JEE (Main) · 2025
JEE (Main) 2025 — 28 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
22 Q
- Sets, Relations and Functions3
- Sequences and Series2
- Probability2
6 easy · 15 medium · 1 hard
Physics
25 Q
- Motion of System of Particles and Rigid Body3
- Electrostatics3
- Optics3
14 easy · 10 medium · 1 hard
Chemistry
22 Q
- Solutions2
- Equilibrium2
- Classification of Elements and Periodicity2
12 easy · 8 medium · 2 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 FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Reflexive: $x + x = 2x$ is even, so $(x, x) \in R$.
Symmetric: $x + y$ even $\Rightarrow y + x$ even.
Transitive: if $x + y$ and $y + z$ are even, then $x + z = (x + y) + (y + z) - 2y$ is even.
($x + y$ is even exactly when $x$ and $y$ have the same parity.)
Answer: (D) an equivalence relation. | ||||
| 2 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy 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 $f(1 - x) = \frac{2^{1-x}}{2^{1-x} + \sqrt2} = \frac{2}{2 + \sqrt2\cdot2^x} = \frac{\sqrt2}{\sqrt2 + 2^x}$, so $f(x) + f(1 - x) = 1$.
The terms $f\left(\frac{k}{82}\right)$ and $f\left(\frac{82 - k}{82}\right)$ pair up for $k = 1, \dots, 40$, giving $40$; the middle term is $f\left(\frac{41}{82}\right) = f\left(\frac12\right) = \frac{\sqrt2}{2\sqrt2} = \frac12$.
Sum $= 40 + \frac12 = \frac{81}{2}$.
Answer: (B) $\frac{81}{2}$. | ||||
| 4 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Case $x \ge \frac32$: $x^2 + 2x - 7 = 0$, $x = -1 \pm 2\sqrt2$; only $x = 2\sqrt2 - 1 \approx 1.83$ satisfies $x \ge \frac32$.
Case $x < \frac32$: $x^2 - 2x - 1 = 0$, $x = 1 \pm \sqrt2$; only $x = 1 - \sqrt2$ satisfies $x < \frac32$.
Sum of squares: $(2\sqrt2 - 1)^2 + (1 - \sqrt2)^2 = (9 - 4\sqrt2) + (3 - 2\sqrt2) = 12 - 6\sqrt2 = 6(2 - \sqrt2)$.
Answer: (A) $6(2 - \sqrt2)$. | ||||
| 5 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The characteristic equation $2r^2 - 5r + 3 = 0$ has roots $r = 1$ and $r = \frac32$, so $a_n = A + B\left(\frac32\right)^n$.
$a_0 = 0$: $A + B = 0$; $a_1 = \frac12$: $A + \frac32B = \frac12$. So $B = 1$, $A = -1$ and $a_n = \left(\frac32\right)^n - 1$.
$\sum_{k=1}^{100} a_k = \sum_{k=1}^{100}\left(\frac32\right)^k - 100 = \frac{\frac32\left(\left(\frac32\right)^{100} - 1\right)}{\frac12} - 100 = 3\left(\left(\frac32\right)^{100} - 1\right) - 100 = 3a_{100} - 100$.
Answer: (D) $3a_{100} - 100$. | ||||
| 6 | 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 $\sum_{r=1}^{25} T_r = \frac{13}{20} = \frac{25}{2}(T_1 + T_{25})$, so $T_1 + T_{25} = \frac{26}{500}$ and $T_1 = \frac{26}{500} - \frac{25}{500} = \frac{1}{500}$.
$d = \frac{T_{25} - T_1}{24} = \frac{24/500}{24} = \frac{1}{500}$, so $T_r = \frac{r}{500}$.
$T_m = \frac{1}{25} = \frac{20}{500}$ gives $m = 20$.
$\sum_{r=20}^{40} T_r = \frac{1}{500}\cdot\frac{21(20 + 40)}{2} = \frac{630}{500}$.
$5m\sum = 100\cdot\frac{630}{500} = 126$.
Answer: (C) 126. | ||||
| 7 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Digits may repeat (with all digits different the count is only $4\cdot120 + 3\cdot120 + 2\cdot120 = 1080$, not an option).
The first digit is 5, 6 or 7. With first + last $\le 8$: first 5 allows last $0$–$3$ (4 ways), first 6 allows $0$–$2$ (3), first 7 allows $0$–$1$ (2): $9$ choices of (first, last).
The three middle digits: $8^3 = 512$ ways. Total $9 \times 512 = 4608$.
This includes $50000$, which is not greater than $50000$: $4608 - 1 = 4607$.
Answer: (A) 4607. | ||||
| 8 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Draws are without replacement from 10 oranges (3 defective).
$P(x = 0) = \frac{\binom72}{\binom{10}2} = \frac{21}{45}$, $P(x = 1) = \frac{3\cdot7}{45} = \frac{21}{45}$, $P(x = 2) = \frac{3}{45}$.
$E(x) = \frac{21 + 6}{45} = \frac35$; $E(x^2) = \frac{21 + 12}{45} = \frac{11}{15}$.
