JEE (Main) · 2025
JEE (Main) 2025 — 24 January, Shift 1
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
25 Q
- Matrices and Determinants2
- Limits, Continuity and Differentiability2
- Three Dimensional Geometry2
8 easy · 14 medium · 3 hard
Physics
24 Q
- Optics4
- Electrostatics3
- Motion of System of Particles and Rigid Body2
18 easy · 6 medium · 0 hard
Chemistry
24 Q
- Aldehydes, Ketones and Carboxylic Acids3
- Chemical Bonding and Molecular Structure2
- Thermodynamics2
13 easy · 10 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 FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Put $t = 2^x$. Then $f(x) = \frac{4t+16}{2t^2+16t+32} = \frac{4(t+4)}{2(t+4)^2} = \frac{2}{t+4}$.
For $4 - x$, $2^{4-x} = \frac{16}{t}$, so $f(4-x) = \frac{2}{\frac{16}{t}+4} = \frac{t}{2(t+4)}$, and $f(x)+f(4-x) = \frac{4+t}{2(t+4)} = \frac{1}{2}$.
Since $4 = \frac{60}{15}$, the terms $f\left(\frac{k}{15}\right)$ and $f\left(\frac{60-k}{15}\right)$ pair up for $k = 1,\dots,29$, giving $29\times\frac{1}{2}$, and the middle term is $f(2) = \frac{2}{4+4} = \frac{1}{4}$.
Sum $= \frac{29}{2}+\frac{1}{4} = \frac{59}{4}$, and $8\times\frac{59}{4} = 118$.
Answer: (A) $118$. | ||||
| 2 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $(x-4)(x-5) = x^2-9x+20$. Let $y = x^2-9x+11$; then $(x-4)(x-5) = y+9$ and the equation is $y^2-(y+9) = 3$, i.e. $y^2-y-12=0$, so $y = 4$ or $y = -3$.
$y = 4$: $x^2-9x+7=0$, discriminant $81-28=53$, not a perfect square: irrational roots.
$y = -3$: $x^2-9x+14=0$, so $x = 2, 7$, both rational.
Product of the rational roots $= 2\times7 = 14$.
Answer: (B) $14$. | ||||
| 3 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctHard | |
A hard single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha+\beta = \frac{3}{2}$ and $\alpha\beta = -i$, so $(\alpha\beta)^4 = 1$, i.e. $\alpha^{-4} = \beta^4$ and $\beta^{-4} = \alpha^4$.
Then $\alpha^{19}+\alpha^{11} = \alpha^{15}(\alpha^4+\alpha^{-4}) = \alpha^{15}(\alpha^4+\beta^4)$, and likewise $\beta^{19}+\beta^{11} = \beta^{15}(\alpha^4+\beta^4)$.
So the ratio is $\alpha^4+\beta^4$. Now $\alpha^2+\beta^2 = \frac{9}{4}+2i$, and $\alpha^4+\beta^4 = \left(\frac{9}{4}+2i\right)^2 - 2(\alpha\beta)^2 = \frac{81}{16}-4+9i+2 = \frac{49}{16}+9i$.
$16\cdot\frac{49}{16}\cdot9 = 441$.
Answer: (D) $441$. | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The first two equations are independent, so for infinitely many solutions the third must be a combination $a(\text{I}) + b(\text{II})$.
$x$: $2a+5b=100$; constants: $4a+12b=212$, i.e. $a+3b=53$. Hence $b=6$, $a=35$.
$y$: $-a+\lambda b=-47 \Rightarrow -35+6\lambda=-47 \Rightarrow \lambda=-2$; $z$: $\mu = a+3b = 53$.
$\mu-2\lambda = 53+4 = 57$.
Answer: (C) $57$. | ||||
| 5 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1}$, so $S_n = 1-\frac{1}{n+1} = \frac{n}{n+1}$ and $2026S_{2025} = 2025$, whose square root is $45$.
Sum of six terms $= \frac{6}{2}\left(2(-p)+5p\right) = 9p = 45$, so $p = 5$.
$|T_{20}-T_{15}| = 5p = 25$.
Answer: (B) $25$. | ||||
| 6 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Let $N = n+4$. The 5th, 6th, 7th coefficients are ${}^NC_4, {}^NC_5, {}^NC_6$; A.P. means $2\,{}^NC_5 = {}^NC_4+{}^NC_6$.
Dividing by ${}^NC_5$: $2 = \frac{5}{N-4}+\frac{N-5}{6}$, i.e. $N^2-21N+98=0$, so $N = 7$ or $N = 14$.
$N = 14$ gives $n = 10$, which is excluded; so $N = 7$ ($n = 3$).
Largest coefficient of $(1+x)^7$ is ${}^7C_3 = {}^7C_4 = 35$.
Answer: (A) $35$. | ||||
| 7 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Recorded sum $= 55$; corrected sum $= 55-4-5+6+8 = 60$, mean $6$.
Corrected $\sum x_i^2 = 371-16-25+36+64 = 430$.
Variance $= \frac{430}{10}-6^2 = 43-36 = 7$.
