JEE (Advanced) · 2019
JEE Advanced 2019, 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
36 Q
- Optics5
- Electrostatics4
- Properties of Bulk Matter4
6 easy · 26 medium · 4 hard
Chemistry
36 Q
- Hydrocarbons5
- p-Block Elements3
- Chemical Kinetics3
16 easy · 17 medium · 3 hard
Mathematics
36 Q
- Integral Calculus5
- Trigonometry5
- Matrices and Determinants4
3 easy · 22 medium · 11 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 11: Thermodynamics
Class 12: Atoms; Class 12: Nuclei
Class 11: Oscillations; Class 11: Waves
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 11: Work, Energy and Power
Class 11: Gravitation
Class 12: Current Electricity
Class 11: Units and Measurement
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 11: Systems of Particles and Rotational Motion
Class 11: Laws of Motion
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Class 12: Electromagnetic Waves
No matching NCERT chapter
Class 11: Hydrocarbons
No matching NCERT chapter
Class 12: Chemical Kinetics
Class 11: Structure of Atom
No matching NCERT chapter
Class 12: The d- and f-Block Elements
Class 11: Chemical Bonding and Molecular Structure
Class 12: Solutions
Class 12: Alcohols, Phenols and Ethers
No matching NCERT chapter
No matching NCERT chapter
Class 12: Aldehydes, Ketones and Carboxylic Acids
Class 11: Thermodynamics
Class 11: Some Basic Concepts of Chemistry
Class 12: Biomolecules
Class 11: Equilibrium
Class 12: Amines
No matching NCERT chapter
Class 11: Redox Reactions
Class 12: Coordination Compounds
Class 11: Organic Chemistry – Some Basic Principles and Techniques
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 12: Matrices; Class 12: Determinants
Class 11: Conic Sections
Class 11: Probability; Class 12: Probability
Class 11: Complex Numbers and Quadratic Equations
Class 11: Sequences and Series
Class 12: Application of Derivatives
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 12: Vector Algebra
Class 11: Conic Sections
Class 12: Differential Equations
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 11: Binomial Theorem
Class 11: Permutations and Combinations
03
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | GravitationOrbital motion | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation A particle at radius $r$ is held in its circular orbit by the gravity of the mass $M(r)$ inside radius $r$ (the shells outside pull nothing net):
$\frac{mv^2}{r} = \frac{GM(r)m}{r^2}$, and $K = \frac{1}{2}mv^2$ gives $\frac{2K}{r} = \frac{GM(r)m}{r^2}$, so $M(r) = \frac{2Kr}{Gm}$.
The mass in a thin shell is $dM = 4\pi r^2\rho(r)\,dr$, so $4\pi r^2\rho = \frac{dM}{dr} = \frac{2K}{Gm}$, giving $\rho(r) = \frac{K}{2\pi r^2 G m}$.
Then $n(r) = \frac{\rho}{m} = \frac{K}{2\pi r^2 m^2 G}$.
Answer: (A). | ||||
| Q.2 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Let the full shell carry charge $Q$, so $V_0 = \frac{kQ}{R}$. Cutting the hole removes a small patch of charge $\alpha Q$, which acts like a point charge at the hole. The holed shell = full shell + a point charge $-\alpha Q$ at the hole.
(A) Potential at the centre: $V_0 - \frac{k\alpha Q}{R} = V_0(1-\alpha)$, reduced by $\alpha V_0$, not $2\alpha V_0$. False.
(B) Field at the centre: the full shell gives zero, the patch gives $\frac{k\alpha Q}{R^2} = \frac{\alpha V_0}{R}$. The field rises from $0$ to $\frac{\alpha V_0}{R}$; it is not reduced by $\frac{\alpha V_0}{2R}$. False.
(C) At the point $\frac{R}{2}$ from the centre towards the hole, the full shell still gives $V_0$ (inside) and the missing patch is at distance $\frac{R}{2}$: $V = V_0 - \frac{k\alpha Q}{R/2} = V_0(1-2\alpha)$. The ratio is $\frac{V_0(1-\alpha)}{V_0(1-2\alpha)} = \frac{1-\alpha}{1-2\alpha}$. True.
(D) At $2R$ from the centre on the line through the hole (hole side), the patch is at distance $R$ and the full shell gives $\frac{kQ}{4R^2} = \frac{V_0}{4R}$. The field is reduced by $\frac{k\alpha Q}{R^2} = \frac{\alpha V_0}{R}$ (or by $\frac{k\alpha Q}{9R^2}$ on the far side), never $\frac{\alpha V_0}{2R}$. False.
Answer: (C). | ||||
| Q.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 | Single correctMedium | |
A medium single 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 no losses, all the power heats the rod: $P = C\frac{dT}{dt}$, so $C = \frac{P}{dT/dt}$.
$\frac{dT}{dt} = \frac{T_0\beta}{4}t^{-3/4}$, so $C = \frac{4P\,t^{3/4}}{\beta T_0}$.
From the law, $t^{1/4} = \frac{T - T_0}{\beta T_0}$, so $t^{3/4} = \frac{(T-T_0)^3}{\beta^3 T_0^3}$.
Therefore $C = \frac{4P(T(t)-T_0)^3}{\beta^4 T_0^4}$.
Answer: (A). | ||||
| Q.4 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Two parallel decay routes add: $\lambda = 4.5\times10^{-10} + 0.5\times10^{-10} = 5\times10^{-10}$ per year.
After time $T$, $N_K = N_0e^{-\lambda T}$ and the stable products total $N_0(1 - e^{-\lambda T})$.
Ratio: $\frac{1 - e^{-\lambda T}}{e^{-\lambda T}} = e^{\lambda T} - 1 = 99$, so $e^{\lambda T} = 100$ and $\lambda T = 2\ln 10 = 4.6$.
$T = \frac{4.6}{5\times10^{-10}} = 9.2\times10^{9}$ years, so $t = 9.2$.
Answer: (B). | ||||
| Q.5 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Multiple correctHard | |
A hard multiple correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Capillary rise: $h = \frac{2T\cos\theta}{r\rho g} = \frac{2(0.075)\cos\theta}{(2\times10^{-4})(1000)(10)} = 0.075\cos\theta$ m.
In T1 ($\theta = 0^\circ$): $h_1 = 7.5$ cm. In T2 ($\theta = 60^\circ$): $h_2 = 3.75$ cm.
Case I has T1 below and T2 above; case II has T2 below and T1 above.
(A) The weight of water in the meniscus depends on the meniscus shape, which is set by the contact angle of the tube the meniscus sits in. In case I the lower tube is T1 (hemispherical meniscus, $\theta = 0^\circ$); in case II it is T2 ($\theta = 60^\circ$, a flatter meniscus). The menisci differ, so the corrections differ. True.
(B) Case II, joint at 5 cm: the lower tube T2 alone would raise water only $3.75$ cm $< 5$ cm, so water never reaches the joint and stands at $3.75$ cm. True.
(C) Case I, joint at 8 cm: the lower tube T1 raises water $7.5$ cm $< 8$ cm, so the column is $7.5$ cm. True.
(D) Case I, joint at 5 cm: water rises through T1 to the joint. Above it, T2 can support a column of at most $3.75$ cm, which is less than 5 cm, so the water cannot go into T2; the meniscus stays at the joint (its angle adjusting there) and the column is 5 cm, not more than 8.75 cm. False.
Answer: (A), (B), (C). | ||||
| Q.6 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Multiple correctMedium | |
A medium multiple correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The motional emf of an element $d\vec l$ is $(\vec V\times\vec B)\cdot d\vec l$. Here $\vec V\times\vec B = V_0B\,(\hat i\times\hat k) = -V_0B\,\hat j$, so only the $dy$ part of each element counts: $|d\phi| = V_0B(y)\,dy$.
The wire runs from $y = 0$ to $y = L$, so
$|\Delta\phi| = V_0B_0\int_0^L\left(1 + \left(\frac{y}{L}\right)^{\beta}\right)dy = V_0B_0L\left(1 + \frac{1}{\beta+1}\right) = B_0V_0L\,\frac{\beta+2}{\beta+1}$.
(A) $\beta = 0$: $|\Delta\phi| = 2B_0V_0L$, not $\frac{1}{2}B_0V_0L$. False.
(B) $\beta = 2$: $|\Delta\phi| = \frac{4}{3}B_0V_0L$. True.
(C) The straight wire $y = x$ of length $\sqrt2L$ also runs from $y = 0$ to $y = L$; since only the $y$-extent matters, $|\Delta\phi|$ is the same. True.
(D) For a given $\beta$, $|\Delta\phi| = B_0V_0L\frac{\beta+2}{\beta+1}$ is proportional to $L$, the wire's projection on the $y$-axis. True.
Answer: (B), (C), (D). | ||||
| Q.7 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Multiple correctHard | |
A hard multiple correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity With S$_1$ closed and S$_2$ open, the single loop holds C$_1$, the 70 Ω resistor, the 100 Ω resistor, C$_3$, the 30 Ω resistor, the 5 V cell and C$_4$ in series.
(A) At $t = 0$ the uncharged capacitors act as short circuits: $I = \frac{5}{70 + 100 + 30} = \frac{5}{200}$ A $= 25$ mA. True.
(B) Fully charged, the three capacitors are in series: $\frac{1}{C} = \frac{1}{10} + \frac{1}{80} + \frac{1}{80} = \frac{1}{8}$, so $C = 8\ \mu$F and $q = 8\times5 = 40\ \mu$C. Then $V_{C_1} = \frac{40}{10} = 4$ V (and $V_{C_3} = V_{C_4} = 0.5$ V). True.
(D) With S$_2$ still open, no current flows, so P and Q differ only by the voltage across C$_1$ (the 70 Ω carries no current): $|V_P - V_Q| = 4$ V, not 10 V. False.
(C) Now close S$_2$. At that instant C$_2$ is uncharged and the other capacitors keep their charges. Take $V_Q = 0$. The 5 V cell has its positive plate towards P, so the charging current flowed P→S$_1$→C$_1$→70 Ω→Q; the top plate of C$_1$ is positive and $V_P = 4$ V when no current flows. Between P and Q there are now three branches:
left (C$_1$, 70 Ω): $V_P = 4 + 70I_1$;
right (C$_4$, 5 V, 30 Ω, C$_3$, 100 Ω): the capacitor and cell voltages add to the same 4 V, so $V_P = 4 + 130I_3$;
middle (C$_2$, 30 Ω, 10 V cell with its positive plate towards P): $V_P = 10 + 30I_2$,
with $I_1 + I_2 + I_3 = 0$ (each current taken from P to Q). The left and right branches together are a 4 V source behind $70\parallel130 = 45.5$ Ω, so the current in the 30 Ω resistor is $\frac{10 - 4}{30 + 45.5} = 0.079$ A $\approx 0.1$ A, not 0.2 A. False.
Answer: (A), (B). | ||||
| Q.8 | 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 By Gauss's law $\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$, and the charge sits uniformly on the sphere of radius $R$.
(A) $h > 2R$, $r > R$: the cylinder contains the whole shell, $\Phi = \frac{Q}{\epsilon_0}$. True.
(B) $h < \frac{8R}{5}$, $r = \frac{3R}{5}$: the farthest point of the cylinder from the centre is a rim corner at $\sqrt{(h/2)^2 + r^2} < \sqrt{\left(\frac{4R}{5}\right)^2 + \left(\frac{3R}{5}\right)^2} = R$. The cylinder lies inside the shell and encloses no charge: $\Phi = 0$. True.
(C) $h > 2R$, $r = \frac{3R}{5}$: the enclosed charge is on the two polar caps where the distance from the axis is below $\frac{3R}{5}$, i.e. $\cos\theta > \frac{4}{5}$. One cap has area fraction $\frac{1 - \cos\theta}{2} = \frac{1}{10}$; two caps give $\frac{1}{5}$. $\Phi = \frac{Q}{5\epsilon_0}$. True.
(D) $r = \frac{4R}{5}$: now $\cos\theta > \frac{3}{5}$, each cap is $\frac{1 - 3/5}{2} = \frac{1}{5}$, both together $\frac{2}{5}$: $\Phi = \frac{2Q}{5\epsilon_0}$, not $\frac{Q}{5\epsilon_0}$. False.
Answer: (A), (B), (C). | ||||
| Q.9 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Read the states from the graph: 1 $(T_0, V_0)$, 2 $(2T_0, 2V_0)$, 3 $(T_0, 2V_0)$, 4 $(\frac{T_0}{2}, V_0)$.
1→2 is a straight line through the origin ($V\propto T$): isobaric at $p = \frac{RT_0}{V_0}$. 2→3: isochoric at $2V_0$. 3→4: again $V\propto T$, isobaric at $p = \frac{RT_0}{2V_0}$. 4→1: isochoric at $V_0$.
(A) Work is done only on the isobars: $W_{12} = \frac{RT_0}{V_0}(2V_0 - V_0) = RT_0$, $W_{34} = \frac{RT_0}{2V_0}(V_0 - 2V_0) = -\frac{RT_0}{2}$. Net $|W| = \frac{1}{2}RT_0$. True.
(B) The cycle is two isobars and two isochores; there is no adiabatic. False.
(C) $Q_{12} = C_p\Delta T = \frac{5}{2}R\,T_0$; $Q_{23} = C_v\Delta T = \frac{3}{2}R(-T_0)$. $\left|\frac{Q_{12}}{Q_{23}}\right| = \frac{5}{3}$. True.
(D) $Q_{34} = \frac{5}{2}R\left(-\frac{T_0}{2}\right) = -\frac{5}{4}RT_0$; $\left|\frac{Q_{12}}{Q_{34}}\right| = 2$, not $\frac{1}{2}$. False.
Answer: (A), (C). | ||||
| Q.10 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The flat interface between the two halves adds no power, so each curved face contributes on its own: $\frac{1}{f} = \frac{n_1 - 1}{R} + \frac{n_2 - 1}{R}$.
With $n_1 = n_2 = n$: $\frac{1}{f} = \frac{2(n-1)}{R}$. Changing $n_2$ by $\Delta n$ changes the power by $\frac{\Delta n}{R}$, and $\Delta\left(\frac{1}{f}\right) = -\frac{\Delta f}{f^2}$, so
$\frac{\Delta f}{f} = -f\frac{\Delta n}{R} = -\frac{\Delta n}{2(n-1)} = -\frac{\Delta n}{n}\cdot\frac{n}{2(n-1)}$.
(A) For $1 < n < 2$, $\frac{n}{2(n-1)} > 1$, so $\left|\frac{\Delta f}{f}\right| > \left|\frac{\Delta n}{n}\right|$. False.
(B) $n = 1.5$: $\Delta f = -\frac{f\,\Delta n}{2(0.5)} = -20\times10^{-3} = -0.02$ cm, so $|\Delta f| = 0.02$ cm. True.
(C) $\frac{\Delta f}{f} = -\frac{\Delta n}{2(n-1)}$: if $\frac{\Delta n}{n} < 0$ then $\frac{\Delta f}{f} > 0$. True.