$\text{Var}(x) = \frac{11}{15} - \frac{9}{25} = \frac{55 - 27}{75} = \frac{28}{75}$.
Answer: (C) $\frac{28}{75}$. | ||||
| 9 | 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 $i^k$ depends only on $k \bmod 4$, and the four remainders are equally likely, so the pair of remainders takes 16 equally likely values.
$i^{k_1} + i^{k_2} = 0$ exactly when $i^{k_1} = -i^{k_2}$, i.e. the remainders differ by 2: $(0, 2), (2, 0), (1, 3), (3, 1)$ — 4 of the 16 cases.
Probability of a non-zero sum $= 1 - \frac{4}{16} = \frac34$.
Answer: (C) $\frac34$. | ||||
| 10 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $(1, 0)$ is the focus, so AC is a focal chord. With $A = (t^2, 2t)$, $C = \left(\frac1{t^2}, -\frac2t\right)$ and length $\left(t + \frac1t\right)^2 = \frac{25}{4}$, $t + \frac1t = \frac52$, $t = 2$.
$A = (4, 4)$, $C = \left(\frac14, -1\right)$. AD is vertical: $D = (4, -4)$; BC is vertical: $B = \left(\frac14, 1\right)$.
Parallel sides $AD = 8$, $BC = 2$; distance between them $4 - \frac14 = \frac{15}{4}$.
Area $= \frac12(8 + 2)\cdot\frac{15}{4} = \frac{75}{4}$.
Answer: (B) $\frac{75}{4}$. | ||||
| 11 | 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 $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n - r + 1}{r} = 2$ and $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n - r}{r + 1} = \frac54$ give $n + 1 = 3r$ and $4n = 9r + 5$, so $r = 3$, $n = 8$.
$C = (9 - 8, 9 - 8 - 1) = (1, 0)$.
Centroid: $3x = 4\cos t + 2\sin t + 1$, $3y = 4\sin t - 2\cos t$.
$(3x - 1)^2 + (3y)^2 = (4\cos t + 2\sin t)^2 + (4\sin t - 2\cos t)^2 = 16 + 4 = 20$.
Answer: (A) 20. | ||||
| 13 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\alpha = \sin^{-1}\frac35$, $\beta = \sin^{-1}\frac5{13}$: $\cos\alpha = \frac45$, $\cos\beta = \frac{12}{13}$.
$\cos(\alpha + \beta) = \frac45\cdot\frac{12}{13} - \frac35\cdot\frac5{13} = \frac{33}{65}$, and $\alpha + \beta \in \left(0, \frac\pi2\right)$, so $\alpha + \beta = \cos^{-1}\frac{33}{65}$.
Then $\alpha + \beta + \sin^{-1}\frac{33}{65} = \cos^{-1}\frac{33}{65} + \sin^{-1}\frac{33}{65} = \frac\pi2$.
$\cos\frac\pi2 = 0$.
Answer: (A) 0. | ||||
| 14 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $A = (x, y, 0)$. Equal distances to $(0, 3, 2)$ and $(0, 0, 1)$: $x^2 + (y - 3)^2 + 4 = x^2 + y^2 + 1 \Rightarrow y = 2$.
To $(2, 0, 3)$ and $(0, 0, 1)$: $(x - 2)^2 + y^2 + 9 = x^2 + y^2 + 1 \Rightarrow x = 3$. So $A = (3, 2, 0)$.
$\vec{AB} = (-2, 2, -1)$, $|\vec{AB}| = 3$; $\vec{AC} = (-1, -2, -2)$, $|\vec{AC}| = 3$; $\vec{AB}\cdot\vec{AC} = 2 - 4 + 2 = 0$.
So the triangle is right-angled and isosceles at A: (S1) is true. Area $= \frac12\cdot3\cdot3 = \frac92 \ne \frac{9\sqrt2}{2}$: (S2) is false.
Answer: (C) only (S1) is true. | ||||
| 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 A point of the line: $F = (1 + 2t, 2 + t, 1 + 3t)$. For the foot of the perpendicular, $(F - P)\cdot(2, 1, 3) = 0$ with $P = (4, 4, 3)$:
$2(2t - 3) + (t - 2) + 3(3t - 2) = 14t - 14 = 0$, so $t = 1$ and $F = (3, 3, 4)$.
Image $= 2F - P = (2, 2, 5)$; $\alpha + \beta + \gamma = 9$.
Answer: (C) 9. | ||||
| 16 | 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 $x < -1$: $f = 1 - 2x$ decreases to $3$; $f(-1) = 3$.
$-1 \le x \le 0$: $f = \frac{7 - 2x}{3}$ decreases from $3$ to $\frac73$; $0 \le x \le 2$: $f = \frac{7 + 2x}{3}$ increases to $\frac{11}{3}$. So $x = 0$ is a local minimum, value $\frac73$, and $x = -1$ is not an extremum ($f$ decreases on both sides).
$x > 2$: $f \to \frac{11}{18}\cdot6 = \frac{11}{3}$ (continuous at 2), decreases to the vertex $x = \frac92$ with $f = \frac{11}{18}\cdot\frac12\cdot\left(-\frac12\right) = -\frac{11}{72}$, then increases. So $x = 2$ is a local maximum and $x = \frac92$ a local minimum.
