JEE (Advanced) · 2026
JEE Advanced 2026, 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.
Mathematics
34 Q
- Trigonometry4
- Conic Sections4
- Matrices and Determinants3
10 easy · 20 medium · 4 hard
Physics
34 Q
- Motion of System of Particles and Rigid Body5
- Optics5
- Thermodynamics3
5 easy · 23 medium · 6 hard
Chemistry
34 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Chemical Bonding and Molecular Structure3
- p-Block Elements3
7 easy · 20 medium · 7 hard
02
Chapters
Every chapter asked, and the NCERT chapters to read for it.
Class 11: Systems of Particles and Rotational Motion
Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics
Class 11: Thermodynamics
Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter
Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter
Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance
Class 12: Electromagnetic Induction; Class 12: Alternating Current
Class 12: Atoms; Class 12: Nuclei
Class 11: Units and Measurement
Class 11: Motion in a Straight Line; Class 11: Motion in a Plane
Class 12: Electromagnetic Waves
Class 11: Oscillations; Class 11: Waves
Class 12: Current Electricity
Class 11: Gravitation
No matching NCERT chapter
Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions
Class 11: Conic Sections
Class 12: Matrices; Class 12: Determinants
Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability
Class 12: Integrals; Class 12: Application of Integrals
Class 11: Probability; Class 12: Probability
Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry
Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions
Class 11: Complex Numbers and Quadratic Equations
Class 12: Vector Algebra
Class 12: Differential Equations
Class 12: Application of Derivatives
Class 11: Conic Sections
Class 11: Permutations and Combinations
Class 11: Binomial Theorem
Class 11: Statistics
Class 11: Organic Chemistry – Some Basic Principles and Techniques
Class 11: Chemical Bonding and Molecular Structure
No matching NCERT chapter
Class 12: Aldehydes, Ketones and Carboxylic Acids
Class 12: Solutions
Class 11: Thermodynamics
Class 12: Chemical Kinetics
Class 11: Structure of Atom
Class 11: Hydrocarbons
Class 12: Coordination Compounds
Class 12: Biomolecules
Class 11: Some Basic Concepts of Chemistry
Class 12: Electrochemistry
Class 11: Classification of Elements and Periodicity in Properties
Class 12: Amines
No matching NCERT chapter
No matching NCERT chapter
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 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Differentiate: $f'(x)=\frac{\log_e x}{2\sqrt x}+\frac{1}{\sqrt x}-1=\frac{\log_e x+2-2\sqrt x}{2\sqrt x}$.
Put $t=\sqrt x$, so that $\log_e x=2\log_e t$. Then $f'(x)=\frac{\log_e t+1-t}{t}$.
For every $t>0$ we have $\log_e t\le t-1$, with equality only at $t=1$. So $f'(x)\le 0$ everywhere and $f'(x)=0$ only at $x=1$.
A function whose derivative keeps one sign, touching zero at a single point, is strictly decreasing, so $f$ has no local maximum and no local minimum. This makes (D) true and (B), (C) false.
(A) is false: $f'(x)=\frac{\log_e t+1-t}{t}\to-\infty$ as $x\to0^+$ and $f'(1)=0$, so $f'$ rises on $(0,1)$ rather than falling.
Answer: (D). | ||||
| Q.2 | CirclesRadius and centre | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Parabola: $\frac{dy}{dx}=2x=4$ gives $x=2$, so $P=(2,4)$.
Circle: $\frac{dy}{dx}=-\frac{x}{y}=-1$ gives $x=y$, and $2x^2=2$ in the first quadrant gives $Q=(1,1)$.
Ellipse: $2x+8y\frac{dy}{dx}=0$ gives $\frac{dy}{dx}=-\frac{x}{4y}=-\frac12$, so $x=2y$; then $8y^2=8$ gives $R=(2,1)$.
$PR$ is vertical and $QR$ is horizontal, so the triangle has a right angle at $R$ and $PQ$ is a diameter of its circumcircle.
$PQ=\sqrt{1^2+3^2}=\sqrt{10}$, so the radius is $\frac{\sqrt{10}}{2}=\sqrt{\frac{5}{2}}$.
Answer: (C). | ||||
| Q.3 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Elementary row operations never change whether a determinant is zero, and the identity has determinant $1$. So only an invertible matrix can be reached, and every invertible matrix can be reached.
(A) has equal rows, so its determinant is $0$.
(B): $1(3-8)-1(2-4)+1(4-3)=-5+2+1=-2\neq0$.
(C): $1(24-20)-1(16-8)+1(10-6)=4-8+4=0$.
(D): $1(3-4)-1(-3-0)+1(-2-0)=-1+3-2=0$.
Only (B) is invertible.
Answer: (B). | ||||
| Q.4 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions The principal range of $\cot^{-1}$ is $(0,\pi)$ and $\cot$ has period $\pi$. Since $4\pi-11\approx1.57$ lies in $(0,\pi)$, $\cot^{-1}(\cot(-11))=4\pi-11$.
$\cos^{-1}\frac{1}{\sqrt2}=\frac{\pi}{4}$, so $10\sin\frac{\pi}{2}=10$.
With $\tan\theta=2$, $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{4}{5}$, so the last term is $8$.
Total: $4\pi-11+10+8=4\pi+7$.
Answer: (C). | ||||
| Q.5 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Multiple correctEasy | |
A easy multiple correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are $35$ balls: $15$ from Box I, $20$ from Box II, $14$ red, $21$ green.
(A) True: $P(E_1)P(F_1)=\frac{15}{35}\cdot\frac{14}{35}=\frac{6}{35}$, and this equals $P(E_1\cap F_1)=\frac{6}{35}$.
(B) False: $P(E_2)P(F_2)=\frac{20}{35}\cdot\frac{21}{35}=\frac{12}{35}=P(E_2\cap F_2)$, so these two are independent too.
(C) True: $P(F_1\mid E_1)=\frac{6}{15}=\frac25$ and $P(F_1\mid E_2)=\frac{8}{20}=\frac25$.
(D) False: $P(F_2\mid E_2)=\frac{12}{20}=\frac35$, which is more than $\frac25$.
Answer: (A), (C). | ||||
| Q.6 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctMedium | |
A medium multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The normal of $P$ is perpendicular to the line's direction $(2,3,1)$ and to the other plane's normal $(1,2,3)$. Their cross product is $(9-2,\ 1-6,\ 4-3)=(7,-5,1)$.
$P$ contains $(1,3,-2)$, so $7x-5y+z=7-15-2=-10$. (A) is true.
$P_1$ is $7x-5y+z=c$ with $c=28-10+2=20$. The distance between the planes is $\frac{|20+10|}{\sqrt{49+25+1}}=\frac{30}{5\sqrt3}=2\sqrt3$, not $30$. (B) is false.
Distance of $P$ from the origin: $\frac{10}{5\sqrt3}=\frac{2}{\sqrt3}$. (C) is false.
Angle with $2x+2y+z=3$: $\cos\theta=\frac{|14-10+1|}{5\sqrt3\cdot3}=\frac{1}{3\sqrt3}$. (D) is true.
Answer: (A), (D). | ||||
| Q.7 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Multiple correctMedium | |
A medium multiple correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Note $g(0)=0$, so the difference quotient at $0$ is $\frac{g(x)-g(0)}{x}=f(x)$ for $x\neq0$. Hence $g$ is differentiable at $0$ exactly when $\lim_{x\to0}f(x)$ exists.
(A) False: take $f(x)=\frac1x$ for $x\neq0$ and $f(0)=0$. Then $g(x)=1$ for $x\neq0$ while $g(0)=0$.
(B) True: if $f$ is continuous at $0$, the limit of $f$ exists (it is $f(0)$), so $g'(0)$ exists.
(C) False: take $f(x)=0$ for $x\neq0$ and $f(0)=1$. Then $g$ is identically $0$, which is differentiable, but $f$ is not continuous at $0$.
(D) True: this is the statement in the first line.
Answer: (B), (D). | ||||
| Q.8 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Multiple correctHard | |
A hard multiple correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let $A=M-I=\begin{bmatrix}1&-1\\1&-1\end{bmatrix}$. Then $A^2=0$, so $M^k=(I+A)^k=I+kA=\begin{bmatrix}1+k&-k\\k&1-k\end{bmatrix}$.
(A) True: we need $N=\begin{bmatrix}u&v\end{bmatrix}$ (columns $u,v$) with $Mu=u$ and $Mv=u+v$, that is $Au=0$ and $Av=u$. Taking $v=\begin{bmatrix}1\\0\end{bmatrix}$ gives $u=Av=\begin{bmatrix}1\\1\end{bmatrix}$ and $Au=0$. This $N=\begin{bmatrix}1&1\\1&0\end{bmatrix}$ is invertible.
(B) False: $a=\sum_{k=1}^{26}(1+k)=26+351=377$.
(C) True: $\det M^{26}=(\det M)^{26}=1$, so $M^{26}$ has an inverse with integer entries, and the system has exactly one solution, which is integral.
(D) True: $a=377$, $b=-351$, $c=351$, $d=26-351=-325$. The determinant of the system is $(377+t)(t-325)+351^2=t^2+52t+676=(t+26)^2$, which is not zero for $t>0$.
Answer: (A), (C), (D). | ||||
| Q.9 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions An equivalence relation is a partition of $S$. A class of size $k$ contributes $k^2$ ordered pairs, so we need class sizes adding to $10$ whose squares add to $42$.
Largest class $6$: the remaining $4$ elements must give squares summing to $6$, which is only $2,1,1$.
Largest class $5$: the remaining $5$ must give $17$, which is only $4,1$.
Largest class $4$ or less: the remaining $6$ must give $26$ using parts at most $4$, and the most available is $16+4=20$, so this fails. Largest class $7$ or more gives more than $42$ or leaves too little: $49>42$.
So the types are $\{6,2,1,1\}$ and $\{5,4,1\}$.
Count: $\binom{10}{6}\binom{4}{2}=210\cdot6=1260$ and $\frac{10!}{5!\,4!\,1!}=1260$.
Total $=2520$. | ||||
| Q.10 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability On this interval $x\sin x$ is even, is $0$ at $x=0$, and increases on $\left(0,\frac{\pi}{2}\right)$ up towards $\frac{\pi}{2}<2$.
So $[x\sin x]$ is $0$ near $0$ and jumps to $1$ only where $x\sin x=1$, at two points $x=\pm x_0$ with $x_0\approx1.11$. The first term is continuous everywhere, so $\alpha=2$.
These two points are also non-differentiable. For the first term $h(x)=(|x|+|x-1|)\sin x$:
At $x=0$: $h=(1-2x)\sin x$ on the left and $h=\sin x$ on the right. Both have derivative $1$ at $0$, so $h$ is differentiable there.