Answer: (C) $7$. | ||||
| 8 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(\text{sum }5) = \frac{4}{36} = \frac{1}{9}$, $P(\text{sum }8) = \frac{5}{36}$.
A wins on his first throw, or after one round in which both fail, and so on: $P = \frac{1}{9}\left(1 + r + r^2 + \dots\right)$ with $r = \frac{8}{9}\cdot\frac{31}{36} = \frac{62}{81}$.
$P = \frac{1/9}{1-62/81} = \frac{1}{9}\cdot\frac{81}{19} = \frac{9}{19}$.
Answer: (B) $\frac{9}{19}$. | ||||
| 9 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines The midpoint of $(1,2)$ and its image is $\left(\frac{35}{13},\frac{-7}{13}\right)$, which lies on the line: $\frac{70}{13}+\frac{21}{13}+\lambda = 0$, so $\lambda = -7$. (The join $(\frac{44}{13},\frac{-66}{13})$ is along $(2,-3)$, the normal, as it should be.)
Intersection of $8x-11y=33$ and $2x-3y=7$: $x = \frac{7+3y}{2}$ gives $4(7+3y)-11y = 33$, so $y = 5$, $x = 11$.
Concurrency: $3(11)-4(5)-\alpha = 0$, so $\alpha = 13$. $|\alpha\lambda| = 91$.
Answer: (C) $91$. | ||||
| 10 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections The given circle has centre $(1,-2)$ and radius $\sqrt{1+4+4} = 3$.
Image of $(1,-2)$ in $2x-3y+5=0$: $\frac{x-1}{2} = \frac{y+2}{-3} = \frac{-2(2+6+5)}{13} = -2$, so $O = (-3, 4)$; C has radius 3.
$A = (0, 4)$. Arc AB $= \frac{1}{6}$ of the circumference means $\angle AOB = 60^\circ$, so $B = (-3+3\cos60^\circ, 4\pm3\sin60^\circ)$; $\beta<4$ takes the minus sign.
$\alpha = -\frac{3}{2}$, $\beta = 4-\frac{3\sqrt3}{2}$, and $\beta-\sqrt3\alpha = 4-\frac{3\sqrt3}{2}+\frac{3\sqrt3}{2} = 4$.
Answer: (D) $4$. | ||||
| 11 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Focal distances of $(x_1,y_1)$ are $a\pm ex_1$; their product is $a^2-e^2x_1^2 = a^2-3e^2 = a^2-3+\frac{3b^2}{a^2} = \frac{7}{4}$.
Let $A = a^2$, $B = b^2$: $A+\frac{3B}{A} = \frac{19}{4}$, so $B = \frac{A(19-4A)}{12}$; and the point lies on the ellipse: $\frac{3}{A}+\frac{1}{4B} = 1$.
Substituting: $\frac{3}{A}+\frac{3}{A(19-4A)} = 1$, i.e. $3(19-4A)+3 = A(19-4A)$, so $4A^2-31A+60 = 0$, $A = 4$ or $A = \frac{15}{4}$.
$A = 4$: $B = 1$, $e^2 = \frac{3}{4}$. $A = \frac{15}{4}$: $B = \frac{5}{4}$, $e^2 = 1-\frac{1}{3} = \frac{2}{3}$. Both have $a>b$.
$\left|\frac{\sqrt3}{2}-\sqrt{\frac{2}{3}}\right| = \frac{3-2\sqrt2}{2\sqrt3}$.
Answer: (A) $\frac{3-2\sqrt{2}}{2\sqrt{3}}$. | ||||
| 12 | 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 Both roots tend to $\sqrt5$, so rationalise: the bracket equals $\frac{(2\cos^2x+3\cos x)-(\cos^2x+\sin x+4)}{\sqrt{\cdot}+\sqrt{\cdot}} = \frac{\cos^2x+3\cos x-4-\sin x}{\sqrt{\cdot}+\sqrt{\cdot}}$.
$\cos^2x+3\cos x-4 = (\cos x-1)(\cos x+4)$, and $\frac{\cos x-1}{\sin x}\to0$, while $\frac{-\sin x}{\sin x} = -1$.
Limit $= \frac{0-1}{2\sqrt5} = -\frac{1}{2\sqrt5}$.
Answer: (C) $-\frac{1}{2\sqrt{5}}$. | ||||
| 13 | 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 Write $\vec c = x\vec a + y\vec b$. $\vec a\cdot\vec a = 14$, $\vec a\cdot\vec b = 3+2-3 = 2$, $\vec b\cdot\vec b = 11$.
$\vec c\cdot\vec b = 2x+11y = 0$ and $\vec a\cdot\vec c = 14x+2y = 5$. So $x = -\frac{11y}{2}$, $-77y+2y = 5$, $y = -\frac{1}{15}$, $x = \frac{11}{30}$.
$\vec c = \frac{11}{30}(1,2,3)-\frac{2}{30}(3,1,-1) = \frac{1}{30}(5,20,35) = \frac{1}{6}(1,4,7)$.
$|\vec c| = \frac{\sqrt{66}}{6} = \sqrt{\frac{11}{6}}$.
