 in warm sunny daytime conditions from nitrogen oxides and hydrocarbons; it contains ozone, PAN and other oxidants, so it is oxidising smog.
Answer: (B). | ||||
| 41 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Chromatographic separation depends on how the compound distributes between the stationary and mobile phases: its solubility (and adsorption), the flow of the solvent, and the length of the column or plate. Whether the pure compound is a solid or a liquid does not matter.
Answer: (D). | ||||
| 42 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | StatementsEasy | |
A easy statements question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A) The ring carbon carrying the exocyclic $=CH_2$ has its p orbital used in the exocyclic double bond, so the ring has no continuous cyclic $\pi$ sextet of its own (it is a cross-conjugated triene); not aromatic.
(B) Benzene: planar, $6\pi$ electrons: aromatic.
(C) Cyclopentadienyl anion: planar, $6\pi$ electrons: aromatic.
(D) Cyclopentadienyl cation: $4\pi$ electrons: antiaromatic.
Aromatic: (B) and (C) only.
Answer: (D). | ||||
| 43 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Lindlar's catalyst is palladium on charcoal (or $CaCO_3$) partially deactivated (poisoned) with lead salts or quinoline; it reduces alkynes to cis-alkenes.
Answer: (A). | ||||
| 44 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Reaction 1: 1-methylcyclohexanol with 20% $H_3PO_4$ at 358 K is acid-catalysed dehydration through the tertiary carbocation; loss of a ring proton gives the more substituted (Saytzeff) alkene, 1-methylcyclohexene, as A.
Reaction 2: 1-chloro-1-methylcyclohexane with the bulky base $(CH_3)_3CO^-K^+$ undergoes E2 elimination; the bulky base removes the less hindered methyl proton, giving the Hofmann product, methylenecyclohexane, as B.
Answer: (C). | ||||
| 45 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids DIBAL-H at $-78^\circ$C reduces an ester (here a lactone) only as far as the aldehyde level: the lactone becomes a lactol (cyclic hemiacetal). On aqueous work-up the hemiacetal opens to the hydroxy aldehyde: the ring oxygen becomes the OH on the carbon it was attached to, and the carbonyl carbon becomes CHO on the $CH_2$ side chain. The ring double bond is untouched.
P is the dihydronaphthalene with OH at C1 and $CH_2CHO$ at C2.
Answer: (C). | ||||
| 46 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Assertion and reasonMedium | |
A medium assertion and reason question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Acetone is almost entirely keto (enol far below 0.1%). In acetylacetone the enol is conjugated with the second C=O and forms a six-membered intramolecular hydrogen-bonded ring, which stabilises it; its enol content is much higher (about 15% in aqueous solution, more in non-polar media). Both statements are true and R explains A.
Answer: (A). | ||||
| 47 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Aniline with $NaNO_2$ and HCl at 273$-$278 K is diazotised to benzenediazonium chloride, X. Warming the diazonium salt with water (hydrolysis, $H_2O/\Delta$) replaces $N_2^+$ by OH, giving phenol with loss of $N_2$.
Answer: (B). | ||||
| 48 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hofmann bromamide degradation is the reaction of a primary amide with $Br_2$ and NaOH, giving an amine with one carbon fewer.
(B) $PhCH_2CONH_2 \xrightarrow{Br_2,\ NaOH} PhCH_2NH_2$: Hofmann.
(C) $PhCOCl \xrightarrow{NH_3} PhCONH_2 \xrightarrow{Br_2,\ NaOH} PhNH_2$: Hofmann.
(D) $PhCN \xrightarrow{KOH,\ H_2O} PhCONH_2$ (partial hydrolysis) $\xrightarrow{Br_2,\ NaOH} PhNH_2$: Hofmann.
(A) $Br_2$/NaOH acts on the methyl ketone (haloform reaction), giving $PhCH_2COOH$ after acidification; $NH_3/\Delta$ gives the amide $PhCH_2CONH_2$, and $LiAlH_4$ reduces the amide to the amine. No amide meets $Br_2$/NaOH, so no Hofmann degradation is involved.
