JEE (Advanced) · 2021
JEE Advanced 2021, 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
38 Q
- Motion of System of Particles and Rigid Body6
- Electromagnetic Induction and Alternating Currents4
- Properties of Bulk Matter4
7 easy · 28 medium · 3 hard
Chemistry
38 Q
- Hydrocarbons4
- Thermodynamics4
- Coordination Compounds3
9 easy · 24 medium · 5 hard
Mathematics
38 Q
- Integral Calculus6
- Conic Sections6
- Probability5
7 easy · 22 medium · 9 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 11: Systems of Particles and Rotational Motion
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 11: Thermodynamics
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 12: Atoms; Class 12: Nuclei
Class 12: Current Electricity
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
No matching NCERT chapter
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Class 11: Laws of Motion
Class 11: Oscillations; Class 11: Waves
Class 11: Units and Measurement
Class 11: Gravitation
Class 12: Dual Nature of Radiation and Matter
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Conic Sections
Class 11: Probability; Class 12: Probability
Class 12: Application of Derivatives
Class 12: Matrices; Class 12: Determinants
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 11: Complex Numbers and Quadratic Equations
Class 12: Vector Algebra
Class 11: Sequences and Series
Class 11: Straight Lines
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 11: Complex Numbers and Quadratic Equations
Class 11: Permutations and Combinations
Class 12: Differential Equations
Class 11: Hydrocarbons
Class 11: Thermodynamics
Class 12: Coordination Compounds
No matching NCERT chapter
Class 12: Electrochemistry
Class 12: Aldehydes, Ketones and Carboxylic Acids
Class 12: Solutions
Class 11: Organic Chemistry – Some Basic Principles and Techniques
No matching NCERT chapter
Class 11: Some Basic Concepts of Chemistry
- Principles Related to Practical Chemistry2
No matching NCERT chapter
No matching NCERT chapter
Class 11: Equilibrium
Class 12: Biomolecules
Class 12: Haloalkanes and Haloarenes
No matching NCERT chapter
No matching NCERT chapter
Class 12: Amines
Class 12: Chemical Kinetics
Class 11: Structure of Atom
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 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count: $10$ VSD $= 9$ MSD, so $1$ VSD $= 0.09$ cm and $LC = 1\text{ MSD} - 1\text{ VSD} = 0.1 - 0.09 = 0.01$ cm.
Zero error (left figure): the Vernier zero lies to the left of the main-scale zero, so the error is negative. The 6th Vernier division coincides with a main-scale line (the 5 mm mark), so the Vernier zero sits at $0.5 - 6\times0.09 = -0.04$ cm. Zero error $= -0.04$ cm (equivalently $-(10 - 6)\times LC$).
Reading with the sphere (right figure): the Vernier zero lies between $3.1$ and $3.2$ cm, so MSR $= 3.1$ cm; the 1st Vernier division coincides, so VSR $= 1\times0.01 = 0.01$ cm. Observed reading $= 3.11$ cm.
Correct diameter $=$ observed $-$ zero error $= 3.11 - (-0.04) = 3.15$ cm.
Answer: (C). | ||||
| Q.2 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Step 1: isobaric expansion at the higher pressure. $V$ rises at constant $P$, so $T$ rises; $\Delta U > 0$ and $W > 0$, hence $Q = \Delta U + W > 0$: heat absorbed.
Step 2: isochoric, pressure falls, so $T$ falls; $W = 0$, $Q = \Delta U < 0$: heat released.
Step 3: isobaric compression at the lower pressure; $T$ falls and $W < 0$, so $Q < 0$: heat released.
Step 4: isochoric, pressure rises, so $T$ rises; $Q = \Delta U > 0$: heat absorbed.
Heat is absorbed in steps 1 and 4.
Answer: (C). | ||||
| Q.3 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Focal lengths (lens maker): $\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} + \frac{1}{20}\right) = \frac{1}{20}$, so $f_1 = +20$ cm (biconvex) and $f_2 = -20$ cm (biconcave).
Lens $L_1$: $u = -10$ cm. $\frac{1}{v} = \frac{1}{f_1} + \frac{1}{u} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20}$, so $v_1 = -20$ cm (virtual, 20 cm to the left of $L_1$). $m_1 = \frac{v_1}{u} = \frac{-20}{-10} = 2$.
Lens $L_2$ is 10 cm to the right of $L_1$, so this image is $30$ cm to the left of $L_2$: $u_2 = -30$ cm. $\frac{1}{v_2} = -\frac{1}{20} - \frac{1}{30} = -\frac{1}{12}$, so $v_2 = -12$ cm and $m_2 = \frac{-12}{-30} = 0.4$.
Total magnification $m = m_1m_2 = 2\times0.4 = 0.8$.
Answer: (B). | ||||
| 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 One hour is $3$ half-lives, so the nuclei left are $1000\times\left(\frac{1}{2}\right)^3 = 125$ and the number that decayed is $1000 - 125 = 875$.
Each decay is an alpha-decay with probability $0.6$, so the number of alpha-decays is $0.6\times875 = 525$.
Answer: (D). | ||||
| Q.5 | 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 Components of the launch velocity: $u_x = u_y = 5\sqrt2\cos45^\circ = 5$ m/s.
Time to the top: $\frac{u_y}{g} = 0.5$ s; height of the top: $\frac{u_y^2}{2g} = 1.25$ m.
At the top the projectile moves horizontally. The part that falls vertically takes $0.5$ s to fall $1.25$ m, which is exactly free fall from rest ($\frac{1}{2}g(0.5)^2 = 1.25$ m), so it has zero velocity just after splitting.
The other part therefore also starts with zero vertical velocity (momentum conservation gives it only a horizontal velocity), so it too falls $1.25$ m from rest: $t = \sqrt{\frac{2\times1.25}{10}} = 0.5$ s.
Answer: $0.5$. | ||||
| Q.6 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | NumericalMedium | |
A medium numerical question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion With $u_x = u_y = 5$ m/s, the top is reached after $0.5$ s at horizontal distance $5\times0.5 = 2.5$ m from O, at height $1.25$ m, moving horizontally at $5$ m/s.
The part that falls vertically down in $0.5$ s falls from rest, so its velocity just after splitting is zero.
Momentum along $x$ during the split: $m\times5 = \frac{m}{2}\times0 + \frac{m}{2}v$, so $v = 10$ m/s (horizontal).
This part falls $1.25$ m from zero vertical velocity in $t = 0.5$ s, covering $10\times0.5 = 5$ m horizontally.
$x = 2.5 + 5 = 7.5$ m.
Answer: $7.5$. | ||||
| Q.7 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Take the bottom rail as $0$ V. Both cells have their positive terminal at the top (longer plate), so with S at P the left end of the $1\ \Omega$ resistor is at $+1$ V and the right end of the $2\ \Omega$ resistor is at $+2$ V.
After a long time no current flows into the capacitor, so one current flows through $1\ \Omega$ and $2\ \Omega$ in series: $I = \frac{2 - 1}{1 + 2} = \frac{1}{3}$ A (from the 2 V side to the 1 V side).
Potential of the capacitor node: $V = 2 - 2\times\frac{1}{3} = \frac{4}{3}$ V (check: $1 + 1\times\frac{1}{3} = \frac{4}{3}$ V).
$q_1 = CV = 1\ \mu\text{F}\times\frac{4}{3}\text{ V} = 1.33\ \mu$C.
Answer: $1.33$. | ||||
| Q.8 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity With S at Q the left end of the $1\ \Omega$ resistor is joined straight to the bottom rail ($0$ V); the 1 V cell is out of the circuit.
In the steady state the capacitor draws no current, so $I = \frac{2 - 0}{1 + 2} = \frac{2}{3}$ A through both resistors.
Capacitor node: $V = 0 + 1\times\frac{2}{3} = \frac{2}{3}$ V.
$q_2 = 1\ \mu\text{F}\times\frac{2}{3}\text{ V} = 0.67\ \mu$C.
Answer: $0.67$. | ||||
| Q.9 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance On the $V = 0$ curve: $\frac{-Q}{r_1} + \frac{Q/\sqrt3}{r_2} = 0$, so $r_1 = \sqrt3\,r_2$, where $r_1$, $r_2$ are the distances from $(0,0)$ and $(2,0)$.
$x^2 + y^2 = 3\left[(x - 2)^2 + y^2\right] \Rightarrow 2x^2 + 2y^2 - 12x + 12 = 0 \Rightarrow x^2 + y^2 - 6x + 6 = 0$.
So $(x - 3)^2 + y^2 = 3$: centre $(3, 0)$, radius $\sqrt3$.
$R = \sqrt3 = 1.732\ldots \approx 1.73$ m.
Answer: $1.73$. | ||||
| Q.10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance On the $V = 0$ curve: $\frac{-Q}{r_1} + \frac{Q/\sqrt3}{r_2} = 0$, so $r_1 = \sqrt3\,r_2$, where $r_1$, $r_2$ are the distances from $(0,0)$ and $(2,0)$.
$x^2 + y^2 = 3\left[(x - 2)^2 + y^2\right] \Rightarrow 2x^2 + 2y^2 - 12x + 12 = 0 \Rightarrow x^2 + y^2 - 6x + 6 = 0$.
So $(x - 3)^2 + y^2 = 3$: centre $(3, 0)$, radius $\sqrt3$.
$b = 3$ m.
Answer: $3$. | ||||
| Q.11 | 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 Let the moment of inertia be $I = kmR^2$ and friction $f$ act backwards. Then $F - f = ma$ and $fR = I\frac{a}{R}$, so $f = kma$ and $a = \frac{F}{m(1 + k)}$.
(A) False: $k = \frac12$ for a solid cylinder and $1$ for a thin hollow one, so $a$ differs for the same $F$.
(B) True: for a solid cylinder $f = \frac{ma}{2}$; static friction can be at most $\mu mg$, so $a_{max} = 2\mu g$.
(C) False: the friction is static and equals $kma$, which is $\mu mg$ only at the limit.
(D) True: for a thin-walled hollow cylinder $k = 1$, so $a = \frac{F}{2m}$.
Answer: (B), (D). | ||||
| Q.12 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctMedium | |
A medium multiple correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Snell's law across parallel layers keeps $n\sin(\text{angle})$ constant: $n_1\sin\theta = n_2\sin\theta_2 = 1\cdot\sin\theta_{air}$. Since $n_1\sin\theta > 1$, no ray can ever exist in air: wherever the ray meets the air boundary it is totally reflected.
(A) False: with $n_2 = n_1$ the ray goes straight to the air boundary and is totally reflected there.
(B) True: if $n_2 < n_1\sin\theta$ it is totally reflected at the $n_1$–$n_2$ boundary; if $n_1\sin\theta < n_2 < n_1$ it enters $n_2$, is totally reflected at the top (air) surface, and by symmetry returns into $n_1$. Either way it finally goes back into $n_1$.
(C) True: with $n_2 > n_1$ it enters $n_2$, is totally reflected at the air surface and refracts back into $n_1$.
(D) True: with $n_2 = 1$, $n_1\sin\theta > n_2$, so it is totally reflected at the $n_1$–$n_2$ boundary back into $n_1$.
Answer: (B), (C), (D). | ||||
| Q.13 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Multiple correctMedium | |
A medium multiple correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Motion: $x = -10 + 5t^2$, $y = -1$, $v_x = 10t$. Force $\vec F = M a\,\hat i = 2\hat i$ N.
(A) True: $x = l = 10$ when $5t^2 = 20$, $t = 2$ s.
(B) True: $\vec\tau = \vec r\times\vec F = (x\hat i - \hat j)\times2\hat i = -2(\hat j\times\hat i) = 2\hat k$, the same at every point of the path.
(C) True: at $t = 2$ s, $\vec p = 0.2\times20\,\hat i = 4\hat i$; $\vec L = (10\hat i - \hat j)\times4\hat i = -4(\hat j\times\hat i) = 4\hat k$.
(D) False: at $(0, -h)$ the torque is still $2\hat k$, not $\hat k$.
Answer: (A), (B), (C). | ||||
| Q.14 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Multiple correctMedium | |
A medium multiple correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.
(A) True: Balmer ($n_1 = 2$): longest $\lambda$ for $n_2 = 3$, $\frac{1}{\lambda} = R\cdot\frac{5}{36}$; shortest for $n_2 \to \infty$, $\frac{1}{\lambda} = \frac{R}{4}$. Ratio $= \frac{1/4}{5/36} = \frac{9}{5}$.
(B) False: Balmer lies between $\frac{4}{R} \approx 365$ nm and $\frac{36}{5R} \approx 656$ nm; Paschen ($n_1 = 3$) starts at $\frac{9}{R} \approx 820$ nm. No overlap.
(C) False: Lyman: $\lambda = \frac{1}{R}\cdot\frac{n^2}{n^2 - 1} = \left(1 + \frac{1}{n^2 - 1}\right)\lambda_0$ with $\lambda_0 = \frac{1}{R}$; $n^2 - 1$ is never a perfect square $m^2$ for $n \ge 2$, so the given form is wrong.
(D) True: Lyman lies between $\frac{1}{R} \approx 91$ nm and $\frac{4}{3R} \approx 122$ nm, far below Balmer's $365$ nm.
Answer: (A), (D). | ||||
| Q.15 | 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 the curved rod equals that of the straight segment joining its ends, which runs perpendicular to the wire from $r = 1$ cm to $r = 4$ cm. With $B = \frac{\mu_0 I}{2\pi r}$:
$\varepsilon = \int_{0.01}^{0.04} vB\,dr = \frac{\mu_0 I v}{2\pi}\ln 4 = 2\times10^{-7}\times2\times3\times(2\times0.7) = 1.68\times10^{-6}$ V.
