JEE (Main) · 2024
JEE (Main) 2024 — 6 April, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Sequences and Series3
- Limits, Continuity and Differentiability3
- Integral Calculus3
9 easy · 15 medium · 6 hard
Physics
30 Q
- Magnetic Effects of Current and Magnetism3
- Laws of Motion2
- Kinematics2
23 easy · 6 medium · 1 hard
Chemistry
30 Q
- Coordination Compounds3
- Solutions2
- Chemical Bonding and Molecular Structure2
17 easy · 12 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct 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 Since $-1\le \sin 5x\le 1$, we get $6\le 7-\sin 5x\le 8$, and every value in $[6,8]$ is attained.
Taking reciprocals (all positive) reverses the inequalities: $\frac{1}{8}\le f(x)\le\frac{1}{6}$.
Range $=\left[\frac{1}{8},\frac{1}{6}\right]$.
Answer: (D). | ||||
| 2 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions List the pairs with $4x\le 5y$, i.e. $y\ge \frac{4x}{5}$:
$x=1$: $y\ge 0.8$ → $y=1,2,3,4,5$ (5 pairs)
$x=2$: $y\ge 1.6$ → $y=2,3,4,5$ (4)
$x=3$: $y\ge 2.4$ → $y=3,4,5$ (3)
$x=4$: $y\ge 3.2$ → $y=4,5$ (2)
$x=5$: $y\ge 4$ → $y=4,5$ (2)
So $m=16$.
For symmetry, each $(x,y)\in R$ needs $(y,x)\in R$. The off-diagonal pairs are $(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5),(5,4)$. Of these only $(4,5)$ and $(5,4)$ already have their reverses; the other 9 do not, so $n=9$.
$m+n=16+9=25$.
Answer: (C). | ||||
| 3 | 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 $|z_1-2z_2|^2=4\left|\tfrac{1}{2}-z_1\bar z_2\right|^2$.
Left: $|z_1|^2+4|z_2|^2-2(z_1\bar z_2+\bar z_1 z_2)$.
Right: $4\left(\tfrac14+|z_1|^2|z_2|^2-\tfrac12(z_1\bar z_2+\bar z_1 z_2)\right)=1+4|z_1|^2|z_2|^2-2(z_1\bar z_2+\bar z_1 z_2)$.
The cross terms cancel: $|z_1|^2+4|z_2|^2-1-4|z_1|^2|z_2|^2=0$, which factors as $(|z_1|^2-1)(1-4|z_2|^2)=0$.
So $|z_1|=1$ or $|z_2|=\frac12$: either $z_1$ lies on a circle of radius 1 or $z_2$ lies on a circle of radius $\frac12$.
Answer: (D). | ||||
| 4 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctHard | |
A hard single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3\times3$ matrix $M$: $\det(\mathrm{adj}\,M)=(\det M)^2$ and $\det(kM)=k^3\det M$.
$\det((2A)^{-1})=\frac{1}{2^3\cdot 3}=\frac{1}{24}$.
$\det(\mathrm{adj}((2A)^{-1}))=\frac{1}{24^2}$; then $\det(3\,\mathrm{adj}(\cdot))=\frac{27}{576}=\frac{3}{2^6}$. Call this $d$.
$\det(\mathrm{adj}(\cdot))=d^2$; $\det(-3\,\mathrm{adj}(\cdot))=-27d^2$.
$\det(\mathrm{adj}(\cdot))=729d^4$; $\det(-4\,\mathrm{adj}(\cdot))=-64\cdot729d^4$.
Outermost adjoint: $(64\cdot 729\,d^4)^2=2^{12}3^{12}d^8=2^{12}3^{12}\cdot\frac{3^8}{2^{48}}=2^{-36}3^{20}$.
So $m=-36$, $n=20$ and $m+2n=-36+40=4$.
Answer: (B). | ||||
| 5 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Letters in alphabetical order: A, G, N, P, R, U (all different).
Words beginning with A: $5!=120$; with G: 120. Total 240.
Words beginning with N occupy positions 241–360. Within them:
NA…: $4!=24$ (241–264), NG…: 24 (265–288), NP…: 24 (289–312).
NR… starts at 313. NRA… with the remaining G, P, U in order: NRAGPU (313), NRAGUP (314), NRAPGU (315).
Answer: (B). | ||||
| 6 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations ${}^{n+1}C_{r+1}=\frac{n+1}{r+1}\,{}^{n}C_{r}$ and ${}^{n}C_{r}=\frac{n}{r}\,{}^{n-1}C_{r-1}$.
So $\frac{n+1}{r+1}=\frac{55}{35}=\frac{11}{7}$ and $\frac{n}{r}=\frac{35}{21}=\frac53$.
From the second, $n=\frac{5r}{3}$. The first gives $7n+7=11r+11$, so $\frac{35r}{3}-11r=4$, $\frac{2r}{3}=4$, $r=6$ and $n=10$.
$2n+5r=20+30=50$.
Answer: (A). | ||||
| 7 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Measure work in system-days. The assignment needs $17m$.
It took $17+8=25$ days, with $m, m-4, m-8,\dots$ systems working on successive days.
Work done $=\sum_{k=0}^{24}(m-4k)=25m-4\cdot\frac{24\cdot25}{2}=25m-1200$.
$25m-1200=17m$ gives $8m=1200$, $m=150$.
Answer: (A). | ||||
| 8 | 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 $\ln f(x)=-2x\ln x$; its derivative $-2(\ln x+1)$ vanishes at $x=\frac1e$, where $f$ has its maximum (the derivative changes from $+$ to $-$).
So for every $x>0$, $x\ne\frac1e$: $\left(\frac1x\right)^{2x}<e^{2/e}$. Put $x=\frac1t$: $t^{2/t}<e^{2/e}$, i.e. $t^{1/t}<e^{1/e}$ for $t\ne e$.
