JEE (Main) · 2023
JEE (Main) 2023 — 11 April, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Three Dimensional Geometry3
- Sets, Relations and Functions2
5 easy · 22 medium · 3 hard
Physics
30 Q
- Electromagnetic Induction and Alternating Currents3
- Oscillations and Waves2
- Gravitation2
25 easy · 5 medium · 0 hard
Chemistry
30 Q
- Coordination Compounds3
- Haloalkanes and Haloarenes3
- Solutions2
16 easy · 12 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 FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions An element of R is a choice of $(a_1, b_2)$ with $a_1 \le b_2$ together with an independent choice of $(b_1, a_2)$ with $b_1 \le a_2$ ($a_1, a_2 \in A$; $b_1, b_2 \in B$).
Pairs $(a_1, b_2)$ with $a_1 \le b_2$: $a_1 = 1$: 5; $a_1 = 3$: 4 (4, 5, 8, 10); $a_1 = 4$: 4; $a_1 = 6$: 2 (8, 10); $a_1 = 9$: 1 (10). Total 16.
Pairs $(b_1, a_2)$ with $b_1 \le a_2$: $b_1 = 2$: 4 (3, 4, 6, 9); $b_1 = 4$: 3 (4, 6, 9); $b_1 = 5$: 2 (6, 9); $b_1 = 8$: 1 (9); $b_1 = 10$: 0. Total 10.
Number of elements $= 16 \times 10 = 160$.
Answer: (C). | ||||
| 2 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions We need $[x]^2 - 3[x] - 10 > 0$, i.e. $([x] - 5)([x] + 2) > 0$, so $[x] > 5$ or $[x] < -2$.
Since $[x]$ is an integer: $[x] \ge 6 \iff x \ge 6$, and $[x] \le -3 \iff x < -2$.
Domain: $(-\infty, -2) \cup [6, \infty)$.
Answer: (A). | ||||
| 3 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | StatementsMedium | |
A medium statements question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $a = p + iq$ and $z = x + iy$. Then $\mathrm{Re}(a + \bar{z}) = p + x$ and $\mathrm{Im}(\bar{a} + z) = y - q$.
For a real number z ($y = 0$): z is in A iff $p + x > -q$, i.e. $x > -(p + q)$; z is in B iff $x < -(p + q)$.
(S1): with $p, q > 0$, A contains only the reals $x > -(p+q)$; e.g. $x = -(p+q) - 1$ is not in A. False.
(S2): with $p, q < 0$, $-(p+q) > 0$ and B contains only the reals $x < -(p+q)$; e.g. $x = -(p+q) + 1$ is not in B. False.
Both statements are false.
Answer: (B). | ||||
| 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 (their x, y coefficients are not proportional), so for infinitely many solutions the third equation must be a combination $m(\text{eq}_1) + k(\text{eq}_2)$.
x: $7m + 5k = 175$; y: $11m + 4k = 194$. Solving: $m = 10$, $k = 21$.
z: $10\alpha + 21 \cdot 7 = 57 \Rightarrow \alpha = -9$.
Constants: $10 \cdot 13 + 21\beta = 361 \Rightarrow \beta = 11$.
$\alpha + \beta + 2 = -9 + 11 + 2 = 4$.
Answer: (B). | ||||
| 5 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Subtract $R_1$ from $R_2$ and $R_3$ and expand (or use $\det(D + xJ) = \det D\,(1 + x\,\mathrm{tr}\,D^{-1})$ with $D = \mathrm{diag}(1, \lambda, \lambda^2)$):
$\Delta = \lambda^3 + x(\lambda^3 + \lambda^2 + \lambda)$.
Comparing with $\frac{9}{8}(103x + 81)$: $\lambda^3 = \frac{729}{8}$, so $\lambda = \frac{9}{2}$; check: $\lambda^3 + \lambda^2 + \lambda = \frac{729 + 162 + 36}{8} = \frac{927}{8} = \frac{9 \cdot 103}{8}$.
Roots $\frac{9}{2}$ and $\frac{3}{2}$: sum 6, product $\frac{27}{4}$. Equation $x^2 - 6x + \frac{27}{4} = 0$, i.e. $4x^2 - 24x + 27 = 0$.
Answer: (D). | ||||
| 6 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Letters in alphabetical order: A, H, M, S, T.
Words beginning with A, H, M or S: $4 \times 4! = 96$.
Beginning with T, second letter before H (A): $3! = 6$.
TH followed by A (the smallest remaining), then M (smaller than S): THAMS is the next word.
Serial number $= 96 + 6 + 1 = 103$.
Answer: (C). | ||||
| 7 | 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 There are 2023 terms, so the $1011^{\text{th}}$ term from the end is $T_{1013}$ from the beginning.
$T_{1011} = \binom{2022}{1010}\left(\frac{4x}{5}\right)^{1012}\left(-\frac{5}{2x}\right)^{1010}$, $T_{1013} = \binom{2022}{1012}\left(\frac{4x}{5}\right)^{1010}\left(-\frac{5}{2x}\right)^{1012}$, and $\binom{2022}{1010} = \binom{2022}{1012}$.
$\frac{T_{1011}}{T_{1013}} = \left(\frac{4x}{5}\right)^2\left(\frac{2x}{5}\right)^2 = \frac{64x^4}{625}$.
Taken literally ($T_{1013} = 1024\,T_{1011}$) this gives $x^4 = \frac{625}{65536}$, $|x| = \frac{5}{16}$, which is not an option. The options fit the relation the other way round, $T_{1011} = 1024\,T_{1013}$:
$\frac{64x^4}{625} = 1024 \Rightarrow x^4 = 10^4 \Rightarrow |x| = 10$.
Answer: (B). | ||||
| 8 | 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 $N = n + 2$ and the coefficients be $\binom{N}{r-1}, \binom{N}{r}, \binom{N}{r+1}$.
$\frac{\binom{N}{r}}{\binom{N}{r-1}} = \frac{N - r + 1}{r} = 3 \Rightarrow N = 4r - 1$.
$\frac{\binom{N}{r+1}}{\binom{N}{r}} = \frac{N - r}{r + 1} = \frac{5}{3} \Rightarrow 3N = 8r + 5$.
So $12r - 3 = 8r + 5$, $r = 2$, $N = 7$. The coefficients are $\binom{7}{1}, \binom{7}{2}, \binom{7}{3} = 7, 21, 35$ (ratio 1:3:5), sum 63.
