JEE (Main) · 2022
JEE (Main) 2022 — 25 July, 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 Calculus4
- Conic Sections3
- Matrices and Determinants2
7 easy · 21 medium · 2 hard
Physics
30 Q
- Magnetic Effects of Current and Magnetism2
- Properties of Bulk Matter2
- Laws of Motion2
25 easy · 5 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- p-Block Elements2
- Aldehydes, Ketones and Carboxylic Acids2
23 easy · 7 medium · 0 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 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|z-z_1|+|z-z_2|$ is the sum of the distances of $z$ from $z_1 = 3\sqrt{2}$ and $z_2 = p\sqrt{2}i$; its minimum is $|z_1 - z_2|$ (taken when $z$ lies on the segment joining them).
$|z_1-z_2| = \sqrt{(3\sqrt{2})^2 + (p\sqrt{2})^2} = \sqrt{18 + 2p^2} = 5\sqrt{2}$, so $18 + 2p^2 = 50$, $p^2 = 16$, $p = \pm 4$.
Of the options, $p = 4$.
Answer: (C). | ||||
| 2 | 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 Let $k = \lambda^2 - |\lambda|$. $\Delta = \begin{vmatrix}2&-3&5\\1&3&-1\\3&-1&k\end{vmatrix} = 2(3k-1) + 3(k+3) + 5(-1-9) = 9k - 43$.
No solution needs $\Delta = 0$, i.e. $k = \frac{43}{9}$.
Then the third row of coefficients is a combination of the first two: $a(2,-3) + b(1,3) = (3,-1)$ gives $a = \frac{10}{9}, b = \frac{7}{9}$, and indeed $5a - b = \frac{43}{9} = k$. For consistency the right side would have to be $9a - 18b = 10 - 14 = -4$, but it is $16$: the system is inconsistent, so there is no solution whenever $k = \frac{43}{9}$.
$\lambda^2 - |\lambda| - \frac{43}{9} = 0$ gives $|\lambda| = \frac{1 + \sqrt{1 + \frac{172}{9}}}{2}$ (the other root is negative), so $\lambda$ takes 2 real values.
Answer: (C). | ||||
| 3 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Both sets have 50 elements. The constrained inputs $3, 9, 15, \ldots, 99$ number $\frac{99-3}{6}+1 = 17$.
Since $f$ is one-one, the inequalities are strict: choose any 17 of the 50 images, ${}^{50}C_{17}$ ways, and they must be assigned in decreasing order (1 way). The other 33 inputs take the remaining 33 values in $33!$ ways.
Total $= {}^{50}C_{17}\cdot 33! = \frac{50!}{17!} = {}^{50}P_{33}$.
Answer: (B). | ||||
| 4 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $11 \equiv 2 \pmod 9$ and $2^6 = 64 \equiv 1 \pmod 9$. $1011 = 6 \cdot 168 + 3$, so $11^{1011} \equiv 2^3 = 8 \pmod 9$.
$1011 = 9 \cdot 112 + 3$, so $1011^{11} \equiv 3^{11} \equiv 0 \pmod 9$.
Remainder $= 8$.
Answer: (D). | ||||
| 5 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{3}{(4n-1)(4n+3)} = \frac{3}{4}\left(\frac{1}{4n-1} - \frac{1}{4n+3}\right)$, and the sum telescopes:
$\frac{3}{4}\left(\frac{1}{3} - \frac{1}{87}\right) = \frac{3}{4}\cdot\frac{28}{87} = \frac{7}{29}$.
Answer: (B). | ||||
| 6 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Put $t = \cos x + \sin x$; then $t^2 = 1 + \sin 2x$ and $t \to \sqrt{2}$.
Denominator $= \sqrt{2}(1 - \sin 2x) = \sqrt{2}(2 - t^2)$; numerator $= 8\sqrt{2} - t^7 = (\sqrt{2})^7 - t^7$.
$\lim_{t\to\sqrt{2}} \frac{(\sqrt{2})^7 - t^7}{\sqrt{2}(2 - t^2)} = \frac{1}{\sqrt{2}}\cdot\frac{7(\sqrt{2})^6}{2\sqrt{2}} = \frac{7 \cdot 8}{4} = 14$ (using $\frac{a^7 - t^7}{a^2 - t^2} \to \frac{7a^6}{2a}$).