The emf is constant, so the R–C circuit charges: $i = \frac{\varepsilon}{R}e^{-t/RC}$, largest at $t = 0$: $i_{max} = \frac{1.68\times10^{-6}}{1.4} = 1.2\times10^{-6}$ A. (A) true, (B) false.
The charge tends to $q_{max} = C_o\varepsilon = 5\times10^{-6}\times1.68\times10^{-6} = 8.4\times10^{-12}$ C. (C) true, (D) false.
Answer: (A), (C). | ||||
| Q.16 | Properties of Bulk MatterFluid mechanics | 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 fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The tube is vertical and its base lies along the $45^\circ$ incline, so point 1 (at the right wall) is $d$ to the right of and $d$ below point 2 (at the left wall).
Take $x$ to the right and $y$ up. Down the incline the acceleration is $\vec a = \left(\frac{a}{\sqrt2}, -\frac{a}{\sqrt2}\right)$. In the tube's frame the water feels effective gravity $\vec g_{eff} = \vec g - \vec a = \left(-\frac{a}{\sqrt2}, -g + \frac{a}{\sqrt2}\right)$, and pressure grows along it: $dP = \rho\,\vec g_{eff}\cdot d\vec r$.
$P_1 - P_2 = \rho\,\vec g_{eff}\cdot(d, -d) = \rho d\left(-\frac{a}{\sqrt2} + g - \frac{a}{\sqrt2}\right) = \rho d\,(g - \sqrt2\,a)$.
So $\beta = 1 - \frac{\sqrt2\,a}{g}$.
(A) True: $a = \frac{g}{\sqrt2}$ gives $\beta = 0$. (B) False for the same reason.
(C) True: $a = \frac{g}{2}$ gives $\beta = 1 - \frac{1}{\sqrt2} = \frac{\sqrt2 - 1}{\sqrt2}$. (D) False.
Answer: (A), (C). | ||||
| Q.17 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter After acceleration $\frac{1}{2}mv^2 = qV$, so $mv = \sqrt{2mqV}$, and in the field $r = \frac{mv}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}} \propto \sqrt{\frac{m}{q}}$.
$\frac{r_S}{r_\alpha} = \sqrt{\frac{32/1}{4/2}} = \sqrt{16} = 4$.
Answer: $4$. | ||||
| Q.18 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rod about O: $L_1 = \frac{Ma^2}{3}\Omega$.
Disc: its centre is at $\frac{3a}{4}$ from O and moves with the rod, giving orbital angular momentum $M\left(\frac{3a}{4}\right)^2\Omega = \frac{9Ma^2\Omega}{16}$. Its spin, as seen by the stationary observer, is $4\Omega$ about its own axis: $\frac{1}{2}M\left(\frac{a}{4}\right)^2\cdot4\Omega = \frac{Ma^2\Omega}{8}$.
Total: $Ma^2\Omega\left(\frac13 + \frac{9}{16} + \frac18\right) = Ma^2\Omega\cdot\frac{16 + 27 + 6}{48} = \frac{Ma^2\Omega}{48}\times49$.
Answer: $49$. | ||||
| Q.19 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The object only radiates (the container at 0 K sends nothing back): $C\frac{dT}{dt} = -\sigma A T^4$.
Integrating: $\frac{1}{T^3} - \frac{1}{T_0^3} = \frac{3\sigma A}{C}t$, so $t \propto \frac{1}{T^3} - \frac{1}{T_0^3}$.
$t_1 \propto \frac{1}{100^3} - \frac{1}{200^3} = \frac{8 - 1}{200^3} = \frac{7}{200^3}$; $t_2 \propto \frac{1}{50^3} - \frac{1}{200^3} = \frac{64 - 1}{200^3} = \frac{63}{200^3}$.
$\frac{t_2}{t_1} = \frac{63}{7} = 9$.
Answer: $9$. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The substrate has a terminal alkyne and an internal alkyne.
NaNH$_2$ first removes the acidic terminal $\equiv$C–H, giving a sodium acetylide. The acetylide anion is not reduced by Na in liquid NH$_3$ (an electron cannot add to the already negative species).
Na/liq. NH$_3$ (dissolving-metal reduction) reduces the internal alkyne to the trans (E) alkene.
On work-up the terminal alkyne is restored. Product: the terminal alkyne kept, the internal triple bond turned into a trans double bond, which is structure (B). (C) has a cis double bond (that would need Lindlar's catalyst), and (A), (D) wrongly reduce the terminal alkyne.
Answer: (B). | ||||
| Q.2 | HydrocarbonsConformations of ethane | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on conformations of ethane, from Hydrocarbons. NCERT: Class 11: Hydrocarbons First decide which projections are meso. Reading each carbon's three groups clockwise:
(A): front OH, H, Me; back OH, H, Me (same sense on both carbons, related by a centre of symmetry): meso, with OH/OH anti and Me/Me anti.
(B): front OH, Me, H; back OH, Me, H (again the same sense on both carbons; it is (A) with both centres mirrored, and the mirror image of a meso form is itself): meso, with OH/OH gauche ($60^\circ$) and Me/Me gauche.
(C) and (D): the two carbons have opposite senses, so these are conformations of the chiral ($\pm$) diol, not of the meso compound.
Between the meso conformers, the one with the two OH groups gauche lets them form an intramolecular O–H···O hydrogen bond, which outweighs the extra gauche interactions in a 1,2-diol. So the most stable conformation is the OH-gauche one, (B).
Answer: (B). | ||||
| Q.3 | Solid StateIonic radii and radius ratio | No matching NCERT chapter | Single correctHard | |
A hard single correct question on ionic radii and radius ratio, from Solid State. NCERT: No matching NCERT chapter The face shown has anions Y at its four corners and the cation X at its centre, all in contact: Y touches Y along the edge and X touches the Y ions along the face diagonal.
Along the edge: $2r_Y = a$, so $r_Y = 0.5a$. Along half the face diagonal: $r_X + r_Y = \frac{a}{\sqrt2}$, so $r_X = \left(\frac{1}{\sqrt2} - \frac12\right)a = 0.2071a$ (radius ratio $0.414$).
Every face is alike, so per unit cell: Y at 8 corners $= 1$ ion; X at 6 face centres $= 3$ ions.
Packing fraction $= \frac{\frac43\pi\left[1\times(0.5a)^3 + 3\times(0.2071a)^3\right]}{a^3} = 4.189\times(0.125 + 0.02665) = 4.189\times0.1517 \approx 0.635$.
Answer: (B) $0.63$. | ||||
| Q.4 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds [Cr(NH$_3$)$_6$]$^{3+}$: Cr$^{3+}$ is $3d^3$; in an octahedral field the three electrons occupy $t_{2g}$ singly, $n = 3$. $\mu = \sqrt{n(n+2)} = \sqrt{15} = 3.87$ BM.
[CuF$_6$]$^{3-}$: Cu$^{3+}$ is $3d^8$; octahedral $d^8$ is $t_{2g}^6e_g^2$ (F$^-$ is a weak-field ligand, and $d^8$ has two unpaired $e_g$ electrons in either case), $n = 2$. $\mu = \sqrt8 = 2.83 \approx 2.84$ BM.
Answer: (A). | ||||
| Q.5 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Mg$_2$C$_3$ + 4H$_2$O $\to$ 2Mg(OH)$_2$ + CH$_3$C$\equiv$CH. P is propyne, $M = 40$: $\frac{4.0}{40} = 0.10$ mol.
NaNH$_2$ then MeI methylates the terminal carbon: Q is but-2-yne, CH$_3$C$\equiv$CCH$_3$. At 75%: $0.075$ mol.
Red hot iron tube at 873 K trimerises an alkyne to a benzene ring: three but-2-yne give hexamethylbenzene, C$_{12}$H$_{18}$ ($M = 162$). At 40%: $\frac{0.075\times0.40}{3} = 0.010$ mol.
$x = 0.010\times162 = 1.62$ g.
Answer: $1.62$. | ||||
| Q.6 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids P is propyne ($0.10$ mol, from $4.0$ g, $M = 40$).
Hg$^{2+}$/H$^+$ hydrates it (Markovnikov) to acetone, S: $0.10$ mol (100%).
Ba(OH)$_2$, heat: aldol condensation of two acetone molecules to mesityl oxide, T = (CH$_3$)$_2$C=CHCOCH$_3$. At 80%: $\frac{0.10}{2}\times0.80 = 0.040$ mol.
NaOCl: haloform reaction of the methyl ketone gives 3-methylbut-2-enoic acid, U = (CH$_3$)$_2$C=CHCOOH ($M = 100$), whose C=C decolourises Baeyer's reagent. At 80%: $0.040\times0.80 = 0.032$ mol.
$y = 0.032\times100 = 3.2$ g.
Answer: $3.20$. | ||||
| Q.7 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium From the graph, $\ln K = -3$ at $\frac{10^4}{T} = 10$ and $\ln K = -7$ at $\frac{10^4}{T} = 12$.
Slope $\frac{d\ln K}{d(1/T)} = \frac{-7 - (-3)}{(12 - 10)\times10^{-4}} = -2\times10^4$ K.
$\Delta H^\ominus = -R\times\text{slope} = 8.314\times2\times10^4 = 166280$ J mol$^{-1}$ $= 166.28$ kJ mol$^{-1}$.
Answer: $166.28$. | ||||
| Q.8 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Slope of the line: $\frac{-7 - (-3)}{2\times10^{-4}} = -2\times10^4$ K, so $\Delta H^\ominus = 8.314\times2\times10^4 = 166280$ J mol$^{-1}$.
At $T = 1000$ K, $\frac{10^4}{T} = 10$, where $\ln K = -3$. $\Delta G^\ominus = -RT\ln K = 8.314\times1000\times3 = 24942$ J mol$^{-1}$.
$\Delta S^\ominus = \frac{\Delta H^\ominus - \Delta G^\ominus}{T} = \frac{166280 - 24942}{1000} = 141.338 \approx 141.34$ J K$^{-1}$ mol$^{-1}$.
Answer: $141.34$. | ||||
| Q.9 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions AgNO$_3$ dissociates into Ag$^+$ + NO$_3^-$, so $i = 2$.
$\Delta T_b = iK_bm = 2\times0.5\times0.1 = 0.1$ K.
Boiling point $x = 100 + 0.1 = 100.1$ $^\circ$C.
Answer: $100.1$. | ||||
| Q.10 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Take 1 kg of water in each solution (densities as water, so equal volumes carry equal water).
Solution A: $0.1$ mol Ag$^+$ and $0.1$ mol NO$_3^-$. Barium chloride solution: $0.1$ mol Ba$^{2+}$ and $0.2$ mol Cl$^-$.
On mixing, Ag$^+$ + Cl$^-$ $\to$ AgCl$\downarrow$ removes $0.1$ mol of each. Ions left in 2 kg water: NO$_3^-$ $0.1$, Ba$^{2+}$ $0.1$, Cl$^-$ $0.1$ = $0.3$ mol, i.e. $0.15$ mol kg$^{-1}$ of particles.
$\Delta T_b(B) = 0.5\times0.15 = 0.075$ K; $\Delta T_b(A) = 0.5\times0.2 = 0.1$ K.
Difference $= 0.1 - 0.075 = 0.025 = 2.5\times10^{-2}$ $^\circ$C, so $|y| = 2.5$.
Answer: $2.5$. | ||||
| Q.11 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Multiple correctHard | |
A hard multiple correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules HNO$_3$ oxidises both ends (CHO and CH$_2$OH) to COOH, giving an aldaric acid. P is D-glucaric acid; the acid with $[\alpha]_D = -52.7^\circ$ is its enantiomer, L-glucaric acid.
Write each sugar's OH positions from C2 to C5 (R = right, L = left). D-glucose: R L R R. L-glucaric acid has L R L L; since an aldaric acid has COOH at both ends, turning its Fischer projection through $180^\circ$ in the plane gives the same acid as R R L R. So any aldohexose with C2–C5 = L R L L (L-glucose) or R R L R (D-gulose) gives L-glucaric acid.
(A) L L R R: D-mannose, gives D-mannaric acid. Not it.
(B) L L R L: L-gulose, gives D-glucaric acid ($+52.7^\circ$). Not it.
(C) L R L L: L-glucose, gives L-glucaric acid ($-52.7^\circ$). Yes.
(D) R R L R: D-gulose, gives L-glucaric acid ($-52.7^\circ$). Yes.
Answer: (C), (D). | ||||
| Q.12 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Multiple correctHard | |
A hard multiple correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Fusion with Na$_2$O$_2$ turns any sulphur in the product into sulphate, which BaCl$_2$ precipitates as BaSO$_4$: the test detects sulphur. So the product must contain S.
(A) 2,4-Dinitrofluorobenzene: F is ortho and para to two NO$_2$ groups, so PhS$^-$ displaces it readily (nucleophilic aromatic substitution) to give 2,4-dinitrophenyl phenyl sulphide, which contains S. Yes.
(B) 1-Iodo-3,5-dinitrobenzene: both NO$_2$ groups are meta to I, so they cannot stabilise the negative intermediate by resonance; no substitution takes place and no sulphur-containing product forms. No.
(C) The Br is meta to NO$_2$ and is not replaced, but the compound already carries an SMe group, so the organic product (the recovered aryl sulphide) contains S. Yes.
(D) The benzylic CH$_2$Cl undergoes S$_N$2 with PhS$^-$ to give a benzyl phenyl sulphide; it also carries SMe. Contains S. Yes.
Answer: (A), (C), (D). | ||||
| Q.13 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter (A) False: precipitating a sol with an electrolyte is coagulation; peptization is the reverse, turning a precipitate into a sol with a small amount of electrolyte.