(D) With both faces concave, $\frac{1}{f} = -\frac{2(n-1)}{R}$ and $\Delta\left(\frac{1}{f}\right) = -\frac{\Delta n}{R}$; then $\frac{\Delta f}{f} = -f\Delta\left(\frac{1}{f}\right) = -\frac{\Delta n}{2(n-1)}$ again. Unchanged. True.
Answer: (B), (C), (D). | ||||
| Q.11 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Multiple correctEasy | |
A easy multiple correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Angular momentum has dimensions $ML^2T^{-1}$. With $M$ and angular momentum dimensionless, $L^2T^{-1}$ is dimensionless, so $T = L^2$.
(A) Linear momentum $MLT^{-1} = L\cdot L^{-2} = L^{-1}$. True.
(B) Energy $ML^2T^{-2} = L^2\cdot L^{-4} = L^{-2}$. True.
(C) Force $MLT^{-2} = L\cdot L^{-4} = L^{-3}$. True.
(D) Power $ML^2T^{-3} = L^2\cdot L^{-6} = L^{-4}$, not $L^{-5}$. False.
Answer: (A), (B), (C). | ||||
| Q.12 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctMedium | |
A medium multiple correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter (A) Voltmeter resistance $= \frac{V}{I_g} = \frac{100\ \text{mV}}{2\ \mu\text{A}} = 50$ kΩ, not 100 kΩ. False.
(B) Ammeter: shunt $S = \frac{I_gG}{I - I_g} = \frac{2\times10^{-6}\times10}{10^{-3} - 2\times10^{-6}} = 0.02004$ Ω; the ammeter is $G\parallel S = \frac{V_{fs}}{I} = \frac{2\times10^{-5}}{10^{-3}} = 0.02$ Ω. True.
(C) In the usual Ohm's law circuit the voltmeter is across $R$ and the ammeter in series with the pair, so the ammeter reads the current through $R$ and the voltmeter together. Measured $R = R\parallel R_V = \frac{1000\times50000}{51000} = 980.4$ Ω, which lies between 978 Ω and 982 Ω. True.
(D) The measured value is $\frac{V}{I}$ of the same combination; an internal resistance in the cell changes both readings in the same proportion and leaves $\frac{V}{I} = 980.4$ Ω, not more than 1000 Ω. False.
Answer: (B), (C). | ||||
| Q.13 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power From the figure: A $(0,1)$, B $(1,1)$, C $(1,0.5)$, D $(0.5,0.5)$, E $(0.5,0)$, F $(0,0)$. $W = \int(\alpha y\,dx + 2\alpha x\,dy)$ along each segment.
AB ($y = 1$, $x$: $0\to1$): $\alpha(1)(1) = \alpha$.
BC ($x = 1$, $y$: $1\to0.5$): $2\alpha(1)(-0.5) = -\alpha$.
CD ($y = 0.5$, $x$: $1\to0.5$): $\alpha(0.5)(-0.5) = -0.25\alpha$.
DE ($x = 0.5$, $y$: $0.5\to0$): $2\alpha(0.5)(-0.5) = -0.5\alpha$.
EF ($y = 0$): $0$. FA ($x = 0$): $0$.
Total $W = \alpha - \alpha - 0.25\alpha - 0.5\alpha = -0.75\alpha = 0.75$ J for $\alpha = -1$.
Check: the force is not conservative ($\partial F_y/\partial x - \partial F_x/\partial y = \alpha$); the loop is clockwise around area $1 - 0.25 = 0.75$ m$^2$, so $W = -\alpha(0.75) = 0.75$ J.
Answer: 0.75. | ||||
| Q.14 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical 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 Equilibrium of the block (angles measured from the ceiling):
horizontal: $T_S\cos60^\circ = T_C\cos30^\circ \Rightarrow T_S = \sqrt3\,T_C$;
vertical: $T_S\sin60^\circ + T_C\sin30^\circ = 100 \Rightarrow \frac{3}{2}T_C + \frac{1}{2}T_C = 100$, so $T_C = 50$ N and $T_S = 50\sqrt3$ N.
Elongation $\Delta l = \frac{Tl}{AY}$ (same $A$):
$\frac{\Delta l_C}{\Delta l_S} = \frac{T_Cl_C/Y_C}{T_Sl_S/Y_S} = \frac{50\sqrt3/(1\times10^{11})}{50\sqrt3\times1/(2\times10^{11})} = 2$.
Answer: 2. | ||||
| Q.15 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Speeds: S1 moves at $30$ m/s towards S2; O moves at $10$ m/s towards S2, perpendicular to the S1–S2 line. S1 is 800 m from S2 and O is 600 m from S2, so S1–O $= 1000$ m.
From S2 (at rest), O moves straight towards it: $f_2 = 120\times\frac{330 + 10}{330} = 123.64$ Hz.
From S1: only velocity components along the line S1–O count. The direction from S1 to O has cosine $\frac{800}{1000} = 0.8$ with S1's motion, so the source approaches at $30\times0.8 = 24$ m/s; O's velocity has component $10\times\frac{600}{1000} = 6$ m/s towards S1.
$f_1 = 120\times\frac{330 + 6}{330 - 24} = 120\times\frac{336}{306} = 131.76$ Hz.
Beats $= f_1 - f_2 = 131.76 - 123.64 = 8.13$ per second.
Answer: 8.13. | ||||
| Q.16 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Layers parallel to the plates are capacitors in series: $\frac{1}{C} = \sum_{m=1}^{N}\frac{\delta}{K_m\epsilon_0A} = \frac{d}{K\epsilon_0A}\cdot\frac{1}{N}\sum_{m=1}^{N}\frac{1}{1 + m/N}$.
For large $N$ the sum becomes an integral: $\frac{1}{N}\sum\frac{1}{1 + m/N}\to\int_0^1\frac{dx}{1 + x} = \ln2$.
So $\frac{1}{C} = \frac{d\ln2}{K\epsilon_0A}$ and $C = \frac{K\epsilon_0A}{d\ln2}$: $\alpha = 1$.
Answer: 1. | ||||
| Q.17 | 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 | NumericalMedium | |
A medium numerical 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 Let the calorimeter's heat capacity be $W$, the liquid's specific heat $s$ and latent heat $L$.
First 5 g: poured slowly, it boils away while the calorimeter is above 80°C, so the calorimeter cools from 110°C to the boiling point 80°C: $W(110 - 80) = 5[s(80 - 30) + L]$, i.e. $30W = 250s + 5L$.
Next 80 g: the calorimeter cools from 80°C to 50°C and the liquid warms from 30°C to 50°C: $W(80 - 50) = 80s(50 - 30)$, i.e. $30W = 1600s$.
Equating: $1600s = 250s + 5L$, so $L = 270s$ and $\frac{L}{s} = 270$°C.
Answer: 270. | ||||
| Q.18 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Inside the core ($n_1$) the ray zigzags at angle $\phi$ to the normal of the core–cladding surface. It is guided only if $\sin\phi \ge \frac{n_2}{n_1} = \frac{1.44}{1.5} = 0.96$.
The path length to cover the axial length $L$ is $\frac{L}{\sin\phi}$, so the time is $t = \frac{n_1L}{c\sin\phi}$, largest when $\sin\phi$ is smallest, i.e. at the critical angle, $\sin\phi = 0.96$ (reachable: it needs $\sin\theta = n_1\cos\phi = 1.5\times0.28 = 0.42$ at the entry face).
$t_{max} = \frac{1.5\times9.6}{3\times10^8\times0.96} = \frac{14.4}{2.88\times10^8} = 5\times10^{-8}$ s $= 50\times10^{-9}$ s.
Answer: 50. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter On heating, borax loses water and gives sodium metaborate and boric anhydride: $\mathrm{Na_2B_4O_7 \to 2NaBO_2 + B_2O_3}$.
The glassy $\mathrm{B_2O_3}$ combines with the metal oxide to form the metal metaborate, whose colour identifies the metal. For chromium(III): $\mathrm{Cr_2O_3 + 3B_2O_3 \to 2Cr(BO_2)_3}$, the green chromium(III) metaborate.
$\mathrm{Cr_2O_3}$ is itself green but is not what the bead holds; $\mathrm{Cr_2(B_4O_7)_3}$ and CrB are not formed.
Answer: (A). | ||||
| Q.2 | 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 Calamine is zinc carbonate, $\mathrm{ZnCO_3}$.
Malachite is basic copper carbonate, $\mathrm{CuCO_3\cdot Cu(OH)_2}$.
Magnetite is $\mathrm{Fe_3O_4}$ ($\mathrm{Fe_2O_3}$ is haematite).
Cryolite is $\mathrm{Na_3AlF_6}$.
Only (C) has all four right.
Answer: (C). | ||||
| Q.3 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter Below the CMC sodium stearate behaves as a strong electrolyte: $\Lambda_m$ falls linearly with $\sqrt c$ (Kohlrausch, $\Lambda_m = \Lambda_m^0 - A\sqrt c$).
At the CMC the stearate ions aggregate into micelles. Many ions are now bound in a few large, slow aggregates (which also hold back part of the counter-ions), so the number of free charge carriers per mole drops and $\Lambda_m$ falls much more steeply after the CMC.
The plot is a falling line that bends down more sharply at the CMC: (C). (A) would mean no further fall, (B) and (D) a rise, none of which happens.
Answer: (C). | ||||
| Q.4 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The acids are I propiolic acid $\mathrm{HC{\equiv}C{-}COOH}$, II acrylic acid $\mathrm{CH_2{=}CH{-}COOH}$, III 4-methoxybenzoic acid, IV propanoic acid $\mathrm{CH_3CH_2COOH}$.
The carbon attached to $-\mathrm{COOH}$ withdraws electrons more the more s character it has: sp (I) > sp$^2$ (II, III) > sp$^3$ (IV). So I is the strongest (pK$_a\approx1.9$) and IV the weakest (pK$_a\approx4.9$).
Between II and III: in III the para methoxy group pushes electrons into the ring by resonance, weakening the acid (pK$_a\approx4.5$, weaker than benzoic acid), while acrylic acid has pK$_a\approx4.25$.
Order: I > II > III > IV.
Answer: (B). | ||||
| Q.5 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Q reduces $\mathrm{Cu^{2+}}$ to $\mathrm{Cu^+}$, so it is the reducing tin(II) chloride, $\mathrm{SnCl_2}$.
X: $\mathrm{SnCl_2 + Cl^- \to SnCl_3^-}$, a monoanion; Sn has three bond pairs and one lone pair, so it is pyramidal.
Y: $\mathrm{SnCl_2 + Me_3N \to Me_3N{\to}SnCl_2}$, the Lewis acid $\mathrm{SnCl_2}$ taking the nitrogen's lone pair.
Z: $\mathrm{SnCl_2 + 2CuCl_2 \to SnCl_4 + 2CuCl}$.
(A) Sn in $\mathrm{SnCl_3^-}$ has four electron pairs: sp$^3$. True.
(B) Y has the coordinate bond N→Sn. True.
(C) Sn in $\mathrm{SnCl_4}$ is +4, not +2. False.
(D) Sn in $\mathrm{SnCl_4}$ uses all four valence electrons in bonds; it has no lone pair. False.
Answer: (A), (B). | ||||
| Q.6 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Multiple correctMedium | |
A medium multiple correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mathrm{2MnO_2 + 4KOH + O_2 \to 2K_2MnO_4 + 2H_2O}$: W is potassium manganate, Y = $\mathrm{MnO_4^{2-}}$ (Mn +6, $d^1$).
Electrolytic oxidation: $\mathrm{MnO_4^{2-} \to MnO_4^- + e^-}$: X is $\mathrm{KMnO_4}$, Z = $\mathrm{MnO_4^-}$ (Mn +7, $d^0$).
(A) In acid, manganate disproportionates: $\mathrm{3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O}$. True.
(B) Manganate is green and permanganate purple; both are tetrahedral. True.
(C) Y ($d^1$, one unpaired electron) is paramagnetic and Z ($d^0$) is diamagnetic, the reverse of the statement. False.
(D) In both ions the Mn–O π bonds are formed by oxygen p orbitals overlapping manganese d orbitals (pπ–dπ). True.
Answer: (A), (B), (D). | ||||
| Q.7 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Multiple correctEasy | |
A easy multiple correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics The standard enthalpy of formation is for forming one mole of a compound from its elements in their standard states.
(A) Forms 2 mol of water, so $\Delta H^\circ = 2\Delta_fH^\circ(\mathrm{H_2O})$. False.
(B) Carbon's standard state is graphite, not C(g). False.
(C) One mole of $\mathrm{O_3}$ from $\mathrm{O_2(g)}$, the standard state of oxygen. True.
(D) One mole of $\mathrm{SO_2}$ from solid $\mathrm{S_8}$ (rhombic sulphur, the standard state) and $\mathrm{O_2(g)}$. True.
Answer: (C), (D). | ||||
| Q.8 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Multiple correctEasy | |
A easy multiple correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $u_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}$ and $\varepsilon_{\mathrm{av}} = \frac{3}{2}k_BT$.
(A) $u_{\mathrm{rms}}\propto\sqrt T$: four times the temperature doubles it. True.
(B) $\varepsilon_{\mathrm{av}}\propto T$: four times the temperature makes it four times, not double. False.
(C) $\varepsilon_{\mathrm{av}} = \frac{3}{2}k_BT$ has no mass in it. True.
(D) $u_{\mathrm{rms}}\propto\frac{1}{\sqrt M}$. True.
Answer: (A), (C), (D). | ||||
| Q.9 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Multiple correctMedium | |
A medium multiple correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) $\mathrm{BeCl_2}$ (linear), $\mathrm{CO_2}$ (linear) and $\mathrm{BCl_3}$ (trigonal planar) are non-polar. False.
(B) $\mathrm{NO_2}$ (bent), $\mathrm{NH_3}$ (pyramidal), $\mathrm{POCl_3}$ (distorted tetrahedron, P=O differs from P–Cl) and $\mathrm{CH_3Cl}$ are all polar. True.
(C) $\mathrm{BF_3}$ (trigonal planar) and $\mathrm{SF_6}$ (octahedral) are non-polar. False.
(D) $\mathrm{SO_2}$ (bent), $\mathrm{C_6H_5Cl}$, $\mathrm{H_2Se}$ (bent) and $\mathrm{BrF_5}$ (square pyramidal) are all polar. True.
Answer: (B), (D). | ||||
| Q.10 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Multiple correctEasy | |
A easy multiple correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics U-238 → Th-234: mass −4, charge −2: $\mathrm{x}_1$ is an α particle.
Th-234 → Pa-234: charge +1: $\mathrm{x}_2$ is $\beta^-$.
Z → Th-230 loses mass 4, so it is an α decay and Z has atomic number $90 + 2 = 92$: Z is U-234.
Pa-234 (91) → U-234 (92): charge +1, so $\mathrm{x}_3$ is $\beta^-$, not a γ-ray.
(A) The α particle is positive and bends towards the negative plate. True.
(B) True. (C) False. (D) Z is ${}^{234}_{92}$U, an isotope of uranium. True.