Answer: (C). | ||||
| 9 | Sequences and SeriesRelation between A.M. and G.M. | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on relation between a.m. and g.m., from Sequences and Series. NCERT: Class 11: Sequences and Series Write the sum as 11 terms: five of $\frac{a}{5}$, three of $\frac{b}{3}$, two of $\frac{c}{2}$ and d. Their A.M. is $\frac{11}{11} = 1$.
A.M. $\ge$ G.M.: $\left(\frac{a}{5}\right)^5\left(\frac{b}{3}\right)^3\left(\frac{c}{2}\right)^2 d \le 1$, with equality at $a = 5, b = 3, c = 2, d = 1$.
Maximum $a^5b^3c^2d = 5^5 \cdot 3^3 \cdot 2^2 = 3125 \cdot 27 \cdot 4 = 337500 = 3750 \times 90$.
$\beta = 90$.
Answer: (B). | ||||
| 10 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $x < -1$: $f(x) = x + 1 < 0$, so $g(f(x)) = x + 2$.
For $-1 \le x < 0$: $f(x) = x + 1 \ge 0$, so $g(f(x)) = 1$.
For $x \ge 0$: $f(x) = |x - 1| \ge 0$, so $g(f(x)) = 1$.
So $g \circ f = x + 2$ for $x < -1$ and 1 for $x \ge -1$. At $x = -1$ both pieces give 1: continuous everywhere. The left derivative at $-1$ is 1 and the right derivative is 0, so it is not differentiable at $x = -1$, and differentiable at every other point (in particular at $x = 1$).
Answer: (D). | ||||
| 11 | 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 On $[0, 1)$: $x - [x] = x \ge x^2$, so $f(x) = e^{x^2}$.
On $[1, 2]$: $h(x) = x - \ln x$ has $h'(x) = 1 - \frac{1}{x} \ge 0$, so $1 = h(1) \le h(x) \le h(2) = 2 - \ln 2 < 2$; hence $[x - \ln x] = 1$ and $f(x) = e$.
$\int_0^1 x e^{x^2} dx = \frac{e - 1}{2}$ and $\int_1^2 e x\, dx = \frac{3e}{2}$.
Total $= \frac{e - 1}{2} + \frac{3e}{2} = 2e - \frac{1}{2}$.
Answer: (C). | ||||
| 12 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $I_1 = \int_0^{\pi/2} f(\sin 2x)\sin x\,dx$. Replacing x by $\frac{\pi}{2} - x$ gives $I_1 = \int_0^{\pi/2} f(\sin 2x)\cos x\,dx$, so $I_1 = \frac{1}{2}\int_0^{\pi/2} f(\sin 2x)(\sin x + \cos x)\,dx$. The integrand is symmetric about $\frac{\pi}{4}$, so $I_1 = \int_0^{\pi/4} f(\sin 2x)(\sin x + \cos x)\,dx$.
Let $I_2 = \int_0^{\pi/4} f(\cos 2x)\cos x\,dx$. Put $x = \frac{\pi}{4} - t$: $\cos 2x = \sin 2t$ and $\cos x = \frac{\cos t + \sin t}{\sqrt{2}}$, so $I_2 = \frac{1}{\sqrt{2}}\int_0^{\pi/4} f(\sin 2t)(\sin t + \cos t)\,dt = \frac{I_1}{\sqrt{2}}$.
$I_1 + \frac{\alpha}{\sqrt{2}} I_1 = 0$ for every such f, so $\alpha = -\sqrt{2}$.
Answer: (A). | ||||
| 13 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\int \frac{5}{x(x^5 + 1)}dx = \int\left(\frac{5}{x} - \frac{5x^4}{x^5 + 1}\right)dx = 5\ln x - \ln(x^5 + 1)$, so the integrating factor is $\frac{x^5}{x^5 + 1}$.
$y\frac{x^5}{x^5 + 1} = \int \frac{(x^5 + 1)^2}{x^7}\cdot\frac{x^5}{x^5 + 1}dx = \int \frac{x^5 + 1}{x^2}dx = \frac{x^4}{4} - \frac{1}{x} + C$.
$x = 1$: $2 \cdot \frac{1}{2} = \frac{1}{4} - 1 + C \Rightarrow C = \frac{7}{4}$.
$x = 2$: $y\cdot\frac{32}{33} = 4 - \frac{1}{2} + \frac{7}{4} = \frac{21}{4}$, so $y(2) = \frac{21 \cdot 33}{128} = \frac{693}{128}$.
Answer: (C). | ||||
| 14 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections r is the least distance from (2, 0) to the ellipse $\frac{x^2}{36} + \frac{y^2}{9} = 1$.
For a point of the ellipse, $y^2 = 9 - \frac{x^2}{4}$, so $d^2 = (x - 2)^2 + 9 - \frac{x^2}{4} = \frac{3}{4}x^2 - 4x + 13$, $-6 \le x \le 6$.
Its minimum is at $x = \frac{8}{3}$: $d^2 = \frac{3}{4}\cdot\frac{64}{9} - \frac{32}{3} + 13 = \frac{23}{3}$.
$r^2 = \frac{23}{3}$, $12r^2 = 92$.
Answer: (C). | ||||
| 15 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium 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 Two vectors in the plane: $(8, 0, -2)$ and $(-4, 3, 2)$. Normal: $(8, 0, -2)\times(-4, 3, 2) = (6, -8, 24) \parallel (3, -4, 12)$.
Plane: $3x - 4y + 12z = 3$ (through (5, 3, 0)); $|n| = 13$.
A: $\frac{|9 - 16 + 12\alpha - 3|}{13} = 2 \Rightarrow 12\alpha - 10 = \pm 26 \Rightarrow \alpha = 3$ (the other root is not natural).
B (2, 3, a): $\frac{|6 - 12 + 12a - 3|}{13} = 3 \Rightarrow 12a - 9 = \pm 39 \Rightarrow a = 4$ or $a = -\frac{5}{2}$.
Positive value: $a = 4$.
Answer: (B). | ||||
| 16 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Line PQ: $(2 + 3t, -1 + 4t, 2 + 2t)$. In $x - y + z = 4$: $5 + t = 4$, $t = -1$, so $R = (-1, -5, 0)$.