With $t=\pi$: $\pi^{1/\pi}<e^{1/e}$; raising to the power $e\pi$: $\pi^{e}<e^{\pi}$. So (A) is true and (B) false.
(C): with $t=2\pi$, $(2\pi)^{e}<e^{2\pi}$, so (C) is false.
(D): compare logarithms: $\pi\ln(2e)=\pi(1+\ln2)\approx5.32$ and $2e\ln\pi\approx6.22$, so $(2e)^\pi<\pi^{2e}$; (D) is false.
Answer: (A). | ||||
| 9 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Numerator $=\sum_{k=1}^{n-1}(k^2-k)(n-k)=n\sum k^2-\sum k^3-n\sum k+\sum k^2$ (sums to $n-1$). Its leading term is $n\cdot\frac{n^3}{3}-\frac{n^4}{4}=\frac{n^4}{12}$; the rest are of lower degree.
Denominator $=\frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2n+1)}{6}$, leading term $\frac{n^4}{4}$.
Limit $=\frac{1/12}{1/4}=\frac13$.
Answer: (B). | ||||
| 10 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $g'(x)=h'(e^x)\,e^x\,e^{h(x)}+h(e^x)\,e^{h(x)}\,h'(x)$.
At $x=0$: $g'(0)=h'(1)\cdot1\cdot e^{h(0)}+h(1)\,e^{h(0)}\,h'(0)=2\cdot1\cdot1+1\cdot1\cdot2=4$.
Answer: (B). | ||||
| 11 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals For $1\le x\le2$ and $0<a<1$, $\frac{a}{x^2}\le\frac1x$ (since $a\le x$), so
Area $=\int_1^2\left(\frac1x-\frac{a}{x^2}\right)dx=\ln2-a\left(1-\frac12\right)=\ln2-\frac a2$.
$\frac a2=\frac17$, so $a=\frac27$ and $7a-3=2-3=-1$.
Answer: (A). | ||||
| 12 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Divide numerator and denominator by $\cos^2x$ and put $t=\tan x$:
$\int\frac{\sec^2x\,dx}{a^2\tan^2x+b^2}=\int\frac{dt}{a^2t^2+b^2}=\frac{1}{ab}\tan^{-1}\left(\frac{a\tan x}{b}\right)+C$.
Comparing: $\frac ab=3$ and $ab=12$, so $a^2=36$, $b^2=4$.
Maximum of $a\sin x+b\cos x$ is $\sqrt{a^2+b^2}=\sqrt{40}$.
Answer: (C). | ||||
| 13 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Write the equation as $\left[(2+\alpha)x-\beta y+2\right]dx+\left[-\beta x+2\alpha y+\beta\gamma-4\alpha\right]dy=0$.
The terms $-\beta y\,dx-\beta x\,dy=-\beta\,d(xy)$ would give an $xy$ term, which a circle cannot have; so $\beta=0$.
With $\beta=0$ the equation is exact: $\left[(2+\alpha)x+2\right]dx+(2\alpha y-4\alpha)dy=0$, which integrates to
$\frac{2+\alpha}{2}x^2+2x+\alpha y^2-4\alpha y=C$.
For a circle the coefficients of $x^2$ and $y^2$ are equal: $\frac{2+\alpha}{2}=\alpha$, so $\alpha=2$.
Then $2x^2+2x+2y^2-8y=C$; through the origin $C=0$: $x^2+y^2+x-4y=0$.
Centre $\left(-\frac12,2\right)$, radius $\sqrt{\frac14+4}=\frac{\sqrt{17}}{2}$.
Answer: (A). | ||||
| 14 | Straight LinesCartesian coordinates and distance formula | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines $4\sqrt{(x-2)^2+(y-1)^2}=5\sqrt{(x-1)^2+(y-3)^2}$.
$16(x^2+y^2-4x-2y+5)=25(x^2+y^2-2x-6y+10)$
$0=9x^2+9y^2+14x-118y+170$.
The constant term is already 170, so $a=9$, $b=9$, $c=0$, $d=14$, $e=-118$.
$a^2+2b+3c+4d+e=81+18+0+56-118=37$.
Answer: (D). | ||||
| 15 | 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 $\overrightarrow{AP}=(1,3)$ is perpendicular to $\overrightarrow{BC}=(h-8,k-3)$: $(h-8)+3(k-3)=0$, i.e. $h+3k=17$.
$\overrightarrow{BP}=(-2,-2)$ is perpendicular to $\overrightarrow{AC}=(h-5,k+2)$: $-2(h-5)-2(k+2)=0$, i.e. $h+k=3$.
Solving: $k=7$, $h=-4$. Then $h^2+k^2=16+49=65$, so C lies on $x^2+y^2-65=0$.
Answer: (C). | ||||
| 16 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the side of ABC be $a$. The sides are $a,\frac a2,\frac a4,\dots$ (ABC included).
$P=3a\left(1+\frac12+\frac14+\cdots\right)=6a$.
$Q=\frac{\sqrt3}{4}a^2\left(1+\frac14+\frac1{16}+\cdots\right)=\frac{\sqrt3}{4}a^2\cdot\frac43=\frac{a^2}{\sqrt3}$.
$P^2=36a^2=36\sqrt3\,Q$.
Answer: (C). | ||||
| 17 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line is $(t,3+t,1-t)$. For the foot F of the perpendicular from Q: $(3-t,-6-t,t)\cdot(1,1,-1)=0$, so $-3-3t=0$, $t=-1$ and $F=(-1,2,2)$.
Image $P=2F-Q=(-5,7,3)$.
$\overrightarrow{QP}=(-8,10,2)$, $\overrightarrow{QR}=(-1,8,-2)$.
$\overrightarrow{QP}\times\overrightarrow{QR}=(10(-2)-2\cdot8,\ -[(-8)(-2)-2(-1)],\ (-8)8-10(-1))=(-36,-18,-54)$.