Answer: (A) $\sqrt{\frac{11}{6}}$. | ||||
| 14 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy 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 The line passes through $P(6,7,7)$ with direction $\vec d = (3,2,-2)$, $|\vec d| = \sqrt{17}$. $\vec{PB} = (-5,-5,-4)$.
$\vec{PB}\times\vec d = (18,-22,5)$, of magnitude $\sqrt{324+484+25} = \sqrt{833}$.
Distance of B from the line $= \sqrt{\frac{833}{17}} = \sqrt{49} = 7$, the height of the triangle on base AC.
Area $= \frac{1}{2}\times6\times7 = 21$.
Answer: (B) $21$. | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy 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 First line: $(-1+2t, 2+3t, 1+4t)$; second: $(-2+3s, 3+2s, 4+s)$.
$2t-3s = -1$, $3t-2s = 1$ give $s = 1$, $t = 1$; check $4t-s = 3$ ✓.
$P = (1,5,5)$, $PQ = \sqrt{3^2+10^2+4^2} = \sqrt{125} = 5\sqrt5$.
Answer: (C) $5\sqrt{5}$. | ||||
| 16 | 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 Replace $x$ by $\frac{1}{x}$: $f\left(\frac1x\right)-6f(x) = \frac{35x}{3}-\frac52$, so $f\left(\frac1x\right) = 6f(x)+\frac{35x}{3}-\frac52$.
Substituting: $f(x)-36f(x)-70x+15 = \frac{35}{3x}-\frac52$, so $-35f(x) = \frac{35}{3x}+70x-\frac{35}{2}$ and $f(x) = -\frac{1}{3x}-2x+\frac12$.
$\frac{1}{\alpha x}+f(x) = \left(\frac1\alpha-\frac13\right)\frac1x-2x+\frac12$ has a finite limit only if $\alpha = 3$; then $\beta = \frac12$.
$\alpha+2\beta = 3+1 = 4$.
Answer: (B) $4$. | ||||
| 17 | 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 For a rectangle $[x_1,x_2]\times[y_1,y_2]$ inside R we need $x_1\ge0$, $y_1\ge x_2$ (from $y\ge x$ at the right side) and $y_2\le 9-\frac{11}{3}x_2^2$ (the parabola falls as $x$ grows).
Best is $x_1 = 0$, $y_1 = x_2 = x$, $y_2 = 9-\frac{11}{3}x^2$: area $A(x) = x\left(9-\frac{11}{3}x^2-x\right) = 9x-x^2-\frac{11}{3}x^3$.
$A'(x) = 9-2x-11x^2 = 0 \Rightarrow (11x-9)(x+1) = 0$, $x = \frac{9}{11}$ (a maximum, $A''<0$).
$A = \frac{9}{11}\left(9-\frac{27}{11}-\frac{9}{11}\right) = \frac{9}{11}\cdot\frac{63}{11} = \frac{567}{121}$.
Answer: (A) $\frac{567}{121}$. | ||||
| 18 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $I(9,14)+I(10,13) = \int_0^1 x^8(1-x)^{13}dx+\int_0^1x^9(1-x)^{12}dx = \int_0^1x^8(1-x)^{12}\left[(1-x)+x\right]dx$.
$= \int_0^1x^8(1-x)^{12}dx = I(9,13)$.
Answer: (C) $I(9, 13)$. | ||||
| 19 | 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 Put $u = x+2$: the region is $u^2-2\le y\le|u|$, symmetric about $u = 0$.
For $u\ge0$: $u^2-2\le u$ for $0\le u\le2$.
Area $= 2\int_0^2\left(u-u^2+2\right)du = 2\left(2-\frac83+4\right) = \frac{20}{3}$.
Answer: (C) $20/3$. | ||||
| 20 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx}+\frac{x}{1+x^2}y = \frac{5x^2}{\sqrt{1+x^2}}$. Integrating factor $e^{\int\frac{x}{1+x^2}dx} = \sqrt{1+x^2}$.
$y\sqrt{1+x^2} = \int5x^2dx = \frac{5x^3}{3}+C$, and $y(0) = 0$ gives $C = 0$.
$y(\sqrt3) = \frac{5\cdot3\sqrt3/3}{2} = \frac{5\sqrt3}{2}$.
Answer: (A) $\frac{5\sqrt{3}}{2}$. | ||||
| 21 | 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 Every element of A is the product of a non-empty set of distinct primes from S (a single prime coming from S itself), and distinct sets give distinct products, so A corresponds to the $2^{10}-1$ non-empty subsets of S.
A prime $p\in S$ divides such a product exactly when $p$ is in the subset: $2^9 = 512$ subsets.
Number of pairs $= 10\times512 = 5120$.
Answer: 5120. | ||||
| 22 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalHard | |
A hard numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $X^TAX = 0$ for all $X$ means A is skew-symmetric: $A = \begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}$.
$A(1,1,1)^T = (a+b, -a+c, -b-c) = (1,4,-5)$ and $A(1,2,1)^T = (2a+b, -a+c, -b-2c) = (0,4,-8)$.
So $a+b = 1$, $2a+b = 0$: $a = -1$, $b = 2$, and $c = 4+a = 3$ (checks: $-b-c = -5$, $-b-2c = -8$).