(B) True: at the same mass concentration a colloid contains far fewer particles (they are huge), so its freezing-point depression is much smaller; it freezes at a higher temperature than the true solution.
(C) True: surfactants aggregate into micelles only above the critical micelle concentration, and the CMC (like the Kraft temperature) depends on temperature.
(D) False: micelles are associated colloids, not macromolecular colloids.
Answer: (B), (C). | ||||
| Q.14 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics I $\to$ II, isothermal expansion: $T$ constant, $V$ up, $p$ down, $H$ constant (ideal gas), $S$ up.
II $\to$ III, reversible adiabatic expansion: $S$ constant, $V$ up, $p$ and $T$ down, so $H$ down.
(A) True: $p$–$V$: a falling isotherm from I to II, then a steeper falling adiabat to III.
(B) True: $p$–$T$: I to II is a vertical drop at constant $T$; II to III goes to lower $T$ and lower $p$ ($p \propto T^{\gamma/(\gamma - 1)}$), as drawn.
(C) False: on the $H$–$S$ plot II to III should go straight down (H falls at constant S), not up.
(D) True: $T$–$S$: horizontal to the right at constant $T$, then straight down at constant $S$.
Answer: (A), (B), (D). | ||||
| Q.15 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter (A) True: PbS + 2PbO $\to$ 3Pb + SO$_2$ (self-reduction of lead).
(B) False: silica is added to remove iron as slag, FeO + SiO$_2$ $\to$ FeSiO$_3$, not to make copper silicate.
(C) True: partial roasting turns some Cu$_2$S into Cu$_2$O, and then 2Cu$_2$O + Cu$_2$S $\to$ 6Cu + SO$_2$ (self-reduction); the escaping SO$_2$ blisters the copper, giving blister copper.
(D) True: 2Na[Au(CN)$_2$] + Zn $\to$ Na$_2$[Zn(CN)$_4$] + 2Au; zinc displaces gold.
Answer: (A), (C), (D). | ||||
| Q.16 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter (A) True: NaOH gives white Pb(OH)$_2$ and white Zn(OH)$_2$; HCl gives white PbCl$_2$ (Zn$^{2+}$ stays in solution).
(B) True: NaOH gives white Pb(OH)$_2$ and white Bi(OH)$_3$; HCl gives white PbCl$_2$, and on dilution Bi$^{3+}$ chloride hydrolyses to white BiOCl.
(C) False: NaOH gives brown Ag$_2$O with AgNO$_3$.
(D) False: NaOH gives yellow HgO with Hg(NO$_3$)$_2$.
Answer: (A), (B). | ||||
| Q.17 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Number the ring C1(CH$_3$)=C2–C3–C4–C5–C6. Replace one H on each sp$^3$ carbon, and allow for allylic radicals to brominate at either end.
Direct substitution:
• CH$_3$: 1-(bromomethyl)cyclohexene, achiral: 1.
• C3: 3-bromo-1-methylcyclohexene, one stereocentre: 2.
• C4: 4-bromo-1-methylcyclohexene: 2.
• C5: 5-bromo-1-methylcyclohexene: 2.
• C6: 6-bromo-1-methylcyclohexene: 2.
Allylic rearrangement:
• The CH$_2^\bullet$ radical is delocalised to C2: 2-bromo-1-methylenecyclohexane: 2.
• The C3 radical is delocalised to C1: 3-bromo-3-methylcyclohexene: 2.
• The C6 radical delocalised to C2 gives back 6-bromo-1-methylcyclohexene (already counted).
Total $= 1 + 2 + 2 + 2 + 2 + 2 + 2 = 13$.
Answer: $13$. | ||||
| Q.18 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The starting compound is a cyclooctadiene with a cyclopentane ring fused at each of its two C=C bonds (14 carbons; both alkenes tetrasubstituted).
Ozonolysis with Zn/H$_2$O cleaves both C=C bonds; each gives two ketone carbonyls. Cutting both ring-fusion bonds turns the tricycle into one 14-membered ring with four C=O groups: cyclotetradecane-1,4,8,11-tetraone.
Excess NH$_2$OH converts all four ketones into oximes, C=N–OH.
sp$^2$ atoms: each oxime has an sp$^2$ carbon and an sp$^2$ nitrogen ($4\times2 = 8$); the oxime O (two single bonds) and all CH$_2$ carbons are sp$^3$.
Total $= 8$.
Answer: $8$. | ||||
| Q.19 | 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 Pt(IV) is octahedral. Two kinds of isomerism apply.
Ionisation isomers (which anions are inside the coordination sphere): [Pt(NH$_3$)$_4$Cl$_2$]Br$_2$, [Pt(NH$_3$)$_4$Br$_2$]Cl$_2$, [Pt(NH$_3$)$_4$ClBr]ClBr.
Each [Pt(NH$_3$)$_4$X$_2$] or [Pt(NH$_3$)$_4$XY] octahedron has cis and trans forms (neither is optically active).
Total $= 3\times2 = 6$.
Answer: $6$. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines The two given sides meet at $A = (-1, 0)$. Let $B$ be the third vertex on the x-axis and $C$ the one on $x + y + 1 = 0$.
The altitude from $C$ is perpendicular to the x-axis, so it is the vertical line through $H(1, 1)$: $x = 1$. Then $C = (1, -2)$.
The altitude from $B$ is perpendicular to $x + y + 1 = 0$ (slope $-1$), so it has slope $1$ through $H$: $y = x$. It meets the x-axis at $B = (0, 0)$.
Circle $x^2 + y^2 + Dx + Ey + F = 0$ through $(0,0)$: $F = 0$; through $(-1, 0)$: $1 - D = 0$, $D = 1$; through $(1, -2)$: $1 + 4 + 1 - 2E = 0$, $E = 3$.
$x^2 + y^2 + x + 3y = 0$.
Answer: (B). | ||||
| Q.2 | 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 Integrate in horizontal strips. For a given $y$, $x$ runs from $\max(3y, 2 - y)$ to $\frac94$.
$3y = 2 - y$ at $y = \frac12$, and $3y \le \frac94$ needs $y \le \frac34$ (so $y \le 1$ is not binding).
$0 \le y \le \frac12$: width $\frac94 - (2 - y) = \frac14 + y$; area $\int_0^{1/2}\left(\frac14 + y\right)dy = \frac18 + \frac18 = \frac14$.
$\frac12 \le y \le \frac34$: width $\frac94 - 3y$; area $\left[\frac94y - \frac32y^2\right]_{1/2}^{3/4} = \frac{9}{16} - \frac32\cdot\frac{5}{16} = \frac{3}{32}$.
Total $= \frac{8}{32} + \frac{3}{32} = \frac{11}{32}$.
Answer: (A). | ||||
| Q.3 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctHard | |
A hard single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability $E_2$ is the one element of $E_1$ not chosen, so $E_3 = E_1$ exactly when $S_3 = S_1$. Each $S_1$ has probability $\frac13$.
$S_1 = \{1,2\}$: $F_2 = \{1,2,3,4\}$. $S_3 = \{1,2\}$ needs $1 \in G_2$, i.e. $1 \in S_2$ (probability $\frac36$); then $G_2 = \{1,2,3,4,5\}$ and $P(S_3 = \{1,2\}) = \frac{1}{10}$. Product $\frac{1}{20}$.
$S_1 = \{1,3\}$: $F_2 = \{1,3,4\}$; $1 \in S_2$ with probability $\frac23$, then $G_2$ has 5 elements: $\frac23\cdot\frac{1}{10} = \frac{1}{15}$.
$S_1 = \{2,3\}$: $F_2 = \{1,2,3,4\}$; $2, 3$ are always in $G_2$. If $1 \in S_2$ (prob. $\frac12$), $|G_2| = 5$: $\frac{1}{10}$; otherwise $|G_2| = 4$: $\frac16$. Total $\frac12\cdot\frac{1}{10} + \frac12\cdot\frac16 = \frac{2}{15}$.
$p = \frac{\frac{1}{20}}{\frac{1}{20} + \frac{1}{15} + \frac{2}{15}} = \frac{3}{3 + 4 + 8} = \frac{1}{5}$.
Answer: (A). | ||||
| Q.4 | 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 $z_k = e^{i(\theta_1 + \cdots + \theta_k)}$ lies on the unit circle and $z_{10} = e^{2\pi i} = 1$.
$|z_k - z_{k-1}| = |e^{i\theta_k} - 1| = 2\sin\frac{\theta_k}{2} \le \theta_k$, and $|z_1 - z_{10}| = |e^{i\theta_1} - 1| \le \theta_1$. The terms cover each $\theta_k$ once, so the sum is $\le 2\pi$ (a chord is shorter than its arc): $P$ is true.
$|z_k^2 - z_{k-1}^2| = |e^{2i\theta_k} - 1| = 2|\sin\theta_k| \le 2\theta_k$, and likewise for the last term, so the sum is $\le 2\cdot2\pi = 4\pi$: $Q$ is true.
Answer: (C). | ||||
| Q.5 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability The maximum is below 81 only if all three numbers are in $\{1, \ldots, 80\}$: probability $\left(\frac{80}{100}\right)^3 = 0.512$.
$p_1 = 1 - 0.512 = 0.488$.
$\frac{625}{4}\times0.488 = 156.25\times0.488 = 76.25$.
Answer: $76.25$. | ||||
| Q.6 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability The minimum exceeds 40 only if all three numbers are in $\{41, \ldots, 100\}$: probability $\left(\frac{60}{100}\right)^3 = 0.216$.
$p_2 = 1 - 0.216 = 0.784$.
$\frac{125}{4}\times0.784 = 31.25\times0.784 = 24.5$.
Answer: $24.50$. | ||||
| Q.7 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The coefficient rows satisfy $R_1 - 2R_2 + R_3 = 0$ (rank 2), so the system is consistent exactly when the right sides satisfy the same relation: $\alpha - 2\beta + (\gamma - 1) = 0$, i.e. $\alpha - 2\beta + \gamma = 1$. This is the plane $P$.
$|M| = \alpha(1\cdot1 - 0) - 2(\beta\cdot1 - 0\cdot(-1)) + \gamma(\beta\cdot0 - 1\cdot(-1)) = \alpha - 2\beta + \gamma = 1$.
Answer: $1$. | ||||
| Q.8 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The coefficient rows satisfy $R_1 - 2R_2 + R_3 = 0$ (rank 2), so the system is consistent exactly when the right sides satisfy the same relation: $\alpha - 2\beta + (\gamma - 1) = 0$, i.e. $\alpha - 2\beta + \gamma = 1$. This is the plane $P$.
Distance of $(0, 1, 0)$ from $x - 2y + z - 1 = 0$: $\frac{|0 - 2 + 0 - 1|}{\sqrt{1 + 4 + 1}} = \frac{3}{\sqrt6}$.
$D = \frac{9}{6} = 1.5$.
Answer: $1.50$. | ||||
| Q.9 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Product of distances: $\frac{|(\sqrt2x + y - 1)(\sqrt2x - y + 1)|}{3} = \lambda^2$, i.e. $C : |2x^2 - (y - 1)^2| = 3\lambda^2$.
On $y = 2x + 1$: $y - 1 = 2x$, so $|2x^2 - 4x^2| = 2x^2 = 3\lambda^2$, $x = \pm\sqrt{\frac{3\lambda^2}{2}}$.
$RS = \sqrt{1 + 2^2}\,|x_R - x_S| = \sqrt5\cdot2\sqrt{\frac{3\lambda^2}{2}}$, so $RS^2 = 5\cdot4\cdot\frac{3\lambda^2}{2} = 30\lambda^2 = 270$, giving $\lambda^2 = 9$.
Answer: $9$. | ||||
| Q.10 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Product of distances: $\frac{|(\sqrt2x + y - 1)(\sqrt2x - y + 1)|}{3} = \lambda^2$, i.e. $C : |2x^2 - (y - 1)^2| = 3\lambda^2$.
On $y = 2x + 1$: $y - 1 = 2x$, so $|2x^2 - 4x^2| = 2x^2 = 3\lambda^2$, $x = \pm\sqrt{\frac{3\lambda^2}{2}}$.
$RS = \sqrt{1 + 2^2}\,|x_R - x_S| = \sqrt5\cdot2\sqrt{\frac{3\lambda^2}{2}}$, so $RS^2 = 5\cdot4\cdot\frac{3\lambda^2}{2} = 30\lambda^2 = 270$, giving $\lambda^2 = 9$.
So $C : |2x^2 - (y - 1)^2| = 27$. The midpoint of $RS$ is $(0, 1)$, and the perpendicular bisector is $y - 1 = -\frac{x}{2}$.
Substituting: $\left|2x^2 - \frac{x^2}{4}\right| = \frac{7x^2}{4} = 27$, so $x^2 = \frac{108}{7}$, $x = \pm\sqrt{\frac{108}{7}}$.
$D = (R'S')^2 = (2x)^2\left(1 + \frac14\right) = 4\cdot\frac{108}{7}\cdot\frac54 = \frac{540}{7} = 77.142\ldots \approx 77.14$.
Answer: $77.14$. | ||||
| Q.11 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Multiple correctMedium | |
A medium multiple correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $P$ swaps rows 2 and 3 when it multiplies on the left and columns 2 and 3 on the right.
(A) True: $PE = \begin{bmatrix}1&2&3\\8&13&18\\2&3&4\end{bmatrix}$, and swapping its columns 2, 3 gives $\begin{bmatrix}1&3&2\\8&18&13\\2&4&3\end{bmatrix} = F$. Swapping twice restores, so $P^2 = I$.