At $x=1$: $h=\sin x$ on the left and $h=(2x-1)\sin x$ on the right. The derivatives are $\cos1$ and $\cos1+2\sin1$, which differ.
So $\beta=3$ (the points $\pm x_0$ and $1$), and $\alpha+\beta=5$. | ||||
| Q.11 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Let the persons get $r_1,r_2,r_3,r_4$ red pens. Each $r_i$ lies between $0$ and $6$, and $r_1+r_2+r_3+r_4=10$. Each person's blue count is then fixed as $6-r_i$, and these add to $14$ automatically.
Without the upper limit: $\binom{10+3}{3}=286$.
Remove the cases with some $r_i\ge7$. Only one person can have that many, and subtracting $7$ leaves a sum of $3$: $4\cdot\binom{3+3}{3}=80$.
$286-80=206$. | ||||
| Q.12 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalHard | |
A hard numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Use $\cos3\phi=\cos\phi\,(4\cos^2\phi-3)=\cos\phi\,(2\cos2\phi-1)$. With $\phi=\frac{\theta}{2}$ this gives $1-2\cos\theta=-\frac{\cos(3\theta/2)}{\cos(\theta/2)}$.
Take $\theta=\frac{3^k\pi}{11}$ for $k=0,1,2,3,4$. Each factor is $-\frac{\cos(3^{k+1}\pi/22)}{\cos(3^k\pi/22)}$, so the product telescopes:
$\alpha=(-1)^5\frac{\cos(243\pi/22)}{\cos(\pi/22)}$.
Now $\frac{243\pi}{22}=11\pi+\frac{\pi}{22}$, so $\cos\frac{243\pi}{22}=-\cos\frac{\pi}{22}$. Hence $\alpha=1$.
Numerical check: the factors are about $-0.919,\ -0.310,\ 2.683,\ 0.715,\ 1.831$, whose product is $1.000$.
$5-\alpha^2=4$. | ||||
| Q.13 | Quadratic EquationsForming equations from given roots | Class 11: Complex Numbers and Quadratic Equations | Matching listMedium | |
A medium matching list question on forming equations from given roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations (P) $\alpha^3=1$ and $\alpha+1=-\alpha^2$. So $(\alpha+1)^{2026}=\alpha^{4052}=\alpha^2$ (as $4052$ leaves remainder $2$ on division by $3$), and its reciprocal is $\alpha$. The new roots are $\alpha,\beta$ again: $x^2+x+1=0$, entry (1).
(Q) $(\alpha+1)^{2027}=-\alpha^{4054}=-\alpha$, with reciprocal $-\alpha^2$. The new roots $-\alpha^2,-\beta^2$ have sum $-(\alpha^2+\beta^2)=1$ and product $1$: $x^2-x+1=0$, entry (2).
(R) $\gamma^2=\gamma-1$ and $\gamma^6=1$. So $(\gamma-1)^{2026}=\gamma^{4052}=\gamma^2$, since $4052$ leaves remainder $2$ on division by $6$. With $\delta=\gamma^{-1}$, the sum is $\gamma^{-2}+\delta^{-2}=\gamma^2+\delta^2=(\gamma+\delta)^2-2\gamma\delta=1-2=-1$, entry (4).
(S) Let $v=\frac{1}{p+1}$ or $\frac{1}{r+1}$. Putting $x=\frac1v-1$ in $x^2+x-1=0$ gives $v^2+v-1=0$. Its roots have sum $-1$ and product $-1$, so the sum of cubes is $(-1)^3-3(-1)(-1)=-4$, entry (5).
P→1, Q→2, R→4, S→5. Answer: (C). | ||||
| Q.14 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Matching listMedium | |
A medium matching list question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions (P) With $s=\sin^2x$: $s^3+(1-s)^2=1$ gives $s(s+2)(s-1)=0$, so $s=0$ or $1$. That gives $x=0,\pm\pi$ and $x=\pm\frac{\pi}{2}$: $5$ values.
(Q) With $c=\cos^2x$: $(1-c)+c^3=1$ gives $c(c^2-1)=0$, so $c=0$ or $1$. That gives $x=\pm\frac{\pi}{2}$ and $x=0$: $3$ values.
(R) $\cos^2\frac x2=\frac{1+\cos x}{2}$. The equation becomes $2\cos^2x+\cos x-2=0$, so $\cos x=\frac{-1+\sqrt{17}}{4}\approx0.78$ (the other root is below $-1$). That gives $2$ values in $[-\pi,\pi]$.
(S) $6\sin^2\frac x2=3-3\cos x$, and $\cos3x=4\cos^3x-3\cos x$. The equation reduces to $4\cos^3x=0$, so $\cos x=0$: $x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2}$, $4$ values.
P→5, Q→3, R→2, S→4. Answer: (B). | ||||
| Q.15 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Matching listMedium | |
A medium matching list question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $MM^T=I$ makes $M$ orthogonal, so its rows are orthonormal and so are its columns. The vectors $\vec u,\vec v,\vec w$ are its columns.
First row: $\alpha^2+\frac12+\frac12=1$ gives $\alpha=0$. Second row: $\frac13+\beta^2+\frac13=1$ gives $\beta^2=\frac13$. Rows 1 and 2 orthogonal: $\frac{\beta}{\sqrt2}-\frac{1}{\sqrt6}=0$, so $\beta=\frac{1}{\sqrt3}$.
(P) Column 1: $\frac13+\gamma^2=1$, so $\gamma^2=\frac23$. Column 2: $\frac12+\frac13+\delta^2=1$, so $\delta^2=\frac16$. The sum is $\frac56$, entry (5).
(Q) Dot the equation with $\vec u$: since the columns are orthonormal, $x=\vec u\cdot\hat j=\frac{1}{\sqrt3}$, entry (4).
(R) This is $|\det M|=1$, entry (2).
(S) $\vec v\times\vec w=\pm\vec u$ for an orthonormal set, and $\vec u\times\vec u=\vec0$, so the value is $0$, entry (1).
P→5, Q→4, R→2, S→1. Answer: (A). | ||||
| Q.16 | Conic SectionsStandard forms of conics | Class 11: Conic Sections | Matching listMedium | |
A medium matching list question on standard forms of conics, from Conic Sections. NCERT: Class 11: Conic Sections (P) Radius $=\frac{|3+8-1|}{5}=2$, so the circle is $(x-1)^2+(y-2)^2=4$. The point $(3,2)$ satisfies it, entry (3).
(Q) A tangent to $y^2=8x$ is $y=mx+\frac2m$. It touches the circle when $\frac{|2/m|}{\sqrt{1+m^2}}=\sqrt2$, that is $m^4+m^2-2=0$, so $m^2=1$ and $m=1$. The line $y=x+2$ passes through $(7,9)$, entry (2).
(R) $\frac{x^2}{16}+\frac{y^2}{12}=1$ has $e=\frac12$, focus at $x=2$, and semi-latus rectum $\frac{12}{4}=3$, so $M=(2,3)$. The normal $\frac{16x}{2}-\frac{12y}{3}=16-12$ reduces to $2x-y=1$, which passes through $(1,1)$, entry (1).
(S) $ae=5$ and $\frac ae=\frac{16}{5}$ give $a^2=16$ and $b^2=25-16=9$, so $\frac{x^2}{16}-\frac{y^2}{9}=1$. The point $(8,3\sqrt3)$ gives $4-3=1$, entry (5).
P→3, Q→2, R→1, S→5. Answer: (B). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single 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 A small disk's centre moves on a circle of radius $R+r=51r$. Rolling without slipping on a fixed surface means the centre's speed is $\omega_{\text{spin}}\,r$.
So the two centres move round the large disk at angular rates $\frac{\omega r}{51r}=\frac{\omega}{51}$ and $\frac{2\omega}{51}$, in opposite directions. They separate at the combined rate $\frac{3\omega}{51}$.
When touching, the centres are $2r$ apart, so $(R+r)\Delta\theta=2r$ for small angles and $\Delta\theta=\frac{2}{51}$.
They start with angular gap $\Delta\theta$ and touch again when they have gone round the rest of the circle, stopping $\Delta\theta$ short on the far side. The angle covered between them is $2\pi-2\Delta\theta=2\pi-\frac{4}{51}$.
$\tau=\frac{2\pi-\frac{4}{51}}{3\omega/51}=51\times\left(2\pi-\frac{4}{51}\right)/3\omega$.
Answer: (C). | ||||
| Q.2 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctHard | |
A hard single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Large coil: $L=\mu_0N^2Sd$.
Small coil: $L_2=\mu_0(2N)^2\frac{S}{2}\cdot\frac{d}{2}=\mu_0N^2Sd=L$.
Mutual inductance: the large coil's field $\mu_0NI$ links $2N\cdot\frac d2=Nd$ turns of area $\frac S2$, so $M=\frac{\mu_0N^2Sd}{2}=\frac L2$.
The small coil is shorted with no resistance, so the flux through it stays zero: $L_2I_2+MI_1=0$, giving $I_2=-\frac{M}{L_2}I_1$.
The flux of the large coil is then $LI_1+MI_2=\left(L-\frac{M^2}{L_2}\right)I_1$, so its effective inductance is $L-\frac{L}{4}=\frac{3L}{4}$.
$\omega=\frac{1}{\sqrt{\frac{3L}{4}C}}=\frac{2}{\sqrt{3LC}}$.
Answer: (C). | ||||
| Q.3 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single 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 While pivoting about the corner the cylinder's moment of inertia about the contact line is $\frac32mR^2$, so its kinetic energy is $\frac12\cdot\frac32mR^2\cdot\frac{v^2}{R^2}=\frac34mv^2$.
Let $\theta$ be the angle of the line from the corner to the centre, measured from the vertical. Energy conservation: $\frac34mv^2=\frac34mv_0^2+mgR(1-\cos\theta)$.
Contact is lost when the normal force is zero, so gravity's radial part alone supplies the centripetal force: $mg\cos\theta=\frac{mv^2}{R}$, that is $v^2=gR\cos\theta$.
Substituting $v_0^2=\frac{gR}{3}$: $\frac34\cos\theta=\frac14+1-\cos\theta$, so $\cos\theta=\frac57$.
$v=\sqrt{\frac{5gR}{7}}$.
Answer: (B). | ||||
| Q.4 | 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 Lens maker's formula in a liquid: $\frac1f=\left(\frac{1.5}{n_L}-1\right)\left(\frac{1}{0.2}+\frac{1}{0.2}\right)=10\left(\frac{1.5}{n_L}-1\right)$ per metre.