$\det(A+I) = 1+a^2+b^2+c^2 = 1+1+4+9 = 15$, so $\det(2(A+I)) = 2^3\cdot15 = 120$.
$\det(\mathrm{adj}\,M) = (\det M)^2 = 120^2 = 14400 = 2^6\cdot3^2\cdot5^2$, so $\alpha = 6$, $\beta = 2$, $\gamma = 2$.
$\alpha^2+\beta^2+\gamma^2 = 36+4+4 = 44$.
Answer: 44. | ||||
| 23 | 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 Divisible by 2 and 3 means divisible by 6: the multiples of 6 from 102 to 996 number $\frac{996-102}{6}+1 = 150$.
\"Not divisible by 4 and 9\" read in the same way as \"divisible by 2 and 3\" means not divisible by both 4 and 9, i.e. not divisible by 36. Multiples of 36 from 108 to 972: $27-3+1 = 25$.
Count $= 150-25 = 125$.
Read instead as divisible by neither 4 nor 9, the count is 50.
Answer: 125 (50 on the 'neither 4 nor 9' reading). | ||||
| 24 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sec^2(\tan^{-1}\alpha) = 1+\alpha^2$ and $\mathrm{cosec}^2(\cot^{-1}\beta) = 1+\beta^2$, so $\alpha^2+\beta^2 = 34$.
With $\alpha+\beta = 8$: $\alpha\beta = \frac{64-34}{2} = 15$, so $\{\alpha,\beta\} = \{3,5\}$ and $\alpha\le\beta$ gives $\alpha = 3$, $\beta = 5$.
$\alpha^2+\beta = 9+5 = 14$.
Answer: 14. | ||||
| 25 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations At $x = 0$: $8f(0)-12 = 0$, so $f(0) = \frac32$.
Differentiate: $4(x+2)f+2(x+2)^2f'-6(x+2) = 10(x+2)f$; divide by $2(x+2)$: $(x+2)f' = 3f+3$.
$\frac{f'}{f+1} = \frac{3}{x+2}$, so $f+1 = C(x+2)^3$; $f(0)+1 = \frac52 = 8C$ gives $C = \frac{5}{16}$.
$f(2)+1 = \frac{5}{16}\cdot64 = 20$, so $f(2) = 19$.
Answer: 19. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Physical World and MeasurementSignificant figures | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on significant figures, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\frac{32.3\times1125}{27.4} = 1326.186\dots$
In multiplication and division the result keeps as many significant figures as the least precise factor: 32.3 and 27.4 have three, so the result has three: $1.33\times10^3 = 1330$.
Answer: (D) $y = 1330$. | ||||
| 27 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion At the top the velocity is horizontal, $v_x = \frac{v_0}{\sqrt2}$, and the height is $H = \frac{v_0^2\sin^245^\circ}{2g} = \frac{v_0^2}{4g}$.
$\vec L = \vec r\times m\vec v$ with $\vec r = (x, H)$, $\vec v = (v_x, 0)$: $L_z = m(x\cdot0-H v_x) = -mHv_x$.
$|L| = m\cdot\frac{v_0^2}{4g}\cdot\frac{v_0}{\sqrt2} = \frac{mv_0^3}{4\sqrt2g}$, along the negative z-axis (x to the right, y upward).
Answer: (C) $\frac{mv_0^3}{4\sqrt{2}g}$ along negative z–axis. | ||||
| 28 | 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 slope: $v_0^2 = rg\frac{\tan\theta+\mu}{1-\mu\tan\theta}$.
$v_0^2-\mu v_0^2\tan\theta = rg\tan\theta+\mu rg$, so $\mu(rg+v_0^2\tan\theta) = v_0^2-rg\tan\theta$.
$\mu = \frac{v_0^2-rg\tan\theta}{rg+v_0^2\tan\theta}$.
Answer: (C). | ||||
| 29 | 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 $W = \int_0^1(\alpha+\beta x^2)dx = \alpha+\frac{\beta}{3} = 5$ (taking the displacement from $x = 0$ to $x = 1$ m).
With $\alpha = 1$: $\frac\beta3 = 4$, $\beta = 12$ N/m$^2$.
Answer: (B) 12 N/m$^2$. | ||||
| 30 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy 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 For rolling without slipping $a = \frac{g\sin\theta}{1+\frac{k^2}{r^2}}$; for a solid cylinder $\frac{k^2}{r^2} = \frac12$.
$a = \frac{2}{3}g\sin45^\circ = \frac{2g}{3\sqrt2} = \frac{\sqrt2g}{3}$.
Answer: (D) $\frac{\sqrt{2}g}{3}$. | ||||
| 31 | GravitationKepler's laws | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation $T\propto R^{3/2}$, so $\frac{\Delta T}{T}\approx\frac32\cdot\frac{\Delta R}{R} = \frac32\times3\% = 4.5\%$ (exactly $1.03^{1.5}-1 = 4.53\%$).
Answer: (D) 4.5%. | ||||
| 32 | 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 Volume is conserved: $n$ drops of radius $r = Rn^{-1/3}$. Work $= T\cdot4\pi(nr^2-R^2) = 4\pi R^2T(n^{1/3}-1)$.