(B) True: $|E| = 1(54 - 52) - 2(36 - 32) + 3(26 - 24) = 2 - 8 + 6 = 0$, so $|F| = |P|^2|E| = 0$ and the right side is $0$. On the left, $PF = P\cdot PEP = EP$, so $EQ + PFQ^{-1} = E(Q + PQ^{-1})$, whose determinant is $|E|\cdot|Q + PQ^{-1}| = 0$.
(C) False: $|EF| = 0$, so both sides are $0$ and the strict inequality fails.
(D) True: trace is additive and $\mathrm{tr}(P^{-1}XP) = \mathrm{tr}(X)$, so both sums equal $\mathrm{tr}E + \mathrm{tr}F$.
Answer: (A), (B), (D). | ||||
| Q.12 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Multiple correctMedium | |
A medium multiple correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The denominator $x^2 + 2x + 4 = (x + 1)^2 + 3 > 0$, so $f$ is defined everywhere.
$f'(x) = \frac{(2x - 3)(x^2 + 2x + 4) - (x^2 - 3x - 6)(2x + 2)}{(x^2 + 2x + 4)^2} = \frac{(2x^3 + x^2 + 2x - 12) - (2x^3 - 4x^2 - 18x - 12)}{(\ldots)^2} = \frac{5x(x + 4)}{(x^2 + 2x + 4)^2}$.
$f' < 0$ on $(-4, 0)$ and $f' > 0$ on $(-\infty, -4)$ and $(0, \infty)$.
(A) True: $(-2, -1) \subset (-4, 0)$. (B) True: $(1, 2) \subset (0, \infty)$.
Extremes: $f(-4) = \frac{16 + 12 - 6}{16 - 8 + 4} = \frac{11}{6}$ (maximum), $f(0) = -\frac{3}{2}$ (minimum), and $f \to 1$ at $\pm\infty$. Range $= \left[-\frac32, \frac{11}{6}\right]$.
(C) False: the range is not $\mathbb{R}$. (D) False: the range is $\left[-\frac32, \frac{11}{6}\right]$, not up to $2$.
Answer: (A), (B). | ||||
| Q.13 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Multiple correctHard | |
A hard multiple correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability (A) True: $P(E\cap F\cap G^c) = P(E\cap F) - P(E\cap F\cap G) \le P(E) - \frac{1}{10} = \frac18 - \frac{1}{10} = \frac{1}{40}$.
(B) True: $P(E^c\cap F\cap G) = P(F\cap G) - \frac{1}{10} \le P(F) - \frac{1}{10} = \frac16 - \frac{1}{10} = \frac{1}{15}$.
(C) True: $P(E\cup F\cup G) = \sum P - \sum P(\text{pairs}) + P(E\cap F\cap G)$. Each pairwise intersection contains $E\cap F\cap G$, so is at least $\frac{1}{10}$. Hence $P(E\cup F\cup G) \le \frac{13}{24} - \frac{3}{10} + \frac{1}{10} = \frac{41}{120} \le \frac{13}{24}$.
(D) False: $P(E^c\cap F^c\cap G^c) = 1 - P(E\cup F\cup G) \ge 1 - \frac{41}{120} = \frac{79}{120} > \frac{5}{12} = \frac{50}{120}$.
Answer: (A), (B), (C). | ||||
| Q.14 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Multiple correctHard | |
A hard multiple correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $G(I - EF) = (I - EF)G = I$ gives $G = I + GEF = I + EFG$.
(B) True: $(I - FE)(I + FGE) = I - FE + FGE - FEFGE = I - FE + F(G - EFG)E = I - FE + F\cdot I\cdot E = I$.
(A) True: $|I - FE| = |I - EF|$ (Sylvester's identity, $|I - AB| = |I - BA|$), and $|FGE| = |F||G||E| = \frac{|FE|}{|I - EF|}$. So $|I - FE|\,|FGE| = |FE|$.
(C) True: $G = I + GEF = I + EFG$ gives $GEF = EFG$ ($G$ commutes with $EF$, as the inverse of $I - EF$ does).
(D) False: $(I - FE)(I - FGE) = I$ together with (B) would force $FGE = -FGE$, i.e. $FGE = 0$, which fails in general (e.g. $E = F = \frac12I$ gives $FGE = \frac13I$).
Answer: (A), (B), (C). | ||||
| Q.15 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Multiple correctMedium | |
A medium multiple correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions For $x > 0$ the argument is positive, so $\cot^{-1}\left(\frac{1 + k(k+1)x^2}{x}\right) = \tan^{-1}\frac{x}{1 + kx\cdot(k+1)x} = \tan^{-1}((k+1)x) - \tan^{-1}(kx)$.
The sum telescopes: $S_n(x) = \tan^{-1}((n+1)x) - \tan^{-1}x = \tan^{-1}\frac{nx}{1 + (n+1)x^2}$, a value in $\left(0, \frac{\pi}{2}\right)$.
(A) True: $S_{10}(x) = \tan^{-1}\frac{10x}{1 + 11x^2} = \frac{\pi}{2} - \tan^{-1}\frac{1 + 11x^2}{10x}$ (both arguments positive).
(B) True: $S_n(x) \to \frac{\pi}{2} - \tan^{-1}x = \cot^{-1}x$, so $\cot(S_n(x)) \to x$.
(C) False: $S_3 = \frac{\pi}{4}$ needs $\frac{3x}{1 + 4x^2} = 1$, i.e. $4x^2 - 3x + 1 = 0$, discriminant $9 - 16 < 0$: no real root.
(D) False: $\tan S_n = \frac{nx}{1 + (n+1)x^2} \le \frac{n}{2\sqrt{n+1}}$ (AM–GM), with equality at $x = \frac{1}{\sqrt{n+1}}$; for $n = 3$ this is $\frac34 > \frac12$.
Answer: (A), (B). | ||||
| Q.16 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Multiple correctHard | |
A hard multiple correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\arg\frac{z - (-\alpha)}{z - (-\beta)} = \frac{\pi}{4}$ is an arc of a circle through the real points $-\alpha$ and $-\beta$. The given circle meets $y = 0$ where $x^2 + 5x + 4 = 0$: $x = -1, -4$. So $\{-\alpha, -\beta\} = \{-1, -4\}$.
The circle has centre $\left(-\frac52, \frac32\right)$ above the real axis and radius $\frac{3}{\sqrt2}$; the chord from $-4$ to $-1$ subtends $\frac{\pi}{4}$ on the upper (major) arc, which is the locus.
To fix the order, take the top point $z = -\frac52 + i\left(\frac32 + \frac{3}{\sqrt2}\right) \approx -2.5 + 3.62i$. Then $\arg(z + 1) \approx 112.5^\circ$ and $\arg(z + 4) \approx 67.5^\circ$, and $\arg\frac{z + 1}{z + 4} = 45^\circ$. So $\alpha = 1$, $\beta = 4$.
(A) False. (B) True: $\alpha\beta = 4$. (C) False. (D) True: $\beta = 4$.
Answer: (B), (D). | ||||
| Q.17 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Case $|x| \ge 1$: $3x^2 - 4x^2 + 4 + x - 1 = 0 \Rightarrow x^2 - x - 3 = 0$, $x = \frac{1 \pm \sqrt{13}}{2} \approx 2.30, -1.30$. Both have $|x| \ge 1$: 2 roots.
Case $|x| < 1$: $3x^2 + 4x^2 - 4 + x - 1 = 0 \Rightarrow 7x^2 + x - 5 = 0$, $x = \frac{-1 \pm \sqrt{141}}{14} \approx 0.777, -0.920$. Both have $|x| < 1$: 2 roots.
Total: $4$.
Answer: $4$. | ||||
| Q.18 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions With $a = BC = 3$, $b = CA = 4$, $c = AB = \sqrt{23}$ and area $\Delta$: $\cot A = \frac{b^2 + c^2 - a^2}{4\Delta}$, and similarly for $B$, $C$.
$\frac{\cot A + \cot C}{\cot B} = \frac{(b^2 + c^2 - a^2) + (a^2 + b^2 - c^2)}{a^2 + c^2 - b^2} = \frac{2b^2}{a^2 + c^2 - b^2} = \frac{32}{9 + 23 - 16} = \frac{32}{16} = 2$.
Answer: $2$. | ||||
| Q.19 | 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 Let $c = \vec u\cdot\vec v$. The square of the volume is the Gram determinant:
$[\vec u\,\vec v\,\vec w]^2 = \begin{vmatrix}1 & c & 1\\ c & 1 & 1\\ 1 & 1 & 4\end{vmatrix} = 1(4 - 1) - c(4c - 1) + 1(c - 1) = 2 + 2c - 4c^2$.
Setting it equal to $2$: $2c - 4c^2 = 0$, so $c = 0$ or $c = \frac12$. They are not perpendicular, so $c = \frac12$.
$|3\vec u + 5\vec v|^2 = 9 + 25 + 30c = 34 + 15 = 49$, so $|3\vec u + 5\vec v| = 7$.
Answer: $7$. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Motion of System of Particles and Rigid BodyEquilibrium of rigid bodies | Class 11: Systems of Particles and Rotational Motion | Multiple correctMedium | |
A medium multiple correct question on equilibrium of rigid bodies, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The rope runs from P (at distance $L$ from the wall) to Q (height $L$ above O), so it makes $45^\circ$ with the beam.
Torques about O: the rope's vertical component $T\sin45^\circ$ acts at distance $L$; the beam's weight $W$ at $L/2$; the block $\alpha W$ at $L$.
$T\frac{1}{\sqrt2}L = W\frac{L}{2} + \alpha WL \Rightarrow T = \sqrt2\,W\left(\frac12 + \alpha\right)$.
Horizontal reaction at O: $H = T\cos45^\circ = W\left(\frac12 + \alpha\right)$.
Vertical reaction at O: $V = W + \alpha W - T\sin45^\circ = W + \alpha W - W\left(\frac12 + \alpha\right) = \frac{W}{2}$.
(A) $V = W/2$ for every $\alpha$: true.
(B) $\alpha = 0.5$: $H = W(0.5 + 0.5) = W$: true.
(C) $\alpha = 0.5$: $T = \sqrt2 W$, not $2W$: false.
(D) The rope breaks when $\sqrt2 W\left(\frac12 + \alpha\right) > 2\sqrt2 W$, i.e. $\alpha > 1.5$: true.
Answer: (A), (B), (D). | ||||
| Q.2 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | Multiple correctMedium | |
A medium multiple correct question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves A pipe closed at one end resonates only at odd multiples of its fundamental: $f_0,\ 3f_0,\ 5f_0, \dots$ with $f_0 = \frac{v}{4L}$.
The source moves towards the stationary pipe, so the frequency reaching the pipe is $f = f_s\frac{v}{v - u}$.
(A) $u = 0.8v$: $f = \frac{f_s}{0.2} = 5f_s = 5f_0$, an odd harmonic: resonance.
(B) $u = 0.8v$, $f_s = 2f_0$: $f = 10f_0$, even multiple: no.
(C) $u = 0.8v$, $f_s = 0.5f_0$: $f = 2.5f_0$: no.
(D) $u = 0.5v$: $f = 2f_s = 3f_0$, an odd harmonic: resonance.
Answer: (A), (D). | ||||
| Q.3 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Multiple correctHard | |
A hard multiple correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics With $n_1 = n_2 = 1.5$ and minimum deviation, the ray inside the prism travels parallel to the base and $r_1 = r_2 = \frac{\theta}{2} = 30^\circ$, so $\sin i = 1.5\sin30^\circ = 0.75$.
Now let the halves differ. At the first face nothing changes ($n_1 = n$), so the ray inside is still parallel to the base. The boundary between the halves is the vertical line from the apex to the base, so the ray meets it at normal incidence and passes straight through, whatever $n_2$ is.
At the second face the angle of incidence is still $30^\circ$, now inside $n_2$: $\sin e = n_2\sin30^\circ = \frac{n + \Delta n}{2}$.
Differentiating about $e = i$: $\cos i\,\Delta e = \frac{\Delta n}{2}$, so $\Delta e = \frac{\Delta n}{2\cos i}$, with $\cos i = \sqrt{1 - 0.5625} = 0.6614$.
$\Delta e = \frac{\Delta n}{1.3229} \approx 0.756\,\Delta n$.
(A) $\Delta e < \Delta n$: false.
(B) $\Delta e$ is proportional to $\Delta n$: true.
(C) and (D): for $\Delta n = 2.8\times10^{-3}$, $\Delta e = 0.756\times2.8\times10^{-3} \approx 2.12\times10^{-3}$ rad $= 2.12$ mrad, between 2.0 and 3.0: (C) true, (D) false.
Answer: (B), (C). | ||||
| Q.4 | 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 $\vec S$ is the Poynting vector: energy flow per unit area per unit time, in W m$^{-2}$. Check: $\frac{EB}{\mu_0} = \frac{E\cdot(E/c)}{\mu_0} = \epsilon_0cE^2$, which is (energy density) $\times$ (speed) = J m$^{-3}\cdot$ m s$^{-1}$ = W m$^{-2}$. Dimensions $[\mathrm{MT^{-3}}]$.
(A) $\frac{\text{Energy}}{\text{Charge}\times\text{Current}} = \frac{\mathrm{ML^2T^{-2}}}{\mathrm{AT\cdot A}} = \mathrm{ML^2T^{-3}A^{-2}}$ (that of resistance): no.
(B) $\frac{\text{Force}}{\text{Length}\times\text{Time}} = \frac{\mathrm{MLT^{-2}}}{\mathrm{LT}} = \mathrm{MT^{-3}}$: yes.
