JEE (Advanced) · 2013
JEE Advanced 2013, question by question
Which chapter each question came from, how the paper was spread, 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. One year is not a forecast.
Physics
40 Q
- Atoms and Nuclei5
- Optics4
- Properties of Bulk Matter4
18 easy · 21 medium · 1 hard
Chemistry
40 Q
- p-Block Elements5
- Aldehydes, Ketones and Carboxylic Acids3
- Principles of Qualitative Analysis3
17 easy · 19 medium · 4 hard
Mathematics
40 Q
- Applications of Derivatives6
- Complex Numbers4
- Probability4
5 easy · 23 medium · 12 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 12: Application of Derivatives
Class 11: Complex Numbers and Quadratic Equations
Class 11: Probability; Class 12: Probability
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 11: Conic Sections
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 12: Vector Algebra
Class 11: Sequences and Series
Class 12: Matrices; Class 12: Determinants
Class 12: Differential Equations
Class 11: Straight Lines
Class 11: Binomial Theorem
Class 11: Complex Numbers and Quadratic Equations
Class 11: Conic Sections
Class 12: Atoms; Class 12: Nuclei
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 11: Work, Energy and Power
Class 11: Oscillations; Class 11: Waves
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 11: Laws of Motion
Class 11: Units and Measurement
Class 12: Electromagnetic Induction; Class 12: Alternating Current
No matching NCERT chapter
Class 11: Kinetic Theory
Class 12: Electromagnetic Waves
Class 12: Dual Nature of Radiation and Matter
Class 11: Systems of Particles and Rotational Motion
Class 11: Gravitation
Class 12: Current Electricity
Class 11: Thermodynamics
No matching NCERT chapter
Class 12: Aldehydes, Ketones and Carboxylic Acids
No matching NCERT chapter
Class 11: Thermodynamics
Class 12: Coordination Compounds
Class 11: Equilibrium
Class 12: Alcohols, Phenols and Ethers
No matching NCERT chapter
Class 11: Hydrocarbons
Class 11: Organic Chemistry – Some Basic Principles and Techniques
Class 11: Structure of Atom
Class 12: Electrochemistry
No matching NCERT chapter
No matching NCERT chapter
Class 12: Chemical Kinetics
Class 12: Haloalkanes and Haloarenes
Class 12: Solutions
Class 12: Biomolecules
Class 11: Chemical Bonding and Molecular Structure
03
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | 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 The zero lies between the marks 5.10 cm and 5.15 cm, so one main-scale division is 1 MSD $= 0.05$ cm.
50 Vernier divisions span 2.45 cm, so 1 VSD $= \frac{2.45}{50} = 0.049$ cm.
Least count $= 1\text{ MSD} - 1\text{ VSD} = 0.05 - 0.049 = 0.001$ cm.
Reading $=$ main-scale reading $+$ (coinciding division) $\times$ least count $= 5.10 + 24\times0.001 = 5.124$ cm.
Answer: (B). | ||||
| 2 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics On reflection the component of the ray along the mirror is unchanged and the component along the normal is reversed.
Incident $\hat{e}_1 = \frac{1}{2}(\hat{i}+\sqrt{3}\hat{j})$, reflected $\hat{e}_2 = \frac{1}{2}(\hat{i}-\sqrt{3}\hat{j})$: the $x$-component is kept and the $y$-component flips, so the normal is along $\hat{j}$ (the mirror lies along $x$).
The angle of incidence $i$ is the angle between the ray and the normal: $\cos i = |\hat{e}_1\cdot\hat{j}| = \frac{\sqrt{3}}{2}$, so $i = 30^\circ$.
Answer: (A). | ||||
| 3 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics With equal slits, $I = I_{max}\cos^2\frac{\phi}{2}$, where $\phi$ is the phase difference.
$I = \frac{I_{max}}{2}$ needs $\cos^2\frac{\phi}{2} = \frac{1}{2}$, so $\frac{\phi}{2} = (2n+1)\frac{\pi}{4}$, i.e. $\phi = (2n+1)\frac{\pi}{2}$.
Path difference $\Delta = \frac{\lambda}{2\pi}\phi = (2n+1)\frac{\lambda}{4}$.
Answer: (B). | ||||
| 4 | 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 For each gas in the mixture, $p_iV = \frac{m_i}{M_i}RT$, so its density is $\rho_i = \frac{m_i}{V} = \frac{p_iM_i}{RT}$.
At the same $T$: $\frac{\rho_1}{\rho_2} = \frac{p_1}{p_2}\cdot\frac{M_1}{M_2} = \frac{4}{3}\cdot\frac{2}{3} = \frac{8}{9}$.
Answer: (D). | ||||
| 5 | Properties of Bulk MatterHeat transfer | 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 heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Take each block of length $L$ along $x$ and cross-section $A$. Thermal resistance of a block: $R = \frac{L}{kA}$.
Configuration I (series): $R_I = \frac{L}{\kappa A} + \frac{L}{2\kappa A} = \frac{3L}{2\kappa A}$.
Configuration II (parallel, both blocks between the same two ends): $\frac{1}{R_{II}} = \frac{\kappa A}{L} + \frac{2\kappa A}{L}$, so $R_{II} = \frac{L}{3\kappa A}$.
Heat current $H = \frac{\Delta T}{R}$, so for the same heat the time is proportional to $R$:
$t_{II} = 9\times\frac{R_{II}}{R_I} = 9\times\frac{1/3}{3/2} = 9\times\frac{2}{9} = 2.0$ s.
Answer: (A). | ||||
| 6 | 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 Energy of the pulse: $E = Pt = 30\times10^{-3}\times100\times10^{-9} = 3\times10^{-9}$ J.
Light carrying energy $E$ carries momentum $\frac{E}{c}$; on complete absorption all of it goes to the object:
$p = \frac{E}{c} = \frac{3\times10^{-9}}{3\times10^{8}} = 1.0\times10^{-17}$ kg m s$^{-1}$.
Answer: (B). | ||||
| 7 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion At its highest point the projectile moves horizontally with speed $u_0\cos\alpha$, at height $H = \frac{u_0^2\sin^2\alpha}{2g}$.
The second particle, thrown up with $u_0$, has at height $H$ the speed $v$ with $v^2 = u_0^2 - 2gH = u_0^2 - u_0^2\sin^2\alpha = u_0^2\cos^2\alpha$, so $v = u_0\cos\alpha$ (vertical).
Momentum is conserved in the collision (impulsive forces only):
horizontal $mu_0\cos\alpha$, vertical $mu_0\cos\alpha$.
The composite moves at $\tan\theta = \frac{mu_0\cos\alpha}{mu_0\cos\alpha} = 1$, so $\theta = \frac{\pi}{4}$.
Answer: (A). | ||||
| 8 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power With $\vec{r} = x\hat{i}+y\hat{j}$, the force is $\vec{F} = \frac{K\vec{r}}{r^3} = \frac{K}{r^2}\hat{r}$: it is radial.
Along a circle about the origin the displacement $d\vec{r}$ is tangential, so $\vec{F}\cdot d\vec{r} = 0$ at every point.
(Equally, $\vec{F} = -\nabla\left(\frac{K}{r}\right)$ is conservative and both end points are at $r = a$.)
Work done $= 0$.
Answer: (D). | ||||
| 9 | 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 in series, so both carry the same tension $F$, and both are copper (same $Y$).
Elongation $\Delta l = \frac{Fl}{AY}$.
Thin: $\Delta l_1 = \frac{FL}{\pi R^2Y}$. Thick: $\Delta l_2 = \frac{F(2L)}{\pi(2R)^2Y} = \frac{FL}{2\pi R^2Y}$.
$\frac{\Delta l_1}{\Delta l_2} = 2.00$.
Answer: (C). | ||||
| 10 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Real image, one-third size: $m = \frac{v}{u} = -\frac{1}{3}$ with $v = +8$ m, so $u = -24$ m.
Lens formula: $\frac{1}{f} = \frac{1}{v}-\frac{1}{u} = \frac{1}{8}+\frac{1}{24} = \frac{4}{24}$, so $f = 6$ m.
Refractive index: $\mu = \frac{\lambda_0}{\lambda} = \frac{3}{2}$.
Plano-convex lens: $\frac{1}{f} = (\mu-1)\frac{1}{R}$, so $R = (\mu-1)f = 0.5\times6 = 3$ m.
Answer: (C). | ||||
| 11 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Multiple correctMedium | |
A medium multiple correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $k = \frac{2\pi}{\lambda} = 62.8$ m$^{-1}$ gives $\lambda = \frac{6.28}{62.8} = 0.1$ m; $\omega = 2\pi f = 628$ s$^{-1}$ gives $f = 100$ Hz.
(A) The fifth harmonic has 5 loops, hence 6 nodes (counting the two fixed ends). False.
(B) Five loops: $L = 5\cdot\frac{\lambda}{2} = 0.25$ m. True.
(C) Midpoint $x = 0.125$ m: $\sin(62.8\times0.125) = \sin(2.5\pi) = 1$, so its amplitude is 0.01 m (an antinode). True.
(D) 100 Hz is the fifth-harmonic frequency; the fundamental is $\frac{100}{5} = 20$ Hz. False.
Answer: (B), (C). | ||||
| 12 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctMedium | |
A medium multiple 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 Let $V = \frac{4}{3}\pi R^3$. Total weight $= \rho Vg + 3\rho Vg = 4\rho Vg$. The largest buoyancy possible (both spheres under) $= 2\times2\rho Vg = 4\rho Vg$.
So the system floats exactly when both spheres are completely submerged: the light sphere is fully under the surface (D true, C false), the heavy sphere hanging below it.
Heavy sphere: weight $3\rho Vg$ down, buoyancy $2\rho Vg$ up, spring tension $T$ up: $T = \rho Vg$.
(Check, light sphere: buoyancy $2\rho Vg$ = weight $\rho Vg$ + $T$.)
Elongation $= \frac{T}{k} = \frac{\rho g}{k}\cdot\frac{4}{3}\pi R^3 = \frac{4\pi R^3\rho g}{3k}$: (A) true, (B) false.
Answer: (A), (D). | ||||
| 13 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctMedium | |
A medium multiple correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $|\vec{u}_2| = 2\sqrt{3+1} = 4$ m s$^{-1}$: the speed is unchanged, as it must be in a magnetic field. The velocity has turned from $+x$ by $\theta$ with $\tan\theta = \frac{1}{\sqrt{3}}$, i.e. $\theta = 30^\circ = \frac{\pi}{6}$, towards $+y$.
Direction: at entry the force $Q\vec{u}_1\times\vec{B}$ must point along $+y$. With $\vec{B} = -B\hat{k}$: $\hat{i}\times(-\hat{k}) = \hat{j}$. So $\vec{B}$ is along $-z$: (A) true, (B) false.
Magnitude: the particle turns through $\frac{\pi}{6}$ at angular speed $\omega = \frac{QB}{M}$ in the 10 ms it spends in the field:
$\frac{QB}{M}\times0.01 = \frac{\pi}{6}$, so $B = \frac{100\pi M}{6Q} = \frac{50\pi M}{3Q}$: (C) true, (D) false.
Answer: (A), (C). | ||||
| 14 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The centres are $3R$ apart. By Gauss's law, outside a uniform sphere $E = \frac{\rho a^3}{3\varepsilon_0 r^2}$, inside $E = \frac{\rho r}{3\varepsilon_0}$ (radially outward for $\rho > 0$).
Point between the centres (inside the big sphere, $R$ from its centre):
small sphere: $E_1 = \frac{\rho_1R^3}{3\varepsilon_0(2R)^2} = \frac{\rho_1R}{12\varepsilon_0}$, pointing towards the big sphere;
big sphere: $E_2 = \frac{\rho_2R}{3\varepsilon_0}$, pointing towards the small sphere.
They are opposite, so $E = 0$ needs $\frac{\rho_1}{12} = \frac{\rho_2}{3}$: $\frac{\rho_1}{\rho_2} = 4$ (D).
Point on the far side of the small sphere ($5R$ from the big centre): both fields point the same way for like charges, so the charges must be opposite:
$\frac{\rho_1R}{12\varepsilon_0} = -\frac{\rho_2(2R)^3}{3\varepsilon_0(5R)^2} = -\frac{8\rho_2R}{75\varepsilon_0}$, so $\frac{\rho_1}{\rho_2} = -\frac{96}{75} = -\frac{32}{25}$ (B).