Answer: (A). | ||||
| 49 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Antihistamines such as brompheniramine and terfenadine block histamine's action at its receptors and are used against allergy; others such as cimetidine and ranitidine block histamine's action on the stomach wall, reducing acid secretion, and serve as antacids.
Answer: (C). | ||||
| 50 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules Fat-soluble vitamins (A, D, E, K) are stored in the liver and adipose tissue; water-soluble ones (B group such as thiamine, and C, ascorbic acid) are excreted in urine and must be supplied regularly. Vitamins A and D are stored longer.
Answer: (B). | ||||
| 51 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 kg of water with 6.50 mol KOH (molar mass 56 g/mol): mass of KOH $= 364$ g, mass of solution $= 1364$ g.
Volume $= \frac{1364}{1.89} = 721.7\ \text{cm}^3 = 0.7217\ \text{dm}^3$.
Molarity $= \frac{6.50}{0.7217} = 9.01 \approx 9$ mol $\text{dm}^{-3}$.
Answer: 9. | ||||
| 52 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter The atomic radius r is the same in both. bcc: $\sqrt{3}a_1 = 4r$; fcc: $\sqrt{2}a_2 = 4r$.
$a_2 = \frac{\sqrt{3}}{\sqrt{2}}a_1 = \frac{1.73\times27}{1.41} = \frac{46.71}{1.41} = 33.1$ Å $\approx 33$.
Answer: 33. | ||||
| 53 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Photon energy: $E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3.0\times10^{8}}{248\times10^{-9}} = 8.02\times10^{-19}$ J $\approx 5.0$ eV.
Kinetic energy of the electrons: $5.0 - 3.0 = 2.0$ eV $= 3.2\times10^{-19}$ J.
$\lambda = \frac{h}{\sqrt{2mK}} = \frac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times3.2\times10^{-19}}} = \frac{6.63\times10^{-34}}{7.63\times10^{-25}} = 8.69\times10^{-10}$ m $\approx 8.7$ Å.
To the nearest integer, 9.
Answer: 9. | ||||
| 54 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\ln K = -\frac{\Delta_rG^\circ}{RT} = \frac{9478}{8.314\times495} = \frac{9478}{4115.4} = 2.303$, so $K = 10$.
At equilibrium, with x mmol of B: $\frac{x}{22 - x} = 10$ (the volume cancels, $\Delta n = 0$), so $x = 20$ mmol.
Answer: 20. | ||||
| 55 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $AB_2 \rightarrow A^{2+} + 2B^-$ gives 3 particles: $i = 1 + (3-1)\alpha = 1 + 2(0.1) = 1.2$.
$\Delta T_b = iK_bm = 1.2\times0.5\times10.0 = 6$ K.
Boiling point $= 100 + 6 = 106$°C.
Answer: 106. | ||||
| 56 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $A_2X$: $K_{sp} = (2s)^2s = 4s^3 = 4.0\times10^{-12}$, so $s^3 = 10^{-12}$ and $S(A_2X) = 10^{-4}$ M.
MX: $K_{sp} = s^2$, so $S(MX) = 2\times10^{-6}$ M.
Ratio $= \frac{10^{-4}}{2\times10^{-6}} = 50$.
Answer: 50. | ||||
| 57 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Each $MnO_4^-$ gains 5 electrons (+7 to +2); each $C_2O_4^{2-}$ loses 2 (two C from +3 to +4). For 2 $MnO_4^-$ (10 electrons) b = 5, so x = 2, y = 10.
Oxygen: $8 + 20 = 20 + z$, so z = 8; hydrogen: c = 2z = 16. Charge check: $-2 - 10 + 16 = +4 = 2\times(+2)$.
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$.