Through R parallel to (2, 2, 1): $(-1 + 2s, -5 + 2s, s)$. In $x + 2y + 3z + 2 = 0$: $-1 + 2s - 10 + 4s + 3s + 2 = 9s - 9 = 0$, $s = 1$.
Distance $= |s|\sqrt{2^2 + 2^2 + 1^2} = 3$.
Answer: (A). | ||||
| 17 | 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 Coplanar points: $[\vec{b} - \vec{a}\ \ \vec{c} - \vec{a}\ \ \vec{d} - \vec{a}] = 0$. Expanding (terms with $\vec{a}$ twice vanish):
$[\vec{b}\,\vec{c}\,\vec{d}] - [\vec{b}\,\vec{c}\,\vec{a}] - [\vec{b}\,\vec{a}\,\vec{d}] - [\vec{a}\,\vec{c}\,\vec{d}] = 0$.
So $[\vec{a}\,\vec{b}\,\vec{c}] = [\vec{b}\,\vec{c}\,\vec{a}] = [\vec{b}\,\vec{c}\,\vec{d}] + [\vec{a}\,\vec{b}\,\vec{d}] + [\vec{c}\,\vec{a}\,\vec{d}]$.
Option (B): $[\vec{d}\,\vec{c}\,\vec{a}] = [\vec{c}\,\vec{a}\,\vec{d}]$, $[\vec{b}\,\vec{d}\,\vec{a}] = [\vec{a}\,\vec{b}\,\vec{d}]$, $[\vec{c}\,\vec{d}\,\vec{b}] = [\vec{b}\,\vec{c}\,\vec{d}]$ (cyclic shifts), which is exactly this sum.
(A) has $[\vec{d}\,\vec{b}\,\vec{a}] = -[\vec{a}\,\vec{b}\,\vec{d}]$ and $[\vec{a}\,\vec{c}\,\vec{d}] = -[\vec{c}\,\vec{a}\,\vec{d}]$; (C) and (D) likewise have wrong signs.
Answer: (B). | ||||
| 18 | StatisticsMean deviation | Class 11: Statistics | Single correctEasy | |
A easy single correct question on mean deviation, from Statistics. NCERT: Class 11: Statistics Mean 5: $12 + x + y = 30$, so $x + y = 18$.
Variance 10: $\frac{46 + x^2 + y^2}{6} - 25 = 10$, so $x^2 + y^2 = 164$. Then $2xy = 324 - 164 = 160$, $xy = 80$: $\{x, y\} = \{8, 10\}$.
Deviations from 5: 4, 3, 1, 0, 3, 5; sum 16.
Mean deviation $= \frac{16}{6} = \frac{8}{3}$.
Answer: (B). | ||||
| 19 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Distance of the southern person from the foot of the tower: $5\cot 45^\circ = 5$ m. Of the western person: $5\cot 30^\circ = 5\sqrt{3}$ m.
The two lines of sight from the foot are perpendicular (south and west), so the distance between them is $\sqrt{25 + 75} = 10$ m.
Answer: (D). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter The converse of $A \Rightarrow B$ is $B \Rightarrow A$: here $r \Rightarrow ((\sim p) \wedge q)$.
None of the options is written in that form, so use the equivalent contrapositive: $\sim((\sim p) \wedge q) \Rightarrow \sim r$, i.e. $(p \vee (\sim q)) \Rightarrow (\sim r)$.
Answer: (A). | ||||
| 21 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalHard | |
A hard numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z^2 + 8iz - 15 = (z + 3i)(z + 5i)$ and $z^2 - 3iz - 2 = (z - i)(z - 2i)$.
With $z = \alpha - \frac{13}{11}i$ and $a = 11\alpha$ (multiply each factor by 11):
$11(z + 3i) = a + 20i$, $11(z + 5i) = a + 42i$, $11(z - i) = a - 24i$, $11(z - 2i) = a - 35i$.
Numerator $\propto (a + 20i)(a + 42i) = (a^2 - 840) + 62ai$; denominator $\propto (a - 24i)(a - 35i) = (a^2 - 840) - 59ai$.
The quotient is real iff $N\bar{D}$ is real: $\mathrm{Im}\,[(a^2 - 840 + 62ai)(a^2 - 840 + 59ai)] = 121a(a^2 - 840) = 0$.
$a \ne 0$, so $a^2 = 840$, $\alpha^2 = \frac{840}{121}$, $242\alpha^2 = 1680$.
Answer: 1680. | ||||
| 22 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $f(1) + f(2) \ge 2$, so $f(4) \ge 3$. For $f(4) = k$ the number of pairs $(f(1), f(2))$ with sum $k - 1$ is $k - 2$: $k = 3, 4, 5, 6$ give $1 + 2 + 3 + 4 = 10$.
$f(3)$ and $f(5)$ are free: $6 \times 6 = 36$.
Number of functions $= 10 \times 36 = 360$.
Answer: 360. | ||||
| 23 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The numerators 1, 4, 8, 13, 19, … have differences 3, 4, 5, 6, … Let $x = \frac{1}{k}$ and $S = \sum a_n x^n$.
$(1 - x)S = 1 + 3x + 4x^2 + 5x^3 + \ldots$
$(1 - x)^2 S = 1 + 2x + x^2 + x^3 + \ldots = 1 + 2x + \frac{x^2}{1 - x}$.
So $S = \frac{(1 + 2x)(1 - x) + x^2}{(1 - x)^3} = \frac{1 + x - x^2}{(1 - x)^3} = \frac{k(k^2 + k - 1)}{(k - 1)^3}$.
$S = 10$: $k = 2$ gives $\frac{2 \cdot 5}{1} = 10$. (For $k \ge 3$, $S \le \frac{3 \cdot 11}{8} < 10$, and $k = 1$ diverges.)
Answer: 2. | ||||
| 24 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $t = e^{2x} > 0$: $t^4 - t^3 - 3t^2 - t + 1 = 0$. Divide by $t^2$: $\left(t + \frac{1}{t}\right)^2 - \left(t + \frac{1}{t}\right) - 5 = 0$.
With $u = t + \frac{1}{t} \ge 2$: $u = \frac{1 \pm \sqrt{21}}{2}$; only $u = \frac{1 + \sqrt{21}}{2} \approx 2.79 > 2$ is allowed.
$t + \frac{1}{t} = u$ with $u > 2$ has two distinct positive roots t, each giving one x.
The curve cuts the x-axis at 2 points.