(A) and (C) give no zero on the line at that distance.
Answer: (B), (D). | ||||
| 15 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Read the batteries from the figure: the $2V_0$ cell has its long (positive) plate at the top; the $V_0$ cell has its long plate at the bottom, so its top terminal is $V_0$ below the common bottom wire.
$S_1$ pressed: $C_1$ charges to $2V_0$, upper plate $+2CV_0$.
$S_1$ released, $S_2$ pressed: $C_1$ and $C_2$ are joined in parallel (top to top, bottom to bottom) and share the charge $2CV_0$ equally: each upper plate $+CV_0$.
$S_2$ released: $C_1$ is now isolated and keeps $+CV_0$ on its upper plate. (B) true, (A) false.
$S_3$ pressed: $C_2$ is connected across the $V_0$ cell, whose top terminal is at $-V_0$ relative to the bottom wire. In the steady state the upper plate of $C_2$ carries $C(-V_0) = -CV_0$. (D) true, (C) false.
Answer: (B), (D). | ||||
| 16 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation: $eV_0 = h\nu - \phi$, so $V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$.
The slope $\frac{h}{e}$ is the same for every metal; the work function only shifts the intercept.
Ratio of slopes $= 1$.
Answer: 1. | ||||
| 17 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\lambda = \frac{\ln2}{T_{1/2}} = \frac{0.693}{1386} = 5\times10^{-4}$ s$^{-1}$.
Fraction decayed in 80 s: $1 - e^{-\lambda t} = 1 - e^{-0.04} \approx 0.04 - \frac{0.04^2}{2} = 0.0392$.
That is 3.92%, which to the nearest integer is 4%. (The activity $10^3$ s$^{-1}$ is not needed.)
Answer: 4. | ||||
| 18 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Work done by the force in time $t$ = $Pt$, all of which becomes kinetic energy (work-energy theorem):
$\frac{1}{2}mv^2 = Pt$, so $v = \sqrt{\frac{2Pt}{m}} = \sqrt{\frac{2\times0.5\times5}{0.2}} = \sqrt{25} = 5$ m s$^{-1}$.
Answer: 5. | ||||
| 19 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion No external torque about the vertical axis, so angular momentum about it is conserved.
Disc: $I_d = \frac{1}{2}MR^2 = \frac{1}{2}\times50\times0.16 = 4$ kg m$^2$.
Each ring lies flat with its rim touching the axis, so its centre is $0.2$ m from the axis. About a vertical axis through its own centre $I = mr^2 = 6.25\times0.04 = 0.25$; by the parallel-axis theorem about the disc's axis $I = 0.25 + 6.25\times0.2^2 = 0.5$ kg m$^2$. Two rings: 1 kg m$^2$.
$4\times10 = (4+1)\omega$, so $\omega = 8$ rad s$^{-1}$.
Answer: 8. | ||||
| 20 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power With the minimum launch speed $\sqrt{5gl_1}$, the first bob reaches the top of its circle with speed $\sqrt{gl_1}$ (tension just zero there).
The second bob hangs at rest at the lowest point of its own circle, at that place. An elastic head-on collision between equal masses exchanges velocities, so the second bob starts with $\sqrt{gl_1}$.
To complete its circle from the lowest point it needs $\sqrt{5gl_2}$:
$gl_1 = 5gl_2$, so $\frac{l_1}{l_2} = 5$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 21 | 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 An acid liberates $CO_2$ from $NaHCO_3$ only if it is stronger than carbonic acid ($pK_a \approx 6.4$).
Benzoic acid ($pK_a \approx 4.2$), salicylic acid ($\approx 3.0$) and benzenesulphonic acid (a strong acid) all are, and give effervescence.
Phenol ($pK_a \approx 10$) is weaker than carbonic acid and does not react with sodium bicarbonate.
Answer: (D). | ||||
| 22 | 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 Concentrated nitric acid slowly decomposes on standing, faster in light:
$4HNO_3 \rightarrow 4NO_2 + 2H_2O + O_2$.
The brown gas $NO_2$ dissolves in the acid and colours it yellow-brown.
Answer: (B). | ||||
| 23 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Adsorption of methylene blue on charcoal is physical adsorption (van der Waals forces).
Adsorption is spontaneous ($\Delta G < 0$) while the adsorbate loses freedom ($\Delta S < 0$), so $\Delta H$ must be negative: adsorption is exothermic. (B) is correct.
(A) Physisorption needs no activation energy. (C) Being exothermic, it decreases as the temperature rises. (D) Physisorption is reversible.
Answer: (B). | ||||
| 24 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Silver: argentite $Ag_2S$. Copper: copper pyrites $CuFeS_2$, copper glance $Cu_2S$. Lead: galena $PbS$.
Tin occurs as cassiterite $SnO_2$, magnesium as carbonate/chloride ores (magnesite, carnallite) and aluminium as bauxite (oxide), so the other sets each contain a metal not found chiefly as sulphide.
Answer: (A). | ||||
| 25 | Solid StateIonic radii and radius ratio | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ionic radii and radius ratio, from Solid State. NCERT: No matching NCERT chapter The figure shows $A^+$ in the void formed by six $X^-$ ions (three in one layer, three staggered in the next): an octahedral void, coordination number 6.
For an octahedral void the limiting radius ratio is $\frac{r_+}{r_-} = \sqrt{2}-1 = 0.414$.
$r_{A^+} = 0.414\times250 = 103.5 \approx 104$ pm.
Answer: (A). | ||||
| 26 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter In ammoniacal medium the sulphide-ion concentration is high, and the group IV cations, whose sulphides have larger $K_{sp}$, precipitate: $Zn^{2+} + S^{2-} \rightarrow ZnS$ (white).
$Al^{3+}$ comes down as $Al(OH)_3$, not as a sulphide ($Al_2S_3$ is hydrolysed by water). $Fe^{3+}$ is removed in group III as $Fe(OH)_3$ by $NH_4OH/NH_4Cl$. $Mg^{2+}$ forms no sulphide precipitate and stays in solution (group VI).
Answer: (D). | ||||
| 27 | ThermodynamicsHess law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics $C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$
$\Delta_cH^\circ = 6(-400) + 6(-300) - (-1300) - 0 = -4200 + 1300 = -2900$ kJ mol$^{-1}$.
Molar mass of glucose $= 180$ g mol$^{-1}$, so per gram: $\frac{-2900}{180} = -16.11$ kJ g$^{-1}$.
Answer: (C). | ||||
| 28 | 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 $P$: $Fe^{3+}$, $3d^5$; $F^-$ is a weak-field ligand, high spin, 5 unpaired electrons: $\mu = \sqrt{35} = 5.92$ B.M.
$Q$: $V^{2+}$, $3d^3$: 3 unpaired, $\mu = \sqrt{15} = 3.87$ B.M.
$R$: $Fe^{2+}$, $3d^6$ with weak-field $H_2O$, high spin: 4 unpaired, $\mu = \sqrt{24} = 4.90$ B.M.
Order: $Q < R < P$.
Answer: (B). | ||||
| 29 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For $P$: $t_{75\%} = 2t_{50\%}$ means the half-life does not depend on concentration (75% is two half-lives), which is first order in $P$.
For $Q$: $[Q]$ falls linearly with time, so its rate does not depend on $[Q]$: zero order in $Q$.
Rate $= k[P]^1[Q]^0$; overall order $= 1 + 0 = 1$.
Answer: (D). | ||||
| 30 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $S$ (phenacyl chloride, $C_6H_5COCH_2Cl$): an $\alpha$-halo ketone. The $C=O$ group's $\pi^*$ overlaps the $S_N2$ transition state and stabilises it; such halides react with iodide many thousands of times faster than simple primary halides. Fastest.
$P$ ($CH_3Cl$): least hindered carbon, very fast.
$R$ (allyl chloride): primary, with the transition state stabilised by the adjacent $\pi$ bond; fast, placed after methyl here.
$Q$ (2-chloropropane): secondary, most hindered, slowest.
Order: $S > P > R > Q$. This is also the only option that has the $\alpha$-halo ketone first and the secondary chloride last.
Answer: (B). | ||||
| 31 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Multiple correctMedium | |
A medium multiple correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) $[Cr(NH_3)_5Cl]Cl_2$ ($Ma_5b$) has no geometrical isomers, and with $Cl^-$ both inside and outside it has no ionisation isomer. $[Cr(NH_3)_4Cl_2]Cl$ ($Ma_4b_2$) shows cis–trans isomerism. Not the same.
(B) $[Co(NH_3)_4Cl_2]^+$ (octahedral $Ma_4b_2$): cis and trans. $[Pt(NH_3)_2(H_2O)Cl]^+$ (square planar $Ma_2bc$): cis and trans. Both geometrical. Same.
(C) $[CoBr_2Cl_2]^{2-}$ is tetrahedral and has no geometrical isomers; $[PtBr_2Cl_2]^{2-}$ is square planar and has cis and trans forms. Not the same.
(D) $[Pt(NH_3)_3(NO_3)]Cl$ has the ionisation isomer $[Pt(NH_3)_3Cl]NO_3$; $[Pt(NH_3)_3Cl]Br$ has the ionisation isomer $[Pt(NH_3)_3Br]Cl$. Both ionisation. Same.
Answer: (B), (D). | ||||
| 32 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Multiple correctMedium | |
A medium multiple correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A) $AlCl_3$ removes $Cl^-$ from 3-chlorocyclopropene, giving the cyclopropenyl cation: planar, cyclic, fully conjugated, $2\pi$ electrons ($4n+2$, $n=0$). Aromatic.
(B) $NaH$ removes the acidic $CH_2$ proton of cyclopentadiene, giving the cyclopentadienyl anion: $6\pi$ electrons. Aromatic.
(C) Hexane-2,5-dione with ammonia (from $(NH_4)_2CO_3$) undergoes the Paal–Knorr condensation to 2,5-dimethylpyrrole: the nitrogen lone pair completes a $6\pi$ ring. Aromatic.
(D) Tropone is protonated on oxygen by $HCl$, giving the hydroxytropylium cation: a seven-membered ring with $6\pi$ electrons. Aromatic.
Answer: (A), (B), (C), (D). | ||||
| 33 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Multiple correctMedium | |
A medium multiple correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Hyperconjugation is the donation of electron density from a filled $C$–$H$ $\sigma$ orbital into an adjacent empty or antibonding orbital.
tert-Butyl cation: the carbocation carbon has an empty $p$ orbital; the $\alpha$ $C$–$H$ bonds donate into it: $\sigma\rightarrow p$ (empty).
2-Butene: the methyl $C$–$H$ bonds donate into the empty antibonding $\pi^*$ orbital of $C=C$: $\sigma\rightarrow\pi^*$.
Donation into a filled orbital ($p$ filled, or bonding $\pi$) gives no stabilisation, which rules out (B), (C) and (D).
Answer: (A). | ||||
| 34 | SolutionsRaoult's law | Class 12: Solutions | Multiple correctEasy | |
A easy multiple correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions For an ideal solution: $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$ (D true).
Mixing increases randomness: $\Delta S_{system} > 0$ (B true).
No heat passes to the surroundings, so $\Delta S_{surroundings} = -\frac{\Delta H}{T} = 0$ (C true).
$\Delta G = \Delta H - T\Delta S = -T\Delta S < 0$: mixing is spontaneous, so (A) is false.
Answer: (B), (C), (D). | ||||
| 35 | EquilibriumIonic equilibrium | Class 11: Equilibrium | Multiple correctMedium | |
A medium multiple correct question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Acid-catalysed ester hydrolysis has rate $\propto [H^+]$.
Strong acid HX (1 M): $[H^+] = 1$ M. The weak acid gives 1/100 of the rate, so its $[H^+] = 0.01$ M.
For HA: $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(0.01)^2}{1-0.01} \approx 1\times10^{-4}$.
Answer: (A). | ||||
| 36 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Step 1 ($H_3O^+$, heat): the cyclic anhydride is hydrolysed to two $-COOH$ groups. Each sits on a carbon next to a ring ketone, so each is a $\beta$-keto acid, and on heating both lose $CO_2$. What remains is the bicyclic dione: a cyclohexene ring fused to a cyclohexane-1,4-dione.