Taking power as $P=\frac1f$: $P=5$ at $n_L=1$, $P=0$ at $n_L=1.5$, $P=-2.5$ at $n_L=2$, and the curve is not straight. This is plot (A).
Taking power in a medium as $P=\frac{n_L}{f}$: $P=10(1.5-n_L)$, a straight line through $5$, $0$ and $-5$. This is plot (B).
The final key accepts either. Plots (C) and (D) diverge at $n_L=1.5$, which would need the power to become infinite where the lens is in fact optically invisible.
Answer: (A) or (B). | ||||
| Q.5 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Multiple correctMedium | |
A medium multiple correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Bohr's condition gives $mv_kr_k=\frac{kh}{2\pi}$, so $mv_k=\frac{kh}{2\pi r_k}$ and $K_k=\frac12v_k(mv_k)=\frac{h}{4\pi}\cdot\frac{kv_k}{r_k}$.
(A) True: $|K_n-K_1|=\frac{h}{4\pi}\left|\frac{nv_n}{r_n}-\frac{v_1}{r_1}\right|$.
(D) False: total energy is $E_k=-K_k$, so its change has the same size as in (A), with $\frac{h}{4\pi}$, not $\frac{h}{2\pi}$.
(C) True: $E_k=-\frac{e^2}{8\pi\epsilon_0r_k}$, so $\nu=\frac{E_n-E_1}{h}=\frac{e^2}{8\pi\epsilon_0h}\left(\frac{1}{r_1}-\frac{1}{r_n}\right)$.
(B) False: $\lambda_k=\frac{2\pi r_k}{k}$ and $r_k=\frac{e^2}{8\pi\epsilon_0K_k}$ give $\lambda_k=\frac{e^2}{4\epsilon_0kK_k}$. The change is $\frac{e^2}{4\epsilon_0}\left|\frac{1}{nK_n}-\frac{1}{K_1}\right|$, which has a factor $n$ the option lacks.
Answer: (A), (C). | ||||
| Q.6 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Multiple correctMedium | |
A medium multiple correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The path is $y=x\tan\theta-\frac{gx^2}{2v^2\cos^2\theta}$, and it must pass through $(5,1)$.
(A) True: at $45^\circ$, $1=5-\frac{25g}{v^2}$, so $v^2=\frac{25g}{4}$ and $v=\frac{5\sqrt g}{2}$.
(B) True: the top is at $x=\frac{v^2\sin2\theta}{2g}=\frac{25}{8}=3.125$ m, less than $5$ m.
(C) False: at $30^\circ$, $1=\frac{5}{\sqrt3}-\frac{50g}{3v^2}$ gives $\frac{v^2}{g}\approx8.83$. The top is at $x=\frac{v^2\sin60^\circ}{2g}\approx3.8$ m, before $P$.
(D) False: at $\tan\theta=\frac15$ the first term is already $5\cdot\frac15=1$, so the gravity term would have to vanish. No finite speed does this.
Answer: (A), (B). | ||||
| Q.7 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Use $\gamma=\frac53$ and write $T$ for the temperature of the isothermal.
(C) True: along the adiabat $TV_1^{2/3}=T_cV_2^{2/3}$, so $\frac{T}{T_c}=8^{2/3}=4$.
(D) False: $a$ and $b$ are at the same temperature, so $\frac{P_a}{P_b}=\frac{V_2}{V_1}=8$.
(A) True: per mole, heat released in $bc$ is $\frac32R(T-T_c)=\frac98RT$, and heat absorbed in $ab$ is $RT\ln8=2.1RT$.
(B) True: $\eta=1-\frac{\frac32\left(1-r^{-2/3}\right)}{\ln r}$ with $r=\frac{V_2}{V_1}$; the temperature cancels.
Answer: (A), (B), (C). | ||||
| Q.8 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Multiple correctMedium | |
A medium multiple correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves The phase is $\vec k\cdot\vec r+\omega t$ with $\vec k=3\hat j+4\hat k$. A phase with $+\omega t$ travels against $\vec k$.
(A) True: the direction is $-\frac15(3\hat j+4\hat k)$.
(B) False: $|\vec k|=5$ m$^{-1}$.
(C) True: $\omega=c|\vec k|=1.5\times10^9$ rad s$^{-1}$.
(D) False: $\vec B=\frac{1}{c}\hat n\times\vec E$ with $\hat n=-\frac15(3\hat j+4\hat k)$. Since $\hat j\times\hat i=-\hat k$ and $\hat k\times\hat i=\hat j$, this gives $-\frac{E_0}{5c}\sin(\ldots)(4\hat j-3\hat k)$, which differs from the option in size and sign.
Answer: (A), (C). | ||||
| Q.9 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalHard | |
A hard 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 Let the rod's cross-section be $A$. An element at distance $s$ from the hinge has a net upward force per unit length of $(\rho_{\text{liquid}}-\rho)Ag$: this is $5\rho Ag$ in the lower half and $\rho Ag$ in the upper half.
For a small tilt $\theta$ the element's force has lever arm $s\theta$, and the height of the boundary along the rod changes only at order $\theta^2$. The restoring torque is
$\tau=\theta\rho Ag\left[5\int_0^{L/2}s\,ds+\int_{L/2}^{L}s\,ds\right]=\theta\rho Ag\left[\frac{5L^2}{8}+\frac{3L^2}{8}\right]=\rho AgL^2\theta$.
The moment of inertia about the hinge is $\frac13(\rho AL)L^2$.
$\omega^2=\frac{\rho AgL^2}{\rho AL^3/3}=\frac{3g}{L}$, so $T=\frac{2\pi}{\sqrt3}\sqrt{\frac Lg}$.
$n=\sqrt3\approx1.73$. | ||||
| Q.10 | ThermodynamicsSecond law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Each engine passes on a fraction $(1-\eta)$ of what it receives, so $Q_5=Q_0(1-\eta)^5$.
$W=Q_0-Q_5$, so $\frac{W}{Q_0}=1-(1-\eta)^5=\frac{211}{243}$.
$(1-\eta)^5=\frac{32}{243}=\left(\frac23\right)^5$, so $\eta=\frac13\approx0.33$. | ||||
| Q.11 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalHard | |
A hard 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 $S_1$ has fixed volume, so its heat capacity is $C_V=\frac32R$. $S_2$ stays at constant pressure, so its heat capacity is $C_P=\frac52R$.
With heat current $q=\frac{KA}{x}\Delta T$ from the hotter side to the colder:
$\frac{d(\Delta T)}{dt}=-q\left(\frac{1}{C_V}+\frac{1}{C_P}\right)=-\frac{KA}{x}\Delta T\cdot\frac{1}{R}\left(\frac23+\frac25\right)=-\frac{16}{15}\frac{KA}{xR}\Delta T$.
The difference decays exponentially and halves at $t=\frac{15}{16}\ln2\cdot\frac{xR}{KA}$.
$n=\frac{15}{16}\times0.7\approx0.66$. | ||||
| Q.12 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Far from the cone it acts as a magnetic dipole, with on-axis field $B=\frac{\mu_0}{4\pi}\frac{2m}{z^3}$.
A ring of charge $dq$ at radius $r$ has current $\frac{\omega\,dq}{2\pi}$ and moment $\frac{\omega\,dq}{2\pi}\pi r^2=\frac{\omega}{2}r^2\,dq$.
On a cone's surface the area, and so the charge, grows in proportion to $r\,dr$. So $dq=Q\frac{2r\,dr}{R^2}$, and $\int r^2\,dq=\frac{2Q}{R^2}\cdot\frac{R^4}{4}=\frac{QR^2}{2}$.
$m=\frac{\omega QR^2}{4}$, which gives $B=\frac{\mu_0}{4\pi}\cdot\frac{QR^2\omega}{2z^3}$.
$n=0.5$. | ||||
| Q.13 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | Matching listMedium | |
A medium matching list question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The detector records a maximum when the two paths differ by a whole number of wavelengths. The smallest $l$ makes the difference equal to $\lambda=0.29$ m.
(P) Semicircle of diameter $l$ has length $\frac{\pi l}{2}$. Difference $=0.571\,l$, so $l=0.51$ m, entry (3).
(Q) The upper path is $0.5l+l+0.5l=2l$. Difference $=l$, so $l=0.29$ m, entry (4).
(R) The dashed line is a diameter of length $\sqrt2\,l$ (the diagonal of a square of side $l$), so the arc is a semicircle of length $\frac{\pi l}{\sqrt2}$. The upper path is $l+\frac{\pi l}{\sqrt2}=3.22\,l$. Difference $=2.22\,l$, so $l=0.13$ m, entry (5).
(S) The triangle has angles $45^\circ$, $105^\circ$ and $30^\circ$, and side $l$ is opposite $105^\circ$. By the sine rule the other two sides total $l\frac{\sin30^\circ+\sin45^\circ}{\sin105^\circ}=l\frac{0.5+0.707}{0.97}=1.244\,l$. Difference $=0.244\,l$, so $l=1.19$ m, entry (2).
P→3, Q→4, R→5, S→2. Answer: (D). | ||||
| Q.14 | OpticsWave optics | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Matching listEasy | |
A easy matching list question on wave optics, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics (P) Aurora: charged particles from the Sun excite atmospheric oxygen and nitrogen, which then emit light, entry (5).
(Q) Sunlight scattered by air molecules is partly polarized, entry (4).
(R) A rainbow comes from light being split by colour on refraction in water drops and reflected inside them, entry (1).
(S) Dark and bright fringes come from diffraction, entry (3).
P→5, Q→4, R→1, S→3. Answer: (A). | ||||
| Q.15 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Matching listMedium | |
A medium matching list question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current A sector of radius $a$ rotating through the field boundary changes its flux at the constant rate $\frac12Ba^2\omega$ while part of it is entering or leaving, and at zero rate otherwise. Entering and leaving give opposite signs.
(P) A semicircle spends $\frac T2$ entering and $\frac T2$ leaving: constant current one way, then the other. Entry (3).
(Q) The sectors are each $\frac T6$ wide with a $\frac T6$ gap. In the first half: on, off, on as the two sectors enter. In the second half: on, off, on the other way as they leave. Entry (2).
(R) One sector: current for $\frac T6$ as it enters, then none until $\frac T2$, then opposite current for $\frac T6$ as it leaves, then none. Entry (1).
(S) The two sectors are diametrically opposite, so one enters just as the other leaves. The total flux stays constant and no current flows. Entry (4).
P→3, Q→2, R→1, S→4. Answer: (C). | ||||
| Q.16 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | Matching listMedium | |
A medium matching list question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion A rod with one end on the axis, at angle $\phi$ to it, has $I=\frac13ml^2\sin^2\phi$.