Answer: 2. | ||||
| 25 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals C: $y = 2x^2 + 1$. Tangent at (1, 3): slope $4x = 4$, $y = 4x - 1$. The tangent meets $x + y = 1$ at $\left(\frac{2}{5}, \frac{3}{5}\right)$, and $x + y = 1$ meets C at (0, 1).
The region is bounded above by C from $x = 0$ to $x = 1$, and below by $x + y = 1$ (from 0 to $\frac{2}{5}$) and the tangent (from $\frac{2}{5}$ to 1).
$A = \int_0^1 (2x^2 + 1)dx - \int_0^{2/5}(1 - x)dx - \int_{2/5}^1 (4x - 1)dx = \frac{5}{3} - \frac{8}{25} - \frac{27}{25} = \frac{5}{3} - \frac{7}{5} = \frac{4}{15}$.
$60A = 16$.
Answer: 16. | ||||
| 26 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines $l_1$ and $l_2$ meet at (0, 1), which must lie on $l_3$: $\beta + 17 = 0$, $\beta = -17$.
$l_3$ is the mirror image of $l_2$ in $l_1$. Take (1, 2) on $l_2$; for $2x - 3y + 3 = 0$ its value is $-1$, so its image is $(1, 2) + \frac{2}{13}(2, -3) = \left(\frac{17}{13}, \frac{20}{13}\right)$.
$l_3$ through (0, 1) and $\left(\frac{17}{13}, \frac{20}{13}\right)$: slope $\frac{7}{17}$, i.e. $7x - 17y + 17 = 0$. So $\alpha = 7$, $\beta = -17$.
$\alpha^2 + \beta^2 - \alpha - \beta = 49 + 289 - 7 + 17 = 348$.
Answer: 348. | ||||
| 27 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $(3, \alpha)$ on the parabola: $\alpha = \pm 6$. The slope of the tangent is $\frac{6}{\alpha}$, and it must be 1 (perpendicular to slope $-1$), so $\alpha = 6$.
Hyperbola: $36x^2 - 9y^2 = 324$, i.e. $\frac{x^2}{9} - \frac{y^2}{36} = 1$; the point is (5, 8) (on it: $\frac{25}{9} - \frac{64}{36} = 1$).
Normal at $(x_1, y_1)$: $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$: $\frac{9x}{5} + \frac{36y}{8} = 45$, i.e. $2x + 5y = 50$.
Distance of (6, −4): $\frac{|12 - 20 - 50|}{\sqrt{29}} = \frac{58}{\sqrt{29}}$; square $= \frac{3364}{29} = 116$.
Answer: 116. | ||||
| 28 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical 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 is $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$: through (0, 1, 3), direction $(1, 2, \lambda)$.
If $\cos\theta = \sqrt{\frac{5}{14}}$ then $\sin\theta = \frac{3}{\sqrt{14}}$, and $\sin\theta = \frac{|1 + 4 + 3\lambda|}{\sqrt{5 + \lambda^2}\sqrt{14}}$.
$(5 + 3\lambda)^2 = 9(5 + \lambda^2) \Rightarrow 30\lambda = 20 \Rightarrow \lambda = \frac{2}{3}$.
Point $\left(t, 1 + 2t, 3 + \frac{2t}{3}\right)$ in P: $t + 2 + 4t + 9 + 2t = 4$, $t = -1$: $(\alpha, \beta, \gamma) = \left(-1, -1, \frac{7}{3}\right)$.
$\alpha + 2\beta + 6\gamma = -1 - 2 + 14 = 11$.
Answer: 11. | ||||
| 29 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalHard | |
A hard numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}\cdot\vec{b} = 1 + 2 - 3 = 0$, so $\vec{a}$, $\vec{b}$ and $\vec{b}\times\vec{a}$ are mutually perpendicular. $|\vec{a}|^2 = 14$, $|\vec{b}|^2 = 3$, $\vec{b}\times\vec{a} = (5, -4, 1)$ with $|\vec{b}\times\vec{a}|^2 = 42$.
Given: $\vec{c}\cdot\vec{a} = 11$, $\vec{c}\cdot\vec{b} = -\sqrt{3}\cdot\sqrt{3} = -3$, $\vec{c}\cdot(\vec{b}\times\vec{a}) = [\vec{b}\,\vec{a}\,\vec{c}] = \vec{b}\cdot(\vec{a}\times\vec{c}) = 27$.
$|\vec{c}|^2 = \frac{11^2}{14} + \frac{(-3)^2}{3} + \frac{27^2}{42} = \frac{121}{14} + 3 + \frac{243}{14} = 29$.
$|\vec{a}\times\vec{c}|^2 = |\vec{a}|^2|\vec{c}|^2 - (\vec{a}\cdot\vec{c})^2 = 14 \cdot 29 - 121 = 285$.
Answer: 285. | ||||
| 30 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(N = n) = \left(\frac{3}{4}\right)^{n-1}\frac{1}{4}$.
No real root: $25N^2 - 256 < 0$, $N^2 < 10.24$, so $N = 1, 2, 3$.
$P = \frac{1}{4}\left(1 + \frac{3}{4} + \frac{9}{16}\right) = \frac{1}{4}\cdot\frac{37}{16} = \frac{37}{64}$.
$q - p = 64 - 37 = 27$.
Answer: 27. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The observer Q (behind) moves towards the source at 40 m/s; the source P moves away from Q at 20 m/s.
$f' = f\frac{v + v_o}{v + v_s} = 400 \times \frac{360 + 40}{360 + 20} = 400 \times \frac{400}{380} \approx 421$ Hz.
Answer: (B). | ||||
| 32 | 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 $v_{rms} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}} = \sqrt{2.74 \times 10^5} \approx 523$ m/s.
Answer: (C). | ||||
| 33 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Orbital speed close to the surface: $v_o = \sqrt{gR} = \sqrt{10 \times 6.4 \times 10^6} = 8$ km/s.
Escape speed from there: $v_e = \sqrt{2gR} = 8\sqrt{2}$ km/s.
Additional velocity $= 8(\sqrt{2} - 1)$ km/s (the mass does not matter).
Answer: (C). | ||||
| 34 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics For an ideal gas the internal energy depends only on temperature, so it stays constant when the temperature is constant: the isothermal process ($\Delta U = 0$, $Q = W$).