(C) $\frac{\text{Energy}}{\text{Volume}} = \mathrm{ML^{-1}T^{-2}}$: no.
(D) $\frac{\text{Power}}{\text{Area}} = \frac{\mathrm{ML^2T^{-3}}}{\mathrm{L^2}} = \mathrm{MT^{-3}}$: yes.
Answer: (B), (D). | ||||
| Q.5 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Multiple correctMedium | |
A medium multiple correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Energy: the mass lost appears as kinetic energy, $E_P + E_Q = \delta c^2$.
Momentum: N was at rest, so P and Q fly apart with equal and opposite momenta of magnitude $p$: $M_Pv_P = M_Qv_Q$.
(A) $E_P + E_Q = c^2\delta$: true.
(B) With $E = \frac{p^2}{2M}$, $\frac{E_P}{E_Q} = \frac{M_Q}{M_P}$, so $E_P = \frac{M_Q}{M_P + M_Q}c^2\delta$, not $\frac{M_P}{M_P + M_Q}c^2\delta$: false.
(C) $\frac{v_P}{v_Q} = \frac{M_Q}{M_P}$: true.
(D) $\frac{p^2}{2M_P} + \frac{p^2}{2M_Q} = \frac{p^2}{2\mu} = c^2\delta$, so $p = c\sqrt{2\mu\delta}$ with $\mu = \frac{M_PM_Q}{M_P + M_Q}$: true.
(Kinetic energies are treated non-relativistically, as the options intend.)
Answer: (A), (C), (D). | ||||
| Q.6 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctMedium | |
A medium multiple correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter (A) For a point in the plane of a loop, each element $d\vec l$ and the vector $\vec r$ to the point both lie in the plane, so $d\vec l\times\vec r$ is perpendicular to the plane. The field of both loops at any point of the xy-plane is therefore along $\pm\hat k$: true.
(B) Both loops are circles about the origin, so the arrangement looks the same from every direction; $|\vec B|$ can depend only on $r$: true.
(C) At the centre: $B = \frac{\mu_0I_1}{2R}$ (out of the plane, anticlockwise current) $- \frac{\mu_0I_2}{2(2R)}$ (into the plane) $= \frac{\mu_0}{4R}(2I_1 - I_2) < 0$, i.e. into the plane, since $I_2 > 2I_1$. As $r \to R$ from inside, the small loop's outward field grows without bound (the point approaches its wire) while the large loop's field stays finite. So the net field changes from into to out of the plane somewhere in $0 < r < R$ and is zero on that circle: false.
(D) Between the loops ($R < r < 2R$): outside the anticlockwise small loop its field in its own plane is reversed, i.e. into the plane; inside the clockwise large loop its field is also into the plane. The net field points into the plane, not outward: false.
Answer: (A), (B). | ||||
| Q.7 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Glass volume: $\frac{5}{2.5} = 2$ cc. The tube with its air sinks once the water it displaces weighs less than the tube (the air's mass is negligible). The displaced volume is glass plus air, so it just begins to sink when $(2 + v)\times1 = 5$, i.e. trapped air $v = 3$ cc.
So $v_0 - \Delta v = 3$ cc and $\Delta v = 3.3 - 3 = 0.3$ cc.
At constant temperature (Boyle's law, with the small hydrostatic pressure differences neglected): $p_0\times3.3 = (p_0 + \Delta p)\times3$, so $\Delta p = 0.1p_0 = 10^4$ Pa $= 10\times10^3$ Pa.
$X = 0.30$. | ||||
| Q.8 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Glass volume: $\frac{5}{2.5} = 2$ cc. The tube with its air sinks once the water it displaces weighs less than the tube (the air's mass is negligible). The displaced volume is glass plus air, so it just begins to sink when $(2 + v)\times1 = 5$, i.e. trapped air $v = 3$ cc.
So $v_0 - \Delta v = 3$ cc and $\Delta v = 3.3 - 3 = 0.3$ cc.
At constant temperature (Boyle's law, with the small hydrostatic pressure differences neglected): $p_0\times3.3 = (p_0 + \Delta p)\times3$, so $\Delta p = 0.1p_0 = 10^4$ Pa $= 10\times10^3$ Pa.
$Y = 10.00$. | ||||
| Q.9 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The impulse gives the bob $u = \frac{P}{m} = \frac{0.2}{0.1} = 2$ m/s along the floor. It slides (frictionless floor) until the string, of length 1.0 m, becomes taut. Then the string makes angle $\theta$ with the vertical, where $\cos\theta = \frac{H}{L} = 0.9$.
Angular momentum about the suspension point while sliding: the velocity is horizontal and its line is at perpendicular distance $H = 0.9$ m from the suspension point, so $J = mu H = 0.1\times2\times0.9 = 0.18$ kg m$^2$/s.
When the string jerks taut, its impulsive tension passes through the suspension point, so angular momentum about that point is conserved (the floor's normal impulse is zero, as the bob then rises). The component of velocity along the string is destroyed; the component perpendicular to it survives: $v = u\cos\theta = 2\times0.9 = 1.8$ m/s (check: $mvL = 0.1\times1.8\times1 = 0.18 = J$).
$J = 0.18$. | ||||
| Q.10 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The impulse gives the bob $u = \frac{P}{m} = \frac{0.2}{0.1} = 2$ m/s along the floor. It slides (frictionless floor) until the string, of length 1.0 m, becomes taut. Then the string makes angle $\theta$ with the vertical, where $\cos\theta = \frac{H}{L} = 0.9$.
Angular momentum about the suspension point while sliding: the velocity is horizontal and its line is at perpendicular distance $H = 0.9$ m from the suspension point, so $J = mu H = 0.1\times2\times0.9 = 0.18$ kg m$^2$/s.
When the string jerks taut, its impulsive tension passes through the suspension point, so angular momentum about that point is conserved (the floor's normal impulse is zero, as the bob then rises). The component of velocity along the string is destroyed; the component perpendicular to it survives: $v = u\cos\theta = 2\times0.9 = 1.8$ m/s (check: $mvL = 0.1\times1.8\times1 = 0.18 = J$).
Just after lift-off: $K = \frac12mv^2 = \frac12\times0.1\times1.8^2 = 0.162$ J.
$K = 0.16$ to two decimal places. | ||||
| Q.11 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The lamp is a resistance: $I = \frac{P}{V_R} = \frac{500}{100} = 5$ A and $R = \frac{100}{5} = 20\ \Omega$.
In a series RC circuit the voltages add as phasors at right angles: $V_C = \sqrt{200^2 - 100^2} = 100\sqrt3$ V.
$X_C = \frac{V_C}{I} = 20\sqrt3\ \Omega$.
$C = \frac{1}{2\pi fX_C} = \frac{1}{2\pi\times50\times20\sqrt3} = \frac{1}{2000\,\pi\sqrt3} = \frac{1}{2000\times5} = 10^{-4}$ F $= 100\ \mu$F.
$C = 100.00$. | ||||
| Q.12 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The lamp is a resistance: $I = \frac{P}{V_R} = \frac{500}{100} = 5$ A and $R = \frac{100}{5} = 20\ \Omega$.
In a series RC circuit the voltages add as phasors at right angles: $V_C = \sqrt{200^2 - 100^2} = 100\sqrt3$ V.
$X_C = \frac{V_C}{I} = 20\sqrt3\ \Omega$.
$\cos\varphi = \frac{V_R}{V} = \frac{100}{200} = \frac12$ (equivalently $\tan\varphi = \frac{X_C}{R} = \sqrt3$), so $\varphi = 60^\circ$.
$\varphi = 60.00$. | ||||
| Q.13 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | ComprehensionMedium | |
A medium comprehension question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The loop started with zero flux (dipole at infinity) and keeps it zero. The dipole's flux through the loop, with $r \gg a$ so its field is nearly uniform over the loop: $\Phi = \frac{\mu_0}{2\pi}\frac{m}{r^3}\cdot\pi a^2$.
The induced current $i$ cancels it with its own flux $L_{\text{loop}}\,i$ ($L_{\text{loop}}$ the loop's self-inductance, a constant): $L_{\text{loop}}\,i = \frac{\mu_0a^2m}{2r^3}$.
So $i \propto \frac{m}{r^3}$.
Answer: (A). | ||||
| Q.14 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | ComprehensionMedium | |
A medium comprehension question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The loop started with zero flux (dipole at infinity) and keeps it zero. The dipole's flux through the loop, with $r \gg a$ so its field is nearly uniform over the loop: $\Phi = \frac{\mu_0}{2\pi}\frac{m}{r^3}\cdot\pi a^2$.
The induced current $i$ cancels it with its own flux $L_{\text{loop}}\,i$ ($L_{\text{loop}}$ the loop's self-inductance, a constant): $L_{\text{loop}}\,i = \frac{\mu_0a^2m}{2r^3}$. So $i \propto \frac{m}{r^3}$.
The loop's moment is $m_2 = i\pi a^2 \propto \frac{m}{r^3}$, and it opposes the dipole (repulsion). At separation $x$ the repulsive force is $F(x) = \frac{k\,m\,m_2(x)}{x^4} \propto \frac{m^2}{x^7}$.
Work done against it in bringing the dipole from infinity to $r$: $W = \int_r^\infty F(x)\,dx \propto m^2\int_r^\infty \frac{dx}{x^7} = \frac{m^2}{6r^6}$.
So $W \propto \frac{m^2}{r^6}$.
Answer: (C). | ||||
| Q.15 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | ComprehensionMedium | |
A medium comprehension question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics The right gas receives no heat and is compressed slowly: a reversible adiabatic process with $\gamma = \frac{C_P}{C_V} = \frac{2R + R}{2R} = \frac32$.
$TV^{\gamma - 1}$ is constant: $T_R = T_0\left(\frac{V_0}{V_0/2}\right)^{1/2} = \sqrt2\,T_0$.
$\frac{T_R}{T_0} = \sqrt2$.
Answer: (A). | ||||
| Q.16 | ThermodynamicsFirst law | Class 11: Thermodynamics | ComprehensionMedium | |
A medium comprehension question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics The right gas receives no heat and is compressed slowly: a reversible adiabatic process with $\gamma = \frac{C_P}{C_V} = \frac{2R + R}{2R} = \frac32$.
$TV^{\gamma - 1}$ is constant: $T_R = T_0\left(\frac{V_0}{V_0/2}\right)^{1/2} = \sqrt2\,T_0$.
Right-side pressure: $p_R = p_0\cdot2^{3/2} = 2\sqrt2\,p_0$, where $p_0V_0 = RT_0$. The partition is frictionless and moves slowly, so the left pressure is the same.
Left volume: $2V_0 - \frac{V_0}{2} = \frac{3V_0}{2}$, so $T_L = \frac{p_LV_L}{R} = \frac{2\sqrt2\,p_0\cdot\frac32V_0}{R} = 3\sqrt2\,T_0$.
The whole cylinder is insulated and does no work outside, so the heat goes into the internal energy of both gases:
$Q = C_V(T_L - T_0) + C_V(T_R - T_0) = 2R\,T_0\left(3\sqrt2 - 1 + \sqrt2 - 1\right) = 2RT_0(4\sqrt2 - 2) = 4(2\sqrt2 - 1)RT_0$.
Answer: (B). | ||||
| Q.17 | Current ElectricityMetre bridge | Class 12: Current Electricity | NumericalHard | |
A hard numerical question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity Reading the figure: the cell $E$ (positive plate towards the left end A of the wire), its internal resistance $r_1$, the resistance $R_0/2$ and the 100 cm wire (resistance $R_0 = 50\ \Omega$) form one series loop. From the junction J between $r_1$ and $R_0/2$, the cell $E/2$ (negative plate at J), $r$ and G go to the jockey at $l$ from A.
At the null point no current flows in the G branch, so the main current is $I = \frac{E}{r_1 + R_0/2 + R_0} = \frac{E}{r_1 + 75}$.
The current leaves the positive plate at A, runs along the wire to B and back through $R_0/2$ to J. So $V_A - V_J = I(50 + 25) = 75I$, and the jockey point is $50I\times0.72 = 36I$ below A: $V_l - V_J = 75I - 36I = 39I$.
Null: this equals the emf of the branch cell, $39I = \frac{E}{2}$, so $I = \frac{E}{78}$.
Then $r_1 + 75 = 78$, $r_1 = 3\ \Omega$.
Answer: 3. | ||||
| Q.18 | GravitationKepler's laws | Class 11: Gravitation | NumericalEasy | |
A easy numerical question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation For two bodies orbiting their common centre of mass at separation $d$, $\text{(period)}^2 = \frac{4\pi^2d^3}{G(m_1 + m_2)}$.
Earth–Sun (Earth's mass negligible): $T^2 = \frac{4\pi^2R^3}{GM_S}$.
Stars: $(nT)^2 = \frac{4\pi^2(9R)^3}{G(9M_S)} = \frac{729}{9}\cdot\frac{4\pi^2R^3}{GM_S} = 81T^2$.
$n = 9$.
Answer: 9. | ||||
| Q.19 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Einstein's equation $E_{max} = h\nu - \phi$.
P and Q share the photon energy $h\nu$: $E_P = h\nu - 4.0$, $E_Q = h\nu - 4.5$.
$E_P = 2E_Q$: $h\nu - 4.0 = 2h\nu - 9.0$, so $h\nu = 5.0$ eV, $E_P = 1.0$ eV and $E_Q = 0.5$ eV.
$E_R = E_Q = 0.5$ eV, so the photon for R has energy $5.5 + 0.5 = 6$ eV.
Answer: 6. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Multiple correctMedium | |
A medium multiple correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons o-Xylene is 1,2-dimethylbenzene: the second carbon on the ring must end as a $\mathrm{CH_3}$.