Steps 2 and 3 ($O_3$, then $H_2O_2$): ozonolysis with oxidative work-up cleaves the ring $C=C$, and each of its two $=CH-$ carbons becomes a $-COOH$.
$P$ is therefore a dione with two carboxylic acid groups, the two $-CH_2COOH$ arms on the cyclohexane-1,4-dione ring.
Answer: 2. | ||||
| 37 | BiomoleculesProteins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The free $-COOH$ is on alanine, so Ala is the C-terminal (fourth) residue.
The other three, Gly, Val and Phe, fill positions 1–3 in $3! = 6$ ways.
The free $-NH_2$ is on residue 1, which must have a chiral $\alpha$-carbon. Glycine ($H_2NCH_2COOH$) is achiral, so Gly cannot be first: that removes $2! = 2$ sequences.
$6 - 2 = 4$ (Val-Gly-Phe-Ala, Val-Phe-Gly-Ala, Phe-Gly-Val-Ala, Phe-Val-Gly-Ala).
Answer: 4. | ||||
| 38 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $EDTA^{4-}$ is hexadentate: it binds through 2 nitrogen atoms and 4 carboxylate oxygen atoms, making an octahedral complex.
Every pair of donor atoms subtends an angle at $Co$. Each of the 2 N atoms forms an angle with each of the 4 O atoms: $2\times4 = 8$ $N$–$Co$–$O$ angles.
(Of these, the two N being cis to each other, each N is cis to three O (90°) and trans to one O (180°): six of 90° and two of 180°.)
Answer: 8. | ||||
| 39 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Melamine is 2,4,6-triamino-1,3,5-triazine, $C_3N_6H_6$: a triazine ring of three C and three N, with an $-NH_2$ on each carbon.
Each of the three ring nitrogens carries one lone pair, and each of the three amino nitrogens carries one lone pair. Carbon and hydrogen carry none.
Total $= 3 + 3 = 6$.
Answer: 6. | ||||
| 40 | 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 For gas molecules with thermal kinetic energy proportional to $T$, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \propto \frac{1}{\sqrt{mT}}$.
$T_{He} = 200$ K, $T_{Ne} = 1000$ K.
$M = \frac{\lambda_{He}}{\lambda_{Ne}} = \sqrt{\frac{m_{Ne}T_{Ne}}{m_{He}T_{He}}} = \sqrt{\frac{20\times1000}{4\times200}} = \sqrt{25} = 5$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 41 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctHard | |
A hard single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations First circle: $|\alpha - z_0|^2 = r^2$, i.e. $|\alpha|^2 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) + |z_0|^2 = r^2$. (1)
Second circle: $\left|\frac{1}{\bar{\alpha}} - z_0\right|^2 = 4r^2$. Multiply by $|\alpha|^2$: $|1 - \bar{\alpha}z_0|^2 = 4r^2|\alpha|^2$, i.e. $1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) + |\alpha|^2|z_0|^2 = 4r^2|\alpha|^2$. (2)
Subtract (1) from (2): $1 - |\alpha|^2 + |z_0|^2(|\alpha|^2 - 1) = r^2(4|\alpha|^2 - 1)$, i.e. $(|\alpha|^2-1)(|z_0|^2-1) = r^2(4|\alpha|^2-1)$.
Given $|z_0|^2 = \frac{r^2+2}{2}$, so $|z_0|^2 - 1 = \frac{r^2}{2}$:
$\frac{r^2}{2}(|\alpha|^2-1) = r^2(4|\alpha|^2-1)$, so $|\alpha|^2 - 1 = 8|\alpha|^2 - 2$, $7|\alpha|^2 = 1$.
$|\alpha| = \frac{1}{\sqrt{7}}$.
Answer: (C). | ||||
| 42 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability By independence, the probability that nobody solves it is
$\left(1-\frac{1}{2}\right)\left(1-\frac{3}{4}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right) = \frac{1}{2}\cdot\frac{1}{4}\cdot\frac{3}{4}\cdot\frac{7}{8} = \frac{21}{256}$.
At least one solves it: $1 - \frac{21}{256} = \frac{235}{256}$.
Answer: (A). | ||||
| 43 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $g(x) = e^{-2x}f(x)$. Then $g'(x) = e^{-2x}(f'(x) - 2f(x)) < 0$, so $g$ is decreasing.
For $x > \frac{1}{2}$: $g(x) < g\left(\frac{1}{2}\right) = e^{-1}$, so $f(x) < e^{2x-1}$.
Hence $0 < \int_{1/2}^{1} f(x)\,dx < \int_{1/2}^{1} e^{2x-1}\,dx = \left[\frac{e^{2x-1}}{2}\right]_{1/2}^{1} = \frac{e-1}{2}$
(the left inequality because $f$ is positive).
Answer: (D). | ||||
| 44 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $f(x) = x^2 - x\sin x - \cos x$. Then $f'(x) = 2x - \sin x - x\cos x + \sin x = x(2-\cos x)$.
Since $2 - \cos x > 0$, $f$ decreases on $(-\infty, 0)$ and increases on $(0, \infty)$.
$f(0) = -1 < 0$, and $f(x) \rightarrow \infty$ as $x \rightarrow \pm\infty$ (the $x^2$ term dominates).
So $f$ crosses zero exactly once on each side of 0: two points.
Answer: (C). | ||||
| 45 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $\left[0, \frac{\pi}{2}\right]$ both $\sin x, \cos x \ge 0$, so $\sin x + \cos x \ge |\cos x - \sin x|$.
On $\left[0, \frac{\pi}{4}\right]$: $|\cos x - \sin x| = \cos x - \sin x$; the gap is $2\sin x$; $\int_0^{\pi/4} 2\sin x\,dx = 2\left(1 - \frac{1}{\sqrt{2}}\right) = 2 - \sqrt{2}$.
On $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$: $|\cos x - \sin x| = \sin x - \cos x$; the gap is $2\cos x$; $\int_{\pi/4}^{\pi/2} 2\cos x\,dx = 2 - \sqrt{2}$.
Area $= 4 - 2\sqrt{2} = 2\sqrt{2}(\sqrt{2}-1)$.
Answer: (B). | ||||
| 46 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} = \frac{y}{x} + \sec\frac{y}{x}$ is homogeneous. Put $y = vx$: $v + x\frac{dv}{dx} = v + \sec v$, so $\cos v\,dv = \frac{dx}{x}$.
Integrating: $\sin v = \log x + C$, i.e. $\sin\frac{y}{x} = \log x + C$.
At $(1, \frac{\pi}{6})$: $\sin\frac{\pi}{6} = 0 + C$, so $C = \frac{1}{2}$.
$\sin\left(\frac{y}{x}\right) = \log x + \frac{1}{2}$.
Answer: (A). | ||||
| 47 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $1 + \sum_{k=1}^{n}2k = 1 + n(n+1)$.
$\cot^{-1}(1 + n(n+1)) = \tan^{-1}\frac{1}{1+n(n+1)} = \tan^{-1}\frac{(n+1)-n}{1+n(n+1)} = \tan^{-1}(n+1) - \tan^{-1}n$.
The sum telescopes: $\sum_{n=1}^{23} = \tan^{-1}24 - \tan^{-1}1 = \tan^{-1}\frac{24-1}{1+24} = \tan^{-1}\frac{23}{25}$.
So $\cot\left(\tan^{-1}\frac{23}{25}\right) = \frac{25}{23}$.
Answer: (B). | ||||
| 48 | Straight LinesCartesian coordinates and distance formula | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines Subtracting the equations: $(a-b)(x-y) = 0$, and $a \ne b$, so $x = y$. Then $(a+b)x + c = 0$: the lines meet at $\left(-\frac{c}{a+b}, -\frac{c}{a+b}\right)$.
Distance from $(1, 1)$: $\sqrt{2}\left(1 + \frac{c}{a+b}\right) < 2\sqrt{2}$, so $\frac{c}{a+b} < 1$, i.e. $a + b - c > 0$: (A).
(Since $a > c$, (A) in fact holds for every $a > b > c > 0$; and since $a > b$, $a - b + c > 0$ holds too, so (C) is also true, while (B) and (D) are false.)
The official key accepts (A) or (C) or both.
Answer: (A). | ||||
| 49 | 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 A point of the line: $(2t-2, -t-1, 3t)$. Its coordinate sum is $4t - 3$.
Foot on $x + y + z = 3$: move along the normal $(1,1,1)$ by $\frac{3 - (4t-3)}{3} = \frac{6-4t}{3}$:
$x = 2t - 2 + \frac{6-4t}{3} = \frac{2t}{3}$, $y = -t - 1 + \frac{6-4t}{3} = 1 - \frac{7t}{3}$, $z = 3t + \frac{6-4t}{3} = 2 + \frac{5t}{3}$.
With $s = \frac{t}{3}$: $(x, y, z) = (0, 1, 2) + s(2, -7, 5)$, i.e. $\frac{x}{2} = \frac{y-1}{-7} = \frac{z-2}{5}$.
Answer: (D). | ||||
| 50 | 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 Let $\vec{a} = \vec{PQ}$, $\vec{b} = \vec{PS}$. Then $\vec{PR} = \vec{a} + \vec{b}$ and $\vec{SQ} = \vec{a} - \vec{b}$.
$\vec{a} = \frac{1}{2}(\vec{PR} + \vec{SQ}) = \frac{1}{2}(4, -2, -6) = (2, -1, -3)$; $\vec{b} = \frac{1}{2}(\vec{PR} - \vec{SQ}) = \frac{1}{2}(2, 4, 2) = (1, 2, 1)$.
Volume $= |[\vec{PT}\ \vec{a}\ \vec{b}]| = \left|\begin{vmatrix}1&2&3\\2&-1&-3\\1&2&1\end{vmatrix}\right| = |1(-1+6) - 2(2+3) + 3(4+1)| = |5 - 10 + 15| = 10$.
Answer: (C). | ||||
| 51 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Multiple correctMedium | |
A medium multiple correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{k(k+1)}{2}$ for $k = 1, 2, 3, 4$ is $1, 3, 6, 10$: odd, odd, even, even, and the pattern repeats with period 4. So the signs run $-, -, +, +$.
Group the terms in fours ($k = 4m-3, \dots, 4m$):
$-(4m-3)^2 - (4m-2)^2 + (4m-1)^2 + (4m)^2 = [(4m-1)^2-(4m-3)^2] + [(4m)^2-(4m-2)^2] = 2(8m-4) + 2(8m-2) = 32m - 12$.
$S_n = \sum_{m=1}^{n}(32m - 12) = 16n(n+1) - 12n = 4n(4n+1)$.
$n = 8$: $32\times33 = 1056$ (A). $n = 9$: $36\times37 = 1332$ (D).
1088 and 1120 lie between 1056 and 1332, which are consecutive values, so they are not values of $S_n$.
Answer: (A), (D). | ||||
| 52 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants (A) $(N^TMN)^T = N^TM^TN = \pm N^TMN$ as $M^T = \pm M$. Correct.
(B) For symmetric $M, N$: $(MN - NM)^T = N^TM^T - M^TN^T = NM - MN = -(MN - NM)$. Correct.
(C) $(MN)^T = N^TM^T = NM$, which equals $MN$ only when $M$ and $N$ commute. Not correct in general.
(D) $\text{adj}(MN) = (\text{adj}\,N)(\text{adj}\,M)$, the reverse order; $(\text{adj}\,M)(\text{adj}\,N)$ differs unless they commute. Not correct in general.
Answer: (C), (D). | ||||
| 53 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Multiple correctMedium | |
A medium multiple correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = \sin\pi x + \pi x\cos\pi x$. Where $\cos\pi x \ne 0$, $f'(x) = 0 \iff \tan\pi x = -\pi x$. (At $x = n + \frac{1}{2}$, $f' = \pm1 \ne 0$; at $x = n$, $f' = \pm\pi n \ne 0$.)
For $x > 0$, $-\pi x < 0$, so a root needs $\tan\pi x < 0$.
On $\left(n, n+\frac{1}{2}\right)$, $\tan\pi x > 0$: no root. (A) false.