Sum of local minimum values $= \frac73 - \frac{11}{72} = \frac{168 - 11}{72} = \frac{157}{72}$.
Answer: (C) $\frac{157}{72}$. | ||||
| 17 | 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 The region is symmetric about the y-axis. For $0 \le x \le 3$ the upper boundary is $\min(2x + 1, x^2 + 1)$, which is $x^2 + 1$ for $0 \le x \le 2$ and $2x + 1$ for $2 \le x \le 3$.
Right half: $\int_0^2(x^2 + 1)\,dx + \int_2^3(2x + 1)\,dx = \left(\frac83 + 2\right) + (12 - 6) = \frac{14}{3} + 6 = \frac{32}{3}$.
Total area $= 2\cdot\frac{32}{3} = \frac{64}{3}$.
Answer: (A) $\frac{64}{3}$. | ||||
| 19 | 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 For $f(x) = px^2 + qx + b$: $f(x + y) - f(x) - f(y) = 2pxy - b$. Matching $1 - \frac27xy$: $2p = -\frac27$ and $-b = 1$, so $p = -\frac17$, $b = -1$.
$2 + 3a = -\frac17 \Rightarrow a = -\frac57$; $q = \frac{a + 2}{a - 1} = \frac{9/7}{-12/7} = -\frac34$.
$f(x) = -\frac{x^2}{7} - \frac{3x}{4} - 1$, negative for $x \ge 1$, so $|f(i)| = \frac{i^2}{7} + \frac{3i}{4} + 1$.
$\sum_{i=1}^5|f(i)| = \frac{55}{7} + \frac{45}{4} + 5$, and $28\sum = 220 + 315 + 140 = 675$.
Answer: (B) 675. | ||||
| 20 | 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 With $x \to -x$: $I = \int \frac{96x^2\cos^2x\,e^x}{1 + e^x}dx$; adding, $2I = \int_{-\pi/2}^{\pi/2}96x^2\cos^2x\,dx$, so $I = \int_0^{\pi/2}96x^2\cos^2x\,dx = 48\int_0^{\pi/2}x^2(1 + \cos2x)\,dx$.
$\int_0^{\pi/2}x^2dx = \frac{\pi^3}{24}$. $\int_0^{\pi/2}x^2\cos2x\,dx = \left[\frac{x^2\sin2x}{2} + \frac{x\cos2x}{2} - \frac{\sin2x}{4}\right]_0^{\pi/2} = -\frac\pi4$.
$I = 48\left(\frac{\pi^3}{24} - \frac\pi4\right) = 2\pi^3 - 12\pi = \pi(2\pi^2 - 12)$.
$\alpha = 2$, $\beta = -12$, $(\alpha + \beta)^2 = 100$.
Answer: (B) 100. | ||||
| 21 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | NumericalHard | |
A hard numerical question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $S_2 = \emptyset$: a skew-symmetric matrix has zero diagonal, and $0 \notin S$.
$n(S_1) = 5^6$ (three diagonal and three upper entries free).
Diagonal triples from S with sum 0: permutations of $(-3, 1, 2)$: 6; of $(-2, 1, 1)$: 3; of $(-1, -1, 2)$: 3; total 12. So $n(S_3) = 12\cdot5^6$ (six off-diagonal entries free).
$n(S_1 \cap S_3) = 12\cdot5^3$.
$n(S_1 \cup S_3) = 5^6 + 12\cdot5^6 - 12\cdot5^3 = 125(13\cdot125 - 12) = 125\cdot1613$.
Answer: 1613. | ||||
| 22 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem In $(1 + i\sqrt3)^{12} = \sum_k {}^{12}C_k(i\sqrt3)^k$, the odd terms are $k = 2r - 1$: $(i\sqrt3)^{2r-1} = i\sqrt3\,(-3)^{r-1}$.
So $\text{Im}(1 + i\sqrt3)^{12} = \sqrt3\sum_{r=1}^6(-3)^{r-1}\,{}^{12}C_{2r-1}$.
$(1 + i\sqrt3)^{12} = \left(2e^{i\pi/3}\right)^{12} = 4096e^{4\pi i} = 4096$, which is real, so the sum is 0 and $\alpha = 1$.
Distance of $(12, \sqrt3)$ from $x - \sqrt3y + 1 = 0$: $\frac{|12 - 3 + 1|}{\sqrt{1 + 3}} = \frac{10}{2} = 5$.
Answer: 5. | ||||
| 23 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Same eccentricity means the same ratio $\frac{b_i}{a_i} = \frac23$. The condition $2b_i = 2a_{i+1}$ gives $a_{i+1} = b_i = \frac23a_i$.
Areas $A_i = \pi a_ib_i = \frac23\pi a_i^2$ form a G.P. with ratio $\frac49$ and $A_1 = \pi\cdot3\cdot2 = 6\pi$.