(A) o-Toluidine $\xrightarrow{\mathrm{NaNO_2/HCl},\ 273\text{ K}}$ diazonium salt $\xrightarrow{\mathrm{CuCN}}$ o-tolunitrile (Sandmeyer) $\xrightarrow{\text{DIBAL-H, then }\mathrm{H_3O^+}}$ o-tolualdehyde $\xrightarrow{\mathrm{N_2H_4,\ KOH,\ heat}}$ (Wolff–Kishner, $\mathrm{CHO}\to\mathrm{CH_3}$) o-xylene. Yes.
(B) o-Bromotoluene $\xrightarrow{\mathrm{Mg,\ CO_2,\ H_3O^+}}$ o-toluic acid $\xrightarrow{\mathrm{SOCl_2}}$ o-toluoyl chloride $\xrightarrow{\mathrm{H_2,\ Pd\text{-}BaSO_4}}$ o-tolualdehyde (Rosenmund) $\xrightarrow{\mathrm{Zn\text{-}Hg,\ HCl}}$ (Clemmensen) o-xylene. Yes.
(C) 2-Methylstyrene: hydroboration–oxidation adds water anti-Markovnikov, giving $\mathrm{ArCH_2CH_2OH}$; $\mathrm{PBr_3}$ gives $\mathrm{ArCH_2CH_2Br}$; Zn/dil. HCl reduces C–Br to C–H, giving $\mathrm{ArCH_2CH_3}$, i.e. 2-ethyltoluene, not o-xylene. No.
(D) Indene: ozonolysis cleaves the C=C of the five-membered ring, giving 2-(2-formylphenyl)acetaldehyde ($\mathrm{OHC{-}C_6H_4{-}CH_2CHO}$, ortho); Wolff–Kishner turns both CHO groups into $\mathrm{CH_3}$, giving $\mathrm{CH_3{-}C_6H_4{-}CH_2CH_3}$, 2-ethyltoluene. No.
Answer: (A), (B). | ||||
| Q.2 | AminesClassification and identification of amines | Class 12: Amines | Multiple correctMedium | |
A medium multiple correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines $\mathrm{PhCH_3}\xrightarrow{\mathrm{Br_2},\ \text{light}}$ P $= \mathrm{PhCH_2Br}$ (side-chain free-radical bromination).
S is foul smelling and is made from R with $\mathrm{CHCl_3}$/KOH (carbylamine reaction), so R is a primary amine and S an isocyanide.
Lower route: $\mathrm{PhCH_3}\xrightarrow{\mathrm{KMnO_4/KOH},\ \mathrm{H_3O^+}}$ T $= \mathrm{PhCOOH}$; with $\mathrm{NH_3}$ then heat, U $= \mathrm{PhCONH_2}$ (benzamide).
R must come both from U (by W) and from $\mathrm{PhCH_2Br}$ (by Q, then $\mathrm{H_2}$, Pd/C). Benzamide with $\mathrm{LiAlH_4}$ gives $\mathrm{PhCH_2NH_2}$; $\mathrm{PhCH_2Br}$ with $\mathrm{AgNO_2}$ gives mainly the nitro compound $\mathrm{PhCH_2NO_2}$ (covalent Ag–O bond, attack through N), which $\mathrm{H_2}$/Pd-C reduces to $\mathrm{PhCH_2NH_2}$. So R = phenylmethanamine (benzylamine), Q = $\mathrm{AgNO_2}$, W = $\mathrm{LiAlH_4}$. (With $\mathrm{KNO_2}$, ionic, the main product is the nitrite ester $\mathrm{PhCH_2ONO}$, which does not lead to the amine; Hofmann degradation of benzamide would give aniline, which cannot come from $\mathrm{PhCH_2Br}$.)
S = $\mathrm{PhCH_2NC}$, which P gives directly with AgCN (attack through N of the covalent cyanide): V = AgCN. KCN would give the nitrile $\mathrm{PhCH_2CN}$.
(A) Q = $\mathrm{KNO_2}$: false. (B) R = benzenamine, V = KCN: false. (C) Q = $\mathrm{AgNO_2}$, R = phenylmethanamine: true. (D) W = $\mathrm{LiAlH_4}$, V = AgCN: true.
Answer: (C), (D). | ||||
| Q.3 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Multiple correctMedium | |
A medium multiple correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Start: [X] = 2 M, [Y] = 1 M. At 50 s, [Y] = 0.5 M, so 0.5 M of P has formed and 1.0 M of X has reacted: [X] = 1 M. X has halved in 50 s.
By the stoichiometry, $-\frac{d[X]}{dt} = 2\frac{d[P]}{dt} = 2k[X]$: X decays by first order with constant $2k$.
Half-life of X: $t_{1/2} = \frac{\ln2}{2k} = 50$ s, so $k = \frac{0.693}{100} = 6.93\times10^{-3}$ s$^{-1}$.
(A) $k = 13.86\times10^{-4}$ s$^{-1}$: false ($k = 69.3\times10^{-4}$ s$^{-1}$).
(B) Half-life of X is 50 s: true.
(C) At 50 s, $-\frac{d[X]}{dt} = 2k[X] = 2\times6.93\times10^{-3}\times1 = 13.86\times10^{-3}$ mol L$^{-1}$ s$^{-1}$: true.
(D) At 100 s (two half-lives), [X] = 0.5 M and $-\frac{d[Y]}{dt} = \frac{d[P]}{dt} = k[X] = 6.93\times10^{-3}\times0.5 = 3.465\times10^{-3}$ mol L$^{-1}$ s$^{-1}$: true.
Answer: (B), (C), (D). | ||||
| Q.4 | ElectrochemistryNernst equation | Class 12: Electrochemistry | Multiple correctMedium | |
A medium multiple correct question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry X dissolves, so X is the anode: $\mathrm{X + Y^{2+} \to X^{2+} + Y}$. This happens when $E_{cell} > 0$.
$E_{cell} = E^\circ_Y - E^\circ_X - \frac{2.303RT}{2F}\log\frac{[\mathrm{X^{2+}}]}{[\mathrm{Y^{2+}}]} = E^\circ_Y - E^\circ_X - 0.0296\log\frac{0.001}{0.1} = E^\circ_Y - E^\circ_X + 0.059$ V.
(A) X = Cd, Y = Ni: $-0.24 + 0.40 + 0.059 = +0.219$ V > 0: yes.
(B) X = Cd, Y = Fe: $-0.44 + 0.40 + 0.059 = +0.019$ V > 0: yes (the concentration ratio tips it).
(C) X = Ni, Y = Pb: $-0.13 + 0.24 + 0.059 = +0.169$ V > 0: yes.
(D) X = Ni, Y = Fe: $-0.44 + 0.24 + 0.059 = -0.141$ V < 0: no.
Answer: (A), (B), (C). | ||||
| Q.5 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Multiple correctMedium | |
A medium multiple correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) $\mathrm{[FeCl_4]^-}$: Fe(III), $d^5$, weak-field $\mathrm{Cl^-}$, $sp^3$: tetrahedral. $\mathrm{[Fe(CO)_4]^{2-}}$: Fe(–II), $d^{10}$, $sp^3$: tetrahedral. Both tetrahedral.
(B) $\mathrm{[Co(CO)_4]^-}$: Co(–I), $d^{10}$: tetrahedral. $\mathrm{[CoCl_4]^{2-}}$: Co(II), $d^7$, weak field, $sp^3$: tetrahedral. Both tetrahedral.
(C) $\mathrm{[Ni(CO)_4]}$: Ni(0), $d^{10}$: tetrahedral. $\mathrm{[Ni(CN)_4]^{2-}}$: Ni(II), $d^8$ with strong-field $\mathrm{CN^-}$, $dsp^2$: square planar. Not both.
(D) $\mathrm{[Cu(py)_4]^+}$ and $\mathrm{[Cu(CN)_4]^{3-}}$: Cu(I), $d^{10}$, no vacant $d$ orbital, $sp^3$: both tetrahedral.
Answer: (A), (B), (D). | ||||
| Q.6 | 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 $\mathrm{H_3PO_3}$ is $\mathrm{HP(O)(OH)_2}$: one P–H bond and two P–OH groups; P is +3.
(A) On heating, $\mathrm{4H_3PO_3 \to 3H_3PO_4 + PH_3}$: P goes from +3 to +5 and –3, a disproportionation: true.
(B) The P–H bond makes $\mathrm{H_3PO_3}$ a reducing agent; in $\mathrm{H_3PO_4}$ P is already +5 and there is no P–H bond: true.
(C) Only the two O–H hydrogens are acidic, so $\mathrm{H_3PO_3}$ is dibasic, not monobasic: false.
(D) The H on P is not ionisable in water: true.
Answer: (A), (B), (D). | ||||
| Q.7 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$: before dilution $\alpha = \frac{100y}{400} = \frac{y}{4}$; after dilution $\alpha' = \frac{300y}{400} = 3\alpha$.
The dissociation constant does not change on dilution (Ostwald): $\frac{c\alpha^2}{1 - \alpha} = \frac{(c/20)(3\alpha)^2}{1 - 3\alpha}$.
$20(1 - 3\alpha) = 9(1 - \alpha) \Rightarrow 20 - 60\alpha = 9 - 9\alpha \Rightarrow 51\alpha = 11 \Rightarrow \alpha = \frac{11}{51} = 0.2157$.
$\alpha = 0.22$. | ||||
| Q.8 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$: before dilution $\alpha = \frac{100y}{400} = \frac{y}{4}$; after dilution $\alpha' = \frac{300y}{400} = 3\alpha$.
The dissociation constant does not change on dilution (Ostwald): $\frac{c\alpha^2}{1 - \alpha} = \frac{(c/20)(3\alpha)^2}{1 - 3\alpha}$.
$20(1 - 3\alpha) = 9(1 - \alpha) \Rightarrow 20 - 60\alpha = 9 - 9\alpha \Rightarrow 51\alpha = 11 \Rightarrow \alpha = \frac{11}{51} = 0.2157$.
$y = 4\alpha = \frac{44}{51} = 0.8627$.
$y = 0.86$. | ||||
| Q.9 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Sn with HCl gives $\mathrm{SnCl_2}$: $\mathrm{Sn + 2HCl \to SnCl_2 + H_2}$.
$\mathrm{SnCl_2}$/HCl reduces nitrobenzene to anilinium chloride, the organic salt:
$\mathrm{C_6H_5NO_2 + 3SnCl_2 + 7HCl \to C_6H_5NH_3^+Cl^- + 3SnCl_4 + 2H_2O}$ (N goes from +3 to –3: six electrons, two from each $\mathrm{Sn^{2+}}$).
Molar mass of $\mathrm{C_6H_5NH_3Cl}$ = 72 + 8 + 14 + 35 = 129 g mol$^{-1}$, so 1.29 g is 0.01 mol.
It needs $3\times0.01 = 0.03$ mol $\mathrm{SnCl_2}$, hence 0.03 mol Sn: $x = 0.03\times119 = 3.57$ g.
$x = 3.57$. | ||||
| Q.10 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Sn with HCl gives $\mathrm{SnCl_2}$: $\mathrm{Sn + 2HCl \to SnCl_2 + H_2}$.
$\mathrm{SnCl_2}$/HCl reduces nitrobenzene to anilinium chloride, the organic salt:
$\mathrm{C_6H_5NO_2 + 3SnCl_2 + 7HCl \to C_6H_5NH_3^+Cl^- + 3SnCl_4 + 2H_2O}$ (N goes from +3 to –3: six electrons, two from each $\mathrm{Sn^{2+}}$).
Molar mass of $\mathrm{C_6H_5NH_3Cl}$ = 72 + 8 + 14 + 35 = 129 g mol$^{-1}$, so 1.29 g is 0.01 mol.
Nitrobenzene used: 0.01 mol; molar mass $\mathrm{C_6H_5NO_2}$ = 72 + 5 + 14 + 32 = 123 g mol$^{-1}$: $y = 0.01\times123 = 1.23$ g.
$y = 1.23$. | ||||
| Q.11 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter $\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O}$: one $\mathrm{MnO_4^-}$ oxidises five $\mathrm{Fe^{2+}}$.
$\mathrm{KMnO_4}$ used: $12.5\times10^{-3}\times0.03 = 3.75\times10^{-4}$ mol, so $\mathrm{Fe^{2+}}$ in 25 mL $= 1.875\times10^{-3}$ mol, and in 250 mL $= 1.875\times10^{-2}$ mol.
$x = 1.875$, i.e. 1.87 or 1.88 to two decimal places. | ||||
| Q.12 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter $\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O}$: one $\mathrm{MnO_4^-}$ oxidises five $\mathrm{Fe^{2+}}$.
$\mathrm{KMnO_4}$ used: $12.5\times10^{-3}\times0.03 = 3.75\times10^{-4}$ mol, so $\mathrm{Fe^{2+}}$ in 25 mL $= 1.875\times10^{-3}$ mol, and in 250 mL $= 1.875\times10^{-2}$ mol.
Mass of iron: $1.875\times10^{-2}\times56 = 1.05$ g. Percentage: $\frac{1.05}{5.6}\times100 = 18.75$.
$y = 18.75$. | ||||
| Q.13 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | ComprehensionMedium | |
A medium comprehension question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Two things raise a C–H BDE: more s-character in the carbon orbital (shorter, stronger bond) and a less stable radical left behind.
(S) $\mathrm{H{-}C{\equiv}CH}$: sp carbon (50% s), unstable ethynyl radical: strongest, 132 → (i).
(R) $\mathrm{H{-}CH{=}CH_2}$: sp$^2$ carbon, unstable vinyl radical: 110 → (ii).