(P) Two rods at $45^\circ$: $2\cdot\frac13ml^2\cdot\frac12=\frac13ml^2$, entry (5).
(Q) CA and CB at $60^\circ$: $2\cdot\frac13ml^2\cdot\frac34=\frac12ml^2$. The rod AB is parallel to the axis at distance $\frac{\sqrt3}{2}l$: $\frac34ml^2$. Total $\frac54ml^2$, entry (1).
(R) Four rods, each with one end on the axis at $45^\circ$: $4\cdot\frac13ml^2\cdot\frac12=\frac23ml^2$, entry (4).
(S) Two rods at $30^\circ$: $2\cdot\frac13ml^2\cdot\frac14=\frac16ml^2$, entry (2).
P→5, Q→1, R→4, S→2. Answer: (A). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Let $nRT=0.5\times R\times600=300R$. The volume at pressure $p$ is $\frac{nRT}{p}$.
Step 1: $W_1=P\left(\frac{nRT}{2}-\frac{nRT}{P}\right)=nRT\left(\frac P2-1\right)$.
Step 2: $W_2=8\left(\frac{nRT}{P}-\frac{nRT}{8}\right)=nRT\left(\frac8P-1\right)$.
$W=nRT\left(\frac P2+\frac8P-2\right)$. By the A.M.–G.M. inequality the bracket is least at $\frac P2=\frac8P$, that is $P=4$, where it equals $2$.
$W_{\min}=300R\times2=600R$.
Answer: (B). | ||||
| Q.2 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics At equilibrium the forward and backward rates are equal: $k_{\text{f}}[\mathbf R]=k_{\text{b}}[\mathbf P]$, so $\frac{[\mathbf P]}{[\mathbf R]}=\frac{k_{\text{f}}}{k_{\text{b}}}=\frac14$.
Since $[\mathbf R]+[\mathbf P]=[\mathbf R]_0$ throughout, the final values are $\frac{[\mathbf R]}{[\mathbf R]_0}=0.8$ and $\frac{[\mathbf P]}{[\mathbf R]_0}=0.2$. Only plot (C) shows these.
Answer: (C). | ||||
| Q.3 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure BF$_3$ (trigonal planar) and NH$_4^+$ (tetrahedral) are symmetric, so their bond dipoles cancel and both have zero dipole moment.
In NH$_3$ the lone pair's moment and the N–H bond moments point the same way, adding up to about $1.47$ D.
In NF$_3$ the N–F bond moments point away from nitrogen, against the lone pair's moment, leaving only about $0.23$ D.
Order: BF$_3$ = NH$_4^+$ < NF$_3$ < NH$_3$.
Answer: (A). | ||||
| Q.4 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctHard | |
A hard single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques LiBH$_4$ reduces the ester arm to $-$CH$_2$CH$_2$OH and leaves the acid. BH$_3$ reduces the acid arm to $-$CH$_2$CH$_2$OH and leaves the ester.
In both cases the product has a $\delta$-hydroxy acid or ester on one ring carbon, and with H$^+$ it closes to the same kind of six-membered spiro-lactone.
The difference is which arm became the ring oxygen's side. With LiBH$_4$ the O–CH$_2$CH$_2$ side is on the wedge and the C(=O)CH$_2$ side on the dash. With BH$_3$ they are the other way round. So the two products differ in the configuration at the spiro carbon while the methyl carbon is unchanged.
First substrate (CH$_3$ on the adjacent carbon): the spiro carbon and the methyl carbon are stereocentres, and only one of them differs, so $\mathbf P$ and $\mathbf Q$ are diastereomers.
Second substrate (CH$_3$ on the opposite carbon): the products are 1,3-disubstituted cyclobutanes with the lactone oxygen arm cis in one and trans in the other to the methyl, so $\mathbf R$ and $\mathbf S$ are diastereomers.
Answer: (C). | ||||
| Q.5 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Multiple correctMedium | |
A medium multiple correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom (A) True: in a many-electron atom $2s$ penetrates the inner shell more than $2p$, so it feels more nuclear charge and lies lower.
(B) True: in hydrogen the energy depends only on $n$, so $2s$ and $2p$ are degenerate.
(C) False: hydrogen's $n=2$ level is at $-3.4$ eV, while lithium's $2s$ electron is at about $-5.4$ eV. So $E_{2p}$(H) is above $E_{2s}$(Li).
(D) True: the same two values give $E_{2s}$(H) $>$ $E_{2s}$(Li), since lithium's greater effective nuclear charge binds its $2s$ electron more strongly.
Answer: (A), (B), (D). | ||||
| Q.6 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Multiple correctEasy | |
A easy multiple correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathbf X$ is Cl$_2$, from the oxidation of HCl by MnO$_2$.
With excess Cl$_2$, ammonia gives NCl$_3$: NH$_3$ + 3Cl$_2$ $\to$ NCl$_3$ + 3HCl. So $\mathbf Y$ is NCl$_3$.
Cl$_2$ with excess F$_2$ at $573$ K gives ClF$_3$. So $\mathbf Z$ is ClF$_3$.
(A) True: chlorine is used to disinfect water.
(B) False: NCl$_3$ has a lone pair on nitrogen and is pyramidal.
(C) True: ClF$_3$ fluorinates uranium to UF$_6$, the form in which $^{235}$U is enriched.
(D) False: electron-withdrawing chlorines make NCl$_3$ a weaker Lewis base than NH$_3$.
Answer: (A), (C). | ||||
| Q.7 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter The product is O$_2^+$[PtF$_6$]$^-$: PtF$_6$ removes an electron from O$_2$.
(A) False: O$_2^+$ has bond order $\frac{10-5}{2}=2.5$.
(B) True: in [PtF$_6$]$^-$ platinum is $+5$. Pt has $10$ valence electrons, so Pt(V) is $d^5$.
(C) True: PtF$_6$ accepts the electron, so it is the oxidant.
(D) False: no fluorine is transferred to oxygen.
Answer: (B), (C). | ||||
| Q.8 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Multiple correctHard | |
A hard multiple correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Kolbe electrolysis of sodium butanoate gives $n$-hexane, and V$_2$O$_5$ at $500^\circ$C and high pressure aromatizes it: $\mathbf Q$ is benzene.
Friedel–Crafts acylation with phthalic anhydride gives $\mathbf R$, 2-benzoylbenzoic acid.
PCl$_5$ makes the acid chloride and Rosenmund reduction gives the aldehyde: $\mathbf S$ is 2-benzoylbenzaldehyde.
Hydrazine condenses with both carbonyl groups, closing a ring with two nitrogens: $\mathbf T$ is 1-phenylphthalazine.
(A) True: $\mathbf S$ has an aldehyde, which reduces Tollens' reagent.
(B) True: benzene adds chlorine under UV to give benzene hexachloride, whose $\gamma$ isomer is gammaxane.
(C) True: the phthalazine ring contains nitrogen.
(D) False: acid cyclization of $\mathbf R$ gives anthraquinone, and Clemmensen reduction removes the carbonyl oxygens altogether. It cannot give a dihydroxy compound.
Answer: (A), (B), (C). | ||||
| Q.9 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry At equal volume and temperature, pressure is proportional to the total moles, so $n_1=5n_2$.
With $M$ the molar mass of He: $n_1=\frac{m_1}{M}+\frac{m_2}{10M}$ and $n_2=\frac{m_2}{M}+\frac{m_1}{10M}$.
$m_1+\frac{m_2}{10}=5m_2+\frac{m_1}{2}$, so $\frac{m_1}{2}=\frac{49}{10}m_2$.
$\frac{m_1}{m_2}=9.8$. | ||||
| Q.10 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Geometric: glycinate has two different donor ends, N and O. A square planar M(N$^\wedge$O)(a)(b) complex has $2$ arrangements, according to which monodentate ligand lies opposite the nitrogen.
Linkage: NCS can bind through N or S ($2$ ways), and NO$_2$ through N or O ($2$ ways).
Total $=2\times2\times2=8$. A square planar complex of this kind has a mirror plane, so there are no optical isomers to add. | ||||
| Q.11 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids First reaction: both acid groups are $\beta$ to the ketone, so both are lost as CO$_2$ on heating. $\mathbf X$ is pentan-3-one, with $1$ carbonyl group.
Second reaction: the two acid groups are $\gamma$ to the ketone, so they do not decarboxylate. They sit on neighbouring carbons on the same face of the ring, as the bold ring bonds show. Heating closes them into a cyclic anhydride with loss of water. $\mathbf Y$ has the ketone and the two anhydride carbonyls, $3$ in all.
Sum $=1+3=4$. | ||||
| Q.12 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons NaNH$_2$ removes both terminal hydrogens. The dianion displaces bromide from two molecules of the allylic bromide, giving
CH$_3$CH=CHCH$_2$–C≡C–C≡C–CH$_2$CH=CHCH$_3$.
The four alkyne carbons are $sp$ and lie in one line, and so do the two CH$_2$ carbons bonded to them.
Collinear carbons: $6$. | ||||
| Q.13 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Matching listMedium | |
A medium matching list question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics (P) Gas molecules settle on a surface: heat is released and freedom is lost, so $\Delta H<0$ and $\Delta S<0$, entry (2).
(Q) Graphite is the more stable form and is less ordered than diamond, so $\Delta H<0$ and $\Delta S>0$, entry (5).
(R) Unfolding breaks many weak bonds and frees the chain, so $\Delta H>0$ and $\Delta S>0$, entry (1).
(S) Forming the strained ring costs energy, and the ring is more rigid than the open chain, so $\Delta H>0$ and $\Delta S<0$, entry (4).
P→2; Q→5; R→1; S→4. Answer: (C). | ||||
| Q.14 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listMedium | |
A medium matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Shapes by electron pairs on the central atom:
SOCl$_2$ and SO$_3^{2-}$: three bonding directions and one lone pair, pyramidal.
ClF$_3$ and XeF$_3^+$: three bonds and two lone pairs, T-shaped.
SF$_4$: four bonds and one lone pair, see-saw.
XeOF$_4$, ClF$_5$ and XeF$_5^+$: five bonds and one lone pair, square pyramidal.
(P) see-saw: one, (1). (Q) T-shaped: two, (2). (R) trigonal planar: zero, (5). (S) square pyramidal: three, (3).
Answer: (A). | ||||
| Q.15 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Matching listHard | |
A hard matching list question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Ozonolysis cuts the ring-fusion C=C, opening each bicycle into one large ring holding two ketones. In base a ketone forms its more substituted enolate, at the carbon carrying CH$_3$, and this adds across the ring to the other ketone. The product has a quaternary C–CH$_3$ next to a C–OH at the new ring fusion.