$n = 27$: factor $3-1 = 2$; $n = 64$: factor $4-1 = 3$.
$W = 10\times\frac32 = 15$ J.
Answer: (C) 15 J. | ||||
| 33 | 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 An air bubble in a liquid has one surface: $P_{in} = P_0+\rho gh+\frac{2T}{r}$.
$\rho gh = 1000\times10\times0.2 = 2000$ N/m$^2$, so $\frac{2T}{r} = 100$ and $T = \frac{100\times0.001}{2} = 0.05$ N/m.
Answer: (A) $0.05$. | ||||
| 34 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics $P\propto T$ for an ideal gas means $\frac{nR}{V}$ is constant: the volume is constant, the process is isochoric (E true, C false).
No volume change, so $W = 0$ (A true) and $Q = \Delta U$ (B false).
Pressure rises with temperature, so T rises and the internal energy increases (D true).
Correct: A, D, E.
Answer: (C) A, D, E Only. | ||||
| 35 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $12.5$ s $= 6T+\frac{T}{4}$. In 6 full periods the particle covers $6\times4A = 24$ cm with no net displacement.
In the remaining quarter period, starting from the mean position (or from an extreme), it moves from mean to extreme (or extreme to mean): distance $1$ cm, displacement of magnitude $1$ cm.
$D = 25$ cm, $d = 1$ cm, $\frac Dd = 25$.
Answer: (B) $25$. | ||||
| 36 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $I = \sqrt{I_A^2+I_B^2}\,\sin(\omega t+\phi)$, a sinusoid of peak $I_0 = \sqrt{I_A^2+I_B^2}$.
$I_{rms} = \frac{I_0}{\sqrt2} = \sqrt{\frac{I_A^2+I_B^2}{2}}$.
Answer: (B). | ||||
| 37 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $C = \frac{\varepsilon_0lb}{d}$, so $C\propto\frac{lb}{d}$; originally $\frac{3\times1}{3} = 1$ (cm$^2$/μm).
A: $\frac{30}{1} = 30$ (×30). B: $\frac{3}{30} = 0.1$. C: $\frac{30}{3} = 10$ ✓. D: $\frac{1}{10} = 0.1$. E: $\frac{10}{1} = 10$ ✓.
Answer: (C) C and E only. | ||||
| 38 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Energy density between the plates is $\frac12\varepsilon_0E^2$; the field fills the volume $Ad$.
$U = \frac12\varepsilon_0E^2Ad$.
Answer: (B) $\frac{1}{2}\varepsilon_0E^2Ad$. | ||||
| 39 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Bright fringes coincide where $n_1\lambda_1 = n_2\lambda_2$: $480n_1 = 600n_2$, i.e. $4n_1 = 5n_2$.
The first coincidence (beyond the central fringe) is $n_1 = 5$, $n_2 = 4$.
Answer: (B) $5$. | ||||
| 40 | 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 In the liquid the lens has $\frac{1}{f_L} = \left(\frac{1.5}{1.2}-1\right)\frac1R = \frac{0.25}{R}$, so $f_L = 4R$.
Light passes the lens, reflects at the silvered plane face (power 0) and passes the lens again: $\frac1F = \frac{2}{f_L}+0$, so $F = \frac{f_L}{2} = 2R$.
$2R = 0.2$ m, $R = 0.10$ m.
Answer: (A) 0.10 m. | ||||
| 41 | 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 $f = \frac1P$, so $\frac{|\Delta f|}{f} = \frac{\Delta P}{P} = \frac{0.1}{2.5} = 0.04$ (exactly, $\frac{1/2.5-1/2.6}{1/2.5} = \frac{0.1}{2.6} = 0.038\approx0.04$).
Answer: (C) $0.04$. | ||||
| 42 | 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 $\frac1{f_1} = (1.5-1)\frac12$, so $f_1 = 4$ cm.
$\frac1{f_2} = \left(\frac{1.5}{1.2}-1\right)\frac13 = \frac{0.25}{3}$, so $f_2 = 12$ cm.
$f_1:f_2 = 1:3$.
Answer: (D) 1 : 3. | ||||
| 43 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Force on the electron $= -e\vec E = eE_0\hat k$, perpendicular to $\vec v_0$; after time t, $v_z = \frac{eE_0t}{m}$ and $v_x = v_0$.
$p = m\sqrt{v_0^2+\frac{e^2E_0^2t^2}{m^2}} = mv_0\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}$.
$\lambda = \frac hp = \frac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}}$.
Answer: (D). | ||||
| 44 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Energies add: $E_{AC} = E_{AB}+E_{BC}$, so $\frac{1}{\lambda_{AB}} = \frac{1}{2000}-\frac{1}{6000} = \frac{2}{6000}$.
$\lambda_{AB} = 3000$ Å.
Answer: (B) 3000 Å. | ||||
| 45 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A false: a solar cell has a large junction area, to collect more light.
B true: a solar cell itself generates the emf; no external bias is applied.
C false: an LED is a heavily doped p-n junction.