Its square magnitude is $1296+324+2916=4536$, so $\lambda^2=\frac{4536}{4}=1134=14K$ and $K=81$.
Answer: (D). | ||||
| 18 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec a\times(\hat i+\hat j)=(2,1,-1)\times(1,1,0)=(1,-1,1)$.
$(1,-1,1)\times\hat i=(0,1,1)$.
$(0,1,1)\times\hat i=(0,1,-1)=\vec b$.
Projection of $\vec a$ on $\vec b$: $\frac{\vec a\cdot\vec b}{|\vec b|}=\frac{1+1}{\sqrt2}=\sqrt2$; its square is 2.
Answer: (B). | ||||
| 19 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\vec c-\vec a|^2=|\vec c|^2+|\vec a|^2-2\vec a\cdot\vec c$: $8=|\vec c|^2+38-12|\vec c|$, so $|\vec c|^2-12|\vec c|+30=0$ and $|\vec c|=6\pm\sqrt6$. As $|\vec c|\ge6$, $|\vec c|=6+\sqrt6$.
$\vec a\times\vec b=(6,1,-1)\times(1,1,0)=(1,-1,5)$, $|\vec a\times\vec b|=\sqrt{27}=3\sqrt3$.
$\left|(\vec a\times\vec b)\times\vec c\right|=3\sqrt3\,(6+\sqrt6)\sin60^\circ=3\sqrt3\cdot\frac{\sqrt3}{2}(6+\sqrt6)=\frac92(6+\sqrt6)$.
Answer: (D). | ||||
| 20 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each letter can go to any of 5 addresses: $5^3=125$ equally likely ways.
Exactly two addresses used: choose them in $\binom52=10$ ways; distribute the 3 letters so that both are used: $2^3-2=6$ ways. Favourable $=60$.
Probability $=\frac{60}{125}=\frac{12}{25}$.
Answer: (C). | ||||
| 21 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Each root satisfies $t^2=-\sqrt2t+8$; multiplying by $t^8$: $t^{10}+\sqrt2t^9=8t^8$.
Adding for $\alpha$ and $\beta$: $U_{10}+\sqrt2U_9=8U_8$.
So $\frac{U_{10}+\sqrt2U_9}{2U_8}=\frac{8U_8}{2U_8}=4$.
Answer: 4. | ||||
| 22 | 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 For infinitely many solutions, $\Delta=\Delta_x=\Delta_y=\Delta_z=0$.
$\Delta_z=\begin{vmatrix}2&7&3\\3&2&4\\1&\mu&-1\end{vmatrix}=2(-2-4\mu)-7(-3-4)+3(3\mu-2)=39+\mu=0$, so $\mu=-39$.
$\Delta_y=\begin{vmatrix}2&3&\lambda\\3&4&5\\1&-1&32\end{vmatrix}=2(133)-3(91)+\lambda(-7)=-7-7\lambda=0$, so $\lambda=-1$.
Check: $\Delta=2(64-5\mu)-7(96-5)+\lambda(3\mu-2)=128+390-637+119=0$ with $\lambda=-1,\mu=-39$, and $\Delta_x=3(64+195)-7(128+5)-(4\mu+2)=777-931+154=0$.
$\lambda-\mu=-1+39=38$.
Answer: 38. | ||||
| 23 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalHard | |
A hard numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Put $y=1+x$: $S=\sum_{k=1}^{60}ky^k$. The standard arithmetico-geometric sum is
$S=\frac{y\left(1-61y^{60}+60y^{61}\right)}{(1-y)^2}$, and $(1-y)^2=x^2$.
At $x=60$, $y=61$: $1-61\cdot61^{60}+60\cdot61^{61}=1+59\cdot61^{61}$.
So $60^2S(60)=61\left(1+59\cdot61^{61}\right)=59\cdot61^{62}+61=(59\cdot61)\,61^{61}+61$.
This is $a(b)^b+b$ with $b=61$ and $a=59\cdot61=3599$.
$a+b=3599+61=3660$.
Answer: 3660. | ||||
| 24 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f(x)=\left[\frac x2\right]+3-[\sqrt x]$.
$\left[\frac x2\right]$ jumps at $x=2,4,6,8$; $[\sqrt x]$ jumps at $x=1,4$ in $[0,8]$.
At $x=4$ both jump up by 1, so $f$ is continuous there: just left of 4, $f=1+3-1=3$; at and right of 4, $f=2+3-2=3$.
At $x=1,2,6$ only one term jumps, so $f$ is discontinuous.
At $x=8$: $f(8)=4+3-2=5$ while the left-hand limit is $3+3-2=4$, so $f$ is discontinuous at 8 (it is defined on $[0,\infty)$).
At $x=0$, $f$ is continuous from the right ($f=3$).
$S=\{1,2,6,8\}$, sum $=17$.
Answer: 17. | ||||
| 25 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\int_0^3[x^2]dx$: $[x^2]=k$ on $[\sqrt k,\sqrt{k+1})$ for $k=0,\dots,8$.
$\sum_{k=1}^{8}k(\sqrt{k+1}-\sqrt k)=8\cdot3-\sum_{k=1}^{8}\sqrt k=24-(1+\sqrt2+\sqrt3+2+\sqrt5+\sqrt6+\sqrt7+2\sqrt2)$
$=21-3\sqrt2-\sqrt3-\sqrt5-\sqrt6-\sqrt7$.
$\int_0^3\left[\frac{x^2}{2}\right]dx$: the value is $k$ on $[\sqrt{2k},\sqrt{2k+2})$; on $[\sqrt8,3]$ it is 4.
$1(2-\sqrt2)+2(\sqrt6-2)+3(2\sqrt2-\sqrt6)+4(3-2\sqrt2)=10-3\sqrt2-\sqrt6$.