Answer: (A), (B), (D). | ||||
| Q.11 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Multiple correctEasy | |
A easy multiple correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules (A) Monosaccharides are the simplest sugars; they are themselves polyhydroxy aldehydes or ketones and cannot be hydrolysed further. True.
(B) Bromine water oxidises the aldehyde group of glucose to give gluconic acid; glutamic acid is an amino acid. False.
(C) Sucrose hydrolyses to D-(+)-glucose (dextrorotatory) and D-(−)-fructose (laevorotatory). True.
(D) The α and β six-membered cyclic hemiacetals of D-(+)-glucose differ only at C-1 and are called anomers. True.
Answer: (A), (C), (D). | ||||
| Q.12 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Multiple correctMedium | |
A medium multiple correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{C_6H_{10}O}$ that adds MeMgBr is cyclohexanone. Q is 1-methylcyclohexan-1-ol.
S: conc. HCl replaces the tertiary OH by Cl through the tertiary carbocation: 1-chloro-1-methylcyclohexane.
R: 20% $\mathrm{H_3PO_4}$ at 360 K dehydrates Q to the more substituted alkene, 1-methylcyclohexene.
T: $\mathrm{H_2}$/Ni gives methylcyclohexane; $\mathrm{Br_2}$/hν brominates the tertiary C–H selectively: 1-bromo-1-methylcyclohexane.
U: HBr with benzoyl peroxide adds anti-Markovnikov (Br on the less substituted alkene carbon): 1-bromo-2-methylcyclohexane.
(A) Its S is an allylic chloride with the double bond kept. False.
(B) Its U is 1-bromo-1-methylcyclohexane (that is T). False.
(C) S is 1-chloro-1-methylcyclohexane and U is 1-bromo-2-methylcyclohexane. True.
(D) U is 1-bromo-2-methylcyclohexane and T is 1-bromo-1-methylcyclohexane. True.
Answer: (C), (D). | ||||
| Q.13 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{B_2H_6}$: the borons are joined only through B–H–B bridges; no B–B bond. No.
$\mathrm{B_3N_3H_6}$ (borazine): the ring alternates B and N; bonds are B–N and B–H, N–H. No.
$\mathrm{N_2O}$: N=N=O (N≡N–O); has an N–N bond. Yes.
$\mathrm{N_2O_4}$: $\mathrm{O_2N{-}NO_2}$; has an N–N bond. Yes.
$\mathrm{H_2S_2O_3}$ (thiosulphuric acid): a terminal S on the central S: S=S bond. Yes.
$\mathrm{H_2S_2O_8}$ (peroxodisulphuric acid): $\mathrm{HO_3S{-}O{-}O{-}SO_3H}$; has an O–O bond. Yes.
Total $= 4$.
Answer: 4. | ||||
| Q.14 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{O_2F_2}$ is a powerful fluorinating agent: $\mathrm{XeF_4 + O_2F_2 \to XeF_6 + O_2}$. So Y is $\mathrm{XeF_6}$.
In $\mathrm{XeF_6}$, Xe has 8 valence electrons, 6 in the six Xe–F bonds and 2 as one lone pair (distorted octahedron). Each F has 3 lone pairs: $6\times3 = 18$.
Total lone pairs $= 18 + 1 = 19$.
Answer: 19. | ||||
| Q.15 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Mixing equal volumes halves each concentration: $[\mathrm{Fe^{2+}}] = 0.03$ M, $[\mathrm{S^{2-}}] = 0.1$ M.
$K_c$ is huge, so almost all the $\mathrm{Fe^{2+}}$ precipitates: $[\mathrm{S^{2-}}]$ left $= 0.1 - 0.03 = 0.07$ M.
$K_c = \frac{1}{[\mathrm{Fe^{2+}}][\mathrm{S^{2-}}]}$, so $[\mathrm{Fe^{2+}}] = \frac{1}{1.6\times10^{17}\times0.07} = \frac{1}{1.12\times10^{16}} = 8.93\times10^{-17}$ M.
Y $= 8.93$.
Answer: 8.93. | ||||
| Q.16 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Raoult's law: $\frac{p^0 - p}{p^0} = x_{\text{solute}} = \frac{10}{650} = \frac{1}{65}$.
So $\frac{n_2}{n_1 + n_2} = \frac{1}{65}$, i.e. $\frac{n_2}{n_1} = \frac{1}{64}$. Benzene: $n_1 = \frac{39}{78} = 0.5$ mol, so $n_2 = \frac{0.5}{64} = 7.81\times10^{-3}$ mol.
Molality $m = \frac{7.81\times10^{-3}}{0.039} = 0.2003$ mol kg$^{-1}$.
$\Delta T_f = K_fm = 5.12\times0.2003 = 1.026$ K, i.e. 1.02 (truncated) or 1.03 (rounded).
Answer: 1.02. | ||||
| Q.17 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Experiments 1 and 2: doubling [B] leaves the rate unchanged, so the order in B is 0.
Experiments 1 and 3: doubling [C] doubles the rate, so the order in C is 1.
Experiments 1 and 4: [A] × 1.5 gives rate × 1.5, so the order in A is 1.
Rate $= k[\mathrm{A}][\mathrm{C}]$, with $k = \frac{6.0\times10^{-5}}{0.2\times0.1} = 3.0\times10^{-3}$ dm$^3$ mol$^{-1}$ s$^{-1}$.
New rate $= 3.0\times10^{-3}\times0.15\times0.15 = 6.75\times10^{-5}$ mol dm$^{-3}$ s$^{-1}$.
Answer: 6.75. | ||||
| Q.18 | AminesDiazonium salts | Class 12: Amines | NumericalMedium | |
A medium numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines Scheme 1 (P = aniline): bromine water brominates all free ortho and para positions: 2,4,6-tribromoaniline. Diazotisation and CuCN/KCN (Sandmeyer) put CN in place of $\mathrm{NH_2}$: 2,4,6-tribromobenzonitrile. Acid hydrolysis gives 2,4,6-tribromobenzoic acid, and $\mathrm{SOCl_2}$ the acid chloride. Q = 2,4,6-tribromobenzoyl chloride (3 Br).
Scheme 2 (R = benzene): oleum gives benzenesulphonic acid; NaOH fusion and acidification give phenol; $\mathrm{Br_2}$ in $\mathrm{CS_2}$ at 273 K (non-polar solvent, cold) gives mainly 4-bromophenol. S = 4-bromophenol (1 Br).
Scheme 3: NaOH makes the phenoxide, which is acylated by Q: T = 4-bromophenyl 2,4,6-tribromobenzoate.
Br atoms in T $= 3 + 1 = 4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $S$ is the region on and outside the circle with centre $2 - i$ and radius $\sqrt5$. Maximising $\frac{1}{|z - 1|}$ means finding the point of $S$ nearest to $1$.
The point $1$ is inside the circle (its distance from $2 - i$ is $\sqrt2 < \sqrt5$), so the nearest point of $S$ is on the circle, along the ray from the centre $(2, -1)$ through $(1, 0)$, i.e. in the direction $\frac{(-1, 1)}{\sqrt2}$:
$z_0 = x_0 + iy_0$ with $x_0 = 2 - \sqrt{5/2}$, $y_0 = -1 + \sqrt{5/2}$.
Numerator: $4 - (z_0 + \overline{z_0}) = 4 - 2x_0 = 2\sqrt{5/2}$.
Denominator: $z_0 - \overline{z_0} + 2i = 2iy_0 + 2i = 2i\sqrt{5/2}$.
The quotient is $\frac{1}{i} = -i$, whose principal argument is $-\frac{\pi}{2}$.
Answer: (A). | ||||
| Q.2 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants A $2\times2$ matrix satisfies $M^2 - (\operatorname{tr}M)M + (\det M)I = 0$. Multiplying by $M^{-1}$: $M = (\operatorname{tr}M)I - (\det M)M^{-1}$. So $\alpha = \operatorname{tr}M$ and $\beta = -\det M$.
Let $t = \sin^2\theta\cos^2\theta = \frac{1}{4}\sin^22\theta \in [0, \frac{1}{4}]$.
$\alpha = \sin^4\theta + \cos^4\theta = 1 - 2t$, smallest at $t = \frac{1}{4}$: $\alpha^* = \frac{1}{2}$.
$\det M = \sin^4\theta\cos^4\theta + (1 + \sin^2\theta)(1 + \cos^2\theta) = t^2 + (1 + 1 + t) = t^2 + t + 2$, so $\beta = -(t^2 + t + 2)$, smallest at $t = \frac{1}{4}$: $\beta^* = -\left(\frac{1}{16} + \frac{1}{4} + 2\right) = -\frac{37}{16}$.
$\alpha^* + \beta^* = \frac{8}{16} - \frac{37}{16} = -\frac{29}{16}$.
Answer: (C). | ||||
| Q.3 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Substitute $y = mx + 1$: $(x - 3)^2 + (mx + 3)^2 = 25$, i.e. $(1 + m^2)x^2 + (6m - 6)x - 7 = 0$.
The midpoint's $x$-coordinate is half the sum of roots: $\frac{-(6m - 6)}{2(1 + m^2)} = -\frac{3}{5}$, i.e. $\frac{3(1 - m)}{1 + m^2} = -\frac{3}{5}$, giving $5(1 - m) = -(1 + m^2)$, i.e. $m^2 - 5m + 6 = 0$: $m = 2$ or $m = 3$.
Both lines do cut the circle: the distance from the centre $(3, -2)$ to the line is $\frac{|3m + 3|}{\sqrt{1 + m^2}}$, which is $\frac{9}{\sqrt5}\approx4.02$ for $m = 2$ and $\frac{12}{\sqrt{10}}\approx3.79$ for $m = 3$, both less than 5.
So $m \in \{2, 3\}$, and $2 \le m < 4$.
Answer: (B). | ||||
| Q.4 | 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 Take the part with $x > 0$ (for $x \le -1$ the strip $1 \le y \le x^2$ is unbounded, so the bounded region meant is in the first quadrant).
The curves $y = x^2$ and $y = \frac{8}{x}$ meet at $x^3 = 8$, $x = 2$; $y = x^2$ meets $y = 1$ at $x = 1$, and $y = \frac{8}{x}$ meets $y = 1$ at $x = 8$.
For $1 \le x \le 2$ the top is $y = x^2$; for $2 \le x \le 8$ it is $y = \frac{8}{x}$; the bottom is $y = 1$.
Area $= \int_1^2(x^2 - 1)\,dx + \int_2^8\left(\frac{8}{x} - 1\right)dx = \left(\frac{7}{3} - 1\right) + (8\ln4 - 6) = 16\log_e2 - \frac{14}{3}$.
Answer: (A). | ||||
| Q.5 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Multiple correctHard | |
A hard multiple correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Since $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$, $a_{n+2} = a_{n+1} + a_n$, with $a_1 = a_2 = 1$: the Fibonacci numbers $1, 1, 2, 3, 5, 8, \dots$
(A) $a_k = a_{k+2} - a_{k+1}$, so the sum telescopes: $\sum_{k=1}^{n}a_k = a_{n+2} - a_2 = a_{n+2} - 1$. True.
(B) $\sum a_nx^n = \frac{1}{\alpha - \beta}\left(\frac{\alpha x}{1 - \alpha x} - \frac{\beta x}{1 - \beta x}\right) = \frac{x}{1 - x - x^2}$ (using $\alpha + \beta = 1$, $\alpha\beta = -1$; it converges since $\alpha/10 < 1$). At $x = \frac{1}{10}$: $\frac{0.1}{0.89} = \frac{10}{89}$. True.
(C) $a_{n-1} + a_{n+1} = \frac{\alpha^{n-1}(1 + \alpha^2) - \beta^{n-1}(1 + \beta^2)}{\alpha - \beta}$. Now $1 + \alpha^2 = \alpha(\alpha - \beta)$ (as $1 = -\alpha\beta$), and likewise $1 + \beta^2 = -\beta(\alpha - \beta)$, so $b_n = \alpha^n + \beta^n$ for $n \ge 2$; and $b_1 = 1 = \alpha + \beta$. True.
(D) $\sum b_nx^n = \frac{\alpha x}{1 - \alpha x} + \frac{\beta x}{1 - \beta x} = \frac{x(\alpha + \beta) - 2\alpha\beta x^2}{1 - (\alpha + \beta)x + \alpha\beta x^2} = \frac{x + 2x^2}{1 - x - x^2}$. At $x = \frac{1}{10}$: $\frac{0.12}{0.89} = \frac{12}{89}$, not $\frac{8}{89}$. False.
Answer: (A), (B), (C). | ||||
| Q.6 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Use $M\,(\operatorname{adj}M) = (\det M)I$.
Row 1 of $M$ with column 2 of adj$M$: $0 - 6 + 3a = 0$, so $a = 2$. Row 1 with column 1: $0 + 8 - 5a = \det M = -2$.
Row 3 with column 1: $-3 + 8b - 5 = 0$, so $b = 1$ (check row 3 with column 3: $-3 + 2b - 1 = -2$ ✓).
(A) $a + b = 3$. True.
(B) $(\operatorname{adj}M)^{-1} = \frac{M}{\det M} = -\frac{M}{2}$, and $\operatorname{adj}(M^{-1}) = (\operatorname{adj}M)^{-1} = -\frac{M}{2}$. Their sum is $-M$. True.
(C) $\det(\operatorname{adj}M^2) = (\det M^2)^{3-1} = (4)^2 = 16$, not 81. False.
(D) $\begin{bmatrix}\alpha\\ \beta\\ \gamma\end{bmatrix} = M^{-1}\begin{bmatrix}1\\2\\3\end{bmatrix} = \frac{1}{-2}\operatorname{adj}M\begin{bmatrix}1\\2\\3\end{bmatrix} = -\frac{1}{2}\begin{bmatrix}-2\\2\\-2\end{bmatrix} = \begin{bmatrix}1\\-1\\1\end{bmatrix}$. Then $\alpha - \beta + \gamma = 3$. True.
Answer: (A), (B), (D). | ||||
| Q.7 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Multiple correctEasy | |
A easy multiple correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(G|B_1) = \frac{5}{10}$, $P(G|B_2) = \frac{5}{8}$, $P(G|B_3) = \frac{3}{8}$.
(A) $P(G|B_3) = \frac{3}{8}$. True.
(B) $P(G) = \frac{3}{10}\cdot\frac{1}{2} + \frac{3}{10}\cdot\frac{5}{8} + \frac{4}{10}\cdot\frac{3}{8} = \frac{12}{80} + \frac{15}{80} + \frac{12}{80} = \frac{39}{80}$. True.
(C) $P(B_3|G) = \frac{12/80}{39/80} = \frac{12}{39} = \frac{4}{13}$, not $\frac{5}{13}$. False.
(D) $P(B_3 \cap G) = \frac{4}{10}\cdot\frac{3}{8} = \frac{3}{20}$, not $\frac{3}{10}$. False.
Answer: (A), (B). | ||||
| Q.8 | 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 Sine rule with circumradius 1: $\sin P = \frac{p}{2} = \frac{\sqrt3}{2}$, $\sin Q = \frac{q}{2} = \frac{1}{2}$. So $Q = 30^\circ$ ($150^\circ$ is impossible alongside $P$), and $P = 60^\circ$ or $120^\circ$. $P = 60^\circ$ would make $R = 90^\circ$, which is excluded; so $P = 120^\circ$, $R = 30^\circ$ and $r = 2\sin30^\circ = 1$.