On $\left(n+\frac{1}{2}, n+1\right)$, $\tan\pi x$ increases from $-\infty$ to 0 while $-\pi x$ decreases; $h(x) = \tan\pi x + \pi x$ is strictly increasing from $-\infty$ to $\pi(n+1) > 0$, so it has exactly one zero. (B) true.
So on $(n, n+1)$ there is exactly one point: (C) true, (D) false.
Answer: (B), (C). | ||||
| 54 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Multiple correctHard | |
A hard multiple correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Sides $8x$ and $15x$ ($x$ fixed by the fixed perimeter); squares of side $s$ cut from the corners. $4s^2 = 100$ gives $s = 5$.
$V(s) = s(8x - 2s)(15x - 2s) = 4s^3 - 46xs^2 + 120x^2s$.
$V'(s) = 12s^2 - 92xs + 120x^2$. The maximum is at $s = 5$:
$300 - 460x + 120x^2 = 0 \Rightarrow 6x^2 - 23x + 15 = 0 \Rightarrow x = \frac{23 \pm 13}{12} = 3$ or $\frac{5}{6}$.
$x = \frac{5}{6}$ gives a side $8x = \frac{20}{3} < 2s = 10$, impossible. So $x = 3$, and $V''(5) = 120 - 276 < 0$ confirms a maximum.
Sides: $8\times3 = 24$ and $15\times3 = 45$.
Answer: (A), (C). | ||||
| 55 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctMedium | |
A medium multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Directions: $l_1$ along $(1, 2, 2)$, $l_2$ along $(2, 2, 1)$. $l$ is along their cross product $(2\cdot1 - 2\cdot2,\ 2\cdot2 - 1\cdot1,\ 1\cdot2 - 2\cdot2) = (-2, 3, -2)$, so $l: \lambda(-2, 3, -2)$.
Meeting $l_1$: $3 + t = -2\lambda$, $-1 + 2t = 3\lambda$, $4 + 2t = -2\lambda$. The first and third give $t = -1$, $\lambda = -1$, and the second checks ($-3 = -3$). Intersection: $(2, -3, 2)$.
Point of $l_2$: $(3+2s, 3+2s, 2+s)$. Distance squared to $(2, -3, 2)$:
$(1+2s)^2 + (6+2s)^2 + s^2 = 9s^2 + 28s + 37 = 17 \Rightarrow 9s^2 + 28s + 20 = 0 \Rightarrow s = -2$ or $-\frac{10}{9}$.
$s = -2$: $(-1, -1, 0)$ (B). $s = -\frac{10}{9}$: $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$ (D).
Answer: (B), (D). | ||||
| 56 | 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 Let $N = n + 5$ and the coefficients be $\binom{N}{r-1}, \binom{N}{r}, \binom{N}{r+1}$.
$\frac{\binom{N}{r}}{\binom{N}{r-1}} = \frac{N-r+1}{r} = 2 \Rightarrow N + 1 = 3r$.
$\frac{\binom{N}{r+1}}{\binom{N}{r}} = \frac{N-r}{r+1} = \frac{14}{10} = \frac{7}{5} \Rightarrow 5N - 5r = 7r + 7 \Rightarrow 5N = 12r + 7$.
With $N = 3r - 1$: $15r - 5 = 12r + 7$, so $r = 4$, $N = 11$ (check: $\binom{11}{3}:\binom{11}{4}:\binom{11}{5} = 165:330:462 = 5:10:14$).
$n = N - 5 = 6$.
Answer: 6. | ||||
| 57 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{n(n+1)}{2} - k - (k+1) = 1224$, so $\frac{n(n+1)}{2} = 1225 + 2k$ with $1 \le k \le n-1$.
Thus $1227 \le \frac{n(n+1)}{2} \le 1223 + 2n$.
$n = 49$: 1225, too small. $n = 50$: 1275, giving $2k = 50$, $k = 25$ (valid). $n = 51$: 1326, $2k = 101$, not an integer. For $n \ge 52$, $\frac{n(n+1)}{2} > 1223 + 2n$.
So $k = 25$ and $k - 20 = 5$.
Answer: 5. | ||||
| 58 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | NumericalHard | |
A hard numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Let $P(E_1) = x$, $P(E_2) = y$, $P(E_3) = z$. By independence
$p = (1-x)(1-y)(1-z)$, $\alpha = x(1-y)(1-z) = \frac{x}{1-x}p$, $\beta = \frac{y}{1-y}p$, $\gamma = \frac{z}{1-z}p$.
Write $a = \frac{x}{1-x}$, $b = \frac{y}{1-y}$, $c = \frac{z}{1-z}$.
First equation: $(a - 2b)p^2 = abp^2 \Rightarrow a - 2b = ab \Rightarrow \frac{1}{b} - \frac{2}{a} = 1$.
With $\frac{1}{a} = \frac{1}{x} - 1$, $\frac{1}{b} = \frac{1}{y} - 1$: $\frac{1}{y} - 1 - \frac{2}{x} + 2 = 1 \Rightarrow \frac{1}{y} = \frac{2}{x}$.
Second equation: $b - 3c = 2bc \Rightarrow \frac{1}{c} - \frac{3}{b} = 2 \Rightarrow \frac{1}{z} - 1 - \frac{3}{y} + 3 = 2 \Rightarrow \frac{1}{z} = \frac{3}{y} = \frac{6}{x}$.
So $\frac{x}{z} = 6$.
Answer: 6. | ||||
| 59 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $P, Q = (h, \pm k)$ with $k = \sqrt{3\left(1 - \frac{h^2}{4}\right)} = \frac{\sqrt{3}}{2}\sqrt{4-h^2}$.
By symmetry the tangents meet on the $x$-axis; the tangent at $P$, $\frac{hx}{4} + \frac{ky}{3} = 1$, meets $y = 0$ at $R = \left(\frac{4}{h}, 0\right)$.
$\Delta(h) = \frac{1}{2}(2k)\left(\frac{4}{h} - h\right) = k\cdot\frac{4-h^2}{h} = \frac{\sqrt{3}}{2}\cdot\frac{(4-h^2)^{3/2}}{h}$,
which decreases as $h$ increases on $\left[\frac{1}{2}, 1\right]$ (numerator falls, denominator rises).
$\Delta_1 = \Delta\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}\cdot2\left(\frac{15}{4}\right)^{3/2} = \sqrt{3}\cdot\frac{15\sqrt{15}}{8} = \frac{45\sqrt{5}}{8}$.
$\Delta_2 = \Delta(1) = \frac{\sqrt{3}}{2}\cdot3\sqrt{3} = \frac{9}{2}$.
$\frac{8}{\sqrt{5}}\cdot\frac{45\sqrt{5}}{8} - 8\cdot\frac{9}{2} = 45 - 36 = 9$.
Answer: 9. | ||||
| 60 | 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 The eight vectors are the four body diagonals of a cube, each with both signs: $\pm(1,1,1)$, $\pm(1,1,-1)$, $\pm(1,-1,1)$, $\pm(-1,1,1)$.
Two vectors of the same $\pm$ pair are collinear, so any three including such a pair are coplanar.
Three vectors from three different diagonals are non-coplanar: e.g. $\begin{vmatrix}1&1&1\\1&1&-1\\1&-1&1\end{vmatrix} = -4 \ne 0$, and by the symmetry of the cube every such triple behaves alike (sign changes only change the sign of the determinant).
Number of ways: choose 3 of the 4 diagonals, $\binom{4}{3} = 4$, and a sign for each, $2^3 = 8$: $4\times8 = 32 = 2^5$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Multiple correctMedium | |
A medium multiple correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $d = \frac{\lambda}{2\sin\theta}$, with $\lambda$ exact and $\Delta\theta$ fixed.
Absolute error: $\Delta d = \left|\frac{dd}{d\theta}\right|\Delta\theta = \frac{\lambda\cos\theta}{2\sin^2\theta}\Delta\theta$. As $\theta$ grows from $0^\circ$ towards $90^\circ$, $\cos\theta$ falls and $\sin^2\theta$ rises, so $\Delta d$ decreases. (A) and (B) are false.
Fractional error: $\frac{\Delta d}{d} = \cot\theta\,\Delta\theta$, which decreases as $\theta$ increases. (C) is false and (D) is true.
Answer: (D). | ||||
| 2 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Inside a uniformly charged sphere, Gauss's law gives $\vec{E} = \frac{\rho}{3\varepsilon_0}\vec{r}$, with $\vec{r}$ measured from its centre.
At a point of the overlap, let $\vec{r}_1$ and $\vec{r}_2$ be its positions from the centres $C_1$ and $C_2$. Then
$\vec{E} = \frac{\rho}{3\varepsilon_0}\vec{r}_1 - \frac{\rho}{3\varepsilon_0}\vec{r}_2 = \frac{\rho}{3\varepsilon_0}(\vec{r}_1 - \vec{r}_2) = \frac{\rho}{3\varepsilon_0}\overrightarrow{C_1C_2}$,
the same vector at every point of the overlap.
(A) False: the field is $\frac{\rho}{3\varepsilon_0}|C_1C_2| \neq 0$.
(B) False: a non-zero uniform field means the potential changes along $\overrightarrow{C_1C_2}$.
(C) True: the magnitude is constant.
(D) True: the direction is everywhere along $C_1 \to C_2$.
Answer: (C), (D). | ||||
| 3 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctMedium | |
A medium multiple correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter With $T$ rising at a constant rate, the rate of heat absorption is $\frac{dQ}{dt} = mC(T)\frac{dT}{dt}$, proportional to $C(T)$. So each statement is read off the graph of $C$.
(A) From 0 to 100 K the curve starts from the origin almost as a straight line, so $C$, and with it the rate of absorption, varies roughly linearly with $T$. True to a reasonable approximation (the key also accepts the answer without it).
(B) Heat absorbed $= m\int C\,dT$. Over 0–100 K, $C$ is well below its value over 400–500 K, so less heat is needed for the same 100 K rise. True.
(C) Over 400–500 K, $C$ is flat (at its saturation value), so the rate of absorption does not change. True.
(D) Over 200–300 K, $C$ is still rising, so the rate of absorption increases. True.
Answer: (A), (B), (C), (D); the key also accepts (B), (C), (D). | ||||
| 4 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Multiple correctMedium | |
A medium multiple correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Angular momentum $\frac{nh}{2\pi} = \frac{3h}{2\pi}$ gives $n = 3$.
Radius $r_n = \frac{n^2a_0}{Z} = \frac{9a_0}{Z} = 4.5a_0$ gives $Z = 2$ (He$^+$).
$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) = 4R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$. From $n = 3$ the possible transitions are:
$3 \to 1$: $\frac{1}{\lambda} = 4R\cdot\frac{8}{9} = \frac{32R}{9}$, so $\lambda = \frac{9}{32R}$ (A).
$3 \to 2$: $\frac{1}{\lambda} = 4R\cdot\frac{5}{36} = \frac{5R}{9}$, so $\lambda = \frac{9}{5R}$ (C).
$2 \to 1$: $\frac{1}{\lambda} = 4R\cdot\frac{3}{4} = 3R$, so $\lambda = \frac{1}{3R}$ (not listed).
$\frac{9}{16R}$ and $\frac{4}{3R}$ match no transition.
Answer: (A), (C). | ||||
| 5 | GravitationEscape velocity | Class 11: Gravitation | Multiple correctEasy | |
A easy multiple correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation At the midpoint the particle is a distance $L$ from each body, so its potential energy is $U = -\frac{GMm}{L} - \frac{GMm}{L} = -\frac{2GMm}{L}$.
To escape it must reach infinity (where $U = 0$) with at least zero kinetic energy:
$\frac{1}{2}mv^2 - \frac{2GMm}{L} = 0 \Rightarrow v = \sqrt{\frac{4GM}{L}} = 2\sqrt{\frac{GM}{L}}$.
So (B) is true and (A), (C) are false.
(D) Gravity is conservative and only it acts, so the total mechanical energy of $m$ stays constant. True.
Answer: (B), (D). | ||||
| 6 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Multiple correctHard | |
A hard multiple correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Let $\omega = \sqrt{\frac{k}{m}}$. Before the collision $x = \frac{u_0}{\omega}\sin\omega t$ and $v = u_0\cos\omega t$.