$\sum A_i = \frac{6\pi}{1 - \frac49} = \frac{54\pi}{5}$; $\frac5\pi\sum A_i = 54$.
Answer: 54. | ||||
| 24 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\vec c - 2\vec a|^2 = |\vec c|^2 - 4\vec a\cdot\vec c + 4|\vec a|^2 = |\vec c|^2 - 4|\vec c| + 12 = 8$, so $(|\vec c| - 2)^2 = 0$, $|\vec c| = 2$ and $\vec a\cdot\vec c = 2$.
$\vec d = \vec a \times \vec b = -\hat i + \hat j$, $|\vec d| = \sqrt2$. $|\vec d \times \vec c|^2 = \left(\sqrt2\cdot2\cdot\frac{1}{\sqrt2}\right)^2 = 4$, and $\vec d\cdot\vec c = 2$.
With $\vec c = (c_1, c_2, c_3)$: $-c_1 + c_2 = 2$, $c_1 + c_2 + c_3 = 2$, $c_1^2 + c_2^2 + c_3^2 = 4$. Then $c_2 = c_1 + 2$, $c_3 = -2c_1$ and $6c_1^2 + 4c_1 = 0$: $c_1 = 0$ or $-\frac23$.
$\vec c = (0, 2, 0)$: $\vec b\cdot\vec c = 4$; $\vec c = \left(-\frac23, \frac43, \frac43\right)$: $\vec b\cdot\vec c = \frac83$. In both cases $|10 - 3\vec b\cdot\vec c| = 2$.
$2 + 4 = 6$.
Answer: 6. | ||||
| 25 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability On $[0, 1)$: $\min(1 + x, x) = x$. On $[1, 2)$: $\min(2 + x, x + 2) = x + 2$. At $x = 2$: $\min(5, 6) = 5$. For $x > 2$: $5$.
$x = 0$: $f(0^-) = 0 = f(0)$, continuous; slopes 3 and 1, not differentiable.
$x = 1$: $f(1^-) = 1$, $f(1) = 3$: discontinuous.
$x = 2$: $f(2^-) = 4$, $f(2) = 5$: discontinuous.
$\alpha = 2$ (at 1, 2); $\beta = 3$ (at 0, 1, 2). $\alpha + \beta = 5$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take strips of width $dx$ parallel to the y-axis: $dm = \sigma b\,dx = \frac{\sigma_0x}{a}dx$.
$x_{cm} = \frac{\int_0^a x\cdot x\,dx}{\int_0^a x\,dx} = \frac{a^3/3}{a^2/2} = \frac{2a}{3}$.
The density does not depend on y, so $y_{cm} = \frac b2$.
Answer: (B) $\left(\frac23a, \frac b2\right)$. | ||||
| 27 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At the top the spring (from the bottom of the hoop) has length $2R$, an extension $R$ over its natural length $R$: stored energy $\frac12kR^2$.
When the spring length is $R$ it is unstretched. A chord of length $R$ from the lowest point of a circle of radius $R$ reaches a height $h = \frac{R^2}{2R} = \frac R2$ above the bottom.
Energy: $\frac12kR^2 + mg(2R) = \frac12mv^2 + mg\frac R2$, so $v^2 = 3gR + \frac{kR^2}{m}$.
Answer: (D) $\sqrt{3Rg + \frac{kR^2}{m}}$. | ||||
| 28 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power A central force whose magnitude depends only on the distance from the centre, $\vec F = f(r)\hat r$, is conservative: $W = \int f(r)\,dr$ depends only on the end points. A is true.
Non-conservative forces such as friction have no associated potential energy, so R is true.
R does not explain why a central force is path-independent.
Answer: (B). | ||||
| 29 | Properties of Bulk MatterViscosity | 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 viscosity, 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: false — surface tension arises because molecules at the surface have extra energy compared with those in the interior.
B: false — the viscosity of a liquid decreases as temperature rises.
C: true — the viscosity of a gas increases with temperature.
D: true — turbulence sets in when the Reynolds number exceeds a critical value.
E: true — two streamlines in steady flow cannot cross, or the velocity at the crossing point would not be unique.
Answer: (B) C, D, E only. | ||||
| 30 | Experimental SkillsProperties of matter experiments | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on properties of matter experiments, from Experimental Skills. NCERT: No matching NCERT chapter Terminal velocity $V = \frac{2R^2(\rho - \sigma)g}{9\eta}$.
A: true — $V \propto R^2$, a parabola.
B: false — V depends on the radius of the ball.
C: true — viscosity, and with it V, depends on temperature.
D: true — with $\eta$ known, the measured V gives the density $\sigma$ of the liquid.
E: false — $\eta$ is a property of the liquid; the initial speed of the ball does not change it.
Answer: (A) A, C and D only. | ||||
| 31 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory From kinetic theory, $\frac12m\overline{v^2} = \frac32k_BT$, so $\overline{v^2} = \frac{3k_BT}{m} \propto T$.
The graph is a straight line through the origin.
Answer: (A). | ||||
| 32 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\eta_{12} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.423$; $\eta_1 = 1 - \frac{373}{473} = \frac{100}{473} \approx 0.211$; $\eta_2 = 1 - \frac{273}{373} = \frac{100}{373} \approx 0.268$.