Answer: 16. | ||||
| 58 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln\frac{k_{200}}{k_{300}} = \frac{E_a}{R}\left(\frac{1}{300} - \frac{1}{200}\right) = \frac{11488}{8.314}\times\left(-\frac{1}{600}\right) = -\frac{1381.8}{600} = -2.303$.
So $\frac{k_{200}}{k_{300}} = \frac{1}{10}$ and $k_{200} = 1.0\times10^{-4}\ \text{s}^{-1} = 10\times10^{-5}\ \text{s}^{-1}$.
Answer: 10. | ||||
| 59 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The trans complex is $trans$-$[Co(NH_3)_4Cl_2]Cl$: the coordination sphere holds 4 neutral $NH_3$ ligands. Ethylenediamine is bidentate, each molecule taking two adjacent positions, so the four $NH_3$ positions (one plane, the two Cl being trans) are filled by 2 en, giving $trans$-$[Co(en)_2Cl_2]Cl$.
So 2 equivalents (moles of en per mole of complex) are needed.
Answer: 2. | ||||
| 60 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Hydrogen: $210\times\frac{2}{18} = 23.3$ g, so %H $= \frac{23.3}{750}\times100 = 3.11 \approx 3$.
(Check carbon: $420\times\frac{12}{44} = 114.5$ g, $\frac{114.5}{750}\times100 = 15.3\%$, as stated.)
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | 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 Write $A = iM$ with $M = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$. Then $M^2 = 2M$, so $M^4 = 8M$, and $A^8 = i^8M^8 = M^8 = 2^7M = 128M$.
The system is $128(x - y) = 8$ and $-128(x - y) = 64$, i.e. $x - y = \frac{1}{16}$ and $x - y = -\frac{1}{2}$: inconsistent.
Answer: (A). | ||||
| 62 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $x < 0$: $g(x) = x^3 < 0$, so $f(g(x)) = x^3 + 2$.
For $0 \le x < 1$: $g(x) = x^3 \in [0, 1)$, so $f(g(x)) = x^6$.
For $x \ge 1$: $g(x) = 3x - 2 \ge 1$, so $f(g(x)) = (3x - 2)^2$.
At $x = 0$: the left limit is 2 but $f(g(0)) = 0$, so $f\circ g$ is discontinuous and not differentiable.
At $x = 1$: both pieces give 1, and the derivatives are $6x^5 = 6$ and $6(3x - 2) = 6$: differentiable.
One point.
Answer: (B). | ||||
| 63 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The line passes through $(1, -4, -2)$ with direction $(-1, 2, 3)$. The plane passes through the origin and contains the line, so its normal is $(1, -4, -2)\times(-1, 2, 3) = (-12 + 4,\ -(3 - 2),\ 2 - 4) = (-8, -1, -2)$, i.e. the plane is $8x + y + 2z = 0$.
At A: $-24 - 6 + 2 = -28$; at B: $16 + 4 - 6 = 14$. The plane divides AB in the ratio $28 : 14 = 2 : 1$, so $k = 2$.
Answer: (A). | ||||
| 64 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctHard | |
A hard single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $\vec{CA} = (a, -a, 4)$ is along the normal, so take $(l, m, n) = (a, -a, 4)$. C lies on the plane: $0 + a^2 - 4 = 0$, so $a = 2$ ($a > 0$).
Plane: $2x - 2y + 4z = 0$, i.e. $x - y + 2z = 0$.
Foot from $B(0, 4, 5)$: $t = \frac{0 - 4 + 10}{1 + 1 + 4} = 1$, so $D = (0, 4, 5) - 1\cdot(1, -1, 2) = (-1, 5, 3)$.
$C = (0, -2, -1)$, $\vec{CD} = (-1, 7, 4)$, $CD = \sqrt{1 + 49 + 16} = \sqrt{66}$.
Answer: (B). | ||||
| 65 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctMedium | |
A medium single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Adding 2 to each observation does not change the standard deviation, so d is the standard deviation of a, b, c.