The speed is $0.5u_0$ when $\cos\omega t = \frac{1}{2}$, i.e. $\omega t_1 = \frac{\pi}{3}$. The elastic collision with the rigid wall only reverses the velocity, so the energy (and amplitude) is unchanged.
(A) Coming back to equilibrium the particle has all the energy as kinetic energy again, so its speed is $u_0$. True.
(B) After reversal the particle retraces its path: it takes the same time $\frac{\pi}{3\omega}$ to return to equilibrium. First passage at $t = \frac{2\pi}{3\omega} = \frac{2\pi}{3}\sqrt{\frac{m}{k}}$, not $\pi\sqrt{\frac{m}{k}}$. False.
(C) It then moves to the other side and reaches maximum compression a quarter period later: $t = \frac{2\pi}{3\omega} + \frac{\pi}{2\omega} = \frac{7\pi}{6}\sqrt{\frac{m}{k}}$, not $\frac{4\pi}{3}\sqrt{\frac{m}{k}}$. False.
(D) It returns to equilibrium half a period after the first passage: $t = \frac{2\pi}{3\omega} + \frac{\pi}{\omega} = \frac{5\pi}{3}\sqrt{\frac{m}{k}}$. True.
Answer: (A), (D). | ||||
| 7 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctMedium | |
A medium multiple correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Two sources add:
Hollow cylinder (current along the axis): by Ampère's law, $B = 0$ for $r < R$ and $B = \frac{\mu_0 I}{2\pi r}$, tangential (circling the axis), for $r > R$.
Solenoid: $B = \mu_0 nI$ along the axis for $r < 2R$, and zero outside.
(A) $0 < r < R$: only the solenoid's axial field $\mu_0 nI$ — non-zero. True.
(B), (C) $R < r < 2R$: both the axial $\mu_0 nI$ and the tangential $\frac{\mu_0 I}{2\pi r}$ are present, so the resultant is neither along the axis nor tangential (it is helical). Both false.
(D) $r > 2R$: the solenoid contributes nothing but the cylinder gives $\frac{\mu_0 I}{2\pi r} \neq 0$. True.
Answer: (A), (D). | ||||
| 8 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | Multiple correctMedium | |
A medium multiple correct question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Wind changes the speed of sound relative to the ground along the line source → observer to $V' = V + w$ (wind from source to observer) or $V' = V - w$ (wind from observer to source).
Source and observer approach each other, each at speed $u$, so
$f_2 = f_1\frac{V' + u}{V' - u}$.
Whichever way the wind blows, $V' > u$ and the fraction is greater than 1, so $f_2 > f_1$ in both cases (only the amount of the rise differs).
(A) and (B) are true; (C) and (D) are false.
Answer: (A), (B). | ||||
| 9 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | ComprehensionEasy | |
A easy comprehension question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The field grows at $\frac{dB}{dt} = \frac{B}{1\ \text{s}}$. By Faraday's law around the orbit,
$\oint \vec{E}\cdot d\vec{l} = E(2\pi R) = \frac{d\Phi}{dt} = \pi R^2\frac{dB}{dt} = \pi R^2 B$ (per second).
$E = \frac{BR}{2}$.
Answer: (B). | ||||
| 10 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | ComprehensionMedium | |
A medium comprehension question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The induced field $E = \frac{BR}{2}$ (per second of change) acts along the orbit, so the torque on the charge about the centre is $\tau = QER = \frac{QBR^2}{2}$ per second. Over the 1 s interval the angular momentum changes by
$|\Delta L| = \tau\,\Delta t = \frac{QBR^2}{2}$.
Direction: by Lenz's law the induced electric field circulates so as to oppose the growth of flux along $+z$, i.e. it pushes a positive charge clockwise seen from $+z$. Its torque is along $-z$, so $\Delta\vec{L} = -\frac{QBR^2}{2}\hat{k}$ whichever way the charge was orbiting.
$\Delta\mu = \gamma\,\Delta L = -\gamma\frac{BQR^2}{2}$.
Answer: (B). | ||||
| 11 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | ComprehensionMedium | |
A medium comprehension question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Test each process with the rule of the passage (atomic masses can be used; the electrons balance).
(A) ${}^{6}_{3}Li \to {}^{4}_{2}He + {}^{2}_{1}H$: $4.002603 + 2.014102 = 6.016705 > 6.015123$. The products are heavier, so it cannot happen.
(B) ${}^{210}_{84}Po \to {}^{209}_{83}Bi + {}^{1}_{1}H$: $208.980388 + 1.007825 = 209.988213 > 209.982876$. Cannot happen.
(C) ${}^{2}_{1}H + {}^{4}_{2}He \to {}^{6}_{3}Li$: $m_3 + m_4 = 6.016705 > 6.015123 = M'$. Complete fusion is possible.
(D) ${}^{70}_{30}Zn + {}^{82}_{34}Se \to {}^{152}_{64}Gd$: $69.925325 + 81.916709 = 151.842034 < 151.919803$. Not possible.
Answer: (C). | ||||
| 12 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | ComprehensionMedium | |
A medium comprehension question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei ${}^{210}_{84}Po \to {}^{206}_{82}Pb + {}^{4}_{2}He$.
Mass defect: $209.982876 - (205.974455 + 4.002603) = 0.005818$ u.
$Q = 0.005818 \times 932 = 5.4224$ MeV.
Momentum is conserved from rest, so the energy is shared inversely as the masses: $K_\alpha = Q\frac{m_{Pb}}{m_{Pb} + m_\alpha} = 5.4224 \times \frac{206}{210} = 5.319$ MeV $= 5319$ keV.
Answer: (A). | ||||
| 13 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | ComprehensionEasy | |
A easy comprehension question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The centre $P$ of the arc is level with the starting point, and $PQ$ makes $30^\circ$ below that horizontal. So $Q$ is $R\sin 30^\circ = 20$ m below the start.
Work-energy theorem from the start to $Q$:
$\frac{1}{2}mv^2 = mgh - W_f = 1 \times 10 \times 20 - 150 = 50$ J.
$v = \sqrt{100} = 10\ m\,s^{-1}$.
Answer: (B). | ||||
| 14 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | ComprehensionMedium | |
A medium comprehension question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion At $Q$ the line $QP$ (towards the centre) is inclined at $30^\circ$ to the horizontal. The component of gravity along $QP$ is $mg\sin 30^\circ$, pointing away from the centre (gravity is downward, the centre is up and to the right).
Radial equation (friction is tangential and does not enter):
$N - mg\sin 30^\circ = \frac{mv^2}{R}$
$N = 1 \times 10 \times \frac{1}{2} + \frac{1 \times 10^2}{40} = 5 + 2.5 = 7.5$ N.
Answer: (A). | ||||
| 15 | Current ElectricityOhm's law | Class 12: Current Electricity | ComprehensionEasy | |
A easy comprehension question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Current: $I = \frac{P}{V} = \frac{600 \times 10^3}{4000} = 150$ A.
Cable resistance: $0.4 \times 20 = 8\ \Omega$.
Loss: $I^2R = 150^2 \times 8 = 1.8 \times 10^5$ W $= 180$ kW.
Fraction: $\frac{180}{600} = 30\%$.
Answer: (B). | ||||
| 16 | Electromagnetic Induction and Alternating CurrentsTransformers | Class 12: Electromagnetic Induction; Class 12: Alternating Current | ComprehensionEasy | |
A easy comprehension question on transformers, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Step-up 1 : 10 raises 4000 V to 40 000 V. With the small line loss of this method neglected, the step-down transformer receives 40 000 V and must deliver 200 V:
$\frac{N_p}{N_s} = \frac{40\,000}{200} = 200 : 1$.
Answer: (A). | ||||
| 17 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement (P) Boltzmann constant: $E = kT$, so $[k] = \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}]$ → 4.
(Q) Viscosity: $F = \eta A\frac{dv}{dx}$, so $[\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]} = [ML^{-1}T^{-1}]$ → 2.
(R) Planck constant: $E = h\nu$, so $[h] = [ML^2T^{-2}][T] = [ML^2T^{-1}]$ → 1.
(S) Thermal conductivity: $\frac{dQ}{dt} = KA\frac{\Delta T}{L}$, so $[K] = \frac{[ML^2T^{-3}][L]}{[L^2][K]} = [MLT^{-3}K^{-1}]$ → 3.
P→4, Q→2, R→1, S→3.
Answer: (C). | ||||
| 18 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Matching listMedium | |
A medium matching list question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The ray enters the block and the prism normally, undeviated, and meets the prism's slanting face at an angle of incidence of $45^\circ$, going from $\mu_1$ into $\mu_2$.
(Q) $e \to g$: the ray goes straight on, which needs no refraction at that face: $\mu_1 = \mu_2$ → 3.
(S) $e \to i$: the ray is turned down and then back (two total internal reflections in the right-angled prism), which needs total internal reflection at $45^\circ$: $\sin 45^\circ > \frac{\mu_2}{\mu_1}$, i.e. $\mu_1 > \sqrt{2}\mu_2$ → 1.
(P) $e \to f$: the ray bends towards the normal of the slanting face (upwards in the figure), so it enters a denser medium: $\mu_2 > \mu_1$; leaving the block it bends further away from the block's normal, $\mu_2 > \mu_3$ → 2.
(R) $e \to h$: the ray bends away from the normal (downwards) without total reflection: $\mu_2 < \mu_1 < \sqrt{2}\mu_2$, with $\mu_2 > \mu_3$ → 4.
P→2, Q→3, R→4, S→1.
Answer: (D). | ||||
| 19 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Matching listEasy | |
A easy matching list question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Balance $A$ and $Z$ for each:
(1) $A$ unchanged, $Z$ down by 1: a positron is emitted — $\beta^+$ decay (Q).
(2) $A$ down by 4, $Z$ down by 2: ${}^{4}_{2}He$ — alpha decay (P).
(3) $A$ down by 1, $Z$ down by 1: ${}^{1}_{1}H$ — proton emission (S).
(4) A heavy nucleus splitting into a medium one (La-140) and the rest — fission (R).
P→2, Q→1, R→4, S→3.
Answer: (C). | ||||
| 20 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Matching listMedium | |
A medium matching list question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $E = (V_0, P_0)$, $F = (V_0, 32P_0)$; $E \to F$ is isochoric and $G \to H \to E$ is isobaric at $P_0$. Of the two curves from $F$, the steeper one ($F \to H$) is the adiabatic and the gentler one ($F \to G$) the isothermal.
Isothermal $F \to G$: $P_0V_G = 32P_0V_0$, so $V_G = 32V_0$. Work $= nRT\ln\frac{V_G}{V_F} = 32P_0V_0\ln 32 = 160P_0V_0\ln 2$ → S→1.
Adiabatic $F \to H$ ($\gamma = \frac{5}{3}$): $32P_0V_0^{5/3} = P_0V_H^{5/3}$, so $V_H = 32^{3/5}V_0 = 8V_0$. Work $= \frac{P_FV_F - P_HV_H}{\gamma - 1} = \frac{32 - 8}{2/3}P_0V_0 = 36P_0V_0$ → R→2.
Isobaric $G \to E$: $|W| = P_0(32V_0 - V_0) = 31P_0V_0$ → P→4.
Isobaric $G \to H$: $|W| = P_0(32V_0 - 8V_0) = 24P_0V_0$ → Q→3.
P→4, Q→3, R→2, S→1.
Answer: (A). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 21 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Ozone is a resonance hybrid of two equivalent structures (O=O–O and O–O=O), so both O–O bonds are equal (about 128 pm). True.
(B) $2O_3 \rightarrow 3O_2$ releases heat ($\Delta H \approx -142$ kJ per mole of $O_3$); ozone is thermodynamically unstable with respect to $O_2$. The decomposition is exothermic, not endothermic. False.
(C) All electrons in $O_3$ are paired, so it is diamagnetic. True.
(D) The central O has a lone pair (sp$^2$, three electron domains), giving a bent molecule with angle about $117^\circ$. True.
Answer: (A), (C), (D). | ||||
| 22 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Multiple correctEasy | |
A easy multiple correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Mass number and atomic number must balance: $9 + A_X = 8 + A_Y$ and $4 + Z_X = 4 + Z_Y$, i.e. $A_Y = A_X + 1$ and $Z_Y = Z_X$.