Answer: (D). | ||||
| 35 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $a = v\frac{dv}{dx} = 10\sqrt{x} \cdot \frac{5}{\sqrt{x}} = 50\ \mathrm{m/s^2}$ (constant).
$F = ma = 0.5 \times 50 = 25$ N.
Answer: (B). | ||||
| 36 | 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 Terminal speed $v = \frac{2r^2(\rho - \sigma)g}{9\eta} \propto r^2$.
Volume conserved: $\frac{4}{3}\pi R^3 = 8 \cdot \frac{4}{3}\pi r^3 \Rightarrow R = 2r$.
New speed $= 10 \times 2^2 = 40$ cm/s.
Answer: (D). | ||||
| 37 | GravitationGravitational potential energy | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation Inside a uniform solid sphere $V(r) = -\frac{GM}{2R^3}(3R^2 - r^2)$. At the surface $V = -\frac{GM}{R}$; at the centre $V_c = -\frac{3GM}{2R} = \frac{3V}{2}$.
Answer: (C). | ||||
| 38 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Read literally, $\vec{B} - \vec{A} = 2\hat{j}$ gives $\vec{B} = 2\hat{i} + 5\hat{j} + 2\hat{k}$ and $|\vec{B}| = \sqrt{33}$, which is not an option.
The options fit $\vec{A} - \vec{B} = 2\hat{j}$: $\vec{B} = 2\hat{i} + \hat{j} + 2\hat{k}$, $|\vec{B}| = \sqrt{4 + 1 + 4} = 3$.
Answer: (A). | ||||
| 39 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $u_x = 40\cos 30^\circ = 20\sqrt{3}$ m/s, $u_y = 40\sin 30^\circ = 20$ m/s.
At $t = 2$ s: $v_y = 20 - 10 \times 2 = 0$ (the top of the path), $v_x = 20\sqrt{3}$.
Velocity $= 20\sqrt{3}\ \mathrm{ms^{-1}}$ (horizontal).
Answer: (B). | ||||
| 40 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $L = VT$ and $M = \frac{F}{\text{acceleration}} = \frac{F}{V/T} = FV^{-1}T$.
Density $= \frac{M}{L^3} = \frac{FV^{-1}T}{V^3T^3} = FV^{-4}T^{-2}$.
Answer: (C). | ||||
| 41 | Communication SystemsSatellite communication | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on satellite communication, from Communication Systems. NCERT: No matching NCERT chapter In satellite communication (C band) the uplink uses 5.925 – 6.425 GHz and the downlink 3.7 – 4.2 GHz. 76 – 88 MHz is FM radio and 420 – 890 MHz is UHF television.
Answer: (D). | ||||
| 42 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The OR gate gives $A + B$ and the AND gate gives $AB$. These feed a NAND gate followed by a NOT gate, which together act as an AND gate:
$Y = (A + B)\cdot AB = AB$ (since $AB = 1$ already means $A + B = 1$).
The circuit is equivalent to AND.
Answer: (A). | ||||
| 43 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Angle of incidence = angle of reflection = $30^\circ$. Deviation $= 180^\circ - 2i = 180^\circ - 60^\circ = 120^\circ$.
Answer: (B). | ||||
| 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 $E_n = -13.6\frac{Z^2}{n^2}$ eV. For $\mathrm{He^+}$, $Z = 2$; first excited state $n = 2$: $E = -13.6 \times \frac{4}{4} = -13.6$ eV.
Answer: (B). | ||||
| 45 | 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 $\lambda = \frac{h}{\sqrt{2mK}}$, so at equal K, $\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{m_p}} = \frac{1}{\sqrt{1849}} = \frac{1}{43}$.
Answer: (D). | ||||
| 46 | 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 $|B| = \frac{E}{c} = \frac{6.6}{3 \times 10^8} = 2.2 \times 10^{-8}$ T.
$\vec{E} \times \vec{B}$ points along the propagation direction $+\hat{i}$: $\hat{j} \times \hat{k} = \hat{i}$, so $\vec{B}$ is along $+\hat{k}$.
$\vec{B} = 2.2 \times 10^{-8}\hat{k}$ T.
Answer: (A). | ||||
| 47 | Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Assertion and reasonEasy | |
A easy assertion and reason question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The falling magnet changes the flux through the rings of the metal pipe and induces eddy currents; by Lenz's law they oppose the motion, so the magnet falls more slowly than a non-magnetic bar, which induces nothing.
A is true, R is true, and R explains A.
Answer: (A). | ||||
| 48 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsEasy | |
A easy statements 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 On the axis the field $\vec{B}$ is along the axis, parallel to the velocity, so $\vec{F} = q\vec{v} \times \vec{B} = 0$.
B: true. C: with no force it keeps moving along the axis with the same velocity: true.
A, D, E: false (no force, no acceleration, a straight path).
Answer: (B). | ||||
| 49 | 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 From the figure: the 8 V cell is across E and A; $R_6 = 4\ \Omega$ is across E–A; $R_5 = 2\ \Omega$ joins E–D; $R_7 = 3\ \Omega$ joins D–A; $R_4 = 3\ \Omega$ joins D–C; $R_3 = 6\ \Omega$ joins C–A; $R_2 = 4\ \Omega$ (C–B) and $R_1 = 2\ \Omega$ (B–A) are in series between C and A.
C to A: $(R_2 + R_1) = 6\ \Omega$ in parallel with $R_3 = 6\ \Omega$: $3\ \Omega$.
D to A: $R_4 + 3 = 6\ \Omega$ in parallel with $R_7 = 3\ \Omega$: $2\ \Omega$.
Branch E–D–A: $R_5 + 2 = 4\ \Omega$ across 8 V, so its current is 2 A and $V_{DA} = 2 \times 2 = 4$ V.
Current through $R_4$: $\frac{4}{6} = \frac{2}{3}$ A, so $V_{CA} = \frac{2}{3} \times 3 = 2$ V.
Current through $R_2$ (and $R_1$): $\frac{2}{6} = \frac{1}{3}$ A.
Answer: (C). | ||||
| 50 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The positive plate carries $Q = CV$. By Gauss's law the flux through a closed surface enclosing only that plate is $\frac{Q}{\varepsilon_0} = \frac{CV}{\varepsilon_0}$.
Answer: (A). | ||||
| 51 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Stress $= Y\frac{\Delta l}{l} = 8 \times 10^{10} \times \frac{3.2 \times 10^{-4}}{0.5} = 5.12 \times 10^7\ \mathrm{N/m^2}$.
Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\text{stress}}{\rho}} = \sqrt{\frac{5.12 \times 10^7}{8 \times 10^3}} = \sqrt{6400} = 80$ m/s.
Fundamental: $f = \frac{v}{2l} = \frac{80}{2 \times 0.5} = 80$ Hz.
Answer: 80. | ||||
| 52 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter A soap bubble has two surfaces: $W = T \times 2 \times 4\pi(r_2^2 - r_1^2) = 3.5 \times 10^{-2} \times 8 \times \frac{22}{7} \times (0.04 - 0.01)$
$= 3.5 \times 10^{-2} \times \frac{176}{7} \times 0.03 = 2.64 \times 10^{-2}$ J $= 264 \times 10^{-4}$ J.
Answer: 264. | ||||
| 53 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion No external torque about the axis: $I\omega = 3I\omega'$, so $\omega' = \frac{\omega}{3}$.
$K' = \frac{1}{2}(3I)\left(\frac{\omega}{3}\right)^2 = \frac{1}{3}\cdot\frac{1}{2}I\omega^2 = \frac{E}{3}$.
$x = 3$.
Answer: 3. | ||||
| 54 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Along the smooth incline: $F - mg\sin 30^\circ = ma \Rightarrow F = 5 \times 1 + 5 \times 10 \times \frac{1}{2} = 30$ N (taken along the incline).
At $t = 10$ s: $v = at = 10$ m/s.
$P = Fv = 30 \times 10 = 300$ W.
Answer: 300. | ||||
| 55 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Nuclear radius $R = R_0A^{1/3}$, so $\frac{A_1}{A_2} = \left(\frac{1}{2^{1/3}}\right)^3 = \frac{1}{2}$: the masses are in the ratio 1 : 2.
The nucleus was at rest, so the momenta are equal and opposite: $m_1v_1 = m_2v_2$, $\frac{v_1}{v_2} = \frac{m_2}{m_1} = 2$.
$n = 2$.
Answer: 2. | ||||
| 56 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Seen from above the water, the bulb appears at a depth $\frac{8}{4/3} = 6$ cm below the surface, i.e. 2 cm above the bottom.
This apparent bulb is $50 - 2 = 48$ cm below the mirror, so the mirror forms its image 48 cm above the mirror: $50 + 48 = 98$ cm from the bottom of the tank.
Answer: 98. | ||||
| 57 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Steady current $I = \frac{10}{4} = 2.5$ A.
$U = \frac{1}{2}LI^2 = \frac{1}{2} \times 2 \times 6.25 = 6.25$ J $= 625 \times 10^{-2}$ J.
Answer: 625. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The free electrons feel $q\vec{v} \times \vec{B}$, along $\pm\hat{x}$ ($\hat{j} \times \hat{k} = \hat{i}$); charge piles up on the two faces normal to x until the electric field balances it: $E = vB$.
Potential difference across the side $l$: $V = Bvl = 0.5 \times 2 \times 0.15 = 0.15$ V $= 150$ mV.
Answer: 150. | ||||
| 59 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Current $I = \frac{1.5}{10} = 0.15$ A.
$3 = 0.15(10 + 2r) \Rightarrow 10 + 2r = 20 \Rightarrow r = 5\ \Omega$.
Answer: 5. | ||||
| 60 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance From the figure: $C_3$, $C_4$, $C_5$ are in series in the right-hand loop: $\frac{1}{C} = \frac{1}{2} + \frac{1}{4} + \frac{1}{2} = \frac{5}{4}$, $C = 0.8\ \mu$F.
This is in parallel with $C_2$: $0.8 + 0.2 = 1\ \mu$F.
In series with $C_1$ and $C_6$: $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{1} + \frac{1}{2} = 2$, $C_{eq} = 0.5\ \mu$F.
Total charge $= 0.5 \times 10 = 5\ \mu$C; voltage across the 1 $\mu$F block $= 5$ V.
Charge on the $C_3C_4C_5$ branch $= 0.8 \times 5 = 4\ \mu$C, which is the charge on $C_4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Solid StateClassification of solids | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on classification of solids, from Solid State. NCERT: No matching NCERT chapter Polar molecular solids are made of polar molecules held by dipole–dipole forces (solid $\mathrm{SO_2}$, HCl, $\mathrm{NH_3}$).
(B) $\mathrm{CO_2}$ is non-polar. (C) AlN is a covalent network solid. (D) MgO is ionic.
Only (A) has two solids built from polar molecules.
Answer: (A). | ||||
| 62 | SolutionsRaoult's law | Class 12: Solutions | Single correctEasy | |
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions For a dilute solution $\frac{\Delta p}{p^\circ} \approx \frac{n_2}{n_1}$.
$\frac{0.20}{54.2} = \frac{w/180}{100/18} \Rightarrow w = \frac{0.20}{54.2} \times \frac{100}{18} \times 180 = \frac{200}{54.2} \approx 3.69$ g.
Answer: (A). | ||||
| 63 | 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 Rate $= k[A][B]$. From the first row: $0.10 = k \times 20 \times 0.5$, $k = 0.01\ \mathrm{L\,mol^{-1}\,s^{-1}}$.
Second row: $0.40 = 0.01 \times x \times 0.5 \Rightarrow x = 80$.
Third row: $0.80 = 0.01 \times 40 \times y \Rightarrow y = 2$.
Answer: (A). | ||||
| 64 | SolutionsConcentration terms | Class 12: Solutions | Single correctEasy | |
A easy single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions 1 mole of water is 18 g, so the solution weighs $2 + 18 = 20$ g.
Mass percent of X $= \frac{2}{20} \times 100 = 10\%$.
Answer: (C). | ||||
| 65 | Isolation of MetalsRoasting and calcination | No matching NCERT chapter | StatementsEasy | |
A easy statements question on roasting and calcination, from Isolation of Metals. NCERT: No matching NCERT chapter Sulphide ores are roasted to oxides before reduction (e.g. $\mathrm{2ZnS + 3O_2 \to 2ZnO + 2SO_2}$), because the formation of the oxide makes the reduction thermodynamically easier: $\Delta_fG^\circ$ of most sulphides is more negative than that of $\mathrm{CS_2}$ or $\mathrm{H_2S}$, so carbon or hydrogen cannot reduce them, whereas oxides are readily reduced by carbon or CO.