Total $=31-6\sqrt2-\sqrt3-\sqrt5-2\sqrt6-\sqrt7$, so $a=31$, $b=-6$, $c=-2$.
$a+b+c=23$.
Answer: 23. | ||||
| 26 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalEasy | |
A easy numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $\frac{e^y}{e^y+1}dy=-\frac{\cos x}{\sin x}dx$, so $\ln(e^y+1)=-\ln\sin x+C$, i.e. $(e^y+1)\sin x=k$.
At $\left(\frac\pi2,0\right)$: $(1+1)\cdot1=2$, so $k=2$.
At $x=\frac\pi6$: $(e^y+1)\cdot\frac12=2$, so $e^{y(\pi/6)}=3$.
Answer: 3. | ||||
| 27 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Take the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Latus rectum: $\frac{2b^2}{a}=9$, so $b^2=\frac{9a}{2}$.
The tangent $y=\sqrt3x-\sqrt3$ has $m=\sqrt3$, $c=-\sqrt3$; tangency needs $c^2=a^2m^2-b^2$: $3=3a^2-\frac{9a}{2}$, i.e. $2a^2-3a-2=0$, so $a=2$ and $b^2=9$.
Then $e^2=1+\frac94=\frac{13}{4}$, $4e^2=13$ (directrices $x=\pm\frac ae=\pm\frac{4}{\sqrt{13}}$).
Point of contact: the tangent at $(x_0,y_0)$ is $\frac{xx_0}{4}-\frac{yy_0}{9}=1$; comparing with $x-\frac{y}{\sqrt3}=1$ gives $x_0=4$, $y_0=3\sqrt3$ (check: $\frac{16}{4}-\frac{27}{9}=1$).
Focal distances: $ex_0\pm a=2\sqrt{13}\pm2$; product $m=4\cdot13-4=48$.
$4e^2+m=13+48=61$.
Answer: 61. | ||||
| 28 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $A=(\lambda,2,1)$, $\vec d_1=(3,-1,1)$; $B=(-2,-5,4)$, $\vec d_2=(-3,2,4)$.
$\vec d_1\times\vec d_2=(-4-2,\ -(12+3),\ 6-3)=(-6,-15,3)$, magnitude $\sqrt{270}=3\sqrt{30}$.
$\overrightarrow{AB}=(-2-\lambda,-7,3)$; $\overrightarrow{AB}\cdot(\vec d_1\times\vec d_2)=12+6\lambda+105+9=6\lambda+126$.
$\frac{|6\lambda+126|}{3\sqrt{30}}=\frac{44}{\sqrt{30}}$ gives $|6\lambda+126|=132$: $\lambda=1$ or $\lambda=-43$.
Largest $|\lambda|=43$.
Answer: 43. | ||||
| 29 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalHard | |
A hard numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability X is hypergeometric: $P(X=k)=\frac{\binom3k\binom9{5-k}}{\binom{12}5}$, $\binom{12}{5}=792$.
$P(0)=\frac{126}{792}$, $P(1)=\frac{3\cdot126}{792}=\frac{378}{792}$, $P(2)=\frac{3\cdot84}{792}=\frac{252}{792}$, $P(3)=\frac{36}{792}$.
$E(X)=\frac{378+504+108}{792}=\frac{990}{792}=\frac54$.
$E(X^2)=\frac{378+1008+324}{792}=\frac{1710}{792}=\frac{95}{44}$.
$\mathrm{Var}(X)=\frac{95}{44}-\frac{25}{16}=\frac{380-275}{176}=\frac{105}{176}$ (the same as $5\cdot\frac14\cdot\frac34\cdot\frac{7}{11}$).
$\gcd(105,176)=1$, so $n-m=176-105=71$.
Answer: 71. | ||||
| 30 | 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 Cosine rule at A: $49=64+\alpha^2-2\cdot8\alpha\cdot\frac23$, so $3\alpha^2-32\alpha+45=0$ and $\alpha=9$ or $\frac53$. As $\alpha\in\mathbb N$, $\alpha=9$.
$\cos C=\frac{7^2+8^2-9^2}{2\cdot7\cdot8}=\frac{32}{112}=\frac27$.
$\cos3C=4\cos^3C-3\cos C=\frac{32}{343}-\frac67=\frac{32-294}{343}=-\frac{262}{343}$.
$49\cos3C+42=-\frac{262}{7}+42=\frac{32}{7}$, so $m+n=39$.
Answer: 39. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion At the maximum speed friction acts down the incline:
$v_{max}^2=rg\,\frac{\tan\theta+\mu}{1-\mu\tan\theta}$.
$\tan30^\circ=\frac{1}{1.73}=0.578$: $\frac{0.578+0.2}{1-0.2\times0.578}=\frac{0.778}{0.884}=0.880$.
$v_{max}^2=300\times10\times0.880\approx2640$, so $v_{max}\approx51.4$ m/s.
Answer: (A). | ||||
| 32 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | StatementsEasy | |
A easy statements question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Specific heat $c=\frac{Q}{m\Delta T}$: $\frac{\mathrm{ML^2T^{-2}}}{\mathrm{M\,K}}=[\mathrm{L^2T^{-2}K^{-1}}]$. Statement (I) is correct.
Gas constant $R=\frac{PV}{nT}$: energy per mole per kelvin, $[\mathrm{ML^2T^{-2}K^{-1}mol^{-1}}]$. The time exponent is $-2$, not $-1$, so Statement (II) is incorrect.
Answer: (C). | ||||
| 33 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctMedium | |
A medium single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Take downward positive, tower height $h$, speed $u$.
Upward throw: $h=-ut_1+\frac12gt_1^2$. Downward throw: $h=ut_2+\frac12gt_2^2$.
Multiply the first by $t_2$, the second by $t_1$ and add: $h(t_1+t_2)=\frac12gt_1t_2(t_1+t_2)$, so $h=\frac12gt_1t_2$.