(A) $(\gamma, n)$: $A$: $9 + 0 = 8 + 1$; $Z$: $0 = 0$. Balanced. True.
(B) $(p, D)$: $A$: $9 + 1 = 8 + 2$; $Z$: $1 = 1$. Balanced. True.
(C) $(n, D)$: $A$ balances ($10 = 10$) but $Z$: $0 \neq 1$. False.
(D) $(\gamma, p)$: $A$ balances ($9 + 0 = 8 + 1$) but $Z$: $0 \neq 1$. False.
Answer: (A), (B). | ||||
| 23 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter (A) Tin: cassiterite $SnO_2$ is reduced by carbon ($SnO_2 + 2C \rightarrow Sn + 2CO$). Carbon is used.
(B) Iron: $Fe_2O_3$ is reduced by coke/CO in the blast furnace. Carbon is used.
(C) Aluminium: $Al_2O_3$ is too stable for carbon below impractically high temperatures (Al forms a carbide); Al is won by electrolysis of alumina in molten cryolite. Carbon reduction is not used.
(D) Magnesium from dolomite $MgCO_3\cdot CaCO_3$: Mg is very electropositive; it is obtained by electrolysis of fused $MgCl_2$ (or by silicothermic reduction), not by carbon. Not used.
Answer: (C), (D). | ||||
| 24 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Multiple correctMedium | |
A medium multiple correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium (A) By Kirchhoff's relation $\frac{d(\Delta H)}{dT} = \Delta C_p$, and $\Delta C_p \neq 0$ here, so $\Delta H$ depends on $T$. True.
(B) Here $K_p = p_{CO_2}$; the solids have unit activity, so $K$ does not depend on how much $CaCO_3$ is taken. True.
(C) At a given $T$, $K$ is a fixed number. The equilibrium pressure of $CO_2$ equals $K$, but $K$ does not depend on the pressure of $CO_2$ applied; changing that pressure only shifts the position of equilibrium. False.
(D) A catalyst changes the path, not the initial and final states, so $\Delta H$ (a state function) is unaffected. True.
Answer: (A), (B), (D). | ||||
| 25 | EquilibriumSolubility product | Class 11: Equilibrium | Multiple correctEasy | |
A easy multiple correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$, $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$.
With solubility $s$ in 0.1 M $AgNO_3$: $[Ag^+] = 0.1 + 2s \approx 0.1$ and $[CrO_4^{2-}] = s$.
$s = \frac{K_{sp}}{(0.1)^2} = \frac{1.1 \times 10^{-12}}{10^{-2}} = 1.1 \times 10^{-10}$ mol/L.
Only one option is correct, although the section allows more.
Answer: (B). | ||||
| 26 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Multiple correctHard | |
A hard multiple correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers This is the Reimer–Tiemann reaction on p-cresol. The phenoxide ion is attacked by dichlorocarbene ($:CCl_2$, from $CHCl_3 + OH^-$).
Attack ortho to O$^-$ gives, after hydrolysis of the $CHCl_2$ group, the ortho-aldehyde: 2-hydroxy-5-methylbenzaldehyde, S. This is the major product. (D) is true.
The para position already carries $CH_3$. Attack there (ipso) gives a cyclohexadienone, 4-methyl-4-(dichloromethyl)cyclohexa-2,5-dienone, Q. It cannot re-aromatise by losing H$^+$, and its $CHCl_2$ group on a quaternary carbon is not hydrolysed, so Q survives as a minor product (the "abnormal" Reimer–Tiemann product). (B) is true.
(A) P, the 2,6-dialdehyde, needs a second formylation of the deactivated ortho-aldehyde; it is not a major product. False.
(C) R is the dienol — the reduction product of Q — and no reducing agent is present. False.
Answer: (B), (D). | ||||
| 27 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Multiple correctMedium | |
A medium multiple correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The OH group strongly activates the ortho and para positions. In aqueous bromine (no deactivation of Br$_2$ by a non-polar solvent) every activated position is brominated:
The two positions ortho to OH take one Br each (2 equivalents).
The para position carries $SO_3H$. Being para to OH, it is attacked by $Br^+$ and the $SO_3H$ group leaves (ipso substitution, desulfonation), using the third equivalent.
The product is 2,4,6-tribromophenol, Q.
P and S keep the $SO_3H$ group and put Br on positions meta to OH, which the OH group does not activate; R has Br at 3, 4 and 5, leaving both ortho positions free. None of these follows the directing effect of OH. Only Q is formed.
Answer: (B). | ||||
| 28 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Multiple correctHard | |
A hard multiple correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Reaction I (base): the enolate is formed and brominated. Each Br on the methyl group makes the remaining H more acidic, so the second and third brominations are faster than the first: once a methyl group starts, it goes to $CBr_3$. The $CBr_3COCH_3$ is then cleaved by $OH^-$ (haloform reaction) to $CH_3COO^-Na^+$ (T) and $CHBr_3$ (U).
Each acetone converted uses 3 mol $Br_2$, so 1.0 mol $Br_2$ converts only $\frac{1}{3}$ mol of acetone; the other $\frac{2}{3}$ mol stays as acetone. Mixture I: T, U and acetone.
Reaction II (acid): bromination goes through the enol, and a Br on the carbon makes the next enolisation slower (it lowers the basicity of the carbonyl oxygen). So the reaction stops at monobromination: 1 mol $Br_2$ gives 1 mol bromoacetone, $CH_3COCH_2Br$ (P).
This is option (C).
(A) misses the haloform course in base; (B) and (D) put polybromination in acid.
Answer: (C). | ||||
| 29 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | ComprehensionEasy | |
A easy comprehension question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics For a fixed mass of gas $T \propto PV$.
K → L: constant pressure, volume increases, so $T$ rises — heating.
L → M: constant volume, pressure falls, so $T$ falls — cooling.
M → N: constant pressure, volume decreases, so $T$ falls — cooling.
N → K: constant volume, pressure rises, so $T$ rises — heating.
Heating, cooling, cooling, heating.
Answer: (C). | ||||
| 30 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | ComprehensionEasy | |
A easy comprehension question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Isochoric means constant volume — vertical lines on a P–V diagram.
K → L and M → N are horizontal (isobaric). L → M (right side) and N → K (left side) are vertical.
Answer: (B). | ||||
| 31 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | ComprehensionEasy | |
A easy comprehension question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Cold, dilute NaOH: $Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O$. The oxo salt is sodium hypochlorite, salt of hypochlorous acid (P).
Hot, concentrated NaOH: $3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$. The oxo salt is sodium chlorate, salt of chloric acid (Q).
Answer: (A). | ||||
| 32 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | ComprehensionMedium | |
A medium comprehension question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter R: $SO_2 + Cl_2 \xrightarrow{\text{charcoal}} SO_2Cl_2$ (sulphuryl chloride).
S: $SO_2Cl_2$ is a chlorinating agent; with white phosphorus $P_4 + 10SO_2Cl_2 \rightarrow 4PCl_5 + 10SO_2$.
T: complete hydrolysis $PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl$.
R, S, T $= SO_2Cl_2$, $PCl_5$, $H_3PO_4$.
Answer: (A). | ||||
| 33 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | ComprehensionEasy | |
A easy comprehension question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Dilute HCl precipitates the group I chlorides: $PbCl_2$, $Hg_2Cl_2$ and $AgCl$.
Of these only $PbCl_2$ dissolves in hot water; $AgCl$ and $Hg_2Cl_2$ do not. ($Hg^{2+}$ gives no precipitate with HCl at all.)
So P contains $Pb^{2+}$.
Answer: (A). | ||||
| 34 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | ComprehensionMedium | |
A medium comprehension question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Q gives nothing with $H_2S$ in acid, so it holds no group II ion — this rules out $Cu^{2+}$.
In ammoniacal medium $H_2S$ ($NH_4OH$/$NH_4Cl$ + $H_2S$) precipitates groups III and IV: $Cr(OH)_3$, $Fe$ (as FeS/Fe(OH)$_3$), $Al(OH)_3$, ZnS, MnS, NiS, CoS. That is R.
$H_2O_2$ in NaOH oxidises chromium(III) to chromate: $2Cr(OH)_3 + 3H_2O_2 + 4NaOH \rightarrow 2Na_2CrO_4 + 8H_2O$, a yellow solution.
Iron would stay as an insoluble brown $Fe(OH)_3$, not a solution; $ZnSO_4$ / zincate is colourless; $CuSO_4$ is excluded above.
So S contains $Na_2CrO_4$.
Answer: (D). | ||||
| 35 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | ComprehensionMedium | |
A medium comprehension question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $C_4H_4O_4$ dicarboxylic acids that add bromine: HOOC–CH=CH–COOH. P forms a cyclic anhydride on heating, so its two COOH groups are on the same side: P is maleic acid (cis); Q is fumaric acid (trans).
Cold dilute alkaline $KMnO_4$ adds two OH groups syn (from the same face).
Syn addition to the cis acid gives meso-tartaric acid, S (both OH on the same side in the Fischer projection; it has an internal mirror plane) — optically inactive.
Syn addition to the trans acid gives (+)- and (−)-tartaric acid, T and U, in equal amounts — a racemic pair, optically inactive as formed.
So P gives optically inactive S and Q gives the optically inactive pair (T, U).
Answer: (B). | ||||
| 36 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | ComprehensionHard | |
A hard comprehension question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Q is fumaric acid. $H_2/Ni$ saturates the double bond to succinic acid, HOOC–CH$_2$–CH$_2$–COOH, which on heating loses water to give succinic anhydride: V.
Benzene + succinic anhydride, anhydrous $AlCl_3$ (Friedel–Crafts acylation): 3-benzoylpropanoic acid, $C_6H_5CO(CH_2)_2COOH$.
Zn-Hg/HCl (Clemmensen) reduces the ketone C=O to $CH_2$: 4-phenylbutanoic acid, $C_6H_5(CH_2)_3COOH$.
$H_3PO_4$ brings about intramolecular Friedel–Crafts acylation, closing a six-membered ring onto the benzene: 1-tetralone (3,4-dihydronaphthalen-1(2H)-one): W.
That is option (A). (C) shows tetralin, which would need a second reduction; (B) and (D) do not come from these reagents.
Answer: (A). | ||||
| 37 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Matching listMedium | |
A medium matching list question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (P) A tertiary chloride with a strong base, $NaOEt$, undergoes E2 elimination to the alkene → 2.
(Q) Williamson synthesis: tert-butoxide (the bulky part as the alkoxide) with a primary halide, $EtBr$, by $S_N2$ → 3.
(R) Markovnikov hydration without rearrangement: oxymercuration–demercuration puts OH on the more substituted carbon, giving the tertiary alcohol → 1.
(S) Hydroboration–oxidation adds H and OH syn and anti-Markovnikov: OH on the less substituted carbon, trans to the methyl group → 4.
P→2, Q→3, R→1, S→4.
Answer: (A). | ||||
| 38 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter (P) $2PbO_2 + 2H_2SO_4 \xrightarrow{\text{warm}} 2PbSO_4 + O_2 + 2H_2O$: $PbO_2$ releases oxygen from sulphuric acid on warming → 3.
(Q) $Na_2S_2O_3 + 4Cl_2 + 5H_2O \rightarrow 2NaHSO_4 + 8HCl$: chlorine oxidises thiosulphate all the way to sulphate → 4.
(R) $N_2H_4 + 2I_2 \rightarrow N_2 + 4HI$: iodine oxidises hydrazine to nitrogen → 2.
(S) $XeF_2 + 2NO \rightarrow Xe + 2NOF$: NO is fluorinated by $XeF_2$ → 1.
P→3, Q→4, R→2, S→1.
Answer: (D). | ||||
| 39 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Matching listHard | |
A hard matching list question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Half-cell potentials do not add; the free energies $\Delta G^\circ = -nFE^\circ$ do.
(P) $Fe^{3+} \to Fe^{2+}$: $-1(0.77)F$; $Fe^{2+} \to Fe$: $-2(-0.44)F$. Total for 3 electrons: $-(0.77 - 0.88)F = +0.11F$, so $E^\circ = \frac{0.77 - 0.88}{3} = -0.037 \approx -0.04$ V → 3.