Mean $= \frac{a + b + c}{3} = \frac{2b}{3}$. $d^2 = \frac{a^2 + b^2 + c^2}{3} - \frac{4b^2}{9}$.
Multiply by 9: $9d^2 = 3(a^2 + c^2) + 3b^2 - 4b^2$, so $b^2 = 3(a^2 + c^2) - 9d^2$.
Answer: (A). | ||||
| 66 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra PR: $(3 + 4t,\ -1 - t,\ 2 + 2t)$; QS: $(1 - 2s,\ 2 + s,\ -4 - 2s)$.
Equating: $-1 - t = 2 + s$ gives $s = -3 - t$; then $3 + 4t = 1 + 6 + 2t$ gives $t = 2$, $s = -5$; the z-coordinates agree ($6 = 6$). $T = (11, -3, 6)$.
A vector perpendicular to both: $(4, -1, 2)\times(-2, 1, -2) = (0, 4, 2)$, of length $2\sqrt{5}$. So $\vec{TA} = \pm(0, 2, 1)$.
$A = (11, -1, 7)$ or $(11, -5, 5)$; in both cases $|\vec{OA}|^2 = 121 + 1 + 49 = 121 + 25 + 25 = 171$.
Answer: (B). | ||||
| 67 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra $\sqrt{3}\hat{i} + \hat{j}$ has length 2 and makes $30^\circ$ with the x-axis. After rotation by $45^\circ$: $\alpha = 2\cos75^\circ$, $\beta = 2\sin75^\circ$.
The triangle is right-angled at $(0, \beta)$ with legs $\alpha$ and $\beta$: area $= \frac{1}{2}\alpha\beta = 2\sin75^\circ\cos75^\circ = \sin150^\circ = \frac{1}{2}$.
Answer: (A). | ||||
| 68 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $t = 81^{\sin^2x}$; then $81^{\cos^2x} = \frac{81}{t}$ and $t + \frac{81}{t} = 30$, so $t^2 - 30t + 81 = 0$, $t = 27$ or $3$.
$81^{\sin^2x} = 3^{4\sin^2x}$: $4\sin^2x = 3$ or $1$, so $\sin x = \frac{\sqrt{3}}{2}$ or $\frac{1}{2}$ (in $[0, \pi]$, $\sin x \ge 0$).
$x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{\pi}{6}, \frac{5\pi}{6}$: four roots.
Answer: (C). | ||||
| 69 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $E_1$: missing card is a spade ($P = \frac{1}{4}$), $E_2$: it is not ($P = \frac{3}{4}$). From the remaining 51 cards, two spades are drawn.
$P(\text{2 spades}\mid E_1) = \frac{^{12}C_2}{^{51}C_2} = \frac{66}{^{51}C_2}$, $P(\text{2 spades}\mid E_2) = \frac{^{13}C_2}{^{51}C_2} = \frac{78}{^{51}C_2}$.
$P(E_2\mid\text{2 spades}) = \frac{\frac{3}{4}\cdot78}{\frac{1}{4}\cdot66 + \frac{3}{4}\cdot78} = \frac{234}{66 + 234} = \frac{234}{300} = \frac{39}{50}$.
Answer: (C). | ||||
| 70 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $2\cot\frac{x}{2}\sin^2\frac{x}{2} = 2\sin\frac{x}{2}\cos\frac{x}{2} = \sin x$, so $f(x) = (4a - 3)(x + \ln5) + (a - 7)\sin x$.
$f'(x) = (4a - 3) + (a - 7)\cos x = 0$ needs $\cos x = \frac{3 - 4a}{a - 7}$ ($a = 7$ gives $f' = 25 \neq 0$).
This has a solution when $\left|\frac{3 - 4a}{a - 7}\right| \le 1$: $(3 - 4a)^2 \le (a - 7)^2$, i.e. $15a^2 - 10a - 40 \le 0$, $3a^2 - 2a - 8 \le 0$, $(3a + 4)(a - 2) \le 0$.