Dropped: $h=\frac12gt^2$, hence $t=\sqrt{t_1t_2}$.
Answer: (A). | ||||
| 34 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion The chain and body are at rest, so the branch's upward pull on the top of the chain balances the total weight below it:
$F=200+10\times10=300$ N.
Answer: (C). | ||||
| 35 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Matching listMedium | |
A medium matching list question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter (A) For a para- or diamagnetic material the susceptibility does not depend on the magnetising field: a horizontal line, graph (III).
(B) Inside a wire with uniform current, Ampere's law gives $B=\frac{\mu_0Ir}{2\pi a^2}\propto r$: a straight line through the origin, graph (I).
(C) Outside the wire $B=\frac{\mu_0I}{2\pi r}\propto\frac1r$: a falling hyperbola-like curve, graph (IV).
(D) Inside a (finite) solenoid the field is nearly uniform about the centre and falls off near the ends: graph (II).
A-III, B-I, C-IV, D-II.
Answer: (B). | ||||
| 36 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $p=\sqrt{2mK}$, so if $K$ becomes $36K$, $p$ becomes $6p$.
Increase $=\frac{6p-p}{p}\times100\%=500\%$.
Answer: (B). | ||||
| 37 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At depth $d$ in a uniform earth, $g_d=g\left(1-\frac dR\right)=\frac34g$ for $d=\frac R4$.
Weight $=\frac34\times300=225$ N.
Answer: (B). | ||||
| 38 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter A soap bubble has two surfaces (inner and outer), each contributing $\frac{2S}{R}$.
Excess pressure $=\frac{4S}{R}$.
Answer: (B). | ||||
| 39 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Helium is monatomic: $\Delta U=nC_v\Delta T=1\times\frac32\times8.3\times2=24.9$ J.
First law: $W=Q-\Delta U=48-24.9=23.1$ J.
Answer: (D). | ||||
| 40 | 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 $=1\text{ MSD}-1\text{ VSD}=1\text{ MSD}\left(1-\frac{49}{50}\right)=\frac{1\text{ MSD}}{50}$.
The zero error is read as the coinciding vernier division times the least count: $4\times LC=0.04$ mm, so $LC=0.01$ mm.
Then $1\text{ MSD}=50\times0.01=0.5$ mm, and 1 cm $=10$ mm holds $\frac{10}{0.5}=20$ main scale divisions.
Answer: (B). | ||||
| 41 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory A non-rigid diatomic molecule has 3 translational, 2 rotational and 2 vibrational (kinetic + potential) terms: 7 quadratic terms, each $\frac12k_BT$.
Energy per molecule $=\frac72k_BT$; for 10 molecules $=35k_BT$.
(35 RT would be the energy of 10 moles.)
Answer: (D). | ||||
| 42 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance R touches P: each gets $\frac Q2$. R (now $\frac Q2$) touches S: each gets $\frac{Q/2+Q}{2}=\frac{3Q}{4}$.
New force $\propto\frac Q2\cdot\frac{3Q}{4}=\frac38Q^2$, so $F=\frac38\times16=6$ N.
Answer: (A). | ||||
| 43 | 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 $I=\frac PV=\frac{110}{220}=0.5$ A.
Electrons per second $=\frac Ie=\frac{0.5}{1.6\times10^{-19}}=3.125\times10^{18}=31.25\times10^{17}$.
Answer: (C). | ||||
| 44 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $|\varepsilon|=L\frac{\Delta I}{\Delta t}$: $0.1=L\cdot\frac{4}{0.2}=20L$, so $L=0.005$ H $=5$ mH.
Answer: (D). | ||||
| 45 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $I=\frac12c\varepsilon_0E_0^2=\frac12\times3\times10^8\times9\times10^{-12}\times600^2$
$=\frac12\times2.7\times10^{-3}\times3.6\times10^5=486$ W/m$^2$.
Answer: (D). | ||||
| 46 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a biconvex lens $R_1=+15$ cm, $R_2=-30$ cm.
$\frac1f=(\mu-1)\left(\frac1{15}+\frac1{30}\right)=\frac{\mu-1}{10}$. With $f=20$ cm: $\mu-1=\frac12$, $\mu=1.5$.
Answer: (B). | ||||
| 47 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon energy $=\frac{1240}{300}=4.13$ eV.
$eV_0=4.13-2.13=2.0$ eV, so $V_0=2$ V.
Answer: (C). | ||||
| 48 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Paschen series ends on $n=3$; the longest wavelength is the $4\to3$ line.
$\frac1\lambda=R\left(\frac19-\frac1{16}\right)=\frac{7R}{144}$.
$\lambda=\frac{144}{7\times1.097\times10^7}=1.875\times10^{-6}$ m $\approx1.876\times10^{-6}$ m.
Answer: (B). | ||||
| 49 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The photon must supply at least 6 eV: $\lambda_{max}=\frac{hc}{E}=\frac{1242}{6}=207$ nm.
Answer: (A). | ||||
| 50 | Experimental SkillsOptics experiments | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on optics experiments, from Experimental Skills. NCERT: No matching NCERT chapter 1 MSD $=\frac{1}{20}$ cm $=0.05$ cm; least count $=\frac{0.05}{50}=0.001$ cm.
Readings: mark $=8.45+26\times0.001=8.476$ cm; mark through slab $=7.12+0.041=7.161$ cm; top surface $=4.05+0.001=4.051$ cm.
Real thickness $=8.476-4.051=4.425$ cm; apparent thickness $=7.161-4.051=3.110$ cm.
$\mu=\frac{4.425}{3.110}=1.42$.
Answer: (B). | ||||
| 51 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Open pipe: $f_n=\frac{nv}{2L}$.
$f_1=\frac{6\times333}{2\times0.6}=1665$ Hz; $f_2=\frac{5\times333}{2\times0.9}=925$ Hz.