(Q) Subtract the acid half-reaction from the basic one: $[O_2 + 2H_2O + 4e^- \to 4OH^-] - [O_2 + 4H^+ + 4e^- \to 2H_2O]$ gives $4H_2O \to 4H^+ + 4OH^-$ with $E^\circ = 0.40 - 1.23 = -0.83$ V → 4.
(S) $Cr^{3+} + 3e^- \to Cr$: $\Delta G = -3(-0.74)F = 2.22F$; minus $Cr^{2+} + 2e^- \to Cr$: $-2(-0.91)F = 1.82F$. For $Cr^{3+} + e^- \to Cr^{2+}$: $\Delta G = 0.40F$, $E^\circ = -0.40$ V → 2.
(R) $E^\circ(Cu^{2+}/Cu^+) = 2(0.34) - 0.52 = 0.16$ V. For $Cu^{2+} + Cu \to 2Cu^+$: $\Delta G^\circ = [-2(0.34) + 2(0.52)]F = 0.36F$. Taken as a two-electron change, $E^\circ = -\frac{0.36}{2} = -0.18$ V → 1 (the remaining value).
P→3, Q→4, R→1, S→2.
Answer: (D). | ||||
| 40 | ElectrochemistryConductance | Class 12: Electrochemistry | Matching listMedium | |
A medium matching list question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry In each case X is added to Y, so before the end point Y's ions are being replaced; after it, excess X accumulates.
(P) Acetic acid is weak and poorly conducting. Adding triethylamine forms the salt $Et_3NH^+CH_3COO^-$, so conductivity increases; after the end point the excess weak base adds few ions — not much change → 3.
(Q) $AgNO_3 + KI \to AgI\downarrow + KNO_3$: $Ag^+$ is replaced by $K^+$ of similar mobility, so conductivity hardly changes; after the end point excess KI adds ions and it increases → 4.
(R) $KOH + CH_3COOH \to CH_3COOK + H_2O$: fast $OH^-$ is replaced by slow $CH_3COO^-$, so conductivity decreases; after the end point excess acetic acid (weak) adds little → 2.
(S) $HI + NaOH \to NaI + H_2O$: fast $H^+$ is replaced by slow $Na^+$, so it decreases; after the end point excess NaOH ($OH^-$ very mobile) makes it increase → 1.
P→3, Q→4, R→2, S→1.
Answer: (A). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 41 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Multiple correctMedium | |
A medium multiple correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $w = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$, so $P = \{e^{in\pi/6}\}$: the twelve points on the unit circle at angles $0, \frac{\pi}{6}, \ldots, \frac{11\pi}{6}$.
$P \cap H_1$ needs $\cos\theta > \frac{1}{2}$: $\theta = -\frac{\pi}{6}, 0, \frac{\pi}{6}$ ($\cos\frac{\pi}{3} = \frac{1}{2}$ exactly is excluded).
$P \cap H_2$ needs $\cos\theta < -\frac{1}{2}$: $\theta = \frac{5\pi}{6}, \pi, \frac{7\pi}{6}$.
The angle between $z_1$ and $z_2$ is the difference of the arguments (taken in $[0, \pi]$): the possible values are $\frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}$, $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$ (and similar pairs), and $\pi$.
So $\frac{2\pi}{3}$ and $\frac{5\pi}{6}$ are possible; $\frac{\pi}{2}$ and $\frac{\pi}{6}$ are not.
Answer: (C), (D). | ||||
| 42 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Multiple correctEasy | |
A easy multiple correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Take logarithms: $x\ln 3 = (x - 1)\ln 4$, so $x(\ln 4 - \ln 3) = \ln 4$ and $x = \frac{\ln 4}{\ln 4 - \ln 3}$.
(A) Divide top and bottom by $\ln 3$: $x = \frac{\log_3 4}{\log_3 4 - 1} = \frac{2\log_3 2}{2\log_3 2 - 1}$. True.
(B) Divide by $\ln 2$: $x = \frac{2}{2 - \log_2 3}$. True.
(C) Divide by $\ln 4$: $x = \frac{1}{1 - \log_4 3}$. True.
(D) $\frac{2\log_2 3}{2\log_2 3 - 1} = \frac{\ln 9}{\ln 9 - \ln 2}$, which is not $\frac{\ln 4}{\ln 4 - \ln 3}$ (numerically 1.46 against 4.82). False.
Answer: (A), (B), (C). | ||||
| 43 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $(P^2)_{ij} = \sum_{k=1}^{n} \omega^{i+k}\omega^{k+j} = \omega^{i+j}\sum_{k=1}^{n}\omega^{2k}$.
The sum $\omega^2 + \omega^4 + \omega^6 + \cdots$ repeats $\omega^2, \omega, 1$, whose sum is 0. So $\sum_{k=1}^{n}\omega^{2k} = 0$ exactly when $n$ is a multiple of 3, and is non-zero otherwise.
$P^2 = 0$ for $n = 57 = 3 \times 19$; $P^2 \neq 0$ for $n = 55, 58, 56$.
Answer: (B), (C), (D). | ||||
| 44 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Multiple correctMedium | |
A medium multiple correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives With $a = |x + 2|$ and $b = 2|x|$, $a + b - |a - b| = 2\min(a, b)$. So $f(x) = 2\min(|x + 2|, 2|x|)$.
The two graphs cross where $|x + 2| = 2|x|$: $x = 2$ or $x = -\frac{2}{3}$.
$x \le -2$: $f = 2|x + 2|$, decreasing to 0 at $x = -2$.
$-2 \le x \le -\frac{2}{3}$: $f = 2(x + 2)$, increasing. So $x = -2$ is a local minimum ($f = 0$).
$-\frac{2}{3} \le x \le 0$: $f = -4x$, decreasing. So $x = -\frac{2}{3}$ is a local maximum ($f = \frac{8}{3}$).
$0 \le x \le 2$: $f = 4x$, increasing; $x = 0$ is a local minimum.
$x \ge 2$: $f = 2(x + 2)$, still increasing, so $x = 2$ is not an extremum. Nor is $\frac{2}{3}$.
Among the options: $-2$ and $-\frac{2}{3}$.
Answer: (A), (B). | ||||
| 45 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctHard | |
A hard multiple correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Denominator: $(na + 1) + \cdots + (na + n) = n^2a + \frac{n(n + 1)}{2}$, and $(n + 1)^{a-1} \sim n^{a-1}$. So the denominator behaves like $n^{a+1}\left(a + \frac{1}{2}\right)$.
Numerator: for $a > -1$, $\frac{1}{n^{a+1}}\sum_{k=1}^{n}k^a = \frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^a \to \int_0^1 x^a\,dx = \frac{1}{a + 1}$.
Limit $= \frac{1/(a + 1)}{a + \frac{1}{2}} = \frac{2}{(a + 1)(2a + 1)} = \frac{1}{60}$.
$(a + 1)(2a + 1) = 120 \Rightarrow 2a^2 + 3a - 119 = 0 \Rightarrow (a - 7)(2a + 17) = 0$, so $a = 7$ or $a = -\frac{17}{2}$.
$a = -\frac{17}{2}$ is not allowed: for $a < -1$ the numerator tends to a finite positive number while the denominator $\sim n^{a+1}(a + \frac{1}{2}) \to 0$, so the limit is not $\frac{1}{60}$ (the Riemann-sum step needs $a > -1$).
Answer: (B). | ||||
| 46 | CirclesStandard and general form | Class 11: Conic Sections | Multiple correctEasy | |
A easy multiple correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Touching the $x$-axis at $(3, 0)$: centre $(3, k)$, radius $|k|$:
$(x - 3)^2 + (y - k)^2 = k^2$, i.e. $x^2 + y^2 - 6x - 2ky + 9 = 0$.
On the $y$-axis ($x = 0$): $y^2 - 2ky + 9 = 0$; the intercept is $2\sqrt{k^2 - 9} = 2\sqrt{7}$, so $k^2 = 16$, $k = \pm 4$.
$k = -4$: $x^2 + y^2 - 6x + 8y + 9 = 0$ (A). $k = 4$: $x^2 + y^2 - 6x - 8y + 9 = 0$ (C).
(The circles touching at $(-3, 0)$ have $+6x$ and are not among the options.) (B) and (D) have the wrong $y$-coefficient: their intercept would be $2\sqrt{\frac{49}{4} - 9} = \sqrt{13}$.
Answer: (A), (C). | ||||
| 47 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctMedium | |
A medium multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $L_1$ passes through $A = (5, 0, 0)$ with direction $(0, 3 - \alpha, -2)$; $L_2$ through $B = (\alpha, 0, 0)$ with direction $(0, -1, 2 - \alpha)$.
Coplanar when $\overrightarrow{AB}\cdot(\vec{d}_1 \times \vec{d}_2) = 0$:
$\begin{vmatrix}\alpha - 5 & 0 & 0\\0 & 3 - \alpha & -2\\0 & -1 & 2 - \alpha\end{vmatrix} = (\alpha - 5)[(3 - \alpha)(2 - \alpha) - 2] = (\alpha - 5)(\alpha^2 - 5\alpha + 4) = (\alpha - 5)(\alpha - 1)(\alpha - 4) = 0$.
$\alpha = 1, 4$ or $5$. Of the options, 1 and 4.
Answer: (A), (D). | ||||
| 48 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Multiple correctHard | |
A hard multiple correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Tangent lengths from the vertices: $PN = s - p$, $QL = s - q$, $RM = s - r$, where $p, q, r$ are the sides opposite $P, Q, R$.
$P$ is the largest angle, so $p$ is the largest side and $s - p$ the smallest tangent length. Let the three be $2k$, $2k + 2$, $2k + 4$ with $s - p = 2k$.
Each side is the sum of the two tangent lengths from its ends: $p = (2k + 2) + (2k + 4) = 4k + 6$, and the other two sides are $4k + 4$ and $4k + 2$.
Write the sides as $2(u + 2)$, $2(u + 1)$, $2u$ with $u = 2k + 1$. Cosine rule at $P$:
$\cos P = \frac{(u + 1)^2 + u^2 - (u + 2)^2}{2u(u + 1)} = \frac{u^2 - 2u - 3}{2u(u + 1)} = \frac{(u - 3)(u + 1)}{2u(u + 1)} = \frac{u - 3}{2u}$.
$\frac{u - 3}{2u} = \frac{1}{3} \Rightarrow u = 9$ (so $k = 4$: tangent lengths 8, 10, 12).
Sides: 22, 20, 18. Of the options, 18 and 22.
Answer: (B), (D). | ||||
| 49 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | ComprehensionMedium | |
A medium comprehension question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Region $S$: with $z = x + iy$, $\frac{1}{1 - \sqrt{3}i} = \frac{1 + \sqrt{3}i}{4}$, so $S_2$ is $\mathrm{Im}[(x - 1 + i(y + \sqrt{3}))(1 + \sqrt{3}i)] > 0$:
$(y + \sqrt{3}) + \sqrt{3}(x - 1) > 0 \iff y > -\sqrt{3}x$, the half-plane above the line through $O$ at $-60^\circ$.
With $x > 0$ and $|z| < 4$, $S$ is the sector of the disc of radius 4 between the arguments $-\frac{\pi}{3}$ and $\frac{\pi}{2}$.
Its angle is $\frac{\pi}{2} + \frac{\pi}{3} = \frac{5\pi}{6}$.
Area $= \frac{1}{2}r^2\theta = \frac{1}{2}\cdot 16\cdot\frac{5\pi}{6} = \frac{20\pi}{3}$.
Answer: (B). | ||||
| 50 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | ComprehensionHard | |
A hard comprehension question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Region $S$: with $z = x + iy$, $\frac{1}{1 - \sqrt{3}i} = \frac{1 + \sqrt{3}i}{4}$, so $S_2$ is $\mathrm{Im}[(x - 1 + i(y + \sqrt{3}))(1 + \sqrt{3}i)] > 0$:
$(y + \sqrt{3}) + \sqrt{3}(x - 1) > 0 \iff y > -\sqrt{3}x$, the half-plane above the line through $O$ at $-60^\circ$.