In general $1 - \eta_{12} = (1 - \eta_1)(1 - \eta_2)$, so $\eta_{12} = \eta_1 + \eta_2 - \eta_1\eta_2 < \eta_1 + \eta_2$.
Answer: (D) $\eta_{12} < \eta_1 + \eta_2$. | ||||
| 33 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v = \sqrt{\frac{E}{\rho}}$; solids have far larger elastic moduli, and sound travels faster in solids than in gases: A is true.
Gases are far more compressible, so their bulk modulus is much smaller than that of solids: R is false.
Answer: (C) A is true but R is false. | ||||
| 34 | 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 Triangle: each side $\frac R3$; between the ends of one side, $\frac R3$ is in parallel with $\frac{2R}3$: $\frac{(R/3)(2R/3)}{R} = \frac{2R}{9}$.
Square: each side $\frac R4$; $\frac R4$ in parallel with $\frac{3R}4$: $\frac{(R/4)(3R/4)}{R} = \frac{3R}{16}$.
Ratio $= \frac{2/9}{3/16} = \frac{32}{27}$.
Answer: (C) $\frac{32}{27}$. | ||||
| 35 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctHard | |
A hard single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The field of the wire is $B = \frac{\mu_0I}{2\pi r}$, circling the wire. The magnetic force does no work, so the speed stays $v_0$.
With $z$ along the current and $r$ the distance from the wire, the force component along the wire is $m\frac{dv_z}{dt} = qv_rB = \frac{q\mu_0I}{2\pi r}\frac{dr}{dt}$ (the sign that makes the particle approach the wire while moving with the current).
Integrating from $r = a$, $v_z = v_0$: $v_z = v_0 + \frac{q\mu_0I}{2\pi m}\ln\frac ra$.
The particle turns round where $v_r = 0$ again, i.e. $|v_z| = v_0$ with $v_z = -v_0$: $-2v_0 = \frac{q\mu_0I}{2\pi m}\ln\frac xa$.
$x = a\,e^{-\frac{4\pi mv_0}{q\mu_0I}}$.
Answer: (A). | ||||
| 36 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Name the nodes along the line: left end L, then $\frac r3$ to node P, $\frac r3$ to node Q, $\frac r3$ to the right end T.
The upper wire joins L to Q, and the lower wire joins P to T. So $Q \equiv L$ and $T \equiv P$.
Each of the three resistors then lies between L and P: three $\frac r3$ resistors in parallel, $\frac{r/3}{3} = \frac r9$.
Answer: (A) $\frac r9$. | ||||
| 37 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The string OA, of length $l$, makes $60^\circ$ with the x-axis, and E points along +x. The force $qE$ has an outward component along the string, so the string stays taut and A moves on a circle of radius $l$ about O.
From A to the x-axis the particle moves along x from $l\cos60^\circ = \frac l2$ to $l$: work by the field $= qE\left(l - \frac l2\right) = \frac{qEl}{2}$. Tension does no work.
$\frac12mv^2 = \frac{qEl}{2}$, so $v = \sqrt{\frac{qEl}{m}}$.
Answer: (B). | ||||
| 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 In parallel both capacitors end at the battery voltage V. The graph shows $C_2$ reaching the larger final charge: $q_2 > q_1$.
$q = CV$ with the same V gives $C_2 > C_1$, and $U = \frac12CV^2$ gives $U_2 > U_1$.
Answer: (C) $C_2 > C_1$, $U_2 > U_1$. | ||||
| 39 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance An infinite line charge gives $V = -\frac{\lambda}{2\pi\varepsilon_0}\ln d + \text{const}$, d the distance from the line.
Distances from the x-, y- and z-axes: $\sqrt{y^2 + z^2}$, $\sqrt{z^2 + x^2}$, $\sqrt{x^2 + y^2}$.
$V = -\frac{\lambda}{4\pi\varepsilon_0}\ln\left[(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)\right] + \text{const}$.
V is constant where the product is constant.
Answer: (C). | ||||
| 40 | 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 Average energy density $u \propto E_0^2$; energy held $= u \times$ volume, and volume $\propto d^2$ for the same length.
Halving the diameter makes the volume $\frac14$, so for the same energy $E_0^2$ must be 4 times larger: $E_0 = 2\times100 = 200$ N C$^{-1}$.
Answer: (B) $200\sin(\omega t - kx)$ NC$^{-1}$. | ||||
| 41 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics A ray reaching E travels along the liquid surface, so it leaves the liquid at grazing emergence ($90^\circ$): $\mu\sin i = 1$ at the point where it leaves.
Rays from O meet the surface at an angle of incidence at most $45^\circ$ (the ray from O to the rim point B, since $OC = CB = R$).
So $\mu = \frac{1}{\sin i} \ge \frac{1}{\sin45^\circ} = \sqrt2$.
Answer: (B) $\sqrt2$. | ||||
| 42 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics No deviation: $(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$.
$0.54\times4 = 0.72\,A_2$, so $A_2 = \frac{2.16}{0.72} = 3^\circ$.