Difference $=740$ Hz.
Answer: 740. | ||||
| 52 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalHard | |
A hard numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The two dielectrics fill the gap side by side (figure), so they act as two capacitors in parallel, each over half the area:
$C=\frac{C_0}{2}K_1+\frac{C_0}{2}K_2=\frac{10}{2}(2+3)=25\ \mu$F.
With the supply connected, the force between the plates is $F=\frac{CV^2}{2d}$ (each part: $\frac{Q^2}{2K\varepsilon_0A}=\frac{CV^2}{2d}$).
$8=\frac{25\times10^{-6}V^2}{2\times0.01}$, so $V^2=6400$ and $V=80$ V.
Answer: 80. | ||||
| 53 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy 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 Ammeter resistance: $\frac{240\times10}{240+10}=9.6\ \Omega$.
Total resistance $=140.4+9.6=150\ \Omega$; current $=\frac{24}{150}=0.16$ A $=160$ mA.
Answer: 160. | ||||
| 54 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $I_{max}=(\sqrt I+\sqrt{4I})^2=9I$, $I_{min}=(\sqrt{4I}-\sqrt I)^2=I$.
Difference $=8I$, so $x=8$.
Answer: 8. | ||||
| 55 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The first excitation energy is 10.2 eV; on returning to the ground state the atom emits a photon of 10.2 eV.
$\lambda=\frac{1245}{10.2}=122.06$ nm $\approx122$ nm.
Answer: 122. | ||||
| 56 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The midpoint of each side is at the inradius from the centroid: $r=\frac{a}{2\sqrt3}=\frac{2}{2\sqrt3}=\frac1{\sqrt3}$ m.
$I=(2+4+6)\times\frac13=4$ kg m$^2$.
Answer: 4. | ||||
| 57 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Differentiate: $2xv=2t$, so $v=\frac tx$.
$a=\frac{dv}{dt}=\frac{x-t\,v}{x^2}=\frac{x-\frac{t^2}{x}}{x^2}=\frac{x^2-t^2}{x^3}=\frac{1}{x^3}$.
So $a=x^{-3}$, $n=3$.
Answer: 3. | ||||
| 58 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Magnetic moment $M=NIA=100\times10^{-3}\times5\times10^{-3}=5\times10^{-4}$ A m$^2$.
Plane perpendicular to B means $\vec M$ is along $\vec B$ ($\theta=0$). Turning to $90^\circ$:
$W=MB(\cos0^\circ-\cos90^\circ)=5\times10^{-4}\times0.2=10^{-4}$ J $=100\ \mu$J.
Answer: 100. | ||||
| 59 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Maximum current means resonance: $f=\frac{1}{2\pi\sqrt{LC}}$.
$LC=0.1\times2.5\times10^{-9}=2.5\times10^{-10}$; $4\pi^2LC=40\times2.5\times10^{-10}=10^{-8}$, so $2\pi\sqrt{LC}=10^{-4}$ s.
$f=10^4$ Hz $=10\times10^3$ Hz.
Answer: 10. | ||||
| 60 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The wire's length is $2L=2$ m, so $L=1$ m.
Equilibrium of the midpoint: $2T\sin\theta=mg$, so $T\approx\frac{20}{2\times0.01}=1000$ N.
Each half stretches from $L$ to $\frac{L}{\cos\theta}$: strain $=\sec\theta-1\approx\frac{\theta^2}{2}=5\times10^{-5}$.
$Y=\frac{T/A}{\text{strain}}$, so $A=\frac{T}{Y\times\text{strain}}=\frac{1000}{2\times10^{11}\times5\times10^{-5}}=10^{-4}$ m$^2$.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | SolutionsConcentration terms | Class 12: Solutions | Single correctEasy | |
A easy single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 L of solution: mass $=1250$ g; NaCl $=3$ mol $=3\times58.5=175.5$ g.
Mass of water $=1250-175.5=1074.5$ g.
Molality $=\frac{3}{1.0745}=2.79$ m.
Answer: (A). | ||||
| 62 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | StatementsEasy | |
A easy statements question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Statement I: PF$_5$ (5 bond pairs) is sp$^3$d, but BrF$_5$ has 5 bond pairs and 1 lone pair on Br (6 electron pairs), so it is sp$^3$d$^2$. False.
Statement II: SF$_6$ is sp$^3$d$^2$, but $[\mathrm{Co(NH_3)_6}]^{3+}$ (Co$^{3+}$, d$^6$, strong-field NH$_3$, low spin) is an inner-orbital complex, d$^2$sp$^3$. False.
Answer: (B). | ||||
| 63 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $K_P=K_C(RT)^{\Delta n_g}$ with $\Delta n_g=1-\left(1+\frac12\right)=-\frac12$.
$\frac{K_P}{K_C}=(RT)^{-1/2}=\frac{1}{\sqrt{RT}}$.
Answer: (A). | ||||
| 64 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Matching listEasy | |
A easy matching list question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions (A) Two elements combine to one product: combination (IV).
(B) One compound breaks into several: decomposition (I).
(C) Na displaces hydrogen from water: displacement (II).
(D) N in NO$_2$ (+4) goes both to +3 (NO$_2^-$) and +5 (NO$_3^-$): disproportionation (III).
Answer: (A). | ||||
| 65 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry When an opposing external potential greater than the cell potential is applied, current flows in the reverse direction and the non-spontaneous reaction is driven: the cell works as an electrolytic cell. A smaller opposing potential only reduces the current; the cell still works galvanically.
Answer: (D). | ||||
| 66 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Flame colours and their characteristic wavelengths:
Li — crimson red, 670.8 nm (III); Na — yellow, 589.2 nm (I); Rb — red-violet, 780.0 nm (IV); Cs — blue, 455.5 nm (II).
A-III, B-I, C-IV, D-II.