Both statements are correct.
Answer: (A). | ||||
| 66 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter LiH is a stable ionic (saline) hydride, and $\mathrm{NH_3}$ and HF are stable covalent hydrides with strong N–H and H–F bonds (HF is the most stable).
$\mathrm{BeH_2}$ is an electron-deficient, polymeric covalent hydride with the weakest bonding of the four; its enthalpy of formation is the least negative, so it is the least stable.
Answer: (B). | ||||
| 67 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $\mathrm{P_4 + 8SOCl_2 \to 4PCl_3 + 4SO_2 + 2S_2Cl_2}$.
Check: P 4 = 4; S 8 = 4 + 4; O 8 = 8; Cl 16 = 12 + 4.
A = $\mathrm{PCl_3}$, B = $\mathrm{S_2Cl_2}$, x = 4.
Answer: (A). | ||||
| 68 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Down the group the atoms get larger and the metallic bond weaker, so melting points fall: Na 371 K, K 336 K, Rb 312 K, Cs 302 K.
Caesium has the lowest melting point.
Answer: (D). | ||||
| 69 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Fe(III) is $d^5$.
$\mathrm{H_2O}$ is a weak-field ligand: high spin $t_{2g}^3e_g^2$, 5 unpaired, $\mu = \sqrt{5 \times 7} = 5.92$ BM.
$\mathrm{CN^-}$ is a strong-field ligand: low spin $t_{2g}^5$, 1 unpaired, $\mu = \sqrt{1 \times 3} = 1.732$ BM.
Answer: (C). | ||||
| 70 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Assertion and reasonMedium | |
A medium assertion and reason question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Both complexes contain Co(III); they differ only in one ligand. $\mathrm{Cl^-}$ is a weaker-field ligand than $\mathrm{H_2O}$, so $\Delta_o$ is smaller for $\mathrm{[CoCl(NH_3)_5]^{2+}}$ and it absorbs light of longer wavelength (it is violet, while the aqua complex is red). A is false.
R is a true general statement: the splitting, and so the wavelength absorbed, depends on the oxidation state of the metal ion (as well as on the ligands). Here it is not the cause, since the oxidation state is the same.
Answer: (D). | ||||
| 71 | Haloalkanes and HaloarenesEnvironmental effects of polyhalogen compounds | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on environmental effects of polyhalogen compounds, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Freons are chlorofluorocarbons, compounds of carbon with chlorine and fluorine only.
Of the options only $\mathrm{C_2Cl_2F_2}$ contains both Cl and F; the others contain no chlorine ($\mathrm{C_2F_4}$ is the monomer of Teflon).
Answer: (A). | ||||
| 72 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | StatementsMedium | |
A medium statements question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[NiCl_2Br_2]^{2-}}$: $\mathrm{Ni^{2+}}$ ($3d^8$) with weak-field halides is $sp^3$, tetrahedral, with 2 unpaired electrons (paramagnetic); a tetrahedral $\mathrm{MA_2B_2}$ shows no geometrical isomerism.
$\mathrm{[PtCl_2Br_2]^{2-}}$: $\mathrm{Pt^{2+}}$ ($5d^8$) is $dsp^2$, square planar and diamagnetic; square planar $\mathrm{MA_2B_2}$ has cis and trans isomers.
So geometry (A), geometrical isomerism (B) and magnetic properties (D) change. Neither complex is optically active (both have mirror planes), so C does not change.
Answer: (D). | ||||
| 73 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter $\mathrm{Mg(NH_4)PO_4}$: white precipitate (test for $\mathrm{Mg^{2+}}$) → II.
$\mathrm{K_3[Co(NO_2)_6]}$: yellow precipitate (test for $\mathrm{K^+}$/$\mathrm{Co^{2+}}$) → III.
$\mathrm{MnO(OH)_2}$: brown precipitate (test for $\mathrm{Mn^{2+}}$) → I.
$\mathrm{Fe_4[Fe(CN)_6]_3}$: Prussian blue → IV.
A-II, B-III, C-I, D-IV.
Answer: (A). | ||||
| 74 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $\mathrm{AgNO_3}$ precipitates AgBr when the C–Br bond ionises to a stable carbocation.
(A) Benzyl bromide gives the resonance-stabilised benzyl cation: precipitate.
(B) $\mathrm{C_6H_5CH{=}CHCH_2Br}$ gives an allylic cation conjugated with the ring: precipitate.
(C) 3-Bromocyclopropene gives the aromatic cyclopropenyl cation ($2\pi$ electrons): precipitate, even more readily.
(D) 1-Bromocyclohexene is a vinylic bromide: the C–Br bond has partial double-bond character and the vinylic cation is very unstable, so no precipitate forms.
Answer: (D). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (i) NaI with $\mathrm{H_3PO_4}$ (HI generated in situ) converts butan-2-ol into 2-iodobutane.
(ii) Mg in dry ether gives the Grignard reagent $\mathrm{CH_3CH_2CH(MgI)CH_3}$.
(iii) $\mathrm{D_2O}$ deuterolyses it: D replaces MgI, giving $\mathrm{CH_3CH_2CH(D)CH_3}$ (2-deuteriobutane).
Answer: (C). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Zn(Hg)/HCl (Clemmensen reduction) reduces the –CHO group to –$\mathrm{CH_3}$, so the chain becomes $\mathrm{C_6H_5{-}CH(OH){-}CH(CH_3){-}CH_2CH_3}$ (structure A).
Under the same hot, strongly acidic conditions this benzylic secondary alcohol is protonated and loses water to a benzylic carbocation, which loses a proton to give the alkene conjugated with the ring and most substituted: $\mathrm{C_6H_5{-}CH{=}C(CH_3){-}C_2H_5}$ (structure B).
C would need the methyl group to migrate onto the benzylic carbon, and D (a cyclic ether) is not the product of a Clemmensen reduction.
Major product: B only.
Answer: (A). | ||||
| 77 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Assertion and reasonMedium | |
A medium assertion and reason question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids R is true: Wolff-Kishner reduction (hydrazine, then KOH in ethylene glycol on heating) converts a carbonyl group into $\mathrm{CH_2}$.
A is false: the substrate, 3-chloro-4-methylpentan-2-one, carries a chlorine on the carbon next to C=O. Under the strongly basic, hot conditions of Wolff-Kishner reduction the C–Cl bond does not survive (it undergoes elimination/substitution), so the chloroalkane shown cannot be obtained this way (a Clemmensen reduction would be used instead).