$a \in \left[-\frac{4}{3}, 2\right]$.
Answer: (C). | ||||
| 71 | 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 $T_{r+1} = {^{60}C_r}\,3^{\frac{60-r}{4}}5^{\frac{r}{8}}$. It is rational when $8 \mid r$ and $4 \mid (60 - r)$; since $4 \mid 60$, this is $8 \mid r$: $r = 0, 8, \ldots, 56$, 8 terms.
There are 61 terms, so $n = 61 - 8 = 53$ and $n - 1 = 52$, divisible by 26 (not by 30, 8 or 7).
Answer: (C). | ||||
| 72 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Let E be the sum of the even-index coefficients and O of the odd-index ones.
$x = 1$: $E + O = (1 - 1 + 1)^n = 1$. $x = -1$: $E - O = (1 + 1 - 1)^n = 1$.
So $E = 1$, $O = 0$, and $E + 4O = 1$.
Answer: (D). | ||||
| 73 | 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 (D): if $p\wedge q$ is true then q is true, so $p\rightarrow q$ is true; an implication with a false antecedent is true. So (D) is always true.
The others fail when p and q are both false (A gives F, C gives F) or when p is true and q false (B: F $\vee$ F).
Answer: (D). | ||||
| 74 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctHard | |
A hard single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Divide numerator and denominator by $2\cdot4^r$ and put $u = \left(\frac{3}{2}\right)^r$: $\frac{6^r}{2\cdot4^r + 3\cdot9^r} = \frac{u}{2 + 3u^2} = \frac{\frac{3}{2}u - u}{1 + \frac{3}{2}u\cdot u}$.
So each term is $\tan^{-1}\left(\frac{3}{2}\right)^{r+1} - \tan^{-1}\left(\frac{3}{2}\right)^r$, and the sum telescopes:
$S_k = \tan^{-1}\left(\frac{3}{2}\right)^{k+1} - \tan^{-1}\frac{3}{2} \rightarrow \frac{\pi}{2} - \tan^{-1}\frac{3}{2} = \cot^{-1}\frac{3}{2}$.
Answer: (B). | ||||
| 75 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $x \ge 0$: $(x - 3)(x + 4) = 6$, $x^2 + x - 18 = 0$, $x = \frac{-1 + \sqrt{73}}{2}$ (the other root is negative).
$-4 \le x < 0$: $(-x - 3)(x + 4) = 6$, $x^2 + 7x + 18 = 0$: discriminant $49 - 72 < 0$, no root.
$x < -4$: $(-x - 3)(-x - 4) = 6$, $x^2 + 7x + 6 = 0$, $x = -1$ or $-6$; only $x = -6$ lies in the range.
Two elements.
Answer: (B). | ||||
| 76 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin x\cos x = \frac{1}{10}$, so $(\sin x + \cos x)^2 = 1 + \frac{2}{10} = \frac{6}{5}$.
The second condition: $2\log_{10}(\sin x + \cos x) = \log_{10}\frac{n}{10}$, so $\frac{6}{5} = \frac{n}{10}$, $n = 12$.
Answer: (B). | ||||
| 77 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor $e^{\int2\tan x\,dx} = \sec^2x$. Then $y\sec^2x = \int\sin x\sec^2x\,dx = \int\sec x\tan x\,dx = \sec x + C$, so $y = \cos x + C\cos^2x$.
$y\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{C}{4} = 0$ gives $C = -2$: $y = \cos x - 2\cos^2x$.
With $t = \cos x \in [-1, 1]$, $t - 2t^2$ is largest at $t = \frac{1}{4}$: $\frac{1}{4} - \frac{1}{8} = \frac{1}{8}$.
Answer: (D). | ||||
| 78 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The chord of $x^2 + y^2 = 25$ with midpoint $(h, k)$ is $hx + ky = h^2 + k^2$, i.e. $y = -\frac{h}{k}x + \frac{h^2 + k^2}{k}$.