Answer: (B). | ||||
| 67 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Inert pair effect: for the heaviest members the lower oxidation state is the stable one.
Pb$^{4+}$ (stable state Pb$^{2+}$) and Tl$^{3+}$ (stable state Tl$^{+}$) readily gain electrons: oxidising agents.
Sn$^{4+}$ is the stable state of tin (Sn$^{2+}$ is a reducing agent); Pb$^{2+}$ and Tl$^{+}$ are already the stable states.
Number $=2$.
Answer: (B). | ||||
| 68 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) False: covalent radius increases down the group (and not regularly, because of d and f electrons).
(B) False: electronegativity falls from C to Si and then stays nearly constant (Si to Pb about 1.8–1.9); it does not decrease gradually.
(C) True: carbon has no d orbitals, so its covalence is limited to 4; Si and heavier elements can expand theirs.
(D) True in the sense taught: heavier members do not form effective p$\pi$–p$\pi$ bonds because their orbitals are too large and diffuse.
(E) True: carbon shows $-4$ (CH$_4$) and other negative states.
Correct: (C), (D), (E).
Answer: (D). | ||||
| 69 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Sc $3d^14s^2$: 1; Ti $3d^24s^2$: 2; V $3d^34s^2$: 3; Mn $3d^54s^2$: 5; Cr $3d^54s^1$: 6.
Increasing order: Sc < Ti < V < Mn < Cr, i.e. (A) < (D) < (C) < (E) < (B).
Answer: (C). | ||||
| 70 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Ligands are named in alphabetical order, ignoring multiplying prefixes: bromo (b) before trimethylphosphine (t).
Two bromo ligands: dibromo. Two trimethylphosphine ligands (a name that already contains a number): bis(trimethylphosphine).
Pt is +2 (2 Br$^-$, neutral phosphines) in a neutral complex: platinum(II).
Name: dibromobis(trimethylphosphine)platinum(II).
Answer: (A). | ||||
| 71 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Matching listMedium | |
A medium matching list question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds In a tetrahedral field the $e$ set lies lower than $t_2$, and $\Delta_t$ is small, so complexes are high spin.
(A) TiCl$_4$: Ti(IV), $d^0$ → $e^0t_2^0$ (III).
(B) $[\mathrm{FeO_4}]^{2-}$: Fe(VI), $d^2$ → $e^2t_2^0$ (I).
(C) $[\mathrm{FeCl_4}]^-$: Fe(III), $d^5$ → $e^2t_2^3$ (IV).
(D) $[\mathrm{CoCl_4}]^{2-}$: Co(II), $d^7$ → $e^4t_2^3$ (II).
Answer: (A). | ||||
| 72 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter With AgNO$_3$ in nitric acid: AgCl is white and readily soluble in NH$_4$OH; AgBr is pale yellow and sparingly soluble; AgI is yellow and insoluble. Carbonate is destroyed by the acid.
A pale yellow precipitate dissolving with difficulty in NH$_4$OH is AgBr: bromide is present.
Answer: (B). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques On a polar adsorbent (silica, alumina) polar compounds are held more strongly.
(A) False: polar compounds are retained near the top; non-polar ones elute first.
(B) False: $R_f$ = distance moved by compound / distance moved by solvent, a fraction less than 1.
(C) False: a compound cannot move faster than the solvent front.
(D) True: a polar compound moves less, so its $R_f$ is smaller.
Answer: (D). | ||||
| 74 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Electron-donating groups activate the ring towards electrophiles, withdrawing groups deactivate it.
$-\mathrm{OCH_3}$ (strong +R) > $-\mathrm{CH_3}$ (+I, hyperconjugation) > H > $-\mathrm{CF_3}$ (strong −I).
Order: (III) > (I) > (II) > (IV).
Answer: (A). | ||||
| 75 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons (A) True: they differ only in spatial arrangement.
(B) True: rotation about the C=C bond needs the π bond to break, which does not happen at room temperature.
(C) True: in the trans isomer the methyl groups are far apart (less steric strain).
(D) False: in trans-2-butene the bond moments cancel (μ ≈ 0), while cis-2-butene has a small dipole moment. So cis has the greater dipole moment.
The incorrect statement is (D).
Answer: (D). | ||||
| 76 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate is 1-chloro-1-cyclohexylpropane, a secondary chloride, treated with NaOH in water (a polar protic solvent).
Ionisation gives the secondary carbocation cyclohexyl–CH$^+$–CH$_2$CH$_3$. A 1,2-hydride shift from the adjacent ring carbon (which carries one H) gives the more stable tertiary carbocation on the ring carbon bearing the propyl chain.
Water/hydroxide then attacks it: 1-propylcyclohexan-1-ol.
Answer: (C). | ||||
| 77 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines (i) NaNO$_2$ + HCl (0–5 °C): aniline → benzenediazonium chloride.
(ii) Cu$_2$Cl$_2$ (Sandmeyer): → chlorobenzene.
(iii) NaOH at 623 K and 300 atm (Dow process): chlorobenzene → sodium phenoxide.
(iv) H$^+$: → phenol.
Answer: (C). | ||||
| 78 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (i) LiAlH$_4$ reduces ethanoic acid to ethanol.
(ii) PCC oxidises ethanol to ethanal, CH$_3$CHO.
(iii) HCN/OH$^-$ adds to the carbonyl: CH$_3$CH(OH)CN.
(iv) Hydrolysis of the nitrile: CH$_3$CH(OH)COOH (lactic acid).
Answer: (A). | ||||
| 79 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers $-\mathrm{OCH_3}$ is ortho/para directing; nitration gives mainly the para product (less steric hindrance): A = 4-nitroanisole.
In A both free positions ortho to OCH$_3$ are also meta to NO$_2$, so both directing effects agree. With excess Br$_2$/Fe both are brominated: B = 2,6-dibromo-4-nitroanisole.