Coordinates: $Q = (0, 0)$, $R = (\sqrt3, 0)$, $P = \left(\frac{\sqrt3}{2}, \frac{1}{2}\right)$ (so $PQ = PR = 1$).
$S$, the midpoint of $PQ$, is $\left(\frac{\sqrt3}{4}, \frac{1}{4}\right)$; $E$, the foot from $P$, is $\left(\frac{\sqrt3}{2}, 0\right)$.
(A) $RS = \sqrt{\left(\frac{3\sqrt3}{4}\right)^2 + \left(\frac{1}{4}\right)^2} = \sqrt{\frac{28}{16}} = \frac{\sqrt7}{2}$. True.
$O$: on $RS$, $x = \sqrt3 - \frac{3\sqrt3}{4}s = \frac{\sqrt3}{2}$ gives $s = \frac{2}{3}$, so $O = \left(\frac{\sqrt3}{2}, \frac{1}{6}\right)$.
(C) $OE = \frac{1}{6}$. True.
(B) $\Delta SOE$ has base $OE = \frac{1}{6}$ on the line $x = \frac{\sqrt3}{2}$ and height $\frac{\sqrt3}{2} - \frac{\sqrt3}{4} = \frac{\sqrt3}{4}$: area $= \frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt3}{4} = \frac{\sqrt3}{48}$, not $\frac{\sqrt3}{12}$. False.
(D) Area of $\Delta PQR = \frac{1}{2}\cdot1\cdot1\cdot\sin120^\circ = \frac{\sqrt3}{4}$; semi-perimeter $s = \frac{2 + \sqrt3}{2}$; inradius $= \frac{\sqrt3/4}{(2 + \sqrt3)/2} = \frac{\sqrt3}{2(2 + \sqrt3)} = \frac{\sqrt3}{2}(2 - \sqrt3)$. True.
Answer: (A), (C), (D). | ||||
| Q.9 | Conic SectionsEllipse | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The largest axis-parallel rectangle in $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ has corner $(a\cos\theta, b\sin\theta)$ maximising $2ab\sin2\theta$: $\theta = 45^\circ$, half-sides $\frac{a}{\sqrt2}, \frac{b}{\sqrt2}$, area $2ab$.
The largest ellipse in that rectangle (centred, axes along the sides) has semi-axes equal to the half-sides. So $a_{n+1} = \frac{a_n}{\sqrt2}$, $b_{n+1} = \frac{b_n}{\sqrt2}$: $a_n = \frac{3}{2^{(n-1)/2}}$, $b_n = \frac{2}{2^{(n-1)/2}}$.
(A) Every $E_n$ has $\frac{b_n}{a_n} = \frac{2}{3}$, so all have the same eccentricity $\frac{\sqrt5}{3}$. False.
(B) Area of $R_n = 2a_nb_n = \frac{12}{2^{n-1}}$; the sum is $12\left(1 + \frac{1}{2} + \dots\right) < 24$ for every $N$. True.
(C) $a_9 = \frac{3}{16}$, $b_9 = \frac{2}{16} = \frac{1}{8}$; latus rectum $= \frac{2b_9^2}{a_9} = \frac{2/64}{3/16} = \frac{1}{6}$. True.
(D) Focal distance $= \sqrt{a_9^2 - b_9^2} = \sqrt{\frac{9 - 4}{256}} = \frac{\sqrt5}{16}$, not $\frac{\sqrt5}{32}$. False.
Answer: (B), (C). | ||||
| Q.10 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Multiple correctHard | |
A hard multiple correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Continuity: at $x = 0$ both sides give 1; at $x = 1$, $1 - 1 + 1 = 1$ and $\frac{2}{3} - 4 + 7 - \frac{8}{3} = 1$; at $x = 3$, $18 - 36 + 21 - \frac{8}{3} = \frac{1}{3}$ and $0 - 3 + \frac{10}{3} = \frac{1}{3}$. So $f$ is continuous.
(A) For $x < 0$, $f(x) = (x + 1)^5 - 2x$, so $f'(x) = 5(x + 1)^4 - 2$, which is negative near $x = -1$ (e.g. $f'(-1) = -2$). $f$ is not increasing on $(-\infty, 0)$. False.
Near $x = 1$: $f'(x) = 2x - 1$ for $0 \le x < 1$ and $f'(x) = 2x^2 - 8x + 7$ for $1 \le x < 3$. Both give $f'(1) = 1$, so $f'$ is continuous at 1.
(B) Just left of 1, $f'' = 2 > 0$ ($f'$ rising to 1); just right, $f'' = 4x - 8 = -4 < 0$ ($f'$ falling from 1). So $f'$ has a local maximum at $x = 1$. True.
(D) The one-sided derivatives of $f'$ at 1 are 2 and $-4$: $f'$ is not differentiable at $x = 1$. True.
(C) $f(x) \to -\infty$ as $x \to -\infty$ (the $x^5$ term) and $f(x) \to +\infty$ as $x \to \infty$ ($(x - 2)\ln(x - 2)$ outgrows $x$). Being continuous, $f$ takes every real value: onto. True.
Answer: (B), (C), (D). | ||||
| Q.11 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Multiple correctHard | |
A hard multiple correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations At $P(x, y)$ the tangent meets the $y$-axis at $Y_P = (0, y - xy')$, so $PY_P^2 = x^2 + x^2y'^2 = 1$, i.e. $y' = \pm\frac{\sqrt{1 - x^2}}{x}$.
The curve is in the first quadrant ($0 < x \le 1$, $y \ge 0$) and reaches $y = 0$ at $x = 1$, so $y$ decreases as $x$ increases: $y' = -\frac{\sqrt{1 - x^2}}{x}$, i.e. $xy' + \sqrt{1 - x^2} = 0$. (B) true, (D) false.
Integrate with $x = \sin t$: $\int\frac{\sqrt{1 - x^2}}{x}dx = \int\frac{\cos^2t}{\sin t}dt = \int(\csc t - \sin t)\,dt = -\ln\left(\frac{1 + \cos t}{\sin t}\right) + \cos t$.
So $y = \ln\left(\frac{1 + \sqrt{1 - x^2}}{x}\right) - \sqrt{1 - x^2} + C$, and $y(1) = 0$ gives $C = 0$.
(A) is this curve. True. (C) is its negative, which is $\le 0$ for $0 < x \le 1$, not in the first quadrant. False.
Answer: (A), (B). | ||||
| Q.12 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctMedium | |
A medium multiple correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Direction of $L_3$: $(-1, 2, 2)\times(2, -1, 2) = (6, 6, -3) \parallel (2, 2, -1)$.
$L_3$ is the common perpendicular. Points $A = (1 - \lambda, 2\lambda, 2\lambda)$ on $L_1$ and $B = (2\mu, -\mu, 2\mu)$ on $L_2$ with $\vec{AB}$ perpendicular to both directions:
$(A - B)\cdot(-1, 2, 2) = -1 + 9\lambda = 0 \Rightarrow \lambda = \frac{1}{9}$; $(A - B)\cdot(2, -1, 2) = 2 - 9\mu = 0 \Rightarrow \mu = \frac{2}{9}$.
$A = \left(\frac{8}{9}, \frac{2}{9}, \frac{2}{9}\right) = \frac{2}{9}(4, 1, 1)$, $B = \left(\frac{4}{9}, -\frac{2}{9}, \frac{4}{9}\right) = \frac{2}{9}(2, -1, 2)$, and their midpoint is $\left(\frac{2}{3}, 0, \frac{1}{3}\right) = \frac{1}{3}(2, 0, 1)$.
(A) passes through $A$ with direction $(2, 2, -1)$. True.
(B) passes through $B$. True.
(C) passes through the midpoint of $AB$, which lies on $L_3$. True.
(D) passes through the origin. Is $O$ on $L_3$? $A + t(2, 2, -1) = 0$ needs $t = -\frac{4}{9}$ from $x$, but then $y = \frac{2}{9} - \frac{8}{9} \ne 0$. False.
Answer: (A), (B), (C). | ||||
| Q.13 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Using $\overline\omega = \omega^2$ and $1 + \omega + \omega^2 = 0$:
$|a + b\omega + c\omega^2|^2 = (a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) = a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]$.
For distinct integers, order them $x < y < z$: the gaps are at least 1 and 1, and $z - x \ge 2$, so the bracket is at least $1 + 1 + 4 = 6$ and the value at least 3.
It is attained by $a, b, c = 1, 2, 3$ (all non-zero).
Answer: 3. | ||||
| Q.14 | 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 A common term $x$ satisfies $x \equiv 1 \pmod 3$, $x \equiv 2 \pmod 5$, $x \equiv 3 \pmod 7$, with $x \ge 3$ (it must be a term of each progression).
$x \equiv 2 \pmod 5$ and $x \equiv 1 \pmod 3$: $x = 7, 22, 37, 52, \dots$ ($x \equiv 7 \pmod{15}$).
Of these, modulo 7: $7 \to 0$, $22 \to 1$, $37 \to 2$, $52 \to 3$. So $x \equiv 52 \pmod{105}$.
The common terms are $52, 157, 262, \dots$: $a = 52$, $d = \operatorname{lcm}(3, 5, 7) = 105$.
$a + d = 157$.
Answer: 157. | ||||
| Q.15 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability A matrix in $E_2$ has seven 1s and two 0s: $\binom{9}{2} = 36$ matrices, all equally likely.
Two 0s in the same row: that row has a single 1, and the other two rows are both $(1, 1, 1)$, equal, so $\det A = 0$. Count $3\times\binom{3}{2} = 9$.
Two 0s in the same column: likewise two equal columns, $\det A = 0$. Count 9.
Two 0s in different rows and columns: up to reordering rows and columns, $A = \begin{bmatrix}0&1&1\\1&0&1\\1&1&1\end{bmatrix}$, with $\det A = 0(0 - 1) - 1(1 - 1) + 1(1 - 0) = 1 \ne 0$. None singular.
$P(E_1|E_2) = \frac{18}{36} = 0.5$.
Answer: 0.5. | ||||
| Q.16 | CirclesStandard and general form | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Reflection: $8(2) - 6(3) - 23 = -25$, so $B = (2, 3) - \frac{2(-25)}{8^2 + 6^2}(8, -6) = (2, 3) + \frac{1}{2}(8, -6) = (6, 0)$. $AB = \sqrt{16 + 9} = 5$.
A tangent with both circles on the same side is a direct common tangent; it meets line $AB$ at the external centre of similitude $C$, which divides $AB$ externally in the ratio of the radii: $\frac{CA}{CB} = \frac{2}{1}$, with $C$ beyond $B$.
So $CA = 2CB = 2(CA - 5)$, giving $CA = 10$.
(Check: from $C$, the tangent makes angle $\phi$ with $CA$ where $\sin\phi = \frac{2}{10} = \frac{1}{5} = \frac{1}{CB}$.)
Answer: 10. | ||||
| Q.17 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replace $x$ by $-x$ (the limits are symmetric): the integrand becomes $\frac{1}{(1 + e^{-\sin x})(2 - \cos2x)} = \frac{e^{\sin x}}{(1 + e^{\sin x})(2 - \cos2x)}$. Adding the two forms of $I$:
$2I = \frac{2}{\pi}\int_{-\pi/4}^{\pi/4}\frac{dx}{2 - \cos2x}$, so $I = \frac{2}{\pi}\int_0^{\pi/4}\frac{dx}{2 - \cos2x}$.
With $t = \tan x$: $2 - \cos2x = \frac{1 + 3t^2}{1 + t^2}$ and $dx = \frac{dt}{1 + t^2}$, so $\int_0^{\pi/4}\frac{dx}{2 - \cos2x} = \int_0^1\frac{dt}{1 + 3t^2} = \frac{1}{\sqrt3}\tan^{-1}\sqrt3 = \frac{\pi}{3\sqrt3}$.
$I = \frac{2}{\pi}\cdot\frac{\pi}{3\sqrt3} = \frac{2}{3\sqrt3}$, $I^2 = \frac{4}{27}$, and $27I^2 = 4$.
Answer: 4. | ||||
| Q.18 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $A$: $\lambda = 1$, $A = (1, 0, 0)$. $B$: $2\mu = 1$, $B = \left(\frac{1}{2}, \frac{1}{2}, 0\right)$. $C$: $3v = 1$, $C = \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$.
$\vec{AB} = \left(-\frac{1}{2}, \frac{1}{2}, 0\right)$, $\vec{AC} = \left(-\frac{2}{3}, \frac{1}{3}, \frac{1}{3}\right)$.
$\vec{AB}\times\vec{AC} = \left(\frac{1}{6}, \frac{1}{6}, \frac{1}{6}\right)$, of length $\frac{\sqrt3}{6}$.
$\Delta = \frac{1}{2}\cdot\frac{\sqrt3}{6} = \frac{\sqrt3}{12}$, $6\Delta = \frac{\sqrt3}{2}$, $(6\Delta)^2 = \frac{3}{4} = 0.75$.
Answer: 0.75. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Multiple correctMedium | |
A medium multiple correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The rod turns about its fixed lower end, with $I = \frac{ML^2}{3}$. Let $\theta$ be its angle with the vertical.
Energy: the centre of mass drops by $\frac{L}{2}(1 - \cos 60^\circ) = \frac{L}{4}$, so $\frac{1}{2}\cdot\frac{ML^2}{3}\omega^2 = \frac{MgL}{4}$, giving $\omega^2 = \frac{3g}{2L}$ and $\omega = \sqrt{\frac{3g}{2L}}$. (A) is true.
Torque about the contact point: $Mg\frac{L}{2}\sin\theta = \frac{ML^2}{3}\alpha$, so $\alpha = \frac{3g\sin\theta}{2L} = \frac{3\sqrt3\,g}{4L}$, not $\frac{2g}{L}$. (B) is false.
Radial (centripetal) acceleration of the centre of mass: $\omega^2\frac{L}{2} = \frac{3g}{2L}\cdot\frac{L}{2} = \frac{3g}{4}$. (C) is true.
Normal reaction: the centre of mass is at height $y = \frac{L}{2}\cos\theta$, so $\ddot y = -\frac{L}{2}(\alpha\sin\theta + \omega^2\cos\theta) = -\frac{L}{2}\left(\frac{3\sqrt3 g}{4L}\cdot\frac{\sqrt3}{2} + \frac{3g}{2L}\cdot\frac{1}{2}\right) = -\frac{L}{2}\cdot\frac{15g}{8L} = -\frac{15g}{16}$.
Then $N - Mg = M\ddot y$ gives $N = Mg - \frac{15Mg}{16} = \frac{Mg}{16}$. (D) is true.
Answer: (A), (C), (D). | ||||
| Q.2 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Multiple correctHard | |
A hard multiple correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion The string from the block on the table passes over the fixed pulley and holds up the movable pulley; over the movable pulley hang $M$ (left) and $2M$ (right). Let the tension be $T_1$ in the first string and $T_2$ in the second; the massless pulley gives $T_1 = 2T_2$.