A line $y = mx + c$ touches $\frac{x^2}{9} - \frac{y^2}{16} = 1$ when $c^2 = 9m^2 - 16$:
$\frac{(h^2 + k^2)^2}{k^2} = \frac{9h^2}{k^2} - 16$, so $(h^2 + k^2)^2 = 9h^2 - 16k^2$.
Locus: $(x^2 + y^2)^2 - 9x^2 + 16y^2 = 0$.
Answer: (C). | ||||
| 79 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections For $y^2 = 4Ax$ with $A = \frac{1}{2}$, the normal of slope m is $y = mx - 2Am - Am^3 = mx - m - \frac{m^3}{2}$.
Through $(a, 0)$: $m\left(a - 1 - \frac{m^2}{2}\right) = 0$, so $m = 0$ or $m^2 = 2(a - 1)$.
Three distinct normals need $m^2 > 0$, i.e. $a > 1$.
Answer: (A). | ||||
| 80 | 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 The base $\frac{1}{\sqrt{2}}$ is less than 1, so the inequality reverses: $\frac{|z| + 11}{(|z| - 1)^2} \ge \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$.
With $r = |z|$: $2(r + 11) \ge (r - 1)^2$, $r^2 - 4r - 21 \le 0$, $(r - 7)(r + 3) \le 0$, so $r \le 7$.
Largest $|z| = 7$.
Answer: (C). | ||||
| 81 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|z + i| = |z - 3i|$: z is equidistant from $-i$ and $3i$, so $\text{Im}\,z = 1$. Let $z = x + i$.
$w = x^2 + 1 - 2x - 2i + 2 = (x^2 - 2x + 3) - 2i$. Re(w) is least at $x = 1$: $w = 2 - 2i = 2\sqrt{2}e^{-i\pi/4}$.
$w^n$ is real when $\frac{n\pi}{4}$ is a multiple of $\pi$: least $n = 4$ ($w^4 = -64$).
Answer: 4. | ||||
| 82 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f(x + 1) + f(x + 2) = 2 = f(x) + f(x + 1)$, so $f(x + 2) = f(x)$: period 2.
Over any interval of length 2: $\int_a^{a+2}f = \int_a^{a+1}[f(t) + f(t + 1)]\,dt = 2$.
$I_1$ covers 4 periods: 8. $I_2$ covers 2 periods: 4.
$I_1 + 2I_2 = 8 + 8 = 16$.
Answer: 16. | ||||
| 83 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $y'(x) = 2x^2 - 15x + 10$. The normal has slope $-\frac{1}{3}$, so the tangent slope is 3: $2a^2 - 15a + 10 = 3$, $2a^2 - 15a + 7 = 0$, $a = 7$ or $\frac{1}{2}$; $a > 1$ gives $a = 7$.
$b = \int_0^7(2t^2 - 15t + 10)\,dt = \frac{2\cdot343}{3} - \frac{15\cdot49}{2} + 70 = \frac{1372 - 4410 + 840}{6} = -\frac{413}{6}$.
$|a + 6b| = |7 - 413| = 406$.
Answer: 406. | ||||
| 84 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Expand: $ae^x - b\cos x + ce^{-x} = (a - b + c) + (a - c)x + \frac{a + b + c}{2}x^2 + \ldots$, and $x\sin x = x^2 + \ldots$.
For a finite limit, $a - b + c = 0$ and $a - c = 0$; the limit is then $\frac{a + b + c}{2} = 2$.
$a + b + c = 4$.
Answer: 4. | ||||
| 85 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series From the set the only AP of four terms is 11, 16, 21, 26 (d = 5), and the only GP is 4, 8, 16, 32 (r = 2).
AP terms are $11 + 5k$, the largest four-digit one being 9996; GP terms are $2^m$ from 4 up to 8192.