Answer: (D). | ||||
| 78 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The starting compound is thymol (2-isopropyl-5-methylphenol).
$\mathrm{NaNO_2/HCl}$ gives $\mathrm{NO^+}$, which attacks the ring at the free position para to the strongly activating OH group: A is 4-nitrosothymol.
$\mathrm{NH_4SH}$ (sulphide) reduces the nitroso group to $\mathrm{NH_2}$: B is 4-amino-2-isopropyl-5-methylphenol, structure (A).
Answer: (A). | ||||
| 79 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Assertion and reasonMedium | |
A medium assertion and reason question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Cl$_2$/red P (Hell–Volhard–Zelinsky) substitutes one α-hydrogen: $\mathrm{H_2N{-}CH_2{-}COOH \to H_2N{-}CHCl{-}COOH}$. The α-carbon now bears H, Cl, $\mathrm{NH_2}$ and COOH, four different groups, so it is a chiral carbon. A is true (the product forms as a racemic mixture, but it does contain a chiral carbon).
R is false: a molecule with two chiral carbons can be optically inactive when it has a plane of symmetry (meso forms, e.g. meso-tartaric acid).
Answer: (C). | ||||
| 80 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | StatementsEasy | |
A easy statements question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Statement I: ethene at 333–343 K and 6–7 atm with the Ziegler–Natta catalyst ($\mathrm{AlEt_3 + TiCl_4}$) gives high density polythene (HDP), not LDP (which needs 350–570 K, 1000–2000 atm and a peroxide initiator). False.
Statement II: heating caprolactam with water at 533–543 K gives nylon 6, a condensation (step-growth) polymer. True.
Answer: (D). | ||||
| 81 | 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 $\mathrm{Mg + 2HCl \to MgCl_2 + H_2}$.
2.4 g Mg = 0.1 mol, giving 0.1 mol $\mathrm{H_2}$ = $0.1 \times 22.4 = 2.24$ L $= 224 \times 10^{-2}$ L.
Answer: 224. | ||||
| 82 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{ClO_3^-}$: Cl has 3 bond pairs and 1 lone pair (pyramidal).
$\mathrm{XeF_4}$: 4 bond pairs, 2 lone pairs (square planar).
$\mathrm{SF_4}$: 4 bond pairs, 1 lone pair (see-saw).
$\mathrm{I_3^-}$: the central I has 2 bond pairs and 3 lone pairs (linear).
Maximum = 3.
Answer: 3. | ||||
| 83 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom A: true; $\psi_{1s}^2$ is greatest at $r = 0$.
B: false; for 2s the probability density is greatest at the nucleus, first decreases sharply to zero (the radial node) and then rises again to a smaller maximum.
C: false; a boundary surface encloses about 90% probability, never 100%.
D: true; the number of angular nodes is l: 1 for p, 2 for d.
E: true; a p orbital has a nodal plane through the nucleus.
Correct: A, D, E = 3.
Answer: 3. | ||||
| 84 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Intensive (independent of the amount): molar heat capacity, molarity, $E^\ominus_{cell}$, molar mass: 4.
Extensive: volume, Gibbs free energy change, mole (amount of substance).
Answer: 4. | ||||
| 85 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium 3 mol HI is formed from 1.5 mol $\mathrm{H_2}$ and 1.5 mol $\mathrm{I_2}$, leaving 3 mol of each.
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(3/10)^2}{(3/10)(3/10)} = 1$ (the volume cancels).
Answer: 1. | ||||
| 86 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry A: true; the cell potential does not depend on the size of the cell (unlike $\Delta G = -nFE$).
B: true; a couple with negative $E^\ominus$ is more easily oxidised than $\mathrm{H_2}$, i.e. a stronger reducing agent.
C: true; e.g. $\mathrm{Ag^+ + e^-}$ needs 1 F per mole, $\mathrm{Cu^{2+} + 2e^-}$ needs 2 F per mole.
D: true; Faraday's first law of electrolysis.
All 4 are correct.
Answer: 4. | ||||
| 87 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter The modern adsorption theory combines the intermediate compound formation theory with the old adsorption theory (D true). Its steps: the reactants diffuse to the surface of the catalyst (so A, which reverses this, is false), are adsorbed on it (B true), react there through an intermediate (C true), and the products desorb and diffuse away.
E is false: the theory explains why the catalyst is unchanged and works in small amounts, but it does not explain the action of catalytic promoters and poisons.
Correct: B, C, D = 3.
Answer: 3. | ||||
| 88 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter The tendency of alkaline earth salts to crystallise with water falls down the group as the cation gets larger and its hydration enthalpy smaller.
Magnesium nitrate crystallises as $\mathrm{Mg(NO_3)_2 \cdot 6H_2O}$ (X = 6), while barium nitrate crystallises anhydrous, $\mathrm{Ba(NO_3)_2}$ (Y = 0).
X + Y = 6.
Answer: 6. | ||||
| 89 | BiomoleculesCarbohydrates | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Benedict's solution gives the orange-red $\mathrm{Cu_2O}$ precipitate with reducing sugars (a free aldehyde/hemiacetal group).
Reducing: glucose, maltose, ribose, 2-deoxyribose, lactose.
Non-reducing: sucrose (both anomeric carbons are in the glycosidic link) and amylose (a polysaccharide that does not give the test).
Number not giving the precipitate = 2.
Answer: 2. | ||||
| 90 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalHard | |
A hard numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{CH_2{=}CH{-}CHCl{-}CH_3}$ adds $\mathrm{H^+}$ at C1 to give the secondary carbocation $\mathrm{CH_3{-}\overset{+}{C}H{-}CHCl{-}CH_3}$.
Capture by $\mathrm{Cl^-}$ gives 2,3-dichlorobutane, which has two similar chiral centres: the (2R,3R) and (2S,3S) enantiomers and the meso form, 3 stereoisomers.
A 1,2-hydride shift gives $\mathrm{CH_3{-}CH_2{-}\overset{+}{C}Cl{-}CH_3}$, stabilised by the lone pair of the chlorine; capture by $\mathrm{Cl^-}$ gives 2,2-dichlorobutane (1 product).
Total isomeric products = 3 + 1 = 4.
Answer: 4. | ||||
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