Constraint: the movable pulley moves down with $a_1$, and the two hanging blocks move relative to it equally and oppositely, so $a_2 + a_3 = 2a_1$, i.e. $a_2 - a_1 = a_1 - a_3$. (D) is true.
Write $a_2 = a_1 + b$, $a_3 = a_1 - b$. For $M$: $Mg - T_2 = M(a_1 + b)$; for $2M$: $2Mg - T_2 = 2M(a_1 - b)$. Subtracting, $b = \frac{a_1 - g}{3}$, and then $T_2 = \frac{4M}{3}(g - a_1)$, $T_1 = \frac{8M}{3}(g - a_1)$.
For the block on the table at extension $x$: $T_1 - kx = 2Ma_1$, so $\frac{8Mg}{3} - kx = \frac{14M}{3}a_1$, i.e. $a_1 = \frac{8Mg - 3kx}{14M}$.
This is SHM about $x_e = \frac{8Mg}{3k}$ with $\omega^2 = \frac{3k}{14M}$. Released from rest at $x = 0$, the amplitude is $x_e$, so $x_0 = 2x_e = \frac{16Mg}{3k}$, not $\frac{4Mg}{k}$. (A) is false.
At $x = \frac{x_0}{2} = x_e$ the speed is the maximum, $v = \omega x_e = \sqrt{\frac{3k}{14M}}\cdot\frac{8Mg}{3k} = g\sqrt{\frac{32M}{21k}} \approx 1.23\,g\sqrt{\frac{M}{k}}$, while $3g\sqrt{\frac{M}{5k}} \approx 1.34\,g\sqrt{\frac{M}{k}}$. (B) is false.
At $x = \frac{x_0}{4} = \frac{4Mg}{3k}$: $a_1 = \frac{8Mg - 4Mg}{14M} = \frac{2g}{7}$, not $\frac{3g}{10}$. (C) is false.
Answer: (D). | ||||
| Q.3 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Multiple correctHard | |
A hard multiple correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power (A) Between two hits on the piston the particle travels to the closed end and back, a distance $2L$, taking $\frac{2L}{v}$. The rate of hitting the piston is $\frac{v}{2L}$, not $\frac{v}{L}$. False.
(B) In the heavy piston's frame the particle approaches at $v + V$ and leaves at $v + V$; back in the ground frame it leaves at $v + 2V$. Each collision adds $2V$. True.
(C) Moving inward by $dL$ takes $dt = \frac{dL}{V}$, during which there are $\frac{v}{2L}\cdot\frac{dL}{V}$ hits, each adding $2V$: $dv = \frac{v}{2L}\cdot\frac{dL}{V}\cdot 2V = v\frac{dL}{L}$, not $2v\frac{dL}{L}$. False.
(D) From $\frac{dv}{v} = -\frac{dL}{L}$ (L decreasing), $vL$ is constant. Halving $L$ doubles $v$, so the kinetic energy becomes $4$ times. True.
Answer: (B), (D). | ||||
| Q.4 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The dipole has magnitude $p_0$ along the $45^\circ$ line OA. At a point at distance $R$ and angle $\phi$ from the dipole's axis its potential is $\frac{p_0\cos\phi}{4\pi\epsilon_0R^2}$. A uniform field adds $-\vec E\cdot\vec r$. For the total to be the same at every $\phi$, the field must lie along the dipole's axis, and with $\vec E = E_0\hat p$ the total is $\left(\frac{p_0}{4\pi\epsilon_0R^2} - E_0R\right)\cos\phi$, constant only if $\frac{p_0}{4\pi\epsilon_0R^2} = E_0R$. (With $\vec E$ opposite to $\vec p$ the two terms would add and could never cancel.)
(A) $R^3 = \frac{p_0}{4\pi\epsilon_0E_0}$, so $R = \left(\frac{p_0}{4\pi\epsilon_0E_0}\right)^{1/3}$. True.
(B) A lies on the dipole's axis: the dipole field there is $\frac{2p_0}{4\pi\epsilon_0R^3} = 2E_0$ along $\hat p$, so the total is $3E_0\hat p = \frac{3E_0}{\sqrt2}(\hat i + \hat j)$, not $\sqrt2E_0(\hat i + \hat j)$. False.
(C) B lies on the dipole's equatorial line: the dipole field is $\frac{p_0}{4\pi\epsilon_0R^3} = E_0$ opposite to $\hat p$, cancelling the uniform field. $\vec E_B = 0$. True.
(D) The field is $3E_0$ at A and $0$ at B, so its magnitude differs around the circle. False.
Answer: (A), (C). | ||||
| Q.5 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat capacities of the mixture: $C_V = 5\cdot\frac{3}{2}R + 1\cdot\frac{5}{2}R = 10R$, $C_P = C_V + 6R = 16R$, so $\gamma = \frac{16}{10} = 1.6$. (D) is true.
Adiabatic: $TV^{\gamma-1}$ constant, so $T = T_0\cdot4^{0.6} = T_0\cdot2^{1.2} = 2.3T_0$.
(A) $|W| = C_V\Delta T = 10R(2.3T_0 - T_0) = 13RT_0$. True.
(B) The kinetic energy of the mixture (its internal energy, all of it kinetic for ideal gases) is $C_VT = 10R\cdot2.3T_0 = 23RT_0$, not between $18RT_0$ and $19RT_0$. False.
(C) $PV^\gamma$ constant: $P = P_0\cdot4^{1.6} = P_0\cdot2^{3.2} = 9.2P_0$, between $9P_0$ and $10P_0$. True.
Answer: (A), (C), (D). | ||||
| Q.6 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Refraction at a single spherical surface, light going from glass ($n_1 = 1.5$) to air ($n_2 = 1$): $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. Take upward as positive with the origin at the top surface, so $u = -30$ cm.
I (flat): $H_1 = \frac{30}{1.5} = 20$ cm.
II (convex top, centre of curvature inside the glass): $R = -300$ cm. $\frac{1}{v} + \frac{1.5}{30} = \frac{-0.5}{-300} = \frac{1}{600}$, so $\frac{1}{v} = \frac{1 - 30}{600}$, $v = -\frac{600}{29} = -20.69$ cm. $H_2 = 20.69$ cm.
III (concave top): $R = +300$ cm. $\frac{1}{v} = -\frac{1}{600} - \frac{30}{600} = -\frac{31}{600}$, $v = -19.35$ cm. $H_3 = 19.35$ cm.
(A) $H_2 > H_1$: true. (B) $H_3 > H_1$: false. (C) $H_2 > H_3$: true. (D) $H_2 - H_1 = 0.69$ cm, not between $0.8$ and $0.9$ cm: false.
Answer: (A), (C). | ||||
| Q.7 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics With oblique incidence the light reaching the two slits already differs in path by $d\sin\alpha$; in the figure the beam slopes downward, so it reaches the lower slit later. A point $y$ above $O$ adds $\frac{yd}{D}$ to the lower slit's path. The total path difference is $\Delta = d\sin\alpha + \frac{yd}{D}$.
$\alpha = \frac{0.36}{\pi}$ degree $= \frac{0.36}{\pi}\cdot\frac{\pi}{180}$ rad $= 0.002$ rad, so $d\sin\alpha = 0.3\times10^{-3}\times0.002 = 6\times10^{-7}$ m $= 600$ nm $= \lambda$.
$\frac{yd}{D}$ at P: $\frac{11\times10^{-3}\times0.3\times10^{-3}}{1} = 3.3\times10^{-6}$ m $= 5.5\lambda$.
(A) At O, $\Delta = \lambda$: constructive, not destructive. False.
(B) $\alpha = 0$: at P, $\Delta = 5.5\lambda$, destructive. False.
(C) $\alpha = \frac{0.36}{\pi}$ degree: at P, $\Delta = 6.5\lambda$ (it would be $4.5\lambda$ had the beam sloped the other way): destructive either way. True.
(D) The fringe width $\beta = \frac{\lambda D}{d}$ does not depend on $\alpha$ (the incidence only shifts the pattern). False.
Answer: (C). | ||||
| Q.8 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Multiple correctEasy | |
A easy multiple correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Absorbed photon: $E_a = 13.6\left(1 - \frac{1}{16}\right) = 12.75$ eV. Since $E \propto \frac{1}{\lambda}$, $E_e = \frac{\lambda_a}{\lambda_e}E_a = \frac{12.75}{5} = 2.55$ eV.
(A) $13.6\left(\frac{1}{m^2} - \frac{1}{16}\right) = 2.55$ gives $\frac{1}{m^2} = 0.1875 + 0.0625 = 0.25$, $m = 2$. True.
(B) $\lambda_e = \frac{1242}{2.55} = 487$ nm, not $418$ nm. False.
(C) The atom's momentum change equals the photon's momentum $\frac{h}{\lambda}$: $\frac{\Delta p_a}{\Delta p_e} = \frac{\lambda_e}{\lambda_a} = 5$, not $\frac{1}{2}$. False.
(D) In the Bohr model the kinetic energy is $13.6\frac{Z^2}{n^2}$ eV, so $\frac{K_2}{K_1} = \frac{1}{4}$. True.
Answer: (A), (D). | ||||
| Q.9 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane For a flight with launch speed $u$: range $\frac{u^2\sin2\theta}{g}$, time $\frac{2u\sin\theta}{g}$. The first flight's average velocity is its range over its time: $V_1 = u_0\cos\theta$.
The $k$-th flight has speed $\frac{u_0}{\alpha^{k-1}}$, so its range is $R_1\alpha^{-2(k-1)}$ and its time $T_1\alpha^{-(k-1)}$. Summing the geometric series:
total displacement $= \frac{R_1}{1 - 1/\alpha^2}$, total time $= \frac{T_1}{1 - 1/\alpha}$.
Average velocity $= V_1\frac{1 - 1/\alpha}{1 - 1/\alpha^2} = \frac{V_1}{1 + 1/\alpha} = \frac{\alpha}{\alpha + 1}V_1$.
$\frac{\alpha}{\alpha + 1} = 0.8$ gives $\alpha = 4$.
Answer: 4. | ||||
| Q.10 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Motional emf: $\varepsilon = B\ell v = 1\times0.1\times0.01 = 10^{-3}$ V, constant since the speed is held constant.
The circuit is a series LR circuit switched on to this emf: $i = \frac{\varepsilon}{R}\left(1 - e^{-tR/L}\right)$ with $\tau = \frac{L}{R} = \frac{10^{-3}}{1} = 1$ ms.
At $t = 1$ ms: $i = 10^{-3}(1 - e^{-1}) = 10^{-3}(1 - 0.37) = 0.63\times10^{-3}$ A.
Answer: 0.63. | ||||
| Q.11 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At the first face: $\sin\theta = \sqrt3\sin r_1$. Inside the prism $r_1 + r_2 = A = 75^\circ$.
A smaller $\theta$ gives a smaller $r_1$ and so a larger $r_2$; total internal reflection for all $\theta \le 60^\circ$ means that $\theta = 60^\circ$ is the limiting case, where $r_2$ equals the critical angle at the prism–film boundary.
At $\theta = 60^\circ$: $\sin r_1 = \frac{\sin60^\circ}{\sqrt3} = \frac{1}{2}$, $r_1 = 30^\circ$, $r_2 = 45^\circ$.
Critical angle condition: $\sqrt3\sin45^\circ = n$, so $n^2 = 3\times\frac{1}{2} = 1.5$.
Answer: 1.5. | ||||
| Q.12 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalMedium | |
A medium numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Each photon reflected at normal incidence gives the mirror momentum $\frac{2h}{\lambda}$. The $N$ photons together give an impulse $\frac{2Nh}{\lambda}$, so the mirror starts from its equilibrium with speed $v = \frac{2Nh}{M\lambda}$.
In SHM the greatest displacement is $A = \frac{v}{\Omega} = \frac{2Nh}{M\Omega\lambda}$.
So $N = \frac{AM\Omega\lambda}{2h} = \frac{A\lambda}{2}\cdot\frac{M\Omega}{h} = \frac{10^{-6}\times8\pi\times10^{-6}}{2}\cdot\frac{10^{24}}{4\pi} = 10^{12}$.
Answer: 1. | ||||
| Q.13 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Energy released to the ground state of Rn: $Q = (226.005 - 222.000 - 4.000)\times931 = 0.005\times931 = 4.655$ MeV.
Momentum conservation: the Rn nucleus recoils with the alpha particle's momentum, so $K_{Rn} = \frac{m_\alpha}{m_{Rn}}K_\alpha = \frac{4}{222}\times4.44 = 0.08$ MeV.
Kinetic energy shared in the decay: $4.44 + 0.08 = 4.52$ MeV. The rest is the excitation of Rn, given off as the gamma photon: $4.655 - 4.52 = 0.135$ MeV $= 135$ keV.
Answer: 135. | ||||
| Q.14 | Experimental SkillsOptics experiments | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on optics experiments, from Experimental Skills. NCERT: No matching NCERT chapter Least count of the scale: $\frac{1}{4}$ cm $= 0.25$ cm.
Object distance $u = 75 - 45 = 30$ cm, image distance $v = 135 - 75 = 60$ cm. $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ gives $f = \frac{uv}{u + v} = \frac{1800}{90} = 20$ cm.
Differentiating, $\frac{\Delta f}{f^2} = \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}$. Taking $\Delta u = \Delta v = 0.25$ cm (the least count):
$\Delta f = 400\left(\frac{0.25}{900} + \frac{0.25}{3600}\right) = 400\times\frac{1.25}{3600} = 0.139$ cm.
Percentage error $= \frac{0.139}{20}\times100 = 0.69\%$.
Answer: 0.69. | ||||
| Q.15 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Matching listEasy | |
A easy matching list question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Fundamental frequency $f = \frac{1}{2L}\sqrt{\frac{T}{\mu'}}$. It is highest at the shortest free length, $L = L_0$.
With $T = T_0$: string $k$ (mass per length $k\mu$) gives $f = \frac{1}{2L_0}\sqrt{\frac{T_0}{k\mu}} = \frac{f_0}{\sqrt k}$.
I: $1$ → P. II: $\frac{1}{\sqrt2}$ → R. III: $\frac{1}{\sqrt3}$ → S. IV: $\frac{1}{2}$ → Q.
Answer: (B). | ||||
| Q.16 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Matching listMedium | |
A medium matching list question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The $n$-th harmonic of a string of length $L$, tension $T$ and mass per length $\mu'$: $f = \frac{n}{2L}\sqrt{\frac{T}{\mu'}}$.
String 1 (first harmonic, length $L_0$) at tension $T_0$ sounds $f_0$; every option except (D) gives it $T_0$, and with that the common frequency is $f_0 = \frac{1}{2L_0}\sqrt{\frac{T_0}{\mu}}$. So I → P.
String 2: $\frac{3}{2\cdot\frac{3L_0}{2}}\sqrt{\frac{T}{2\mu}} = \frac{1}{L_0}\sqrt{\frac{T}{2\mu}}$. Setting it equal to $f_0$: $4\cdot\frac{T}{2} = T_0$, $T = \frac{T_0}{2}$ → Q.