With $x > 0$ and $|z| < 4$, $S$ is the sector of the disc of radius 4 between the arguments $-\frac{\pi}{3}$ and $\frac{\pi}{2}$.
We need the distance from the point $A = (1, -3)$ to $S$.
$A$ is not in $S$: $y + \sqrt{3}x = -3 + \sqrt{3} < 0$. It lies just below the boundary line $\sqrt{3}x + y = 0$.
Distance from $A$ to that line: $\frac{|\sqrt{3}(1) + (-3)|}{2} = \frac{3 - \sqrt{3}}{2} \approx 0.63$.
The foot of the perpendicular is $A + \frac{3 - \sqrt{3}}{4}(\sqrt{3}, 1) \approx (1.55, -2.68)$: it has $x > 0$ and modulus $\approx 3.10 < 4$, so it lies on the edge of $S$ and the infimum is attained in the limit.
$\min|1 - 3i - z| = \frac{3 - \sqrt{3}}{2}$.
Answer: (C). | ||||
| 51 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | ComprehensionEasy | |
A easy comprehension question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability The boxes hold 6, 9 and 12 balls; $6 \times 9 \times 12 = 648$ equally likely triples.
All white: $1 \times 2 \times 3 = 6$.
All red: $3 \times 3 \times 4 = 36$.
All black: $2 \times 4 \times 5 = 40$.
Probability $= \frac{6 + 36 + 40}{648} = \frac{82}{648}$.
Answer: (A). | ||||
| 52 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | ComprehensionMedium | |
A medium comprehension question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each box is chosen with probability $\frac{1}{3}$. Probability of one white and one red:
$B_1$: $\frac{1 \times 3}{\binom{6}{2}} = \frac{3}{15} = \frac{1}{5}$.
$B_2$: $\frac{2 \times 3}{\binom{9}{2}} = \frac{6}{36} = \frac{1}{6}$.
$B_3$: $\frac{3 \times 4}{\binom{12}{2}} = \frac{12}{66} = \frac{2}{11}$.
Bayes' theorem (the $\frac{1}{3}$ cancels):
$P(B_2 \mid WR) = \frac{1/6}{1/5 + 1/6 + 2/11} = \frac{55/330}{(66 + 55 + 60)/330} = \frac{55}{181}$.
Answer: (D). | ||||
| 53 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | ComprehensionHard | |
A hard comprehension question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Put $g(x) = e^{-x}f(x)$. Then $g'(x) = e^{-x}(f'(x) - f(x))$ and $g''(x) = e^{-x}(f''(x) - 2f'(x) + f(x)) \ge 1 > 0$.
So $g$ is strictly convex on $[0, 1]$, with $g(0) = g(1) = 0$.
A strictly convex function lies below the chord joining its end values, and that chord is $y = 0$. So $g(x) < 0$ for $0 < x < 1$, and since $e^{-x} > 0$, $f(x) < 0$ there.
Only (D), $-\infty < f(x) < 0$, is guaranteed. (A) is false. (B) and (C) give lower bounds the condition does not provide: for example $f(x) = Ke^x x(x - 1)$ with $K \ge \frac{1}{2}$ satisfies it (here $g'' = 2K \ge 1$), and for large $K$ it goes far below $-\frac{1}{2}$.
Answer: (D). | ||||
| 54 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | ComprehensionHard | |
A hard comprehension question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Put $g(x) = e^{-x}f(x)$. Then $g'(x) = e^{-x}(f'(x) - f(x))$ and $g''(x) = e^{-x}(f''(x) - 2f'(x) + f(x)) \ge 1 > 0$.
So $g$ is strictly convex on $[0, 1]$, with $g(0) = g(1) = 0$.
$g'$ is strictly increasing, and $g$ has its minimum at $x = \frac{1}{4}$, so $g'(\frac{1}{4}) = 0$, $g' < 0$ on $(0, \frac{1}{4})$ and $g' > 0$ on $(\frac{1}{4}, 1)$.
$g' = e^{-x}(f' - f)$ has the sign of $f' - f$. So $f'(x) < f(x)$ for $0 < x < \frac{1}{4}$ and $f'(x) > f(x)$ for $\frac{1}{4} < x < 1$.
(C) is true; (A), (B) and (D) contradict this.
Answer: (C). | ||||
| 55 | Conic SectionsParabola | Class 11: Conic Sections | ComprehensionMedium | |
A medium comprehension question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Let $P = (at_1^2, 2at_1)$ and $Q = (at_2^2, 2at_2)$. For a focal chord $t_1t_2 = -1$.
The tangents meet at $(at_1t_2, a(t_1 + t_2)) = (-a, a(t_1 + t_2))$ (on the directrix). This lies on $y = 2x + a$: $a(t_1 + t_2) = -2a + a$, so $t_1 + t_2 = -1$.
Then $(t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1t_2 = 1 + 4 = 5$.
Length of a focal chord: $PQ = a(t_1 - t_2)^2$ (with $t_2 = -\frac{1}{t_1}$ this is the familiar $a\left(t_1 + \frac{1}{t_1}\right)^2$).
$PQ = 5a$.
Answer: (B). | ||||
| 56 | Conic SectionsParabola | Class 11: Conic Sections | ComprehensionHard | |
A hard comprehension question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Let $P = (at_1^2, 2at_1)$ and $Q = (at_2^2, 2at_2)$. For a focal chord $t_1t_2 = -1$.
The tangents meet at $(at_1t_2, a(t_1 + t_2)) = (-a, a(t_1 + t_2))$ (on the directrix). This lies on $y = 2x + a$: $a(t_1 + t_2) = -2a + a$, so $t_1 + t_2 = -1$.
Then $(t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1t_2 = 1 + 4 = 5$.
Slopes of $OP$ and $OQ$: $m_1 = \frac{2}{t_1}$, $m_2 = \frac{2}{t_2}$.
$\tan\theta = \frac{m_1 - m_2}{1 + m_1m_2} = \frac{2(t_2 - t_1)/(t_1t_2)}{1 + 4/(t_1t_2)} = \frac{-2(t_2 - t_1)}{1 - 4} = \frac{2(t_2 - t_1)}{3}$, of magnitude $\frac{2\sqrt{5}}{3}$.
The angle is obtuse: $\overrightarrow{OP}\cdot\overrightarrow{OQ} = a^2t_1^2t_2^2 + 4a^2t_1t_2 = a^2 - 4a^2 < 0$. So $\tan\theta$ is negative:
$\tan\theta = -\frac{2}{3}\sqrt{5}$.
Answer: (D). | ||||
| 57 | Conic SectionsParabola | Class 11: Conic Sections | Matching listMedium | |
A medium matching list question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Take $F = (4t^2, 8t)$ on $y^2 = 16x$; $0 \le y_0 \le 6$ means $0 \le t \le \frac{3}{4}$.
Tangent at $F$: $ty = x + 4t^2$; at $x = 0$, $y_1 = 4t$. So $G = (0, 4t)$.
Triangle $EFG$ has base $EG = |3 - 4t| = 3 - 4t$ on the $y$-axis and height $x_0 = 4t^2$:
Area $A(t) = \frac{1}{2}(3 - 4t)(4t^2) = 6t^2 - 8t^3$.
$A'(t) = 12t - 24t^2 = 0 \Rightarrow t = \frac{1}{2}$, and $A''(\frac{1}{2}) = 12 - 24 < 0$: a local maximum.
At $t = \frac{1}{2}$: $F = (1, 4)$, so $y_0 = 4$; $y_1 = 2$; area $= \frac{6}{4} - 1 = \frac{1}{2}$; slope of $EF$: $m = \frac{4 - 3}{1 - 0} = 1$.
P→4, Q→1, R→2, S→3.
Answer: (A). | ||||
| 58 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Matching listHard | |
A hard matching list question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions (P) $\cos(\tan^{-1}y) + y\sin(\tan^{-1}y) = \frac{1}{\sqrt{1 + y^2}} + \frac{y^2}{\sqrt{1 + y^2}} = \sqrt{1 + y^2}$.
$\cot(\sin^{-1}y) + \tan(\sin^{-1}y) = \frac{\sqrt{1 - y^2}}{y} + \frac{y}{\sqrt{1 - y^2}} = \frac{1}{y\sqrt{1 - y^2}}$.
The ratio is $y\sqrt{1 - y^4}$; squared and divided by $y^2$: $1 - y^4$. Adding $y^4$ gives 1, whose square root is 1 → 4.
(Q) $e^{ix} + e^{iy} + e^{iz} = 0$: three unit vectors summing to zero are $120^\circ$ apart, so $x - y = \pm\frac{2\pi}{3} + 2k\pi$ and $\cos\frac{x - y}{2} = \pm\frac{1}{2}$. Possible value $\frac{1}{2}$ → 3.
(R) Collect the $\cos 2x$ terms: $\cos 2x\left[\cos\left(\frac{\pi}{4} - x\right) - \cos\left(\frac{\pi}{4} + x\right)\right] = \sqrt{2}\sin x\cos 2x$. The other side: $\sin 2x\sec x(\cos x - \sin x) = 2\sin x(\cos x - \sin x)$.
So $\sin x\left[\sqrt{2}(\cos x - \sin x)(\cos x + \sin x) - 2(\cos x - \sin x)\right] = 0$, i.e. $\sin x = 0$, or $\cos x = \sin x$, or $\cos x + \sin x = \sqrt{2}$.
The last two give $x = \frac{\pi}{4}$ (or $\frac{5\pi}{4}$), so $\sec x = \pm\sqrt{2}$; $\sin x = 0$ gives $\sec x = \pm 1$. $\sqrt{2}$ is a possible value → 2 (1 is also possible, but the codes with the other matches fixed leave 2).
(S) $\theta = \sin^{-1}\sqrt{1 - x^2} \in [0, \frac{\pi}{2}]$ has $\cos\theta = |x|$, so $\cot\theta = \frac{|x|}{\sqrt{1 - x^2}}$. Also $\sin(\tan^{-1}(x\sqrt{6})) = \frac{x\sqrt{6}}{\sqrt{1 + 6x^2}}$.
The left side is positive, so $x > 0$: $\frac{1}{1 - x^2} = \frac{6}{1 + 6x^2} \Rightarrow 1 + 6x^2 = 6 - 6x^2 \Rightarrow x^2 = \frac{5}{12}$, $x = \frac{1}{2}\sqrt{\frac{5}{3}}$ → 1.
P→4, Q→3, R→2, S→1.
Answer: (B). | ||||
| 59 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Matching listMedium | |
A medium matching list question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Intersection: $L_1$: $(1 + 2s, -s, -3 + s)$; $L_2$: $(4 + u, -3 + u, -3 + 2u)$.
From $z$: $s = 2u$; from $y$: $-2u = -3 + u$, so $u = 1$, $s = 2$; check $x$: $1 + 4 = 4 + 1$. The point is $(5, -2, -1)$.
A plane perpendicular to both $P_1$ and $P_2$ has its normal along $(7, 1, 2) \times (3, 5, -6) = (1\cdot(-6) - 2\cdot 5,\ 2\cdot 3 - 7\cdot(-6),\ 7\cdot 5 - 1\cdot 3) = (-16, 48, 32) \parallel (1, -3, -2)$.
Plane: $x - 3y - 2z = 5 + 6 + 2 = 13$.
$a = 1$, $b = -3$, $c = -2$, $d = 13$: P→3, Q→2, R→4, S→1.
Answer: (A). | ||||
| 60 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Matching listMedium | |
A medium matching list question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra (P) $[\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2 = 4$. Volume $= 2 \times 3 \times 1 \times 4 = 24$ → 3.
(Q) $[\vec{a} + \vec{b}\ \ \vec{b} + \vec{c}\ \ \vec{c} + \vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}] = 10$. Volume $= 3 \times 1 \times 2 \times 10 = 60$ → 4.
(R) $(2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b}) = -2\,\vec{a}\times\vec{b} + 3\,\vec{b}\times\vec{a} = -5\,\vec{a}\times\vec{b}$. The area is 5 times the old one: $5 \times 20 = 100$ → 1.
(S) $(\vec{a} + \vec{b}) \times \vec{a} = \vec{b} \times \vec{a}$, the same magnitude: area 30 → 2.
P→3, Q→4, R→1, S→2.
Answer: (C). | ||||
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