(P) Six-six: a ten-membered 1,6-dione. The enolate carbon reaches the far ketone through a three-carbon chain, closing a five-membered ring and leaving a seven-membered one. The original second CH$_3$ is next to the C–OH on the seven-membered ring, and the remaining C=O is next to the C–CH$_3$ on that same ring. Entry (2).
(Q) Seven-five: a ten-membered 1,5-dione. The only enolate with a CH$_3$ that closes a ring of workable size attacks through the four-carbon chain, giving a six-six skeleton. The second CH$_3$ and the C=O are in the same ring. Entry (1).
(R) Seven-five with both CH$_3$ on the seven-membered ring: again a 1,5-dione, giving a six-six skeleton. One ring holds the C=O and the other holds the second CH$_3$ next to C–OH. Entry (5).
(S) Six-six with both CH$_3$ on one ring: a 1,6-dione, giving a five-seven skeleton. The second CH$_3$ lies on the five-membered ring next to C–OH, and the C=O lies on the seven-membered ring. Entry (3).
P→2; Q→1; R→5; S→3. Answer: (C). | ||||
| Q.16 | Organic Chemistry: Some Basic Principles and TechniquesTypes of organic reactions | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Matching listHard | |
A hard matching list question on types of organic reactions, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The nitro group para to bromine activates it to attack by nucleophiles.
(P) With NaOH the oxime oxygen displaces bromide, closing a benzisoxazole with H at C-3. Base removes that hydrogen and the N–O bond breaks, opening to the 2-cyanophenol: entry (1).
(Q) Acetic anhydride converts the aldoxime into its acetate, and base eliminates acetic acid to give the nitrile. Bromine stays: entry (2).
(R) The ketoxime's oxygen also displaces bromide, but C-3 now carries CH$_3$ and has no hydrogen to lose, so the benzisoxazole survives: entry (4).
(S) Mild aqueous carbonate only hydrolyses the acetate ester, returning the oxime: entry (5).
P→1; Q→2; R→4; S→5. Answer: (B). | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Squaring: $a^2+b^2+2\vec a\cdot\vec b=21$ and $a^2+b^2-2\vec a\cdot\vec b=9$. So $a^2+b^2=15$ and $\vec a\cdot\vec b=3$.
Perpendicularity: $\vec a\cdot(\vec a-\vec b)=0$ gives $a^2=\vec a\cdot\vec b=3$, so $b^2=12$.
Area of $OPR=\frac12|\vec a\times(\vec a+\vec b)|=\frac12|\vec a\times\vec b|$, and $|\vec a\times\vec b|^2=a^2b^2-(\vec a\cdot\vec b)^2=36-9=27$.
Area $=\frac{3\sqrt3}{2}$.
Answer: (C). | ||||
| Q.2 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Here $a=4$, and a point on the parabola is $(4t^2,8t)$. The point $(64,32)$ has $t=4$.
Tangents at $t_1,t_2$ are perpendicular when $t_1t_2=-1$, so the other point has $t=-\frac14$: $(x_1,y_1)=\left(\frac14,-2\right)$.
The distance from a point on a parabola to the focus equals its distance from the directrix: $x_1+a=\frac14+4=\frac{17}{4}$.
Answer: (C). | ||||
| Q.3 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $\frac{1+y^4}{y^3}\,dy=\frac{e^{5x}+1}{e^x}\,dx=(e^{4x}+e^{-x})\,dx$.
Integrate: $\frac{y^2}{2}-\frac{1}{2y^2}=\frac{e^{4x}}{4}-e^{-x}+C$.
At $x=0$, $y^2=\frac12$: the left side is $\frac14-1=-\frac34$ and the right side is $\frac14-1+C$, so $C=0$.
At $x=\log_e2$ the right side is $4-\frac12=\frac72$. So $y^2-\frac{1}{y^2}=7$, and with $u=y^2>0$ we get $u=\frac{7+\sqrt{53}}{2}$.
$y=\sqrt{\frac{7+\sqrt{53}}{2}}$.
Answer: (B). | ||||
| Q.4 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $f(x)=\frac{1}{3^x+3}$ and $t=3^x$. Then $f(2-x)=\frac{1}{9/t+3}=\frac{t}{3(t+3)}$.
$f(x)+f(2-x)=\frac{3+t}{3(t+3)}=\frac13$.
Using $\int_0^2f(x)\,dx=\int_0^2f(2-x)\,dx$: $2I=\int_0^2\frac13\,dx=\frac23$, so $I=\frac13$.
Answer: (B). | ||||
| Q.5 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Multiple correctHard | |
A hard multiple correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Expand: $(x^2-1)^{10}=\sum_k\binom{10}{k}(-1)^{10-k}x^{2k}$. Differentiating ten times sends $x^m$ to $\frac{m!}{(m-10)!}x^{m-10}$.
(C) True: the top term $x^{20}$ becomes a multiple of $x^{10}$.
(A) True: $x^8$ comes from $x^{18}$ ($k=9$), whose coefficient is $-10$. The result is $-10\cdot\frac{18!}{8!}$.
(D) False: the constant comes from $x^{10}$ ($k=5$), whose coefficient is $-\binom{10}{5}=-252$. The constant is $-252\cdot10!$.
(B) True: write $(x^2-1)^{10}=(x-1)^{10}(x+1)^{10}$. By Leibniz's rule, at $x=1$ only the term that differentiates $(x-1)^{10}$ all ten times survives, giving $10!\cdot(1+1)^{10}=10!\,2^{10}$. The function $f$ is even, so $f(-1)$ has the same value. The sum is $10!\,2^{11}$.
Answer: (A), (B), (C). | ||||
| Q.6 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Multiple correctMedium | |
A medium multiple correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let the roots be integers $p,q$. Then $\frac ba=-(p+q)$ and $\frac ca=pq$ are integers, and $2b=a+c$ gives $-2(p+q)=1+pq$.
So $pq+2p+2q+4=3$, that is $(p+2)(q+2)=3$.
The positive coefficients force both roots to be negative, so $p+2=-1$ and $q+2=-3$ (in some order). The roots are $-3$ and $-5$, and so $b=8a$, $c=15a$.
(A) True: $c-b=7a$.
(B) True: $-3$ and $-5$ are odd.
(C) True: $c=15$ gives $a=1$, $b=8$, so $ab=8$.
(D) False: $b=8$ gives $a=1$, with roots $-3$ and $-5$. $x=3$ is not one of them.
Answer: (A), (B), (C). | ||||
| Q.7 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Multiple correctMedium | |
A medium multiple correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $L$: $(1+t,\ 2+t,\ -1+2t)$. Perpendicularity of $RS$ to $(1,1,2)$: $(t-3)+(t+3)+2(2t-6)=0$, so $t=2$ and $S=(3,4,3)$.
$\overrightarrow{PS}=(2,2,4)$, and $T$ lies beyond $S$ at twice that distance: $T=(7,8,11)$. So $|PT|=6\sqrt6$.
$\overrightarrow{SR}=(1,-5,2)$, so $|RS|=\sqrt{30}$, and $RS$ is the altitude on $PT$.
Area $=\frac12\cdot6\sqrt6\cdot\sqrt{30}=18\sqrt5$. So (D) is true and (C) is false.
The orthocentre $H$ lies on the altitude: $H=S+s(1,-5,2)$. Also $\overrightarrow{PH}\perp\overrightarrow{RT}=(3,9,6)$: $(2+s,\ 2-5s,\ 4+2s)\cdot(3,9,6)=48-30s=0$, so $s=\frac85$.
$H=\left(\frac{23}{5},-4,\frac{31}{5}\right)$. So (A) is true and (B) is false.
Answer: (A), (D). | ||||
| Q.8 | 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 Rewrite as $\frac{dy}{dx}-\frac yx=-x^2$. The integrating factor is $\frac1x$: $\frac{d}{dx}\left(\frac yx\right)=-x$, so $\frac yx=-\frac{x^2}{2}+C$. The condition $y(1)=0$ gives $C=\frac12$.
$f(x)=\frac{x-x^3}{2}$ and $f'(x)=\frac{1-3x^2}{2}$.
(A) False and (B) True: $f'$ changes from positive to negative at $x=\frac{1}{\sqrt3}$, which is a local maximum.
(C) False: $f'<0$ for $x>\frac{1}{\sqrt3}$.
(D) True: $f=g$ reduces to $x(9x^2-10x+2)=0$. The quadratic has discriminant $28>0$ and both its roots are positive, so there are $2$ solutions in $(0,\infty)$.
Answer: (B), (D). | ||||
| Q.9 | 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 $ST=\begin{bmatrix}0&-1\\1&1\end{bmatrix}$, so $a=0$, $b=-1$, $c=1$, $d=1$.
(A) False: $\frac{-1}{1+i}=\frac{-1+i}{2}\neq i$.
(B) True: $\frac{-1}{\omega+1}=\frac{-1}{-\omega^2}=\frac{1}{\omega^2}=\omega$, using $\omega^3=1$ and $1+\omega+\omega^2=0$.
(C) False: $(ST)^3=-I$ and $(ST)^6=I$, and $ST$ is invertible. So $(ST)^2=(ST)^m$ holds exactly when $m\equiv2\pmod6$. This includes $m=14$, which is not a multiple of $8$.
(D) True: for real coefficients, $\operatorname{Im}\frac{az+b}{cz+d}=\frac{(ad-bc)\operatorname{Im}z}{|cz+d|^2}$, and $ad-bc=1>0$.
Answer: (B), (D). | ||||
| Q.10 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions The values $2^x$ are all different, so $g\circ f$ is one-one, which forces $f$ to be one-one. Conversely, for any one-one $f$ we can define $g$ on the image of $f$ by $g(f(x))=2^x$ and arbitrarily elsewhere.
So $T$ is the set of one-one functions with $f(2)\neq2$ and $f(4)\neq4$.
All one-one functions: $7\cdot6\cdot5\cdot4\cdot3=2520$.
With $f(2)=2$: $6\cdot5\cdot4\cdot3=360$. With $f(4)=4$: also $360$. With both: $5\cdot4\cdot3=60$.
$2520-360-360+60=1860$. | ||||
| Q.11 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability With $k$ Mathematics books chosen, $X=|2k-6|$, and the number of ways is $\binom6k\binom{5}{6-k}$. The total is $\binom{11}{6}=462$.
$k=1$: $6$ ways, $X=4$. $k=2$: $75$ ways, $X=2$. $k=3$: $200$ ways, $X=0$. $k=4$: $150$ ways, $X=2$. $k=5$: $30$ ways, $X=4$. $k=6$: $1$ way, $X=6$.
$\alpha=\frac{24+150+0+300+120+6}{462}=\frac{600}{462}=\frac{100}{77}$.