String 3: $\frac{5}{2\cdot\frac{5L_0}{4}}\sqrt{\frac{T}{3\mu}} = \frac{2}{L_0}\sqrt{\frac{T}{3\mu}}$. Setting it equal: $16\cdot\frac{T}{3} = T_0$, $T = \frac{3T_0}{16}$ → T.
String 4: $\frac{14}{2\cdot\frac{7L_0}{4}}\sqrt{\frac{T}{4\mu}} = \frac{4}{L_0}\cdot\frac{1}{2}\sqrt{\frac{T}{\mu}} = \frac{2}{L_0}\sqrt{\frac{T}{\mu}}$. Setting it equal: $16T = T_0$, $T = \frac{T_0}{16}$ → U.
I→P, II→Q, III→T, IV→U. Answer: (B). | ||||
| Q.17 | ThermodynamicsFirst law | Class 11: Thermodynamics | Matching listMedium | |
A medium matching list question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics $T\,dX = \frac{3}{2}R\,dT + RT\frac{dV}{V} = C_V\,dT + P\,dV$: the quantity is ordinary heat, so the first law applies as usual.
States: 1 $(P_0, V_0)$, 2 $(P_0, 2V_0)$, 3 $(\frac{3P_0}{2}, 2V_0)$; $P_0V_0 = \frac{1}{3}RT_0$.
(I) Work: only 1→2 at constant pressure, $W = P_0V_0 = \frac{1}{3}RT_0$ → Q.
(II) $\Delta U = \frac{3}{2}(P_3V_3 - P_1V_1) = \frac{3}{2}(3P_0V_0 - P_0V_0) = 3P_0V_0 = RT_0$ → R.
(III) $Q = \Delta U + W = RT_0 + \frac{1}{3}RT_0 = \frac{4}{3}RT_0$ → S.
(IV) 1→2 is isobaric: $Q = \frac{5}{2}P_0\Delta V = \frac{5}{2}P_0V_0 = \frac{5}{6}RT_0$ → U.
I→Q, II→R, III→S, IV→U. Answer: (B). | ||||
| Q.18 | ThermodynamicsFirst law | Class 11: Thermodynamics | Matching listMedium | |
A medium matching list question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics States: 1 $(V_0, \frac{T_0}{3})$, 2 $(2V_0, \frac{T_0}{3})$, 3 $(2V_0, T_0)$. 1→2 is isothermal, 2→3 isochoric.
(I) Work: only the isothermal step, $W = R\frac{T_0}{3}\ln\frac{2V_0}{V_0} = \frac{1}{3}RT_0\ln2$ → P.
(II) $\Delta U = \frac{3}{2}R\left(T_0 - \frac{T_0}{3}\right) = RT_0$ → R.
(III) $Q = \Delta U + W = RT_0 + \frac{1}{3}RT_0\ln2 = \frac{1}{3}RT_0(3 + \ln2)$ → T.
(IV) 1→2 is isothermal, so $Q = W = \frac{1}{3}RT_0\ln2$ → P.
I→P, II→R, III→T, IV→P. Answer: (C). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Leaching: $4\mathrm{Au} + 8\mathrm{CN^-} + \mathrm{O_2} + 2\mathrm{H_2O} \rightarrow 4[\mathrm{Au(CN)_2}]^- + 4\mathrm{OH^-}$. So $\mathbf{Q}$ is $\mathrm{O_2}$ (air) and $\mathbf{R}$ is $[\mathrm{Au(CN)_2}]^-$, gold in the $+1$ state.
Recovery: zinc, the more electropositive metal, displaces gold: $2[\mathrm{Au(CN)_2}]^- + \mathrm{Zn} \rightarrow [\mathrm{Zn(CN)_4}]^{2-} + 2\mathrm{Au}$. So $\mathbf{T}$ is Zn and $\mathbf{Z}$ is $[\mathrm{Zn(CN)_4}]^{2-}$.
(A) true. (B) true. (C) true. (D) false: R is $[\mathrm{Au(CN)_2}]^-$, not $[\mathrm{Au(CN)_4}]^-$.
Answer: (A), (B), (C). | ||||
| Q.2 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Aqua regia is three volumes of concentrated HCl to one of concentrated $\mathrm{HNO_3}$. True.
(B) Gold dissolves as the tetrachloroaurate ion: $\mathrm{Au} + 4\mathrm{H^+} + \mathrm{NO_3^-} + 4\mathrm{Cl^-} \rightarrow [\mathrm{AuCl_4}]^- + \mathrm{NO} + 2\mathrm{H_2O}$. Au is $+3$ in this anion. True.
(C) The reduction product of nitrate here is NO; it turns to $\mathrm{NO_2}$ only on meeting the oxygen of air. In the absence of air, $\mathrm{NO_2}$ is not produced. False.
(D) $3\mathrm{HCl} + \mathrm{HNO_3} \rightarrow \mathrm{NOCl} + \mathrm{Cl_2} + 2\mathrm{H_2O}$: the orange-yellow colour comes from nitrosyl chloride and chlorine. True.
Answer: (A), (B), (D). | ||||
| Q.3 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Multiple correctMedium | |
A medium multiple correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mathrm{Zn} + 2\mathrm{H_2SO_4}\ (\text{hot, conc.}) \rightarrow \mathrm{ZnSO_4} + \mathrm{SO_2} + 2\mathrm{H_2O}$: $\mathbf{G} = \mathrm{ZnSO_4}$, $\mathbf{R} = \mathrm{SO_2}$, $\mathbf{X} = \mathrm{H_2O}$.
$\mathrm{Zn} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2ZnO_2} + \mathrm{H_2}$: $\mathbf{T} = \mathrm{Na_2ZnO_2}$, $\mathbf{Q} = \mathrm{H_2}$.
$\mathrm{ZnSO_4} + \mathrm{H_2S} + 2\mathrm{NH_4OH} \rightarrow \mathrm{ZnS}\downarrow + 2\mathrm{H_2O} + (\mathrm{NH_4})_2\mathrm{SO_4}$: $\mathbf{Z} = \mathrm{ZnS}$, $\mathbf{Y} = (\mathrm{NH_4})_2\mathrm{SO_4}$.
(A) ZnS comes down as a white (dirty white) precipitate, the group IV test for zinc. True.
(B) In $\mathrm{Na_2ZnO_2}$ zinc is $+2$, not $+1$. False.
(C) $\mathrm{SO_2}$: S has two bonding domains and one lone pair, so the molecule is bent (V-shaped). True.
(D) $\mathrm{H_2}$: two electrons in $\sigma_{1s}$, bond order $\frac{2 - 0}{2} = 1$. True.
Answer: (A), (C), (D). | ||||
| Q.4 | 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 For $\mathrm{He^+}$ ($Z = 2$): $E_n = -13.6\frac{Z^2}{n^2} = -\frac{54.4}{n^2}$ eV. Setting this to $-3.4$ eV gives $n^2 = 16$, $n = 4$. With $l = 2$ the state is $4d$.
(A) It is a $4d$ state. True.
(B) Angular nodes $= l = 2$. True.
(C) Radial nodes $= n - l - 1 = 4 - 2 - 1 = 1$, not 3. False.
(D) $\mathrm{He^+}$ has a single electron, so nothing screens the nucleus: the electron feels the full charge $2e$, not less. False.
Answer: (A), (B). | ||||
| Q.5 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Multiple correctEasy | |
A easy multiple correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A) Sodium butanoate, $\mathrm{CH_3CH_2CH_2COONa}$, heated with soda lime loses $\mathrm{CO_2}$ (decarboxylation): $\mathrm{CH_3CH_2CH_2COONa} + \mathrm{NaOH} \rightarrow \mathrm{CH_3CH_2CH_3} + \mathrm{Na_2CO_3}$. Propane. True.
(B) Kolbe electrolysis of sodium butanoate joins two propyl radicals: $2\mathrm{CH_3CH_2CH_2COO^-} \rightarrow \mathrm{CH_3(CH_2)_4CH_3} + 2\mathrm{CO_2}$. The product is hexane. False.
(C) 1-Chloropropane is reduced by Zn and dilute HCl (nascent hydrogen) to propane. True.
(D) 1,2-Dibromopropane with Zn undergoes dehalogenation to propene, $\mathrm{CH_3CH{=}CH_2}$. False.
Answer: (A), (C). | ||||
| Q.6 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Multiple correctMedium | |
A medium multiple correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A) 1,2-Dibromopropane: alcoholic KOH then $\mathrm{NaNH_2}$ remove two HBr to give propyne, $\mathrm{CH_3C{\equiv}CH}$. Passed through a red-hot iron tube at 873 K, three molecules cyclise to 1,3,5-trimethylbenzene (mesitylene), which is aromatic. True.
(B) Cyclopentadiene is unusually acidic for a hydrocarbon (pKa about 16) because its conjugate base, the cyclopentadienyl anion, has $6\pi$ electrons in a planar ring and is aromatic. Methoxide removes the $\mathrm{CH_2}$ proton to give sodium cyclopentadienide. True.
(C) 3-Bromocyclobutene with ethoxide would eliminate HBr to cyclobutadiene, a $4\pi$ antiaromatic ring; the product is not aromatic. False.
(D) Benzene with excess chlorine under UV light at 500 K undergoes free-radical addition to benzene hexachloride, $\mathrm{C_6H_6Cl_6}$, which has no ring $\pi$ system and is not aromatic. False.
Answer: (A), (B). | ||||
| Q.7 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Multiple correctHard | |
A hard multiple correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Start: 4-(4-methoxyphenyl)but-3-ynal, $\mathrm{Ar{-}C{\equiv}C{-}CH_2{-}CHO}$ with Ar = 4-methoxyphenyl.
i) $\mathrm{Hg^{2+}}$, dil. $\mathrm{H_2SO_4}$ hydrates the triple bond to a ketone (on the carbon next to the ring, which best stabilises the positive charge): $\mathrm{Ar{-}CO{-}CH_2CH_2{-}CHO}$.
ii) Tollens' reagent oxidises only the aldehyde: $\mathrm{Ar{-}CO{-}CH_2CH_2{-}COOH}$.
iii) Clemmensen reduction removes the ketone oxygen: $\mathbf{Q} = \mathrm{Ar{-}CH_2CH_2CH_2{-}COOH}$, 4-(4-methoxyphenyl)butanoic acid. (Whichever carbon the hydration put the oxygen on, this step removes it, so Q is the same.)
$\mathrm{SOCl_2}$/pyridine makes the acid chloride; $\mathrm{AlCl_3}$ closes a six-membered ring by intramolecular Friedel–Crafts acylation at the ring carbon next to the side chain (meta to OMe, the only position that can close a six-membered ring): $\mathbf{R}$ = 7-methoxy-1-tetralone, the OMe para to the ring-fusion carbon that carries the $\mathrm{CH_2}$ chain.
Clemmensen again: $\mathbf{S}$ = 6-methoxy-1,2,3,4-tetrahydronaphthalene.
(A) Q there keeps an OH and S is a dihydronaphthalene: false.
(B) R there still has a ring C=C (a naphthalenone): false.
(C) Q is the butanoic acid and R the methoxytetralone: true.
(D) R the methoxytetralone and S the methoxytetralin: true.
Answer: (C), (D). | ||||
| Q.8 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter (A) Natural rubber is cis-1,4-polyisoprene; the trans polymer is gutta-percha. False.
(B) Nylon-6, from caprolactam, is a polyamide: its units are joined by $\mathrm{{-}CO{-}NH{-}}$ (amide) links. True.
(C) Teflon is made by heating tetrafluoroethene with a free-radical (persulphate) catalyst at high pressure. True.
(D) Cellulose is a chain of $\beta$-D-glucose units joined by $\beta$-1,4 glycosidic links; $\alpha$-D-glucose units make starch. False.
Answer: (B), (C). | ||||
| Q.9 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions The highest oxidation state of sulphur is $+6$, in $\mathrm{H_2SO_4}$; concentrated nitric acid is reduced to $\mathrm{NO_2}$.
Rhombic sulphur is made of $\mathrm{S_8}$ molecules. Each S goes from 0 to $+6$ (6 electrons), so $\mathrm{S_8}$ gives 48 electrons and needs 48 $\mathrm{HNO_3}$ (each N from $+5$ to $+4$):
$\mathrm{S_8} + 48\mathrm{HNO_3} \rightarrow 8\mathrm{H_2SO_4} + 48\mathrm{NO_2} + 16\mathrm{H_2O}$.
(Hydrogen: 48 on the left, $16 + 32 = 48$ on the right; oxygen: 144 on the left, $32 + 96 + 16 = 144$ on the right.)
One mole of $\mathrm{S_8}$ gives 16 mol of water: $16\times18 = 288$ g.
Answer: 288. | ||||
| Q.10 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The complex is octahedral: four N atoms from the two en ligands and two Cl. In the cis isomer the two Cl are at $90^\circ$ to each other.
Each position in an octahedron has four cis neighbours and one trans partner. For each Cl, one cis neighbour is the other Cl, so the other three cis neighbours are N atoms; its trans partner is the fourth N.
Cis N–Mn–Cl angles: $3$ for each Cl, $2\times3 = 6$.
Answer: 6. | ||||
| Q.11 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let $2p$ atm of $\mathrm{N_2O_5}$ decompose. Then $p(\mathrm{N_2O_5}) = 1 - 2p$, $p(\mathrm{N_2O_4}) = 2p$, $p(\mathrm{O_2}) = p$, and the total is $1 + p = 1.45$, so $p = 0.45$ and $p(\mathrm{N_2O_5}) = 1 - 0.9 = 0.1$ atm.
The unit $\mathrm{s^{-1}}$ makes the reaction first order. The rate constant belongs to the reaction as written, with rate $= -\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt} = k[\mathrm{N_2O_5}]$, so $\mathrm{N_2O_5}$ is used up with constant $2k$:
$\ln\frac{p_0}{p} = 2kt$, $\ln\frac{1}{0.1} = 2\times5\times10^{-4}\,t$, $t = \frac{2.303}{10^{-3}} = 2303$ s $= 2.30\times10^3$ s.
Answer: 2.30. | ||||
| Q.12 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Water: $\frac{900}{18} = 50$ mol. With $n$ mol urea, $\frac{n}{n + 50} = 0.05$, so $0.95n = 2.5$ and $n = 2.632$ mol.
Mass of urea $= 2.632\times60 = 157.9$ g; mass of solution $= 900 + 157.9 = 1057.9$ g.
Volume $= \frac{1057.9}{1.2} = 881.6\ \mathrm{cm^3} = 0.8816$ L.
Molarity $= \frac{2.632}{0.8816} = 2.985$ M, that is 2.98 to two decimal places.
Answer: 2.98. | ||||
| Q.13 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The substrate has three multiple bonds: the ring C=C of the cyclopentene, the ring C=C of the cyclohexene, and the C≡C in the chain joining them.
i) $\mathrm{H_2}$, Pd–$\mathrm{BaSO_4}$, quinoline (Lindlar's catalyst) reduces only the alkyne, to a cis alkene. Now there are three C=C bonds.
ii) Cold dilute $\mathrm{KMnO_4}$ in excess (Baeyer's reagent at 273 K) adds two OH groups (syn) across every C=C, without cleaving them.