Answer: (A) 3. | ||||
| 43 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Photon: $E = \frac{hc}{\lambda}$, so $\lambda = \frac{hc}{E}$.
Proton with kinetic energy E: $p = \sqrt{2m_pE}$, $\lambda_p = \frac{h}{\sqrt{2m_pE}}$.
$\frac{\lambda_p}{\lambda} = \frac{h}{\sqrt{2m_pE}}\cdot\frac{E}{hc} = \frac1c\sqrt{\frac{E}{2m_p}}$.
Answer: (B). | ||||
| 44 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Charge: $0 = +1 - 1$, so the electron $e^-$ is emitted (rules out (B) and (D)).
Lepton number: $0 = 1 + L_\nu$, so the third particle has lepton number $-1$: the antineutrino.
$n \to p + e^- + \bar\nu$ is $\beta^-$ decay.
Answer: (C). | ||||
| 45 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits From the circuit: the upper AND gate gets A and $\bar B$ (B through the NOT gate); the lower AND gate gets A (the vertical line from A, which B's line hops over) and B.
NOR output: $Y = \overline{A\bar B + AB} = \overline{A(\bar B + B)} = \bar A$.
A NAND gate with both inputs tied to A gives $\overline{A\cdot A} = \bar A$.
Answer: (B). | ||||
| 46 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | NumericalEasy | |
A easy numerical question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Modulus of elasticity: $[ML^{-1}T^{-2}]$; torque: $[ML^2T^{-2}]$.
$\frac{[ML^{-1}T^{-2}]}{[ML^2T^{-2}]} = [M^0L^{-3}T^0]$.
The time exponent is $c = 0$ (the mass and time dimensions cancel whichever way the ratio is taken).
Answer: 0. | ||||
| 47 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $= \frac{0.75}{15} = 0.05$ mm, the error in each reading.
$\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b} = \frac{0.05}{5} + \frac{0.05}{2.5} = 0.01 + 0.02 = \frac{3}{100}$.
Answer: 3. | ||||
| 48 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Both discs are of iron with the same (negligible) thickness t: $M = \rho\pi R^2t \propto R^2$.
About the central axis $I = \frac12MR^2 \propto R^4$.
$\frac{I_1}{I_2} = \left(\frac{R_1}{R_2}\right)^4 = \frac1{16}$, so $x = 16$.
Answer: 16. | ||||
| 49 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion About a diameter: disc $\frac14MR^2$, sphere $\frac25MR^2$ (same mass and radius), ring of radius r: $\frac12Mr^2$.
$\frac{I_{sphere}}{I_{disc}} = \frac{2/5}{1/4} = \frac85$.
$n = \frac{I_{sphere}}{I_{ring}} = \frac{I_{sphere}}{I_{disc}}\cdot\frac{I_{disc}}{I_{ring}} = \frac85\times2.5 = 4$.
(The ring's radius is fixed by the given ratio: $\frac14MR^2 = 2.5\cdot\frac12Mr^2$, $r^2 = \frac{R^2}{5}$.)
Answer: 4. | ||||
| 50 | 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 $y_n = \frac{n\lambda D}{d} \propto \lambda$ for the same set-up.
$y' = 10\times\frac{660}{600} = 11$ mm.
Answer: 11. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 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 $\mathrm{BrF_5}$: 5 bond pairs + 1 lone pair on Br (octahedral arrangement): square pyramidal.
$\mathrm{XeOF_4}$: 5 σ-bonded atoms + 1 lone pair on Xe: square pyramidal.
$\mathrm{PCl_5}$ and $\mathrm{SbF_5}$: 5 bond pairs, no lone pair: trigonal bipyramidal.
Answer: (C) $\mathrm{BrF_5}$ & $\mathrm{XeOF_4}$. | ||||
| 53 | SolutionsColligative properties | Class 12: Solutions | Single correctEasy | |
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions Molality $m = \frac{1/256}{0.050} = 0.078125$ mol kg$^{-1}$.
$K_f = \frac{\Delta T_f}{m} = \frac{0.40}{0.078125} = 5.12$ K kg mol$^{-1}$.
Answer: (C) 5.12 K kg mol$^{-1}$. | ||||
| 54 | EquilibriumLe Chatelier's principle | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on le chatelier's principle, from Equilibrium. NCERT: Class 11: Equilibrium At 1 atm, 273.15 K is the melting point, and ice and water coexist.
Ice is less dense than water, so by Le Chatelier's principle a rise in pressure favours the denser liquid, and the melting point falls below 273.15 K (Clausius–Clapeyron: about $-0.0075$ K per atm).
At 2 atm and a temperature held at 273.15 K, which is now above the melting point, liquid is the only stable phase: the ice melts completely.
The volume of the system decreases as ice turns to water, so (D) is wrong, and (A) is the reverse of what happens.
Answer: (C) strictly; (B), the Le Chatelier statement that the amount of ice decreases, is the textbook phrasing. | ||||
| 55 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{HA \rightleftharpoons H^+ + A^-}$; with degree of dissociation x, $\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = \frac{cx}{c(1 - x)} = \frac{x}{1 - x}$.