(P) $\mathrm{H{-}CH(CH_3)_2}$: sp$^3$, gives a secondary (isopropyl) radical, stabilised by hyperconjugation: 95 → (iii) (below the 105 of methane).
(Q) $\mathrm{H{-}CH_2Ph}$: gives the resonance-stabilised benzyl radical: weakest, 88 → (iv).
P – iii, Q – iv, R – ii, S – i.
Answer: (A). | ||||
| Q.14 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | ComprehensionEasy | |
A easy comprehension question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Use the BDEs in the paragraph (bond broken: +BDE; bond formed: –BDE).
Initiation: $\mathrm{Cl_2 \to 2Cl^\bullet}$, $\Delta H^\circ = +58$ kcal mol$^{-1}$, endothermic. (A) is false.
Propagation 1: $\mathrm{CH_4 + Cl^\bullet \to {}^\bullet CH_3 + HCl}$, $\Delta H^\circ = 105 - 103 = +2$ kcal mol$^{-1}$, endothermic. (B) is false.
Propagation 2: $\mathrm{{}^\bullet CH_3 + Cl_2 \to CH_3Cl + Cl^\bullet}$, $\Delta H^\circ = 58 - 85 = -27$ kcal mol$^{-1}$, exothermic. (C) is false.
Overall: $\Delta H^\circ = (105 + 58) - (85 + 103) = -25$ kcal mol$^{-1}$, exothermic. (D) is true.
Answer: (D). | ||||
| Q.15 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | ComprehensionMedium | |
A medium comprehension question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter $\mathrm{Fe^{2+}}$ with $\mathrm{[Fe(CN)_6]^{4-}}$ (both iron(II)) gives no Fe(II)–Fe(III) charge transfer, so the precipitate is white: potassium iron(II) hexacyanoferrate(II),
$\mathrm{K_4[Fe(CN)_6] + FeSO_4 \to K_2Fe[Fe(CN)_6]\downarrow + K_2SO_4}$.
In air the outer Fe(II) is oxidised to Fe(III) and mixed-valence Prussian blue forms, which is why it turns blue.
$\mathrm{Fe_4[Fe(CN)_6]_3}$ and $\mathrm{KFe[Fe(CN)_6]}$ are blue Fe(III)/Fe(II) compounds; $\mathrm{Fe[Fe(CN)_6]}$ (Berlin green) is Fe(III)/Fe(III).
Answer: (C). | ||||
| Q.16 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | ComprehensionEasy | |
A easy comprehension question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Brown-ring test for nitrate: $\mathrm{Fe^{2+}}$ reduces $\mathrm{NO_3^-}$ to NO in the acid layer,
$\mathrm{NO_3^- + 3Fe^{2+} + 4H^+ \to NO + 3Fe^{3+} + 2H_2O}$,
and NO replaces one water of the aqua complex of the remaining iron(II):
$\mathrm{[Fe(H_2O)_6]^{2+} + NO \to [Fe(NO)(H_2O)_5]^{2+} + H_2O}$, the brown ring at the junction of the layers.
Answer: (D). | ||||
| Q.17 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics For one mole of ideal gas, $U = C_{V,m}T = \frac52RT$, so $\frac{U}{R} = \frac52T$: state 1 ($\frac{U}{R} = 2250$) is at 900 K, as stated; state 2 ($\frac{U}{R} = 450$) is at 180 K.
Process I is vertical on the U–S plot: entropy constant, a reversible adiabatic. Work done by the gas: $W_I = -\Delta U = R(2250 - 450) = 1800R$.
Process II is horizontal: U, hence T, constant at 180 K, with entropy rising: a reversible isothermal expansion. $W_{II} = RT\ln\frac{V_3}{V_2} = 180R\ln\frac{V_3}{V_2}$.
Equal work: $180R\ln\frac{V_3}{V_2} = 1800R$, so $\ln\frac{V_3}{V_2} = 10$.
Answer: 10. | ||||
| Q.18 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Photon momentum: $p = \frac{h}{\lambda} = \frac{6.6\times10^{-34}}{330\times10^{-9}} = 2\times10^{-27}$ kg m s$^{-1}$.
Mass of one He atom: $\frac{4\times10^{-3}}{6\times10^{23}} = \frac{2}{3}\times10^{-26}$ kg.
The atom takes up the photon's momentum: $\Delta v = \frac{2\times10^{-27}}{\frac23\times10^{-26}} = 0.3$ m s$^{-1}$ = 30 cm s$^{-1}$.
Answer: 30. | ||||
| Q.19 | 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 Chlorine dioxide reacts with ozone to give dichlorine hexoxide:
$\mathrm{2ClO_2 + 2O_3 \to Cl_2O_6 + 2O_2}$.
In $\mathrm{Cl_2O_6}$ (in the solid, $\mathrm{[ClO_2]^+[ClO_4]^-}$, with Cl at +5 and +7), the average oxidation state of Cl is $\frac{2\times6}{2} = +6$.
Answer: 6. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Multiple correctEasy | |
A easy multiple correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $n_1$: each of $i, j, k$ has 10 choices: $10^3 = 1000$. (A) true.
$n_2$: $j + 2 \leq 10$ forces $j \leq 8$, and $1 \leq i < j + 2$ means $i \in \{1, \dots, j + 1\}$: $j + 1$ choices. $n_2 = \sum_{j=1}^{8}(j + 1) = 2 + 3 + \dots + 9 = 44$. (B) true.
$n_3$: an increasing quadruple is just a choice of 4 numbers: $\binom{10}{4} = 210$, not 220. (C) false.
$n_4$: ordered quadruples of distinct numbers: $10\cdot9\cdot8\cdot7 = 5040$, and $\frac{5040}{12} = 420$. (D) true.
Answer: (A), (B), (D). | ||||
| Q.2 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Multiple correctMedium | |
A medium multiple correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions (A) $\cos P = \frac{q^2 + r^2 - p^2}{2qr} \geq \frac{2qr - p^2}{2qr} = 1 - \frac{p^2}{2qr}$, since $q^2 + r^2 \geq 2qr$. True.
(B) Multiply by $p + q > 0$: the claim is $(p + q)\cos R \geq (q - r)\cos P + (p - r)\cos Q$, i.e. $(p + q)\cos R + r(\cos P + \cos Q) \geq q\cos P + p\cos Q$.
Projection formulae: $q\cos P + p\cos Q = r$, $p\cos R + r\cos P = q$, $q\cos R + r\cos Q = p$. So the left side is $p + q$ and the right side is $r$, and $p + q > r$ (triangle inequality). True.
(C) By the sine rule $\frac{\sqrt{\sin Q\sin R}}{\sin P} = \frac{\sqrt{qr}}{p}$, so the claim is $q + r < 2\sqrt{qr}$, which contradicts AM ≥ GM. False.
(D) $\cos Q > \frac{p}{r} \Leftrightarrow p^2 + r^2 - q^2 > 2p^2 \Leftrightarrow r^2 - q^2 > p^2$, and $\cos R > \frac{p}{q} \Leftrightarrow q^2 - r^2 > p^2$. Adding the two would give $0 > 2p^2$, so they can never both hold. False.
Answer: (A), (B). | ||||
| Q.3 | 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 Let $F(x) = \int_0^x f(t)\,dt$: $F(0) = 0$, $F\left(\frac{\pi}{3}\right) = 0$, $F' = f$.
(A) $g(x) = F(x) - \sin3x$: $g(0) = 0$, $g\left(\frac{\pi}{3}\right) = 0 - \sin\pi = 0$. By Rolle's theorem $g'(c) = f(c) - 3\cos3c = 0$ for some $c \in \left(0, \frac{\pi}{3}\right)$. True.
(B) $h(x) = F(x) + \cos3x + \frac{6x}{\pi}$: $h(0) = 1$, $h\left(\frac{\pi}{3}\right) = 0 + \cos\pi + 2 = 1$. Rolle: $h'(c) = f(c) - 3\sin3c + \frac{6}{\pi} = 0$, i.e. $f(c) - 3\sin3c = -\frac{6}{\pi}$. True.
(C) As $x \to 0$, $F(x) \approx f(0)x = x$ (by L'Hospital, $\frac{F(x)}{x} \to f(0) = 1$) and $1 - e^{x^2} \approx -x^2$. So $\frac{xF(x)}{1 - e^{x^2}} = \frac{F(x)}{x}\cdot\frac{x^2}{1 - e^{x^2}} \to 1\cdot(-1) = -1$. True.
(D) $\frac{\sin x\,F(x)}{x^2} = \frac{\sin x}{x}\cdot\frac{F(x)}{x} \to 1\cdot1 = 1$, not $-1$. False.
Answer: (A), (B), (C). | ||||
| Q.4 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Multiple correctMedium | |
A medium multiple correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations A function is in $S$ if, for some $\alpha, \beta$, $f' + \alpha f = xe^{\beta x}$ exactly (coefficient 1) and $f(1) = 1$. Each option is a particular part plus a multiple of $e^{-x}$, which suggests $\alpha = 1$ (the multiple of $e^{-x}$ then drops out of $f' + f$).
(A) $f = \left(\frac{x^2}{2} + e - \frac12\right)e^{-x}$: $f' + f = xe^{-x}$, so $\alpha = 1$, $\beta = -1$. $f(1) = \left(\frac12 + e - \frac12\right)e^{-1} = 1$. In $S$.
(B) $f = \left(-\frac{x^2}{2} + e + \frac12\right)e^{-x}$: $f' + f = -xe^{-x}$, coefficient $-1$; for any other $\alpha$, $f' + \alpha f = -xe^{-x} + (\alpha - 1)f$ keeps an $x^2e^{-x}$ term. Not in $S$.
(C) $f = \frac{e^x}{2}\left(x - \frac12\right) + \left(e - \frac{e^2}{4}\right)e^{-x}$: for the first part $P' = \frac{e^x}{2}\left(x + \frac12\right)$, so $P' + P = xe^x$; with $\alpha = 1$, $\beta = 1$, $f' + f = xe^x$. $f(1) = \frac{e}{4} + 1 - \frac{e}{4} = 1$. In $S$.
(D) $f = \frac{e^x}{2}\left(\frac12 - x\right) + \left(e + \frac{e^2}{4}\right)e^{-x}$: $f' + f = -xe^x$, wrong sign, and (as in B) no other $\alpha$ works; also $f(1) = -\frac{e}{4} + 1 + \frac{e}{4} = 1$ but the equation fails. Not in $S$.
Answer: (A), (C). | ||||
| Q.5 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Multiple correctMedium | |
A medium multiple correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\overrightarrow{OA}\times\overrightarrow{OB} = \begin{vmatrix}\hat i&\hat j&\hat k\\2&2&1\\1&-2&2\end{vmatrix} = 6\hat i - 3\hat j - 6\hat k$, of length 9.
$\overrightarrow{OB}\times\overrightarrow{OC} = -\frac{\lambda}{2}\overrightarrow{OB}\times\overrightarrow{OA} = \frac{\lambda}{2}\overrightarrow{OA}\times\overrightarrow{OB}$, length $\frac{9\lambda}{2} = \frac92$, so $\lambda = 1$ and $\overrightarrow{OC} = \frac12(-\hat i - 4\hat j + \hat k)$.
(A) Projection of $\overrightarrow{OC}$ on $\overrightarrow{OA}$: $\frac{\overrightarrow{OC}\cdot\overrightarrow{OA}}{|\overrightarrow{OA}|} = \frac{\frac12(-2 - 8 + 1)}{3} = -\frac32$. True.
(B) Area $OAB = \frac12|\overrightarrow{OA}\times\overrightarrow{OB}| = \frac92$. True.
(C) $\overrightarrow{AB} = -\hat i - 4\hat j + \hat k$, $\overrightarrow{AC} = -\frac52\hat i - 4\hat j - \frac12\hat k$. $\overrightarrow{AB}\times\overrightarrow{AC} = 6\hat i - 3\hat j - 6\hat k$, length 9, area $\frac92$. True.
(D) Diagonals: $\overrightarrow{OA} + \overrightarrow{OC} = \frac32\hat i + \frac32\hat k$ and $\overrightarrow{OA} - \overrightarrow{OC} = \frac52\hat i + 4\hat j + \frac12\hat k$. Dot product $\frac{15}{4} + \frac34 = \frac92$; lengths $\frac{3}{\sqrt2}$ and $\sqrt{\frac{90}{4}} = \frac{3\sqrt{10}}{2}$. $\cos\theta = \frac{9/2}{\frac{3}{\sqrt2}\cdot\frac{3\sqrt{10}}{2}} = \frac{1}{\sqrt5}$, not $\frac12$. False.
Answer: (A), (B), (C). | ||||
| Q.6 | Conic SectionsParabola | Class 11: Conic Sections | Multiple correctMedium | |
A medium multiple correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = 8x$ has $a = 2$: focus $F = (2, 0)$, directrix $x = -2$. So $P = (-2, 4)$ lies on the directrix.
Chord of contact $QQ'$ from $P$: $yy_1 = 4(x + x_1)$ gives $4y = 4(x - 2)$, i.e. $y = x - 2$. It passes through $F(2, 0)$. (D) true.
Tangents from a point of the directrix are perpendicular: the slopes $m$ of tangents $y = mx + \frac{2}{m}$ through $(-2, 4)$ satisfy $2m^2 + 4m - 2 = 0$, product of roots $-1$. So $\angle QPQ' = 90^\circ$. (B) true.
Slope of $PF$ $= \frac{0 - 4}{2 + 2} = -1$ and slope of $QQ'$ is 1: $PF \perp QQ'$, so $\angle PFQ = 90^\circ$ and triangle $PFQ$ is right-angled. (A) true.