D false: the intensity rises with forward current only up to a point, then falls.
E true: light is emitted when electrons and holes recombine under forward bias.
Answer: (B) B, E Only. | ||||
| 47 | ThermodynamicsFirst law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics At constant pressure $E_1 = nC_p\Delta T$ and $E_2 = nC_v\Delta T$, so $\frac{E_1}{E_2} = \frac{C_p}{C_v} = \frac53$ for a monatomic gas.
$\frac{x}{9} = \frac53$, $x = 15$.
Answer: 15. | ||||
| 48 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Each side has $3\,\Omega$. Between two vertices one side ($3\,\Omega$) is in parallel with the other two in series ($6\,\Omega$).
$R = \frac{3\times6}{3+6} = 2\,\Omega$.
Answer: 2. | ||||
| 49 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Each side is at distance $d = \frac{a}{2} = \frac{1}{2\sqrt2}$ m from the centre and subtends $45^\circ$ on each side of the perpendicular.
One side: $B_1 = \frac{\mu_0I}{4\pi d}(\sin45^\circ+\sin45^\circ) = 10^{-7}\times5\times2\sqrt2\times\sqrt2 = 2\times10^{-6}$ T.
Four sides add: $B = 8\times10^{-6}$ T, so $p = 8$.
Answer: 8. | ||||
| 50 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The charge is on the axis through the centre of the square. Taking it at distance $\frac a2$ from the square (the reading the figure intends), the square is one face of a cube of side $a$ centred on q, so the flux through the whole square is $\frac{q}{6\varepsilon_0}$.
The square splits into 8 congruent right triangles about its centre (by its two midlines and two diagonals), each receiving the same flux, $\frac{q}{48\varepsilon_0}$, by symmetry about the axis.
The shaded region is the two lower quarters (4 triangles) and half of the upper-left quarter (1 triangle): 5 triangles.
Flux $= \frac{5q}{48\varepsilon_0} = \frac{5}{48}\cdot\frac{1}{\varepsilon_0}$ (q = 1 C), so $p = 48$.
Answer: 48. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Two atomic orbitals combine only if they have the same symmetry about the internuclear (z) axis, so that their overlap is not zero.
A: $2p_z$ ($\sigma$) and $2p_x$ ($\pi$) — net overlap zero. B: $2s$ ($\sigma$) and $2p_x$ ($\pi$) — zero.
C: $3d_{xy}$ and $3d_{x^2-y^2}$ are both $\delta$-type but rotated by $45^\circ$ about z — net overlap zero.
D: $2s$ and $2p_z$ are both symmetric about z — they combine ($\sigma$ overlap).
E: $2p_z$ ($\sigma$) and $3d_{x^2-y^2}$ ($\delta$) — zero.
Answer: (C) D only. | ||||
| 52 | 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 A true: O, N and C each have four electron pairs, $sp^3$.
B true: lone pairs compress the angles: 104.5° (two lone pairs), 107.5° (one), 109.5° (none).
C true: $\mathrm{CH_4}$ is non-polar (0); $\mathrm{NH_3}$ (1.47 D) < $\mathrm{H_2O}$ (1.85 D).
D false: $\mathrm{H_2O}$ and $\mathrm{NH_3}$ have lone pairs and act as Lewis bases; $\mathrm{CH_4}$ is neither.
E false: in $\mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-}$, $\mathrm{NH_3}$ accepts a proton (base) and $\mathrm{H_2O}$ donates it (acid).
Correct: A, B, C.
Answer: (A) A, B and C Only. | ||||
| 53 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Endothermic: $\Delta H>0$. $\Delta G = \Delta H-T\Delta S$ is positive at 273 K and negative at 373 K, so it falls as T rises: $\Delta S>0$ (spontaneous above $T = \frac{\Delta H}{\Delta S}$).
Answer: (B) Both $\Delta H$ and $\Delta S$ are (+ve). | ||||
| 54 | SolutionsColligative properties | Class 12: Solutions | Single correctMedium | |
A medium single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions A solution has a lower vapour pressure than the pure liquid solvent at every temperature, so its curve lies below the liquid-solvent curve (this rules out the first graph).
The freezing point is where the vapour pressure of the liquid equals that of the solid solvent: the steep solid curve meets the liquid-solvent curve at $T_f^\circ$ and the lower solution curve at a lower temperature $T_f$. $\Delta T_f = T_f^\circ-T_f$.
The graphs with a horizontal dashed line compare temperatures at equal vapour pressure, which describes boiling, not freezing.
Only the fourth graph shows this (its steep curve is labelled \"Frozen Solution\" in the paper; it stands for the solid solvent that separates on freezing).
Answer: (D). | ||||
| 55 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{Cr(OH)_3\rightleftharpoons Cr^{3+}+3OH^-}$: $K_{sp} = s(3s)^3 = 27s^4$.
$s = \sqrt[4]{\frac{1.6\times10^{-30}}{27}}$.