$77\alpha=100$. | ||||
| Q.12 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Sums: all ten add to $50$ and the first eight add to $32$, so $x_9+x_{10}=18$.
Squares: all ten give $10(7+25)=320$ and the first eight give $8(3.5+16)=156$, so $x_9^2+x_{10}^2=164$.
$(x_9+x_{10})^2=324$ then gives $x_9x_{10}=80$. With $x_9<x_{10}$: $x_9=8$ and $x_{10}=10$.
$3x_9+2x_{10}=24+20=44$. | ||||
| Q.13 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse: $e^2=1-\frac{12}{18}=\frac13$ and the foci are at $(\pm\sqrt6,0)$.
Hyperbola: $e=\sqrt3$ with the same foci, so its semi-axis is $\sqrt2$ and $b^2=6-2=4$. It is $\frac{x^2}{2}-\frac{y^2}{4}=1$.
Put $x^2=\sqrt5\,y$: $y^2-2\sqrt5\,y+4=0$, so $y=\sqrt5\pm1$ and $x^2=5\pm\sqrt5$. Both give points in the first quadrant.
$d^2=(x_1-x_2)^2+(y_1-y_2)^2=\left(10-2\sqrt{25-5}\right)+4=14-4\sqrt5$.
$a=14$, $b=-4$, so $a-b=18$. | ||||
| Q.14 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $f$: $\sin^2(\pi(x-[x]))=\sin^2\pi x$ is continuous, so the logarithm is continuous and is zero exactly at integers. The factor $[x^3]$ jumps wherever $x^3$ is an integer.
On $(-3,3)$, $x^3$ takes the integer values $-26,\ldots,26$, which is $53$ points. At those that are integers ($x=0,\pm1,\pm2$) the logarithm is $0$, so $f$ is continuous there. Hence $|A|=53-5=48$.
For $g$: within each interval between integers $g$ is continuous. At an integer $n$, the right value is $n^3\sin^2(0)=0$ and the left limit is $n^3\sin^2(\pi\log_e2)$. These differ unless $n=0$. So $B=\{\pm1,\pm2\}$ and $|B|=4$.
$A$ contains no integers, so $A\cap B$ is empty.
$48+8-0=56$. | ||||
| Q.15 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions The curves meet where $\sin x+\cos x=1$, that is $\sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt2}$. This gives $x=2k\pi$ or $x=\frac{\pi}{2}+2k\pi$.
In $[0,10\pi]$: $x=2k\pi$ for $k=0,\ldots,5$ gives $6$ points, and $x=\frac{\pi}{2}+2k\pi$ for $k=0,\ldots,4$ gives $5$ points.
$n=11$. | ||||
| Q.16 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The first four intersections are $\alpha_1=0$, $\alpha_2=\frac{\pi}{2}$, $\alpha_3=2\pi$, $\alpha_4=\frac{5\pi}{2}$.
An antiderivative of $e^{-x}(\sin x+\cos x-1)$ is $F(x)=e^{-x}(1-\cos x)$, as differentiating confirms.
On $\left(0,\frac{\pi}{2}\right)$ the integrand is positive: area $F\left(\frac{\pi}{2}\right)-F(0)=e^{-\pi/2}$.
On $\left(\frac{\pi}{2},2\pi\right)$ it is negative: area $F\left(\frac{\pi}{2}\right)-F(2\pi)=e^{-\pi/2}$.
On $\left(2\pi,\frac{5\pi}{2}\right)$ it is positive: area $F\left(\frac{5\pi}{2}\right)-F(2\pi)=e^{-5\pi/2}$.
$\beta-2e^{-\pi/2}=e^{-5\pi/2}$, so the value asked for is $\frac52=2.5$. | ||||
| Q.17 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Subtracting the equations gives $x^2=y^2$, and then $5x^2=1$. So $P$ has $x=y=\frac{1}{\sqrt5}$.
Slopes at $P$: for the first ellipse $-\frac{x}{4y}=-\frac14$, and for the second $-\frac{4x}{y}=-4$.
$\tan\theta=\left|\frac{-\frac14+4}{1+1}\right|=\frac{15}{8}$.
$4\tan\theta=7.5$. | ||||
| Q.18 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals By symmetry, take four times the part in the first quadrant. There the region lies under $y=\frac12\sqrt{1-x^2}$ for $0\le x\le\frac{1}{\sqrt5}$ and under $y=\sqrt{1-4x^2}$ for $\frac{1}{\sqrt5}\le x\le\frac12$.
$\int_0^{1/\sqrt5}\frac12\sqrt{1-x^2}\,dx=\frac14\left[\frac25+\sin^{-1}\frac{1}{\sqrt5}\right]$.
With $u=2x$: $\int_{1/\sqrt5}^{1/2}\sqrt{1-4x^2}\,dx=\frac14\left[\frac{\pi}{2}-\frac25-\sin^{-1}\frac{2}{\sqrt5}\right]$.
Adding, and using $\sin^{-1}\frac{2}{\sqrt5}=\frac{\pi}{2}-\sin^{-1}\frac{1}{\sqrt5}$: the first-quadrant area is $\frac12\sin^{-1}\frac{1}{\sqrt5}$.
$\alpha=2\sin^{-1}\frac{1}{\sqrt5}=2\tan^{-1}\frac12$, so $\tan\alpha=\frac{2\cdot\frac12}{1-\frac14}=\frac43$.
$\cot\alpha=\frac34=0.75$. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Resistance: $R=\frac{l}{\sigma A}=\frac{100}{2\times10^8\times0.5\times10^{-6}}=1\ \Omega$.
Current: $I=\frac{2}{1+1}=1$ A.
Electron density: $n=\frac{\rho N_A}{M}=\frac{6.35\times10^3\times6\times10^{23}}{63.5\times10^{-3}}=6\times10^{28}$ m$^{-3}$.
Drift speed: $v_d=\frac{I}{neA}=\frac{1}{6\times10^{28}\times1.6\times10^{-19}\times0.5\times10^{-6}}\approx2.08\times10^{-4}$ m s$^{-1}=0.208$ mm s$^{-1}$.
Answer: (C). | ||||
| Q.2 | 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 Number of nuclei: $\frac{dN}{dt}=\alpha-\lambda N$ with $N(0)=0$ gives $N=\frac{\alpha}{\lambda}\left(1-e^{-\lambda t}\right)$.
Decays per second: $\lambda N=\alpha\left(1-e^{-\lambda t}\right)$, so the heating power is $\alpha E_0\left(1-e^{-\lambda t}\right)$.
Temperature rise: $ms\frac{dT}{dt}=\alpha E_0\left(1-e^{-\lambda t}\right)$.
Answer: (A). | ||||
| Q.3 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a thin prism $D_m=(n-1)A$, so $D_m$ is smallest where $n$ is smallest.
$\frac{dn}{d\lambda}=\alpha-\frac{2\beta}{\lambda^3}=0$ gives $\lambda^3=\frac{2\times0.096}{3}=0.064$, so $\lambda=0.4\ \mu$m.
$n=3\times0.4+\frac{0.096}{0.16}=1.2+0.6=1.8$.
$D_m=0.8\times6^\circ=4.8^\circ$.
Answer: (B). | ||||
| Q.4 | GravitationOrbital motion | Class 11: Gravitation | Single correctHard | |
A hard single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Radial motion with fixed $\ell$: $m\ddot r=\frac{\ell^2}{mr^3}-\frac{k}{r^2}$.
The circular orbit has $\frac{\ell^2}{mr_0^3}=\frac{k}{r_0^2}$, so $r_0=\frac{\ell^2}{mk}$.
Put $r=r_0+x$ and keep the first order in $x$: $m\ddot x=\left(-\frac{3\ell^2}{mr_0^4}+\frac{2k}{r_0^3}\right)x=-\frac{k}{r_0^3}x$.
So $\omega^2=\frac{k}{mr_0^3}=\frac{m^2k^4}{\ell^6}$, which gives $\omega=\frac{mk^2}{\ell^3}$.
$T=\frac{2\pi\ell^3}{mk^2}$.
Answer: (A). | ||||
| Q.5 | 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 Faces $a_1b_1$ and $a_2b_2$ are vertical and the mirror is horizontal. Reflection at the mirror reverses only the ray's vertical component, so the ray makes the same angle with the horizontal normals of both faces: $i_2=e_1$ always.
At minimum deviation a prism has $i=e$ and $\sin i=n\sin\frac A2$.
(A) True: both at minimum deviation means $i_1=e_1=i_2=e_2$, so $n_1\sin\frac{A_1}{2}=n_2\sin\frac{A_2}{2}$.
(B) False: prism 2 at minimum deviation gives $\sin i_2=\sin e_1=n_2\sin\frac{A_2}{2}$. This speaks of $e_1$, and $i_1$ equals $e_1$ only if prism 1 is also at minimum deviation.
(D) True: prism 1 at minimum deviation gives $i_1=e_1$ and $\sin e_1=n_1\sin\frac{A_1}{2}$, and $i_2=e_1$.
(C) True: each dashed axis bisects its prism's apex angle, so it leans from the vertical face by $\frac{A_1}{2}$ and $\frac{A_2}{2}$. The axes meet at $\theta=\frac{A_1}{2}+\frac{A_2}{2}$. For a thin prism $\delta_m=(n-1)A$, so $A=\frac{\delta_m}{n-1}$.
Answer: (A), (C), (D). | ||||
| Q.6 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Multiple correctMedium | |
A medium multiple correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $\frac qm=1$ C kg$^{-1}$, so the electric acceleration is $1$ ms$^{-2}$ along $\hat i$.
Up to $t=0.2$ s: $v_x=1+t$, $v_y=2-10t$ and $y=2t-5t^2$. At $0.2$ s: $v_x=1.2$, $v_y=0$ and $y=0.2$ m.
After that the field is along $\hat j$. It bends only the horizontal velocity, with radius $\frac{mv_x}{qB}=\frac{1.2}{6}=0.2$ m $=20$ cm. So (C) is true.
Vertically the particle falls freely from rest: $y=0.2-5(t-0.2)^2$.
(A) True: at $t=0.3$ s, $y=0.2-0.05=0.15$ m.
(B) False: at $t=0.4$ s, $y=0.2-0.2=0$.
(D) False: $y=0$ happens at $t=0.4$ s, not $0.35$ s.
Answer: (A), (C). | ||||
| Q.7 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Let $\vec p=(a,b)$ and let $\vec r$ be the point. $|V_1|=|V_2|$ means $\frac{1}{|\vec r-\vec p|}=\frac{|m|}{|\vec r-m\vec p|}$, that is $|\vec r-m\vec p|^2=m^2|\vec r-\vec p|^2$.