A common term is a power of 2, at least 11, with $2^m \equiv 1 \pmod 5$, which needs $4 \mid m$: 16, 256, 4096 (65536 is too large).
Three common terms.
Answer: 3. | ||||
| 86 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $P^{-1}AP - I_3 = P^{-1}(A - I_3)P$, so $\det(P^{-1}AP - I_3)^2 = [\det(A - I_3)]^2$.
$A - I_3 = \begin{bmatrix} 1 & 7 & \omega^2 \\ -1 & -\omega - 1 & 1 \\ 0 & -\omega & -\omega \end{bmatrix}$, and $-\omega - 1 = \omega^2$.
$\det = 1(\omega^2\cdot(-\omega) + \omega) - 7(\omega - 0) + \omega^2(\omega - 0) = (-1 + \omega) - 7\omega + 1 = -6\omega$.
Square: $36\omega^2$, so $\alpha = 36$.
Answer: 36. | ||||
| 87 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The limit is $2\int_0^1\log_2\left(1 + \tan\frac{\pi x}{4}\right)dx$. Put $t = \frac{\pi x}{4}$: $= \frac{8}{\pi}\int_0^{\pi/4}\log_2(1 + \tan t)\,dt$.
With $t \rightarrow \frac{\pi}{4} - t$, $1 + \tan\left(\frac{\pi}{4} - t\right) = \frac{2}{1 + \tan t}$, so $J = \int_0^{\pi/4}\log_2(1 + \tan t)\,dt$ satisfies $2J = \int_0^{\pi/4}\log_2 2\,dt = \frac{\pi}{4}$, $J = \frac{\pi}{8}$.
Limit $= \frac{8}{\pi}\cdot\frac{\pi}{8} = 1$.
Answer: 1. | ||||
| 88 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalHard | |
A hard numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $\text{tr}(AA^T)$ is the sum of the squares of all nine entries. Each square is 0, 1, 4 or 9, and they must total 9:
One 9, rest 0: 9 ways.
Two 4s and one 1: $\frac{9!}{2!\,1!\,6!} = 252$.
One 4 and five 1s: $\frac{9!}{1!\,5!\,3!} = 504$.
Nine 1s: 1.
Total $= 9 + 252 + 504 + 1 = 766$.
Answer: 766. | ||||
| 89 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Put $A = (0, 0)$, $B = (1, 0)$, $C = (1, 1)$, $D = (0, 1)$. $C_2$ touches both axes: centre $(r, r)$; it touches $C_1$ internally: $r\sqrt{2} = 1 - r$, so $r = \sqrt{2} - 1$.
Let $E = (1 - s, 0)$ with $s = EB$. Line CE: $x - sy - (1 - s) = 0$. Its distance from $(r, r)$ equals r: $\frac{(1 - s)(1 - r)}{\sqrt{1 + s^2}} = r$.
With $(1 - r)^2 = 6 - 4\sqrt{2} = 2r^2$: $2(1 - s)^2 = 1 + s^2$, so $s^2 - 4s + 1 = 0$, $s = 2 \pm \sqrt{3}$. E is on side AB, so $s = 2 - \sqrt{3}$.
$\alpha = 2$, $\beta = -1$, $\alpha + \beta = 1$.
Answer: 1. | ||||
| 90 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $y = (x + 1)^2 + c$. A finite region with the x-axis needs the parabola to cut the axis: write $y = (x + 1)^2 - k$, $k > 0$, roots $x = -1 \pm \sqrt{k}$.
Area $= \int_{-\sqrt{k}}^{\sqrt{k}}(k - u^2)\,du = \frac{4}{3}k^{3/2}$. Setting $\frac{4}{3}k^{3/2} = \frac{4\sqrt{8}}{3}$: $k^{3/2} = 8^{1/2}$, $k = 2$.
$y = (x + 1)^2 - 2$, so $y(1) = 4 - 2 = 2$.
Answer: 2. | ||||
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