Three C=C bonds, two OH each: $\mathbf{P}$ has 6 hydroxyl groups.
Answer: 6. | ||||
| Q.14 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Count by ring size (the ring must contain the O).
Five-membered ring: tetrahydrofuran (oxolane): 1.
Four-membered ring (oxetane) with one methyl: 2-methyloxetane (one stereocentre, a pair of enantiomers: 2) and 3-methyloxetane (achiral: 1). Total 3.
Three-membered ring (oxirane) with two more carbons: 2-ethyloxirane (chiral: 2); 2,2-dimethyloxirane (1); 2,3-dimethyloxirane (cis, a meso form: 1; trans, a pair of enantiomers: 2). Total 6.
All together: $1 + 3 + 6 = 10$.
Answer: 10. | ||||
| Q.15 | Structure of AtomBohr model | Class 11: Structure of Atom | Matching listEasy | |
A easy matching list question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In Bohr's model: $r_n = \frac{n^2h^2\epsilon_0}{\pi mZe^2} \propto n^2$; $L = \frac{nh}{2\pi} \propto n$; $K = \frac{13.6Z^2}{n^2}$ eV $\propto n^{-2}$; $U = -2K \propto n^{-2}$.
So I → T, II → S, III → P, IV → P.
(A) (I), (P): no. (B) (I), (T): yes. (C) (II), (Q): no. (D) (II), (R): no.
Answer: (B). | ||||
| Q.16 | Structure of AtomBohr model | Class 11: Structure of Atom | Matching listEasy | |
A easy matching list question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Kinetic energy $K = \frac{1}{2}mv^2$ with $v \propto \frac{1}{n}$, so $K \propto n^{-2}$. Potential energy $U = -\frac{Ze^2}{4\pi\epsilon_0 r}$ with $r \propto n^2$, so $U \propto n^{-2}$ (it is $-2K$).
III → P and IV → P.
(A) (III), (P): yes. (B) (III), (S): no. (C) (IV), (Q): no. (D) (IV), (U): no.
Answer: (A). | ||||
| Q.17 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Matching listHard | |
A hard matching list question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (I) DIBAL-H reduces the nitrile to an imine (as its aluminium salt); dilute HCl hydrolyses that to an aldehyde and also opens the acetal to a second aldehyde: $o$-$\mathrm{OHC{-}C_6H_4{-}CH_2CHO}$. $\mathrm{NaBH_4}$ reduces both to the diol Q; conc. $\mathrm{H_2SO_4}$ closes the cyclic ether R (isochroman). I gives Q and R (P, S, T, U are not formed: none of the steps makes an acid).
(II) Ozonolysis with Zn/$\mathrm{H_2O}$ turns the allyl group into $\mathrm{CH_2CHO}$: $o$-$\mathrm{OHCCH_2{-}C_6H_4{-}CO_2H}$ = P. $\mathrm{NaBH_4}$ reduces the aldehyde but not the acid: S. Conc. $\mathrm{H_2SO_4}$ closes the lactone U. II gives P, S, U.
(A) (I), (Q), (T), (U): I gives neither T nor U. False.
(B) (II), (P), (S), (T): II does not give T. False.
(C) (II), (P), (S), (U): true.
(D) (I), (S), (Q), (R): I does not give S. False.
Answer: (C). | ||||
| Q.18 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Matching listHard | |
A hard matching list question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (III) KCN replaces Cl by CN; hot aqueous acid hydrolyses both the nitrile and the methyl ester: homophthalic acid T. $\mathrm{LiAlH_4}$ reduces both acids to the diol Q; conc. $\mathrm{H_2SO_4}$ gives isochroman R. III gives T, Q, R.
(IV) $\mathrm{LiAlH_4}$ reduces both esters to the diol Q; conc. $\mathrm{H_2SO_4}$ dehydrates it to the cyclic ether R. IV gives Q, R.
(A) (III), (S), (R): III does not give S. False.
(B) (IV), (Q), (U): IV does not give the lactone U. False.
(C) (III), (T), (U): III does not give U. False.
(D) (IV), (Q), (R): true.
Answer: (D). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | 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 Let $A = \begin{bmatrix}2&1&3\\1&0&2\\3&2&1\end{bmatrix}$; it is symmetric. The six $P_k$ are all the $3\times3$ permutation matrices.
(B) $(P_kAP_k^T)^T = P_kA^TP_k^T = P_kAP_k^T$, so each term, and hence $X$, is symmetric. True.
(A) Let $u = (1,1,1)^T$. A permutation matrix leaves $u$ unchanged, so $P_k^Tu = u$. $Au = (6, 3, 6)^T$ (the row sums). Then $P_kAu$ is a rearrangement of $(6, 3, 6)$, and summing over all six permutations each entry of $Xu$ receives each of $6, 3, 6$ exactly twice: $Xu = 2(6 + 3 + 6)u = 30u$. So $\alpha = 30$. True.
(C) $\operatorname{tr}(P_kAP_k^T) = \operatorname{tr}(AP_k^TP_k) = \operatorname{tr}A = 3$, so $\operatorname{tr}X = 6\times3 = 18$. True.
(D) $(X - 30I)u = 0$ with $u \ne 0$, so $X - 30I$ is singular. False.
Answer: (A), (B), (C). | ||||
| Q.2 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants (A) Compare the second rows. $PQ$: row 2 of $P$ is $(0, 2, 2)$, giving $(2x,\ 8 + 2x,\ 12)$. $QP$: row 2 of $Q$ is $(0, 4, 0)$, giving $4\times(0, 2, 2) = (0, 8, 8)$. The third entries are $12$ and $8$ for every $x$, so $PQ \ne QP$ always. False.
(B) $\det R = \det P\det Q\det P^{-1} = \det Q = 2(24) - x(0) + x(0 - 4x) = 48 - 4x^2$. And $\det\begin{bmatrix}2&x&x\\0&4&0\\x&x&5\end{bmatrix} = 2(20) + x(-4x) = 40 - 4x^2$; adding 8 gives $48 - 4x^2$. True for all $x$.
(C) For $x = 0$, $Q = \operatorname{diag}(2, 4, 6)$. $R\vec v = 6\vec v$ means $Q(P^{-1}\vec v) = 6(P^{-1}\vec v)$, so $P^{-1}\vec v$ is a multiple of $(0, 0, 1)^T$ and $\vec v$ a multiple of $P(0,0,1)^T = (1, 2, 3)^T$. With first entry 1: $a = 2$, $b = 3$, $a + b = 5$. True.
(D) A nonzero vector with $R\vec v = 0$ needs $\det R = 0$. For $x = 1$, $\det R = 48 - 4 = 44 \ne 0$. False.
Answer: (B), (C). | ||||
| Q.3 | 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 Put $N = n + 2$, $\theta = \frac{\pi}{N}$ and $j = k + 1$, running from 1 to $N - 1$.
Denominator: $\sum_{j=1}^{N-1}\sin^2 j\theta = \frac{1}{2}\sum_{j=1}^{N-1}(1 - \cos2j\theta)$. Since $\sum_{j=0}^{N-1}\cos\frac{2\pi j}{N} = 0$, $\sum_{j=1}^{N-1}\cos2j\theta = -1$, and the denominator is $\frac{1}{2}(N - 1 + 1) = \frac{N}{2}$.
Numerator: $\sin j\theta\sin(j+1)\theta = \frac{1}{2}[\cos\theta - \cos(2j+1)\theta]$. The angles $\frac{(2j+1)\pi}{N}$, $j = 0, \dots, N-1$, are the arguments of the $N$-th roots of $-1$, so their cosines add to 0; hence $\sum_{j=1}^{N-1}\cos(2j+1)\theta = -\cos\theta$. The numerator is $\frac{1}{2}[(N-1)\cos\theta + \cos\theta] = \frac{N}{2}\cos\theta$.
So $f(n) = \cos\frac{\pi}{n+2}$.
(A) $f(4) = \cos\frac{\pi}{6} = \frac{\sqrt3}{2}$. True.
(B) $\lim f(n) = \cos0 = 1$, not $\frac{1}{2}$. False.
(C) $\cos^{-1}f(6) = \frac{\pi}{8}$, $\alpha = \tan\frac{\pi}{8} = \sqrt2 - 1$. Then $\alpha^2 + 2\alpha - 1 = (3 - 2\sqrt2) + (2\sqrt2 - 2) - 1 = 0$. True.
(D) $\cos^{-1}f(5) = \frac{\pi}{7}$, and $\sin\left(7\cdot\frac{\pi}{7}\right) = \sin\pi = 0$. True.
Answer: (A), (C), (D). | ||||
| Q.4 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctEasy | |
A easy multiple correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability (A) $f(x) = |x|$: $\frac{|h|}{\sqrt{|h|}} = \sqrt{|h|} \to 0$. Property 1 holds. True.
(B) $f(x) = x^{2/3}$: $\frac{|h|^{2/3}}{|h|^{1/2}} = |h|^{1/6} \to 0$. Property 1 holds. True.
(C) $f(x) = x|x|$: $\frac{h|h|}{h^2} = \frac{h}{|h|}$, which is $+1$ for $h > 0$ and $-1$ for $h < 0$: no limit. False.
(D) $f(x) = \sin x$: $\frac{\sin h}{h^2} = \frac{\sin h}{h}\cdot\frac{1}{h}$, unbounded as $h \to 0$. False.
Answer: (A), (B). | ||||
| Q.5 | 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 $f'(x) = \frac{\pi x\cos\pi x - 2\sin\pi x}{x^3}$, so critical points satisfy $\tan t = \frac{t}{2}$ with $t = \pi x$.
Where is each root? On $(m\pi, m\pi + \frac{\pi}{2})$, $m \ge 1$, $\tan t$ rises from 0 to $+\infty$ while $\frac{t}{2}$ is positive and slowly rising, so there is exactly one root. On $(m\pi - \frac{\pi}{2}, m\pi)$, $\tan t < 0 < \frac{t}{2}$: none. On $(0, \frac{\pi}{2})$, $\tan t > t > \frac{t}{2}$: none. So the critical points are $t_m = m\pi + \delta_m$ with $0 < \delta_m < \frac{\pi}{2}$, i.e. one in each $(m, m + \frac{1}{2})$.
Type: on $(m, m+1)$, $\sin\pi x$ has the sign of $(-1)^m$, and $f$ is 0 at both ends. So for even $m$ the critical point is a maximum, for odd $m$ a minimum. (On $(0, 1)$ $f$ falls from $+\infty$ with no critical point.) Hence $x_n \in (2n, 2n + \frac{1}{2})$ and $y_n \in (2n - 1, 2n - \frac{1}{2})$.
Also $\tan\delta_m = \frac{m\pi + \delta_m}{2}$ grows with $m$, so $\delta_m$ increases with $m$.
(A) $x_1 \in (2, 2.5)$, $y_1 \in (1, 1.5)$: $x_1 > y_1$. False.
(B) $x_{n+1} - x_n = 2 + \frac{\delta_{2n+2} - \delta_{2n}}{\pi} > 2$. True.
(C) $x_n \in (2n, 2n + \frac{1}{2})$. True.
(D) $x_n - y_n = 1 + \frac{\delta_{2n} - \delta_{2n-1}}{\pi} > 1$. True.
Answer: (B), (C), (D). | ||||
| Q.6 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Numerator: $\sum_{k=1}^n k^{1/3} = n^{4/3}\cdot\frac{1}{n}\sum\left(\frac{k}{n}\right)^{1/3} \sim n^{4/3}\int_0^1 x^{1/3}dx = \frac{3}{4}n^{4/3}$.
Denominator: $n^{7/3}\sum_{k=1}^n\frac{1}{(an + k)^2} = n^{7/3}\cdot\frac{1}{n^2}\sum\frac{1}{(a + k/n)^2} = n^{4/3}\cdot\frac{1}{n}\sum\frac{1}{(a + k/n)^2} \sim n^{4/3}\int_0^1\frac{dx}{(a + x)^2} = n^{4/3}\left(\frac{1}{a} - \frac{1}{a+1}\right) = \frac{n^{4/3}}{a(a+1)}$.
(With $|a| > 1$, $a + x$ never vanishes on $[0, 1]$.)
Limit $= \frac{3}{4}a(a + 1) = 54$, so $a^2 + a - 72 = 0$, $(a + 9)(a - 8) = 0$, $a = 8$ or $a = -9$.
Answer: (A), (D). | ||||
| Q.7 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Multiple correctMedium | |
A medium multiple correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $F'(x) = f(x)$, whose sign is: negative on $(0, 1)$, positive on $(1, 2)$, negative on $(2, 5)$, positive after 5.
So $F$ has local minima at $x = 1$ and $x = 5$ and a local maximum at $x = 2$.
(A) True. (B) True. (C) There are two minima and one maximum, not the other way round. False.
(D) $f(t) = t^3 - 8t^2 + 17t - 10$, so $F(x) = \frac{x^4}{4} - \frac{8x^3}{3} + \frac{17x^2}{2} - 10x$. $F(0) = 0$ and $F$ decreases on $(0, 1)$, so $F < 0$ there. On $[1, 5)$ the largest value is the local maximum $F(2) = 4 - \frac{64}{3} + 34 - 20 = -\frac{10}{3} < 0$. Hence $F(x) < 0$, never 0, on $(0, 5)$. True.
Answer: (A), (B), (D). | ||||
| Q.8 | 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 Take $P = (\lambda, 0, 0)$, $Q = (0, \mu, 1)$, $R = (1, 1, v)$. They are collinear when $\vec{PQ} = t\,\vec{PR}$ for some $t$ (the three points are always distinct):
$(-\lambda,\ \mu,\ 1) = t(1 - \lambda,\ 1,\ v)$.
Then $\mu = t$, $1 = tv$ (so $t \ne 0$), and $-\lambda = t - t\lambda$, i.e. $\lambda(t - 1) = t$, which needs $t \ne 1$ and then gives $\lambda = \frac{t}{t - 1}$.
So a suitable $P$ and $R$ exist exactly when $\mu \ne 0$ and $\mu \ne 1$.
(A) $\mu = -\frac{1}{2}$: yes. (B) $\mu = 0$: no. (C) $\mu = \frac{1}{2}$: yes. (D) $\mu = 1$: no.
Answer: (A), (C). | ||||
| Q.9 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The four sums: $\sum k = \frac{n(n+1)}{2}$; $\sum{}^nC_k\,k = n2^{n-1}$; $\sum{}^nC_k\,k^2 = n(n+1)2^{n-2}$; $\sum{}^nC_k\,3^k = 4^n$.
Determinant: $\frac{n(n+1)}{2}4^n - n(n+1)2^{n-2}\cdot n2^{n-1} = n(n+1)4^n\left(\frac{1}{2} - \frac{n}{8}\right)$. For a positive integer $n$ this is zero only when $n = 4$.