Expanding, the terms in $p^2$ cancel: $r^2-2m\,\vec r\cdot\vec p=m^2r^2-2m^2\vec r\cdot\vec p$, which is $(1-m^2)r^2=2m(1-m)\vec r\cdot\vec p$. With $m\neq1$: $(1+m)r^2=2m\,\vec r\cdot\vec p$.
(A) True: $m=-1$ gives $\vec r\cdot\vec p=0$, that is $ax+by=0$.
(B) True: $m=2$ gives $r^2=\frac43\vec r\cdot\vec p$, a circle through the origin with centre $\frac23\vec p$ and radius $\frac23|\vec p|$.
(C) True: $m=-2$ gives $r^2=4\vec r\cdot\vec p$, with centre $2\vec p$ and radius $2|\vec p|$.
(D) False: $m=-3$ gives $r^2=3\vec r\cdot\vec p$, which is a circle, not a straight line.
Answer: (A), (B), (C). | ||||
| Q.8 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Multiple correctMedium | |
A medium multiple correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance (A) False: the forces $+qE\hat j$ and $-qE\hat j$ cancel, so the centre of mass stays at rest.
The field turns the dipole with no energy lost. Potential energy $-pE\cos\phi$ with $\phi$ the angle to the field falls by $pE\sin\theta_f=qdE\sin\theta_f$ as the dipole turns through $\theta_f$ from $\hat i$. This all becomes rotational kinetic energy.
The moment of inertia about the centre is $I=2m\left(\frac d2\right)^2=\frac{md^2}{2}$, so $\frac{md^2}{4}\omega_f^2=qdE\sin\theta_f$.
(B) True on the first swing: $\omega_f^2=\frac{2qE}{md}$ gives $\sin\theta_f=\frac12$, so $\theta_f=\frac{\pi}{6}$ is reached first.
(C) False: at $\theta_f=\frac{\pi}{3}$ the gain is $\frac{\sqrt3}{2}qEd$.
(D) True: once the field is off no torque acts, so the angular velocity stays constant about the stationary centre of mass.
Answer: (B), (D); the final key also accepts (D) alone. | ||||
| Q.9 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Multiple correctMedium | |
A medium multiple correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics The sudden compression leaves no time for heat exchange. Treat it as adiabatic with $\gamma=\frac53$: $T_b=300\times3^{2/3}=300\times9^{1/3}=624$ K, and $P_b=3^{5/3}P_a=3\times2.08\,P_a=6.24\,P_a$.
(D) False: the temperature is right but the pressure is $6.24$ times atmospheric, not $2.08$ times.
(B) True: $\Delta U_{ab}=10\times\frac32R\times(624-300)=4860R$.
(C) True: the final temperature is $284$ K, so the net $\Delta U=15R\times(284-300)=-240R$.
(A) True: $a\to b$ is a steep rising curve, $b\to c$ is a vertical drop at $V_0/3$, and $c\to f$ is an isothermal expansion back to $V_0$, ending below $a$ because $284\text{ K}<300\text{ K}$.
Answer: (A), (B), (C). | ||||
| Q.10 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $=\frac{0.5}{100}=0.005$ mm.
Wire-1 reads $0.5+42\times0.005=0.710$ mm, but its true diameter is $0.650$ mm. So the zero error is $+0.060$ mm.
Wire-2 reads $1.5+95\times0.005=1.975$ mm, so its true diameter is $1.975-0.060=1.915$ mm.
$d=1915\ \mu$m. | ||||
| Q.11 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | NumericalMedium | |
A medium numerical question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement First minimum: $a\sin\theta=\lambda$.
$\frac{\Delta\lambda}{\lambda}=\frac{\Delta a}{a}+\frac{\cos\theta\,\Delta\theta}{\sin\theta}$.
$\frac{\Delta a}{a}=\frac{0.002}{0.016}=0.125$.
$\Delta\theta=40'=\frac23^\circ=0.0116$ rad, and with $\cos2^\circ\approx1$ the second term is $\frac{0.0116}{0.035}\approx0.333$.
Fractional error $\approx0.125+0.333\approx0.46$. | ||||
| Q.12 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The reflected ray is fully polarized when the reflection at the water–medium interface occurs at Brewster's angle: $\tan\theta_B=\frac{n_p}{n_w}=\frac{4/\sqrt3}{4/3}=\sqrt3$, so $\theta_B=60^\circ$.
All the interfaces are parallel, so $n\sin(\text{angle})$ is the same in every layer. The ray in the lower water and the ray in the upper water have the same index, so they make the same angle with the normal.
$i=60^\circ$. | ||||
| Q.13 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter With $K$ open: $I_1=\frac{10}{R_1+6}$.
With $K$ closed: $G$ ($6\ \Omega$) and $R_2$ ($4\ \Omega$) are in parallel, which is $2.4\ \Omega$. The total current is $I=\frac{10}{R_1+2.4}$, and $G$ carries the fraction $\frac{4}{10}$ of it.
Half deflection: $\frac{0.4\times10}{R_1+2.4}=\frac12\cdot\frac{10}{R_1+6}$, so $0.4(R_1+6)=0.5(R_1+2.4)$ and $R_1=12\ \Omega$.
In this condition the current through $R_1$ is $\frac{10}{14.4}$ A $\approx694.44$ mA. | ||||
| Q.14 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | NumericalMedium | |
A medium numerical question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\sqrt{\mu_0/\epsilon_0}$ has the dimensions of resistance, $[ML^2T^{-3}A^{-2}]$.
One new unit is $5\cdot5^2\cdot5^{-3}\cdot5^{-2}=5^{-2}=\frac{1}{25}$ of the SI unit.
So one SI unit equals $25$ new units. | ||||
| Q.15 | 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 Each chamber has floor area $A=1$ m$^2$ and the hole has area $a=\sqrt{10}\times10^{-4}$ m$^2$. Let the left and right heights be $h_L,h_R$ with $h_L+h_R=2$, and let $\Delta=h_L-h_R$.
Both surfaces are at atmospheric pressure, so the efflux speed is $\sqrt{2g\Delta}$.
$\frac{d\Delta}{dt}=-\frac{2a}{A}\sqrt{2g\Delta}$, so $\frac{d\sqrt\Delta}{dt}=-\frac aA\sqrt{2g}=-\sqrt{10}\times10^{-4}\times\sqrt{20}=-\sqrt{2}\times10^{-3}$ s$^{-1}$.
$\sqrt\Delta$ starts at $\sqrt2$ and falls by $500\times\sqrt2\times10^{-3}=\frac{\sqrt2}{2}$. So $\sqrt\Delta=\frac{1}{\sqrt2}$ and $\Delta=0.5$ m.
$h_L=\frac{2+0.5}{2}=1.25$ m. | ||||
| Q.16 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The two chambers are capacitors in parallel, each of area $1$ m$^2$ with plate gap $2$ m. Within a chamber a liquid layer of height $h$ and an air layer are in series: $C=\frac{\epsilon_0}{\frac{h}{15}+(2-h)}$.
At $t=0$: left $h=2$ gives $7.5\epsilon_0$, right $h=0$ gives $0.5\epsilon_0$. Total $8\epsilon_0$.
At $t=500$ s: $h_L=1.25$ gives $\frac{\epsilon_0}{0.0833+0.75}=1.2\epsilon_0$, and $h_R=0.75$ gives $\frac{\epsilon_0}{0.05+1.25}=0.769\epsilon_0$. Total $1.969\epsilon_0$.
The difference is $(8-1.969)\epsilon_0$, so $n\approx1.97$. | ||||
| Q.17 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalHard | |
A hard numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take the pivot as origin. $P$ is at $(0.2\sin45^\circ,\ -0.2-0.2\cos45^\circ)=(0.1414,\ -0.3414)$ m.
Angular momentum about $C$ is conserved in the collision, since the pivot's force has no torque about $C$.
Before: momentum $(-2,0)$ gives $L_z=xp_y-yp_x=-(-0.3414)(-2)=-0.683$.
After: momentum $(0,-1.8)$ gives $L_z=0.1414\times(-1.8)=-0.255$.
The disk receives $0.428$ kg m$^2$ s$^{-1}$. Its moment of inertia about $C$ is $\frac32\times1\times0.04=0.06$ kg m$^2$, so $\omega=7.14$ rad s$^{-1}$.
Kinetic energy $=\frac12\times0.06\times7.14^2=1.53$ J $=Mgh$, so $h\approx0.15$ m. | ||||
| 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 Particle before: $\frac12\times0.02\times100^2=100$ J.
Particle after: $\frac12\times0.02\times90^2=81$ J.
The disk just after the collision has $\frac12I\omega^2=\frac12\times0.06\times7.14^2\approx1.53$ J (from angular momentum conservation about $C$, as in the previous part).
Energy lost $=100-81-1.53\approx17.47$ J. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| Q.1 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry For strong electrolytes $\Lambda=\Lambda^\circ-A\sqrt c$. The two concentrations have $\sqrt c=0.1$ and $0.2$, so $\Lambda^\circ$ is the first value plus the drop between the two.
NaNO$_3$: $111+10=121$. NaCl: $117+10=127$. AgNO$_3$: $125+9=134$.
Kohlrausch's law: $\Lambda^\circ(\text{AgCl})=134+127-121=140$ S cm$^2$ mol$^{-1}$.
Solubility: $X=\frac{1000\kappa}{\Lambda^\circ}=\frac{1000\times1.40\times10^{-6}}{140}=10^{-5}$ mol L$^{-1}$.
$\log_{10}(X^{-1})=5$.
Answer: (C). | ||||
| Q.2 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure NO$_2^+$ has no lone electrons on nitrogen and is linear: $180^\circ$.
NO$_2$ has one unpaired electron, which repels the bonds only weakly: about $134^\circ$.
NO$_3^-$ is trigonal planar: $120^\circ$.
NO$_2^-$ has a full lone pair on nitrogen, which squeezes the bonds below the trigonal angle: about $115^\circ$.
Order: NO$_2^-$ < NO$_3^-$ < NO$_2$ < NO$_2^+$.
Answer: (B). | ||||
| Q.3 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Natural rubber is cis-polyisoprene, $-[\text{CH}_2\text{C(CH}_3)=\text{CHCH}_2]-$. Cutting every C=C gives $\mathbf X$, CH$_3$COCH$_2$CH$_2$CHO (4-oxopentanal). It has a methyl ketone (iodoform) and an aldehyde (Tollens').
In base the methyl group's enolate attacks the aldehyde, closing a five-membered ring, and heating removes water. The result is cyclopent-2-en-1-one.
Answer: (A). | ||||
| Q.4 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctHard | |
A hard single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules The sweetener is sucralose. Three things fix the drawing.