$K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$ gives $\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$.
$\mathrm{pH} - \mathrm{p}K_a = \log\frac{x}{1 - x}$.
Answer: (D). | ||||
| 56 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions A: methane combines with oxygen (combustion is classed as a combination redox reaction): II.
B: NaH breaks into its elements: decomposition, III.
C: Ca displaces V from its oxide: displacement, IV.
D: in $\mathrm{H_2O_2}$ oxygen ($-1$) goes both to $-2$ (water) and to $0$ ($\mathrm{O_2}$): disproportionation, I.
Answer: (C) A-II, B-III, C-IV, D-I. | ||||
| 57 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For order n, $t_{1/2} \propto [A]_0^{1-n}$.
$\frac{200}{100} = \left(\frac{0.100}{0.025}\right)^{1-n}$: $2 = 4^{1-n}$, $1 - n = \frac12$, $n = \frac12$. A is true.
$t_{1/2} = k'\sqrt{[A]_0}$ with $k' = \frac{200}{\sqrt{0.1}} = 200\sqrt{10}$. At 1 M, $t_{1/2} = 200\sqrt{10}$ min: B is true.
At 1.6 M: $t_{1/2} = 200\sqrt{10}\cdot\sqrt{1.6} = 200\sqrt{16} = 800$ min: D is true.
The order is a property of the rate law and does not change with the starting concentration: C is false.
Answer: (C) A, B and D only. | ||||
| 58 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Atomic radius increases down a group: Mg (period 3) is larger than Be (period 2), so \"Be > Mg\" is wrong.
The others follow the trends: Al > B (down the group), B > N > F (across period 2); Mg > Al (across period 3), and the period-2 atoms C > O are smaller still; Si > P > Cl across period 3, and F is the smallest.
Answer: (C) Be > Mg > Al > Si. | ||||
| 59 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties First ionisation enthalpies (kJ mol$^{-1}$, approximately): Al 577, In 558, Tl 589, Ge 762, Sn 709, Pb 716.
Highest: Ge, whose most stable oxidation state is +4.
Lowest: In, whose most stable oxidation state is +3 (the inert-pair effect makes +1 the stable state only for Tl).
Answer: (D) +4 and +3. | ||||
| 60 | 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 In acidic medium both dichromate and permanganate oxidise iodide to iodine, sulphide ($\mathrm{H_2S}$) to sulphur, and iron(II) to iron(III).
Iodide to iodate and thiosulphate to sulphate are the permanganate reactions in neutral or faintly alkaline solution, not dichromate reactions.
Answer: (A) A, B and C only. | ||||
| 61 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mathrm{ClF_3}$: 3 bond pairs + 2 lone pairs on Cl (trigonal bipyramidal arrangement); both lone pairs take equatorial positions, giving the T-shape. So $n = 2$.
Unpaired electrons: $\mathrm{V^{3+}}$ ($3d^2$) 2; $\mathrm{Ti^{3+}}$ ($3d^1$) 1; $\mathrm{Cu^{2+}}$ ($3d^9$) 1; $\mathrm{Ni^{2+}}$ ($3d^8$) 2; $\mathrm{Ti^{2+}}$ ($3d^2$) 2.
Ions with 2 unpaired electrons: A, D and E.
Answer: (C) A, D and E only. | ||||
| 63 | 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 In 2-chloro-N,N-diethylethanamine the nitrogen lone pair displaces chloride intramolecularly (neighbouring-group participation), forming a strained aziridinium ion; hydroxide then opens this ring rapidly. Statement II is correct.
This anchimeric assistance makes the hydrolysis much faster than that of the carbon analogue, a primary chloride that can react only by ordinary $S_N2$. Statement I is correct.
Answer: (A) Both statements are correct. | ||||
| 64 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques C, the tropylium ion, is aromatic ($6\pi$ electrons in a planar ring): the most stable.
A, the triphenylmethyl cation, has its charge delocalised over three rings; B (diphenylmethyl) over two: A > B.
D, the sec-butyl cation, has only inductive and hyperconjugative stabilisation: the least stable.
Answer: (A) C > A > B > D. | ||||
| 65 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Work back from R, 1-acetyl-2-methylcyclopentene. It is the intramolecular aldol condensation product of octane-2,7-dione, $\mathrm{CH_3CO(CH_2)_4COCH_3}$ (the enolate of one methyl ketone's $\alpha$-$\mathrm{CH_2}$ attacks the other carbonyl, closing a five-membered ring, then dehydration).
Octane-2,7-dione is the ozonolysis product of 1,2-dimethylcyclohexene, so Q is 1,2-dimethylcyclohexene.
P must give Q by an acid-catalysed rearrangement. Isopropenylcyclopentane (D) is protonated (Markovnikov) to the tertiary cation $\mathrm{(CH_3)_2C^+}$–cyclopentyl; ring expansion of the strained cyclopentane gives the 2,2-dimethylcyclohexyl cation, a 1,2-methyl shift gives the tertiary 1,2-dimethylcyclohexyl cation, and loss of $\mathrm{H^+}$ gives the tetrasubstituted alkene Q.