Then $\sum_{k=0}^{4}\frac{{}^4C_k}{k+1} = \frac{1}{5}\sum_{k=0}^4{}^5C_{k+1} = \frac{2^5 - 1}{5} = \frac{31}{5} = 6.2$.
Answer: 6.20. | ||||
| Q.10 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The seats are fixed (the persons are named), so we count colourings of a 5-cycle with 3 colours, adjacent ones different.
On a 5-cycle no three seats are pairwise non-adjacent, so each colour is used at most twice. Five hats in three colours, each at most twice, forces the pattern $2 + 2 + 1$.
Choose the single colour (3 ways) and its seat (5 ways). The remaining four seats form a path; it must alternate the other two colours, which can be done in 2 ways.
Total $= 3\times5\times2 = 30$. (The general formula $(k - 1)^n + (-1)^n(k - 1)$ gives $2^5 - 2 = 30$ as well.)
Answer: 30. | ||||
| Q.11 | 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 Independence: $\frac{|A\cap B|}{6} = \frac{|A|}{6}\cdot\frac{|B|}{6}$, i.e. $|A\cap B| = \frac{|A||B|}{6}$ must be a whole number. Let $a = |A|$, $b = |B|$ with $1 \le b < a \le 6$.
$a = 6$ ($A = S$): every $B$ works; $b = 1, \dots, 5$ gives $6 + 15 + 20 + 15 + 6 = 62$ pairs.
$a = 5$: $\frac{5b}{6}$ is whole only for $b = 6$, not allowed. 0.
$a = 4$: $\frac{2b}{3}$ whole needs $b = 3$, with $|A\cap B| = 2$. Choose $A$: $\binom64 = 15$; $B$: 2 of the 4 in $A$ and 1 of the 2 outside, $6\times2 = 12$. $180$ pairs.
$a = 3$: $\frac{b}{2}$ whole needs $b = 2$, with $|A\cap B| = 1$. $A$: $\binom63 = 20$; $B$: 1 of 3 inside and 1 of 3 outside, 9. $180$ pairs.
$a = 2$: $b = 1$ gives $\frac{1}{3}$, not whole. 0.
Total $= 62 + 180 + 180 = 422$.
Answer: 422. | ||||
| Q.12 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta_k = \frac{7\pi}{12} + \frac{k\pi}{2}$. Since $\cos\left(\theta + \frac{\pi}{2}\right) = -\sin\theta$:
$\sec\theta_k\sec\theta_{k+1} = \frac{1}{\cos\theta_k(-\sin\theta_k)} = -\frac{2}{\sin2\theta_k}$.
$\sin2\theta_k = \sin\left(\frac{7\pi}{6} + k\pi\right) = (-1)^k\sin\frac{7\pi}{6} = (-1)^{k+1}\cdot\frac{1}{2}$, so each term is $-2\cdot\frac{2}{(-1)^{k+1}} = 4(-1)^k$.
$\sum_{k=0}^{10}4(-1)^k = 4$ (eleven terms, alternating, starting and ending with $+$). Then $\frac{1}{4}\times4 = 1$ and $\sec^{-1}1 = 0$, which lies in the interval.
Answer: 0. | ||||
| Q.13 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals With $\theta \to \frac{\pi}{2} - \theta$, $I = \int_0^{\pi/2}\frac{3\sqrt{\sin\theta}}{(\sqrt{\cos\theta} + \sqrt{\sin\theta})^5}d\theta$. Adding the two forms:
$2I = 3\int_0^{\pi/2}\frac{d\theta}{(\sqrt{\cos\theta} + \sqrt{\sin\theta})^4}$.
Divide top and bottom by $\cos^2\theta$: $\int_0^{\pi/2}\frac{\sec^2\theta\,d\theta}{(1 + \sqrt{\tan\theta})^4}$. With $\tan\theta = u^2$, $\sec^2\theta\,d\theta = 2u\,du$:
$\int_0^\infty\frac{2u\,du}{(1 + u)^4} = 2\int_0^\infty\left[\frac{1}{(1+u)^3} - \frac{1}{(1+u)^4}\right]du = 2\left(\frac{1}{2} - \frac{1}{3}\right) = \frac{1}{3}$.
So $2I = 3\times\frac{1}{3} = 1$ and $I = 0.5$.
Answer: 0.50. | ||||
| Q.14 | 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 $\vec a\cdot\vec a = 6$, $\vec b\cdot\vec b = 6$, $\vec a\cdot\vec b = 2 + 2 - 1 = 3$; $\vec a + \vec b = 3\hat i + 3\hat j$, of length $3\sqrt2$.
Projection condition: $\vec c\cdot(\vec a + \vec b) = 3\sqrt2\times3\sqrt2 = 18$. And $\vec c\cdot(\vec a + \vec b) = \alpha(6 + 3) + \beta(3 + 6) = 9(\alpha + \beta)$, so $\alpha + \beta = 2$.
$\vec c$ lies in the plane of $\vec a$ and $\vec b$, so $(\vec a\times\vec b)\cdot\vec c = 0$ and the expression is $|\vec c|^2 = 6\alpha^2 + 6\beta^2 + 6\alpha\beta = 6[(\alpha + \beta)^2 - \alpha\beta] = 6(4 - \alpha\beta)$.
With $\alpha + \beta = 2$, $\alpha\beta \le 1$ (equality at $\alpha = \beta = 1$), so the minimum is $6(4 - 1) = 18$.
Answer: 18. | ||||
| Q.15 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Matching listHard | |
A hard matching list question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $X$: $\pi\cos x = m\pi$ means $\cos x \in \{-1, 0, 1\}$, i.e. $x$ is a multiple of $\frac{\pi}{2}$: $X = \left\{\frac{\pi}{2}, \pi, \frac{3\pi}{2}, \dots\right\}$, an AP containing $\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi$. So I → P, Q.
$Y$: $f'(x) = -\pi\sin x\cos(\pi\cos x) = 0$ means $\sin x = 0$ or $\cos x = \pm\frac{1}{2}$, i.e. $x$ is a multiple of $\frac{\pi}{3}$: an AP containing $\frac{\pi}{3}, \frac{2\pi}{3}, \pi$. So II → Q, T.
$Z$: $2\pi\sin x = \frac{\pi}{2} + m\pi$ means $\sin x \in \left\{\pm\frac{1}{4}, \pm\frac{3}{4}\right\}$. These $x$ are not equally spaced (for instance $\sin^{-1}\frac{1}{4}$, $\sin^{-1}\frac{3}{4}$, $\pi - \sin^{-1}\frac{3}{4}$ have unequal gaps), and $\frac{\pi}{6}$ ($\sin = \frac{1}{2}$) is not in $Z$. So III → R only.
$W$: $g'(x) = -2\pi\cos x\sin(2\pi\sin x) = 0$ means $\cos x = 0$ or $2\sin x \in \mathbb{Z}$, i.e. $\sin x \in \left\{0, \pm\frac{1}{2}, \pm1\right\}$: $W = \left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \dots\right\}$, with gaps $\frac{\pi}{3}, \frac{\pi}{3}, \frac{\pi}{6}, \dots$: not an AP. It contains $\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi$ and $\frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6}$, but not $\frac{\pi}{3}$ or $\frac{3\pi}{4}$. So IV → P, R, S.
(A) (I), (P), (R): X is an AP. False. (B) (II), (Q), (T): true. (C) (I), (Q), (U): $\frac{\pi}{6} \notin X$. False. (D) (II), (R), (S): Y is an AP. False.
Answer: (B). | ||||
| Q.16 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Matching listHard | |
A hard matching list question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $X$: $\pi\cos x = m\pi$ means $\cos x \in \{-1, 0, 1\}$, i.e. $x$ is a multiple of $\frac{\pi}{2}$: $X = \left\{\frac{\pi}{2}, \pi, \frac{3\pi}{2}, \dots\right\}$, an AP containing $\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi$. So I → P, Q.
$Y$: $f'(x) = -\pi\sin x\cos(\pi\cos x) = 0$ means $\sin x = 0$ or $\cos x = \pm\frac{1}{2}$, i.e. $x$ is a multiple of $\frac{\pi}{3}$: an AP containing $\frac{\pi}{3}, \frac{2\pi}{3}, \pi$. So II → Q, T.
$Z$: $2\pi\sin x = \frac{\pi}{2} + m\pi$ means $\sin x \in \left\{\pm\frac{1}{4}, \pm\frac{3}{4}\right\}$. These $x$ are not equally spaced (for instance $\sin^{-1}\frac{1}{4}$, $\sin^{-1}\frac{3}{4}$, $\pi - \sin^{-1}\frac{3}{4}$ have unequal gaps), and $\frac{\pi}{6}$ ($\sin = \frac{1}{2}$) is not in $Z$. So III → R only.
$W$: $g'(x) = -2\pi\cos x\sin(2\pi\sin x) = 0$ means $\cos x = 0$ or $2\sin x \in \mathbb{Z}$, i.e. $\sin x \in \left\{0, \pm\frac{1}{2}, \pm1\right\}$: $W = \left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \dots\right\}$, with gaps $\frac{\pi}{3}, \frac{\pi}{3}, \frac{\pi}{6}, \dots$: not an AP. It contains $\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi$ and $\frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6}$, but not $\frac{\pi}{3}$ or $\frac{3\pi}{4}$. So IV → P, R, S.
(A) (III), (R), (U): $\frac{\pi}{6} \notin Z$. False. (B) (IV), (P), (R), (S): true. (C) (III), (P), (Q), (U): Z is not an AP. False. (D) (IV), (Q), (T): W is not an AP. False.
Answer: (B). | ||||
| Q.17 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Matching listHard | |
A hard matching list question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections $C_1$: centre $O(0, 0)$, radius 3. $C_2$: centre $(3, 4)$, radius 4. The centres are 5 apart along the unit vector $\hat e = \left(\frac{3}{5}, \frac{4}{5}\right)$.
$C_3$ contains both and touches each internally, with its centre on the line of centres, so $M$ and $N$ are the far ends: $M = -3\hat e = \left(-\frac{9}{5}, -\frac{12}{5}\right)$, $N = (3, 4) + 4\hat e = \left(\frac{27}{5}, \frac{36}{5}\right)$. $MN$ is a diameter: $2r = 3 + 5 + 4 = 12$, $r = 6$, centre $(h, k) = \left(\frac{9}{5}, \frac{12}{5}\right)$.
(I) $2h + k = \frac{18}{5} + \frac{12}{5} = 6$ → P.
(II) Common chord: $C_1 - C_2$ gives $6x + 8y - 18 = 0$, i.e. $3x + 4y = 9$. Its distance from $O$ is $\frac{9}{5}$, so $XY = 2\sqrt{9 - \frac{81}{25}} = \frac{24}{5}$. Its distance from the centre of $C_3$ is $\frac{|27/5 + 48/5 - 9|}{5} = \frac{6}{5}$, so $ZW = 2\sqrt{36 - \frac{36}{25}} = \frac{24\sqrt6}{5}$. Ratio $= \sqrt6$ → Q.
(III) $ZW$ is perpendicular to the diameter $MN$. Triangle $MZN$: base $MN = 12$, height $\frac{ZW}{2} = \frac{12\sqrt6}{5}$, area $\frac{72\sqrt6}{5}$. Triangle $ZMW$: base $ZW = \frac{24\sqrt6}{5}$, height = distance from $M$ to $3x + 4y = 9$, $\frac{|-27/5 - 48/5 - 9|}{5} = \frac{24}{5}$; area $\frac{288\sqrt6}{25}$. Ratio $= \frac{72\sqrt6/5}{288\sqrt6/25} = \frac{5}{4}$ → R.
(IV) $C_1$ and $C_3$ touch at $M$, so their common tangent is the tangent at $M$, perpendicular to $\hat e$: $3x + 4y = -15$. Substituting $y = -\frac{15 + 3x}{4}$ into $x^2 = 8\alpha y$: $x^2 + 6\alpha x + 30\alpha = 0$. Tangency: $36\alpha^2 - 120\alpha = 0$, so $\alpha = \frac{10}{3}$ (as $\alpha \ne 0$ for a parabola) → U.
So I → P, II → Q, III → R, IV → U.
(A) (I), (S): no. (B) (I), (U): no. (C) (II), (Q): yes. (D) (II), (T): no.
Answer: (C). | ||||
| Q.18 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Matching listHard | |
A hard matching list question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections $C_1$: centre $O(0, 0)$, radius 3. $C_2$: centre $(3, 4)$, radius 4. The centres are 5 apart along the unit vector $\hat e = \left(\frac{3}{5}, \frac{4}{5}\right)$.
$C_3$ contains both and touches each internally, with its centre on the line of centres, so $M$ and $N$ are the far ends: $M = -3\hat e = \left(-\frac{9}{5}, -\frac{12}{5}\right)$, $N = (3, 4) + 4\hat e = \left(\frac{27}{5}, \frac{36}{5}\right)$. $MN$ is a diameter: $2r = 3 + 5 + 4 = 12$, $r = 6$, centre $(h, k) = \left(\frac{9}{5}, \frac{12}{5}\right)$.
(I) $2h + k = \frac{18}{5} + \frac{12}{5} = 6$ → P.
(II) Common chord: $C_1 - C_2$ gives $6x + 8y - 18 = 0$, i.e. $3x + 4y = 9$. Its distance from $O$ is $\frac{9}{5}$, so $XY = 2\sqrt{9 - \frac{81}{25}} = \frac{24}{5}$. Its distance from the centre of $C_3$ is $\frac{|27/5 + 48/5 - 9|}{5} = \frac{6}{5}$, so $ZW = 2\sqrt{36 - \frac{36}{25}} = \frac{24\sqrt6}{5}$. Ratio $= \sqrt6$ → Q.
(III) $ZW$ is perpendicular to the diameter $MN$. Triangle $MZN$: base $MN = 12$, height $\frac{ZW}{2} = \frac{12\sqrt6}{5}$, area $\frac{72\sqrt6}{5}$. Triangle $ZMW$: base $ZW = \frac{24\sqrt6}{5}$, height = distance from $M$ to $3x + 4y = 9$, $\frac{|-27/5 - 48/5 - 9|}{5} = \frac{24}{5}$; area $\frac{288\sqrt6}{25}$. Ratio $= \frac{72\sqrt6/5}{288\sqrt6/25} = \frac{5}{4}$ → R.
(IV) $C_1$ and $C_3$ touch at $M$, so their common tangent is the tangent at $M$, perpendicular to $\hat e$: $3x + 4y = -15$. Substituting $y = -\frac{15 + 3x}{4}$ into $x^2 = 8\alpha y$: $x^2 + 6\alpha x + 30\alpha = 0$. Tangency: $36\alpha^2 - 120\alpha = 0$, so $\alpha = \frac{10}{3}$ (as $\alpha \ne 0$ for a parabola) → U.
So I → P, II → Q, III → R, IV → U.
(A) (I), (P): correct. (B) (IV), (U): correct. (C) (III), (R): correct. (D) (IV), (S): incorrect, since $\alpha = \frac{10}{3}$, not $\frac{21}{5}$.
The only incorrect combination is (D). Answer: (D). | ||||
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