Chlorine at C-4 of galactose: the C-4 hydroxyl of D-galactose (in the Fischer projection, on the left at C-4) is axial in the chair, so chlorine must be axial, pointing up. That leaves (A) and (D); in (B) and (C) it is equatorial, as in a glucose ring.
$\alpha$-Glycoside: the anomeric oxygen of the pyranose must be axial, pointing down. All the options show this.
$\beta$-Fructose: the furanose ring is drawn turned over, as in the usual picture of sucrose, so up and down are swapped compared with the standard Haworth form. In $\beta$-D-fructofuranose the anomeric oxygen is up and C-1 is down in the standard form, so here the linking oxygen goes down and CH$_2$Cl (C-1) goes up. This is (A); (D) has them the other way round.
Answer: (A). | ||||
| Q.5 | 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 $[\mathbf R]=[\mathbf R]_0e^{-kt}$ and $[\mathbf P]=[\mathbf R]_0(1-e^{-kt})$.
(A) False: $[\mathbf P]$ rises steeply at first and then levels off, so the curve bends the other way.
(B) False: $\frac{d[\mathbf R]}{dt}=-k[\mathbf R]$ is negative and has slope $-k$; the plot shows a positive slope.
(C) True: $\frac{d[\mathbf P]}{dt}=k[\mathbf R]_0e^{-kt}$ is an exponential decay.
(D) True: $k$ does not change with time at a fixed temperature.
Answer: (C), (D). | ||||
| Q.6 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Multiple correctMedium | |
A medium multiple correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Xe with F$_2$ in the ratio $1:5$ at $873$ K and $7$ bar gives $\mathbf P$, XeF$_4$. With O$_2$F$_2$ it gives $\mathbf Q$, XeF$_6$. Complete hydrolysis gives $\mathbf R$, XeO$_3$.
(A) True: XeF$_4$ has two lone pairs on xenon.
(B) False: XeF$_6$ has a lone pair, which makes it a distorted octahedron.
(C) True: XeF$_6$ readily gives up fluorine.
(D) True: XeO$_3$ has three bonding directions and one lone pair, so it is trigonal pyramidal.
Answer: (A), (C), (D). | ||||
| Q.7 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Multiple correctEasy | |
A easy multiple correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties (A) True: C$^+$ loses a $2p$ electron, while B$^+$ must lose a $2s$ electron from a filled subshell. So B has the larger second ionization enthalpy.
(B) True: these ions all have the same electron count, and more nuclear charge pulls them in, so Al$^{3+}$ is smallest and Na$^+$ largest.
(C) False: potassium ($0.86$ g cm$^{-3}$) is less dense than sodium ($0.97$ g cm$^{-3}$).
(D) False: H–H (about $436$ kJ mol$^{-1}$) is much stronger than F–F (about $155$ kJ mol$^{-1}$).
Answer: (A), (B). | ||||
| Q.8 | AminesDiazonium salts | Class 12: Amines | Multiple correctMedium | |
A medium multiple correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Acetylation then nitration gives $\mathbf P$, $p$-nitroacetanilide, with one nitro group. Hydrolysis and diazotization give $\mathbf Q$, $p$-nitrobenzenediazonium chloride.
Cumene gives phenol by the air-oxidation route, and the Kolbe reaction then gives $\mathbf S$, salicylic acid. Coupling with $\mathbf Q$ in base gives $\mathbf T$, an azo compound.
(A) False: ethanol reduces a diazonium salt to the arene (nitrobenzene here), not to an aldehyde.
(B) True: $\mathbf S$ is a phenol, which gives the phthalein test.
(C) False: $\mathbf P$ has only one nitro group.
(D) True: azo compounds are coloured.
Answer: (B), (D). | ||||
| Q.9 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Multiple correctMedium | |
A medium multiple correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules (A) False: HNO$_3$ oxidizes both the aldehyde and the primary alcohol end of glucose, so glucose also gives saccharic acid.
(B) True: in base fructose rearranges through an enediol to glucose and mannose, both aldoses that reduce Fehling's solution.
(C) True: hydrolysis of sucrose gives one mole each of D-glucose and D-fructose.
(D) False: the D forms have rotations opposite to the given L forms: $+52.5^\circ$ and $-92.5^\circ$. The equimolar mixture has $\frac{52.5-92.5}{2}=-20^\circ$.
Answer: (B), (C). | ||||
| Q.10 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom $\frac1\lambda=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$.
$\mathbf X$: $\frac1\lambda=RZ_1^2\cdot\frac34$. $\mathbf Y$: $\frac{1}{9\lambda}=RZ_2^2\cdot\frac{3}{16}$.
Dividing: $9=\frac{Z_1^2\cdot\frac34}{Z_2^2\cdot\frac{3}{16}}=\frac{4Z_1^2}{Z_2^2}$, so $Z_2=\frac23Z_1$.
The smallest whole values are $Z_1=3$ and $Z_2=2$. A hydrogen-like ion of nuclear charge $Z$ has charge $Z-1$, so $a=2$ and $b=1$.
$a+b=3$. | ||||
| Q.11 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter From pH $3$: $[\text{H}^+]=10^{-3}$ M. Then $K_a=\frac{[\text{H}^+]^2}{C}$ gives $C=\frac{10^{-6}}{10^{-5}}=0.1$ M.
Acid left in $50$ mL: $0.1\times0.05=0.005$ mol, which is $0.30$ g.
Adsorbed: $x=0.45-0.30=0.15$ g, with $m=1.0$ g.
A slope of $1$ means $\frac1n=1$, so $\frac xm=kC$ and $k=\frac{0.15}{0.1}=1.5$ L mol$^{-1}$. | ||||
| Q.12 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalHard | |
A hard numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions With $M$ the molar mass of $\mathbf S$ in kg mol$^{-1}$: $K_b=\frac{RT_b^2M}{\Delta H_{\text{vap}}}=\frac{R(400)^2M}{10R}=16000M$.
In $100$ g of solution there are $0.25$ g of $\mathbf B$, in $0.09975$ kg of $\mathbf S$. The moles of $\mathbf B$ are $\frac{0.25\times10^{-3}}{10M}$, so the molality is $\frac{2.506\times10^{-4}}{M}$.
$\Delta T_b=iK_bm$: $8=i\times16000M\times\frac{2.506\times10^{-4}}{M}=4.01\,i$, so $i=1.995$.
One $\mathbf B$ gives four particles when it dissociates, so $i=1+3\alpha$. This gives $\alpha=0.332$, about $33\%$. | ||||
| Q.13 | 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 An octahedron's vertices form three opposite pairs, and each of its $8$ faces takes one vertex from each pair.
$cis$-[Co(NH$_3$)$_4$Cl$_2$]$^+$: the pairs are (Cl, N), (Cl, N), (N, N). A face with two Cl must take Cl from the first two pairs and either N from the third: $2$ faces.
$mer$-[Co(NH$_3$)$_3$Cl$_3$]: the pairs are (Cl, Cl), (Cl, N), (N, N). A face with exactly two Cl takes either Cl from the first pair, the Cl from the second, and either N from the third: $2\times1\times2=4$ faces.
Total $=2+4=6$. | ||||
| Q.14 | PolymersBiodegradable polymers | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on biodegradable polymers, from Polymers. NCERT: No matching NCERT chapter CH$_3$I $\to$ CH$_3$CN $\to$ CH$_3$COOH $\to$ BrCH$_2$COOH (Hell–Volhard–Zelinsky) $\to$ H$_2$NCH$_2$COOH. So $\mathbf X$ is glycine, molar mass $75$.
Acid hydrolysis of caprolactam gives $\mathbf Y$, 6-aminocaproic acid, molar mass $131$.
They condense into the acyclic copolymer nylon-2,6. Joining $1000$ monomers in a chain makes $999$ amide bonds, each losing one H$_2$O.
Mass $=500\times75+500\times131-999\times18=37500+65500-17982=85018$ g. | ||||
| Q.15 | SolutionsRaoult's law | Class 12: Solutions | NumericalMedium | |
A medium numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions A $5$ molal solution has $5$ mol of $\mathbf B$ in $1000$ g of $\mathbf A$, which is $20$ mol of $\mathbf A$. So $x_{\mathbf B}=\frac{5}{25}=0.2$ and $x_{\mathbf A}=0.8$.
Raoult's law: $100=0.8\times105+0.2\,P^\circ_{\mathbf B}$, so $P^\circ_{\mathbf B}=80$ mm Hg.
Vapour molar volume at that pressure: $\frac{RT}{P}=\frac{0.08\times300}{80/760}=228$ L mol$^{-1}$.
Liquid molar volume: $\frac{57}{0.5}=114$ mL mol$^{-1}=0.114$ L mol$^{-1}$.
Ratio $=\frac{228}{0.114}=2000$. | ||||
| Q.16 | SolutionsRaoult's law | Class 12: Solutions | NumericalEasy | |
A easy numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions From Raoult's law, as in the previous part, $x_{\mathbf B}=0.2$ and $P^\circ_{\mathbf B}=\frac{100-0.8\times105}{0.2}=80$ mm Hg.
The partial pressure of $\mathbf B$ is $0.2\times80=16$ mm Hg.
By Dalton's law, $y_{\mathbf B}=\frac{16}{100}=0.16$. | ||||
| Q.17 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Friedel–Crafts acylation of $m$-xylene gives ArCOCH$_2$Cl, with Ar = 2,4-dimethylphenyl. NaI gives ArCOCH$_2$I, and the phenoxide displaces iodide: $\mathbf J$ is ArCOCH$_2$OC$_6$H$_4$NO$_2$.
NaBH$_4$ then PBr$_3$ turns the C=O into CHBr: $\mathbf K$ is ArCHBrCH$_2$OC$_6$H$_4$NO$_2$. Its molar mass is $105+93+14+16+122=350$, as stated.
Excess NH$_3$ replaces Br by NH$_2$: $\mathbf L$ has molar mass $350-80+16=286$.
$5.72$ g is $0.02$ mol. Kjeldahl's method does not convert nitro nitrogen to ammonia, so only the amine nitrogen counts: $0.02$ mol of NH$_3$.
$2$NH$_3$ + H$_2$SO$_4$ needs $0.01$ mol of acid, which is $10$ mL of $1$ M solution. | ||||
| Q.18 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $\mathbf K$ is ArCHBrCH$_2$OC$_6$H$_4$NO$_2$ (molar mass $350$), from the sequence shown.
PhSNa displaces bromide: $\mathbf M$ has molar mass $350-80+109=379$, with one sulphur atom.
$3.79$ g is $0.01$ mol, which gives $0.01$ mol of BaSO$_4$ (molar mass $233$).
Mass $=2.33$ g. | ||||
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