(C) is Q itself and needs no rearrangement; (B) would expand to a cyclopentene, which ozonolysis and aldol turn into a six-membered enone, not R.
Answer: (D). | ||||
| 66 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Silver salts are largely covalent, so the ambident ion attacks through the atom that carries the free lone pair in the covalent salt.
$\mathrm{AgNO_2}$ (Ag–O–N=O) attacks through nitrogen: nitroalkane $\mathrm{CH_3CH_2CH_2NO_2}$ (A).
$\mathrm{AgCN}$ (Ag–C≡N) attacks through nitrogen: isocyanide $\mathrm{CH_3CH_2CH_2NC}$ (B).
Answer: (D). | ||||
| 67 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids A: both contain $\mathrm{CH_3CO-}$ and give iodoform. Yes.
B: both have $\alpha$-hydrogens, so neither undergoes the Cannizzaro reaction. No.
C: both have $\alpha$-hydrogens and undergo aldol condensation. Yes.
D: Tollens' reagent is reduced by the aldehyde only. No.
E: both carbonyl groups are reduced to $\mathrm{CH_2}$ by Zn–Hg/HCl. Yes.
Answer: (A) A, C and E only. | ||||
| 69 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Statement I: glucose pentaacetate has no free –OH at C-1 and cannot open to the free aldehyde, so it does not react with 2,4-DNP (or hydroxylamine). False.
Statement II: glucose is made from starch by boiling it with dilute $\mathrm{H_2SO_4}$ at 393 K under 2–3 atm pressure. Concentrated sulphuric acid at 100 °C chars and dehydrates the carbohydrate instead. False as worded.
Answer: (B) Both statements are false. | ||||
| 70 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Statement I: the permanganate–oxalate reaction is slow at room temperature, so the oxalic acid solution is warmed to about 60 °C; the Fe²⁺ titration is done cold. True.
Statement II: the first $\mathrm{Mn^{2+}}$ ($\mathrm{MnSO_4}$) formed in the warm solution catalyses the rest of the reaction (autocatalysis); heating an Fe²⁺ solution lets air oxidise some Fe²⁺ to Fe³⁺, which would lower the titre. True, and it explains I.
Answer: (A) Both statements are true. | ||||
| 71 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions 1 L of solution has mass $1000\times1.25 = 1250$ g, of which $0.70\times1250 = 875$ g is acid.
Moles $= \frac{875}{70} = 12.5$, so the molarity is $12.5$ M $= 125\times10^{-1}$ M.
Answer: 125. | ||||
| 72 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\mathrm{H_2O(g) \to 2H(g) + O(g)}$: $\Delta H = 2(220.0) + 250.0 - (-242.0) = 932.0$ kJ mol$^{-1}$.
This breaks two O–H bonds: average O–H bond enthalpy $= \frac{932}{2} = 466$ kJ mol$^{-1}$.
Answer: 466. | ||||
| 73 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry From the graph: $\Lambda_m = 150$ S cm² mol⁻¹ at $\sqrt C = 0.1$ ($C = 0.01$ mol L⁻¹) and $\Lambda_m = 100$ at $\sqrt C = 0.15$ ($C = 0.0225$ mol L⁻¹).
$\kappa \propto \Lambda_mC$: higher concentration $100\times0.0225 = 2.25$; dilute $150\times0.01 = 1.5$ (same units).
For the same cell $R \propto \frac1\kappa$: $x = 100\times\frac{2.25}{1.5} = 150\ \Omega$.
Answer: 150. | ||||
| 74 | Some Basic Concepts of ChemistryEmpirical and molecular formula | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry The remainder is oxygen: $100 - (14.5 + 1.8 + 64.46) = 19.24$ %.
Moles per 100 g: C $\frac{14.5}{12} = 1.208$; H $1.8$; Cl $\frac{64.46}{35.5} = 1.816$; O $\frac{19.24}{16} = 1.2025$.
Dividing by 1.2025: C 1.00, H 1.50, Cl 1.51, O 1.00; doubling: $\mathrm{C_2H_3Cl_3O_2}$.
Empirical formula mass $= 24 + 3 + 106.5 + 32 = 165.5 = 1655\times10^{-1}$.
Answer: 1655. | ||||
| 75 | 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 $\mathrm{C_6H_5Cl \xrightarrow{Mg} C_6H_5MgCl \xrightarrow{CO_2, H_3O^+} C_6H_5COOH \xrightarrow{NH_3, \Delta} C_6H_5CONH_2}$ (A).
Hofmann bromamide degradation: $\mathrm{C_6H_5CONH_2 \xrightarrow{Br_2, NaOH} C_6H_5NH_2}$ (B, aniline).
Molar mass of chlorobenzene $= 112.5$ g mol⁻¹: $\frac{11.25}{112.5} = 0.1$ mmol, giving 0.1 mmol aniline (93 g mol⁻¹) $= 9.3$ mg $= 93\times10^{-1}$ mg.
Answer: 93. | ||||
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