Answer: (C). | ||||
| 56 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Add free energies: $\mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}}$ is ($\mathrm{Fe^{3+}+3e^-\rightarrow Fe}$) minus ($\mathrm{Fe^{2+}+2e^-\rightarrow Fe}$): $-1\cdot FE^\ominus = -3Fz+2Fy$, so $E^\ominus_{\mathrm{Fe^{3+}/Fe^{2+}}} = 3z-2y$.
In the cell Ag$^+$ is reduced (cathode) and Fe$^{2+}$ oxidised (anode): $E^\ominus_{cell} = x-(3z-2y) = x+2y-3z$.
Answer: (C) $x+2y-3z$. | ||||
| 57 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Start: $\mathrm{N_2O_5}$ at $P_0$. At 50% completion: $\mathrm{N_2O_5} = 0.5P_0$, $\mathrm{NO_2} = P_0$, $\mathrm{O_2} = 0.25P_0$.
Total $= 1.75P_0 = \frac74P_0$.
Answer: (A) 7/4 times of initial pressure. | ||||
| 58 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties A not true: in the modern periodic law properties are a periodic function of atomic number, not atomic weight (B is true).
C not true: elements with similar outer configurations fall in the same group, not the same period.
D true: the period and block of an element follow from n and l of the last filled orbital.
E not true: the number of elements in a period is twice the number of orbitals being filled (e.g. period 2: 4 orbitals, 8 elements).
Not true: A, C, E.
Answer: (A) A, C and E Only. | ||||
| 59 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Oxygen exists as small diatomic $\mathrm{O_2}$ molecules held by weak van der Waals forces, while sulphur exists as $\mathrm{S_8}$ rings, much larger molecules with much stronger dispersion forces.
This difference in atomicity explains why oxygen is a gas and sulphur a solid with far higher melting and boiling points.
Answer: (C) Atomicity. | ||||
| 60 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The stable oxidation state of the lanthanoids is +3. $\mathrm{Ce^{3+}}$ and $\mathrm{Lu^{3+}}$ are already in it; $\mathrm{Eu^{2+}}$ ($4f^7$) tends to lose an electron and is a reducing agent.
$\mathrm{Tb^{4+}}$ ($4f^7$) readily gains an electron to become $\mathrm{Tb^{3+}}$, so it is the oxidising agent.
Answer: (C) $\mathrm{Tb^{4+}}$. | ||||
| 61 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Fusion of $\mathrm{MnO_2}$ with KOH and an oxidant such as $\mathrm{KNO_3}$ gives green potassium manganate: $\mathrm{MnO_2+2KOH+KNO_3\rightarrow K_2MnO_4+KNO_2+H_2O}$.
The manganate is then oxidised (electrolytically or by disproportionation in acid) to $\mathrm{KMnO_4}$.
Answer: (D) $\mathrm{K_2MnO_4}$. | ||||
| 62 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Two chlorides are precipitated by $\mathrm{AgNO_3}$, so two are outside the coordination sphere; three ions in solution means one complex cation and two $\mathrm{Cl^-}$.
Octahedral: five $\mathrm{NH_3}$ and one Cl in the sphere: $\mathrm{[Co(NH_3)_5Cl]^{2+}+2Cl^-}$.
Answer: (D) $\mathrm{[Co(NH_3)_5Cl]Cl_2}$. | ||||
| 63 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Statement I is true: Dumas method estimates nitrogen, by heating the compound with CuO and measuring the $\mathrm{N_2}$ evolved.
Statement II is false: heating with conc. $\mathrm{H_2SO_4}$ to form ammonium sulphate is Kjeldahl's method.
Answer: (C) Statement I is true but Statement II is false. | ||||
| 64 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques All four are allyl cations; the substituent on the far vinyl carbon decides the extra stabilisation, since the positive charge is delocalised onto that carbon.
$-\mathrm{OCH_3}$ (option B) donates its lone pair strongly (+R), giving an oxocarbenium form $\mathrm{CH_2=CH-CH=\overset{+}{O}-CH_3}$ in which every atom has an octet.
$-\mathrm{OCOCH_3}$ (A) donates less, its oxygen lone pair being drawn into the C=O; $-\mathrm{CH_2OCH_3}$ (C) cannot donate by resonance and withdraws inductively; $-\mathrm{F}$ (D) donates weakly by resonance but withdraws strongly by induction.
Answer: (B). | ||||
| 65 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Electrophilic addition of HBr goes through a carbocation; the most stable cation forms fastest.
Q is an enol ether (the C=C is on the carbon next to the ring oxygen): protonation gives a cation on the carbon attached to O, stabilised by the oxygen lone pair (oxocarbenium ion) — far more stable than any alkyl cation.
P's double bond is one carbon removed from O, so the oxygen only withdraws inductively; R and S give tertiary carbocations but without lone-pair stabilisation.
Answer: (B) Q. | ||||
| 66 | 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 Both are $S_N2$ reactions of a primary chloride; the solvent effect depends on how the charge changes on going to the transition state.
Statement I: $\mathrm{HO^-}$ + neutral substrate → transition state $[\mathrm{HO^{\delta-}\cdots C\cdots Cl^{\delta-}}]$: the charge is spread out in the transition state, so a more polar solvent stabilises the reactant ion more and slows the reaction; a less polar medium favours it. True.