$PF = \sqrt{4^2 + 4^2} = 4\sqrt2$, not $5\sqrt2$. (C) false.
Answer: (A), (B), (D). | ||||
| Q.7 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A circle with centre $(h, 0)$ and radius $r$ lies in $R$ when it stays right of the $y$-axis, $r \leq h$, and inside the parabola $y^2 = 4 - x$.
Squared distance from $(h, 0)$ to a point $(x, 4 - x)$ of the parabola: $D(x) = (x - h)^2 + 4 - x$. $D'(x) = 2(x - h) - 1 = 0$ at $x = h + \frac12$, giving the least distance $D = \frac14 + 4 - h - \frac12 = \frac{15}{4} - h$ (valid while $h + \frac12 \leq 4$).
So $r^2 \leq \frac{15}{4} - h$ and $r \leq h$. The first bound falls and the second rises with $h$; the largest $r$ is where they meet: $h^2 = \frac{15}{4} - h$, $4h^2 + 4h - 15 = 0$, $(2h - 3)(2h + 5) = 0$, $h = \frac32$, $r = \frac32$.
The circle touches the parabola where $x = h + \frac12 = 2$ (then $y^2 = 2$, $y = \pm\sqrt2$).
Radius $= 1.50$. | ||||
| Q.8 | Conic SectionsParabola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A circle with centre $(h, 0)$ and radius $r$ lies in $R$ when it stays right of the $y$-axis, $r \leq h$, and inside the parabola $y^2 = 4 - x$.
Squared distance from $(h, 0)$ to a point $(x, 4 - x)$ of the parabola: $D(x) = (x - h)^2 + 4 - x$. $D'(x) = 2(x - h) - 1 = 0$ at $x = h + \frac12$, giving the least distance $D = \frac14 + 4 - h - \frac12 = \frac{15}{4} - h$ (valid while $h + \frac12 \leq 4$).
So $r^2 \leq \frac{15}{4} - h$ and $r \leq h$. The first bound falls and the second rises with $h$; the largest $r$ is where they meet: $h^2 = \frac{15}{4} - h$, $4h^2 + 4h - 15 = 0$, $(2h - 3)(2h + 5) = 0$, $h = \frac32$, $r = \frac32$.
The circle touches the parabola where $x = h + \frac12 = 2$ (then $y^2 = 2$, $y = \pm\sqrt2$).
$\alpha = 2.00$. | ||||
| Q.9 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f_1'(x) = \prod_{j=1}^{21}(x - j)^j$. The sign of $f_1'$ changes only at $x = j$ with $j$ odd (an even power does not change sign): at $x = 1, 3, 5, \dots, 21$, eleven points, and these are the only extrema.
For $x > 21$ every factor is positive, so $f_1' > 0$. Going left past 21 the sign flips to negative, past 19 back to positive, and so on, alternating.
So at 21 $f_1'$ goes from $-$ to $+$: a minimum; at 19 from $+$ to $-$: a maximum; alternating down to 1. (Check at $x = 1$: for $0 < x < 1$ all 21 factors are negative and the total power is $1 + 2 + \dots + 21 = 231$, odd, so $f_1' < 0$; after 1 it is positive: a minimum.)
Minima at $1, 5, 9, 13, 17, 21$: $m_1 = 6$. Maxima at $3, 7, 11, 15, 19$: $n_1 = 5$.
$2m_1 + 3n_1 + m_1n_1 = 12 + 15 + 30 = 57$.
Answer: 57.00. | ||||
| Q.10 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f_2'(x) = 4900(x - 1)^{49} - 29400(x - 1)^{48} = 4900(x - 1)^{48}(x - 7)$.
At $x = 1$ the factor $(x - 1)^{48}$ does not change sign (and $x - 7 < 0$ on both sides), so $x = 1$ is not an extremum.
At $x = 7$, $f_2'$ goes from negative to positive: a local minimum.
So $m_2 = 1$, $n_2 = 0$ and $6m_2 + 4n_2 + 8m_2n_2 = 6$.
Answer: 6.00. | ||||
| Q.11 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $S_1 = \int_{\pi/8}^{3\pi/8}\sin^2x\,dx = \int_{\pi/8}^{3\pi/8}\frac{1 - \cos2x}{2}\,dx = \frac12\cdot\frac{\pi}{4} - \frac14\left[\sin2x\right]_{\pi/8}^{3\pi/8} = \frac{\pi}{8} - \frac14\left(\frac{1}{\sqrt2} - \frac{1}{\sqrt2}\right) = \frac{\pi}{8}$.
(Alternatively, $x \to \frac{\pi}{2} - x$ turns $\sin^2$ into $\cos^2$ on this interval, so $2S_1 = \frac{\pi}{4}$.)
$\frac{16S_1}{\pi} = 2$.
Answer: 2.00. | ||||
| Q.12 | 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 Substitute $x \to \frac{\pi}{2} - x$ (the interval maps to itself): $\sin^2x \to \cos^2x$ and $|4x - \pi| \to |\pi - 4x|$, unchanged. So $S_2 = \int\cos^2x\,|4x - \pi|\,dx$ too, and adding,
$2S_2 = \int_{\pi/8}^{3\pi/8}|4x - \pi|\,dx = 2\int_{\pi/8}^{\pi/4}(\pi - 4x)\,dx = 2\left[\pi x - 2x^2\right]_{\pi/8}^{\pi/4} = 2\left(\frac{\pi^2}{8} - \frac{3\pi^2}{32}\right) = \frac{\pi^2}{16}$.
$S_2 = \frac{\pi^2}{32}$, so $\frac{48S_2}{\pi^2} = \frac{48}{32} = 1.5$.
Answer: 1.50. | ||||
| Q.13 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | ComprehensionHard | |
A hard comprehension question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_n = 2\left(1 - \frac{1}{2^n}\right) = 2 - \frac{1}{2^{n-1}}$, so $S_{n-1} = 2 - \frac{1}{2^{n-2}}$.
$C_n$ is inside $M$ when its farthest point from the origin, at distance $S_{n-1} + a_n = 2 - \frac{1}{2^{n-2}} + \frac{1}{2^{n-1}} = 2 - \frac{1}{2^{n-1}}$, is at most $r = \frac{1025}{513} = 2 - \frac{1}{513}$.
$\frac{1}{2^{n-1}} \geq \frac{1}{513} \Leftrightarrow 2^{n-1} \leq 513 \Leftrightarrow n \leq 10$. So $k = 10$.
Neighbours $C_n$, $C_{n+1}$: centres $a_n$ apart, radii $a_n$ and $\frac{a_n}{2}$; since $\frac{a_n}{2} < a_n < \frac{3a_n}{2}$ the circles cross.
$C_n$ and $C_{n+2}$: centres $a_n + a_{n+1} = \frac32a_n$ apart, radii sum $a_n + \frac{a_n}{4} = \frac54a_n < \frac32a_n$: they do not meet (and the same holds for any larger gap).
So no two chosen circles may be neighbours: from 10 in a row, at most 5 (e.g. $C_1, C_3, C_5, C_7, C_9$). $l = 5$.
Check: $k + 2l = 20$, $2k + l = 25$, $2k + 3l = 35$, $3k + 2l = 40$. Only (D) holds.
Answer: (D). | ||||
| Q.14 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | ComprehensionHard | |
A hard comprehension question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $r = \sqrt2\left(2 - \frac{1}{2^{198}}\right)$. The centre of $D_n$ is at distance $\sqrt2S_{n-1} = \sqrt2\left(2 - \frac{1}{2^{n-2}}\right)$ from the origin, so $D_n$ is inside $M$ when
$\sqrt2\left(2 - \frac{1}{2^{n-2}}\right) + \frac{1}{2^{n-1}} \leq \sqrt2\left(2 - \frac{1}{2^{198}}\right)$,
i.e. $\frac{\sqrt2}{2^{198}} \leq \frac{2\sqrt2 - 1}{2^{n-1}}$, i.e. $2^{n-199} \leq \frac{2\sqrt2 - 1}{\sqrt2} = 2 - \frac{1}{\sqrt2} \approx 1.29$.
This holds for $n - 199 \leq 0$ and fails for $n = 200$ (it would need $2 \leq 1.29$). So $n = 1, 2, \dots, 199$: 199 circles.
Answer: (B). | ||||
| Q.15 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | ComprehensionHard | |
A hard comprehension question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Rewrite $f$ and $g$. The integrand of $f$ is even: $f(x) = 2\int_0^x(t - t^2)e^{-t^2}\,dt$. In $g$ put $t = u^2$: $g(x) = \int_0^x u\,e^{-u^2}\,2u\,du = 2\int_0^x u^2e^{-u^2}\,du$.
Also $\psi_1' = 1 - e^{-x} > 0$ for $x > 0$, and $\psi_2' = 2x - 2 + 2e^{-x} = 2(\psi_1(x) - 1)$, with $\psi_1(0) = 1$, $\psi_2(0) = 0$.
(A) $f + g = 2\int_0^x te^{-t^2}\,dt = 1 - e^{-x^2}$. At $x = \sqrt{\ln3}$: $1 - \frac13 = \frac23$, not $\frac13$. False.
(B) $\psi_1(x) = 1 + \alpha x$ gives $\alpha = 1 - \frac{1 - e^{-x}}{x}$. For $x > 0$, $0 < 1 - e^{-x} < x$, so $0 < \alpha < 1$: $\alpha$ is never in $(1, x)$. False.
(C) Mean value theorem for $\psi_2$ on $[0, x]$: $\psi_2(x) - \psi_2(0) = x\,\psi_2'(\beta)$ for some $\beta \in (0, x)$, i.e. $\psi_2(x) = 2x(\psi_1(\beta) - 1)$. True.
(D) $f'(x) = 2(x - x^2)e^{-x^2} < 0$ for $x > 1$, so $f$ decreases on $\left(1, \frac32\right]$. False.
Answer: (C). | ||||
| Q.16 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | ComprehensionHard | |
A hard comprehension question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Rewrite $f$ and $g$. The integrand of $f$ is even: $f(x) = 2\int_0^x(t - t^2)e^{-t^2}\,dt$. In $g$ put $t = u^2$: $g(x) = \int_0^x u\,e^{-u^2}\,2u\,du = 2\int_0^x u^2e^{-u^2}\,du$.
Also $\psi_1' = 1 - e^{-x} > 0$ for $x > 0$, and $\psi_2' = 2x - 2 + 2e^{-x} = 2(\psi_1(x) - 1)$, with $\psi_1(0) = 1$, $\psi_2(0) = 0$.
(A) $\psi_1$ is increasing from $\psi_1(0) = 1$, so $\psi_1(x) > 1$ for $x > 0$. False.
(B) Hence $\psi_2' = 2(\psi_1 - 1) > 0$ and $\psi_2$ increases from $\psi_2(0) = 0$: $\psi_2(x) > 0$ for $x > 0$. False.
(C) $f(x) = 2\int_0^x te^{-t^2}dt - 2\int_0^x t^2e^{-t^2}dt = 1 - e^{-x^2} - g(x)$. Since $e^{-u} > 1 - u$ for $u > 0$, $g(x) = 2\int_0^x t^2e^{-t^2}dt > 2\int_0^x(t^2 - t^4)dt = \frac23x^3 - \frac25x^5$. So $f(x) < 1 - e^{-x^2} - \frac23x^3 + \frac25x^5$ for $x > 0$: the stated $\geq$ is false.
(D) Since $e^{-u} \leq 1 - u + \frac{u^2}{2}$ for $u \geq 0$, $g(x) \leq 2\int_0^x\left(t^2 - t^4 + \frac{t^6}{2}\right)dt = \frac23x^3 - \frac25x^5 + \frac17x^7$. True.
Answer: (D). | ||||
| Q.17 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Multiples of 3 up to 2000: $\lfloor2000/3\rfloor = 666$. Multiples of 7: $\lfloor2000/7\rfloor = 285$. Multiples of 21 (both): $\lfloor2000/21\rfloor = 95$.
Multiples of 3 or 7: $666 + 285 - 95 = 856$.
$p = \frac{856}{2000}$, so $500p = \frac{856}{4} = 214$.
Answer: 214. | ||||
| Q.18 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $M(P, Q) - M(P, Q') = \frac{P + Q}{2} - \frac{P + Q'}{2} = \frac{Q - Q'}{2}$, so the distance is half the chord $QQ'$ (midpoint theorem).
The longest chord of the ellipse is the major axis, $2\times4 = 8$, taken with $Q, Q' = (\pm4, 0)$ and $P$ any third point.
Maximum distance $= \frac82 = 4$.
Answer: 4. | ||||
| Q.19 | 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 $\frac{10x}{x + 1}$ increases from 0 to $\frac{100}{11} \approx 9.09$ on $[0, 10]$, so $\sqrt{\frac{10x}{x+1}}$ runs from 0 to about 3.02 and the integrand takes the values 0, 1, 2, 3.
$\sqrt{\frac{10x}{x+1}} \geq 1 \Leftrightarrow 10x \geq x + 1 \Leftrightarrow x \geq \frac19$.
$\geq 2 \Leftrightarrow 10x \geq 4x + 4 \Leftrightarrow x \geq \frac23$.
$\geq 3 \Leftrightarrow 10x \geq 9x + 9 \Leftrightarrow x \geq 9$.
$I = 0\cdot\frac19 + 1\cdot\left(\frac23 - \frac19\right) + 2\left(9 - \frac23\right) + 3(10 - 9) = \frac59 + \frac{50}{3} + 3 = \frac{5 + 150 + 27}{9} = \frac{182}{9}$.
$9I = 182$.
Answer: 182. | ||||
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