Answer: (D). | ||||
| 80 | BiomoleculesProteins | Class 12: Biomolecules | StatementsEasy | |
A easy statements question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules (A) Correct: enzymes are biological catalysts.
(B) Incorrect: enzymes are highly specific.
(C) Correct: most enzymes are globular proteins.
(D) Incorrect: maltose is hydrolysed to glucose by maltase; oxidases catalyse oxidation.
Incorrect statements: (B) and (D).
Answer: (A). | ||||
| 81 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom In the Bohr model $KE=-E$, so $KE=3.4$ eV $=34\times10^{-1}$ eV.
Answer: 34. | ||||
| 82 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure NH$_3$ sp$^3$; SO$_2$ sp$^2$; SiO$_2$ (network) sp$^3$; BeCl$_2$ sp; C$_2$H$_2$ sp; C$_2$H$_4$ sp$^2$; BCl$_3$ sp$^2$; HCHO sp$^2$; C$_6$H$_6$ sp$^2$; BF$_3$ sp$^2$; C$_2$H$_4$Cl$_2$ (dichloroethane, saturated) sp$^3$.
sp$^2$: SO$_2$, C$_2$H$_4$, BCl$_3$, HCHO, C$_6$H$_6$, BF$_3$ = 6.
Answer: 6. | ||||
| 83 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Spontaneous when $\Delta G=\Delta H-T\Delta S<0$, i.e. $T>\frac{\Delta H}{\Delta S}=\frac{400}{0.2}=2000$ K (assuming $\Delta H$ and $\Delta S$ do not change with temperature).
Answer: 2000. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalMedium | |
A medium numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions From the diagram the solution freezes at 270.65 K: $\Delta T_f=273.15-270.65=2.5$ K.
$m=\frac{\Delta T_f}{K_f}=\frac{2.5}{1.86}=1.344$ mol kg$^{-1}$.
Water $=100$ g $=0.1$ kg, so methanol $=0.1344$ mol $=0.1344\times32=4.301$ g.
Volume $=\frac{4.301}{0.792}=5.431$ mL $=543.1\times10^{-2}$ mL.
$x\approx543$.
Answer: 543. | ||||
| 85 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For a first-order reaction $t=\frac{t_{1/2}}{\ln2}\ln\frac{[A]_0}{[A]}$.
$\frac23$ complete leaves $\frac13$: $t_1=\frac{t_{1/2,1}}{\log2}\log3$. $\frac45$ complete leaves $\frac15$: $t_2=\frac{t_{1/2,2}}{\log2}\log5$.
$\frac{t_1}{t_2}=\frac52\times\frac{0.477}{0.699}=2.5\times0.6824=1.706=17.06\times10^{-1}$.
Nearest integer: 17.
Answer: 17. | ||||
| 86 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Oxidising power: MnO$_4^-$ > Cr$_2$O$_7^{2-}$ > VO$_2^+$ (standard reduction potentials about 1.51, 1.33 and 1.00 V). The least oxidising is VO$_2^+$.
V is +5 in VO$_2^+$: $3d^0$, no unpaired electrons, $\mu=0$ BM. (In fact all three are $d^0$.)
Answer: 0. | ||||
| 87 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds NiS dissolves in aqua regia: A = NiCl$_2$ (with NO, S and water).
With NH$_4$OH and dimethylglyoxime (dmgH$_2$), Ni$^{2+}$ gives the red chelate B = $[\mathrm{Ni(dmgH)_2}]$.
B has two dmgH$^-$ ligands, each with two CH$_3$ groups (6 H) and one O–H. The two O–H protons form the intramolecular O–H···O hydrogen bonds; the 12 methyl protons do not.
Answer: 12. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Hyperconjugation needs C–H bonds on an sp$^3$ carbon next to the cationic carbon.
1. 2-Methylbutan-2-yl cation (tertiary): has α C–H bonds — stabilised.
2. Di-tert-butylmethyl cation, (CH$_3$)$_3$C–CH$^+$–C(CH$_3$)$_3$: both α carbons are quaternary, no α-H — not stabilised.
3. CH$_3^+$: no α carbon — not stabilised.
4. Cyclopentadienyl cation: the neighbours are sp$^2$ carbons, no α sp$^3$ C–H — not stabilised.
5. $^+$CH$_2$–OCH$_3$: stabilised by resonance with oxygen's lone pair, not by hyperconjugation — not stabilised by it.
6. tert-Butyl cation: nine α C–H bonds — stabilised.
7. (CH$_3$)$_2$N–CH$_2^+$: stabilised by resonance with nitrogen's lone pair — not by hyperconjugation.
Not stabilised by hyperconjugation: 5.
Answer: 5. | ||||
| 89 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | NumericalMedium | |
A medium numerical question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers $-\mathrm{OC_2H_5}$ is ortho/para directing; nitration gives mainly P = 4-nitrophenetole.
With 2Br$_2$/Fe, both positions ortho to the ethoxy group (also meta to NO$_2$) are brominated: Q = 2,6-dibromo-4-nitrophenetole, C$_8$H$_7$Br$_2$NO$_3$.
O atoms: 1 (ether) + 2 (NO$_2$) = 3; Br atoms: 2. Ratio $=\frac32=1.5=15\times10^{-1}$.
Answer: 15. | ||||
| 90 | AminesClassification and identification of amines | Class 12: Amines | NumericalMedium | |
A medium numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Ammonolysis of benzyl chloride gives a mixture of primary, secondary and tertiary amines (and the quaternary salt).
With p-toluenesulphonyl chloride (Hinsberg-type test) a tertiary amine does not react, so the solution stays clear. X is tribenzylamine, $(\mathrm{C_6H_5CH_2})_3\mathrm{N}$ = C$_{21}$H$_{21}$N.
Molar mass $=21\times12+21\times1+14=287$ g mol$^{-1}$.
Answer: 287. | ||||
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