Statement II: neutral amine + neutral substrate → transition state with developing charges $[\mathrm{R_3N^{\delta+}\cdots C\cdots Cl^{\delta-}}]$ and ionic products: a more polar medium stabilises the charge and speeds it. True.
Answer: (A) Both Statement I and Statement II are true. | ||||
| 67 | 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 x is 1-acetylcyclopentene, an $\alpha,\beta$-unsaturated ketone: the product of an intramolecular aldol condensation of a 1,6-dicarbonyl compound, $\mathrm{CH_3CO(CH_2)_4CHO}$ (6-oxoheptanal).
Ozonolysis of 1-methylcyclohexene cleaves the ring C=C to give exactly $\mathrm{CH_3CO-CH_2CH_2CH_2CH_2-CHO}$.
The enolate of the ketone's ring-side $\mathrm{CH_2}$ attacks the aldehyde to close a five-membered ring; dehydration gives the C=C in the ring conjugated with the acetyl group: 1-acetylcyclopentene.
The other alkenes give the wrong carbon skeleton (3- and 4-methylcyclohexene give dialdehydes with a methyl branch; methylenecyclohexane gives cyclohexanone).
Answer: (A). | ||||
| 68 | 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 Ketones are less reactive than aldehydes (more crowding, two electron-donating groups): acetophenone is least reactive.
Among the aldehydes, an electron-donating p-$\mathrm{CH_3}$ lowers the carbonyl's electrophilicity and an electron-withdrawing p-$\mathrm{NO_2}$ raises it.
Order: acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde.
Answer: (B). | ||||
| 69 | 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) Hydration of propyne (Markovnikov, $\mathrm{Hg^{2+}/H_2SO_4}$) gives acetone, $\mathrm{CH_3COCH_3}$.
ii) HCN adds to the carbonyl: the cyanohydrin $\mathrm{(CH_3)_2C(OH)CN}$.
iii) $\mathrm{H_2/Ni}$ reduces $-\mathrm{CN}$ to $-\mathrm{CH_2NH_2}$: $\mathrm{(CH_3)_2C(OH)CH_2NH_2}$.
Answer: (B) $\mathrm{CH_3-C(OH)(CH_3)-CH_2-NH_2}$. | ||||
| 70 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules The sugar of nucleic acids (2-deoxyribose in DNA, ribose in RNA) is a pentose (A true), of D-configuration (C true), present as a five-membered furanose ring, not pyranose (B false), and joined to the base as the $\beta$-anomer (E false).
When free, it has a hemiacetal/aldehyde group and reduces Tollens' and Fehling's reagents (D true).
Correct: A, C, D.
Answer: (A) A, C and D Only. | ||||
| 71 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{Fe_3O_4}$: $\frac{2.32\times10^6}{232} = 10^4$ mol. CO: $\frac{2.8\times10^5}{28} = 10^4$ mol.
$10^4$ mol $\mathrm{Fe_3O_4}$ would need $4\times10^4$ mol CO, so CO is limiting.
Fe $= \frac34\times10^4 = 7500$ mol $= 7500\times56$ g $= 420$ kg.
Answer: 420. | ||||
| 72 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta S^\ominus = 110-\frac12(70)-\frac52(50) = 110-35-125 = -50$ J K$^{-1}$ mol$^{-1}$.
At equilibrium $\Delta G = \Delta H-T\Delta S = 0$: $T = \frac{-35000}{-50} = 700$ K.
Answer: 700. | ||||
| 73 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $n(\mathrm{N_2O_5}) = \frac{37.8}{108} = 0.35$ mol; $P_0 = \frac{nRT}{V} = 0.35\times0.082\times500 = 14.35$ bar.
If $2p$ bar of $\mathrm{N_2O_5}$ reacts: $\mathrm{N_2O_5} = 14.35-2p$, $\mathrm{N_2O_4} = 2p$, $\mathrm{O_2} = p$; total $= 14.35+p = 18.65$, so $p = 4.3$.
Partial pressures: $\mathrm{N_2O_5} = 5.75$, $\mathrm{N_2O_4} = 8.6$, $\mathrm{O_2} = 4.3$ bar.
$K_p = \frac{(8.6)^2(4.3)}{(5.75)^2} = \frac{318.03}{33.06} = 9.619 = 961.9\times10^{-2}$, nearest integer 962.
Answer: 962. | ||||
| 75 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter In the standard confirmatory tests, potassium ferrocyanide gives:
$\mathrm{Cu^{2+}}$: chocolate-brown $\mathrm{Cu_2[Fe(CN)_6]}$; $\mathrm{Fe^{3+}}$: Prussian blue $\mathrm{Fe_4[Fe(CN)_6]_3}$; $\mathrm{Zn^{2+}}$: bluish-white $\mathrm{K_2Zn_3[Fe(CN)_6]_2}$.
$\mathrm{Ba^{2+}}$, $\mathrm{Ca^{2+}}$, $\mathrm{Mg^{2+}}$ and $\mathrm{NH_4^+}$ are identified by other reagents (chromate, oxalate, phosphate, Nessler's reagent).
Number $= 3$.
Answer: 3. | ||||
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