Answer: (A). | ||||
| 7 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals With $N = 2^n \to \infty$, the expression is $\frac{1}{N}\sum_{r=1}^{N-1} \frac{1}{\sqrt{1 - \frac{r}{N}}}$, a Riemann sum for
$\int_0^1 \frac{dx}{\sqrt{1-x}} = \left[-2\sqrt{1-x}\right]_0^1 = 2$.
Answer: (C). | ||||
| 8 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(A\cap B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{2} = \frac{1}{30}$.
$P(A|B') = \frac{P(A) - P(A\cap B)}{1 - P(B)} = \frac{9/30}{4/5} = \frac{3}{8}$; $P(B|A') = \frac{P(B) - P(A\cap B)}{1 - P(A)} = \frac{5/30}{2/3} = \frac{1}{4}$.
Sum $= \frac{3}{8} + \frac{1}{4} = \frac{5}{8}$.
Answer: (B). | ||||
| 9 | 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 The interval has length 104.
$[\sin \pi x]$ has period 2: it is 0 on $(0,1)$ and $-1$ on $(1,2)$, so each period contributes $-1$; 52 periods give $-52$.
$e^{[\cos 2\pi x]}$ has period 1: $\cos 2\pi x \ge 0$ on $[0,\frac14]\cup[\frac34,1]$ (value $e^0 = 1$, length $\frac12$) and negative on $(\frac14,\frac34)$ (value $e^{-1}$, length $\frac12$); each period contributes $\frac12 + \frac{1}{2e}$, and 104 periods give $52 + \frac{52}{e}$.
Total $= -52 + 52 + \frac{52}{e} = \frac{52}{e}$.
Answer: (B). | ||||
| 10 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines Below $L_1$ (smaller $y$) means $3\alpha - 4\beta + 12 > 0$, so $\frac{3\alpha - 4\beta + 12}{5} = 1$: $3\alpha - 4\beta = -7$.
Above $L_2$ (larger $y$) means $8\alpha + 6\beta + 11 > 0$, so $\frac{8\alpha + 6\beta + 11}{10} = 1$: $8\alpha + 6\beta = -1$.
Solving: $25\alpha = -23$, $\alpha = -\frac{23}{25}$; $\beta = \frac{3\alpha + 7}{4} = \frac{53}{50}$.
$100(\alpha + \beta) = 100\left(-\frac{46}{50} + \frac{53}{50}\right) = 14$.
Answer: (D). | ||||
| 11 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} = -k\frac{y}{x}$ gives $\ln|y| = -k\ln x + c$, i.e. $y = Cx^{-k}$.
$(1,2)$: $C = 2$. $(8,1)$: $1 = 2\cdot 8^{-k}$, so $8^k = 2$, $k = \frac13$.
$y\left(\frac18\right) = 2\left(\frac18\right)^{-1/3} = 2 \cdot 2 = 4$.
Answer: (B). | ||||
| 12 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The first line meets the $x$-axis at $(7, 0)$, so $a = 7$. The second meets the $y$-axis at $(0, -2\sqrt{6})$, so $b = 2\sqrt{6}$.
$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{24}{49} = \frac{25}{49}$, $e = \frac{5}{7}$.
Answer: (A). | ||||
| 13 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Tangent at $(x_1, y_1)$ to $y^2 - 2x - 2y - 1 = 0$: $yy_1 - (x + x_1) - (y + y_1) - 1 = 0$.
At $A(1,3)$: $3y - x - 1 - y - 3 - 1 = 0$, i.e. $2y - x = 5$. At $B(1,-1)$: $-y - x - 1 - y + 1 - 1 = 0$, i.e. $x + 2y = -1$.
They meet at $P(-3, 1)$. $AB$ lies on $x = 1$ with length 4, and $P$ is at distance 4 from it.
Area $= \frac12 \cdot 4 \cdot 4 = 8$.
Answer: (D). | ||||
| 14 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse: $c^2 = 16 - 7 = 9$, foci $(\pm 3, 0)$.
Hyperbola: $\frac{x^2}{144/25} - \frac{y^2}{\alpha/25} = 1$, so $a^2 = \frac{144}{25}$, $b^2 = \frac{\alpha}{25}$ and $a^2 + b^2 = 9$ gives $b^2 = \frac{81}{25}$.
Latus rectum $= \frac{2b^2}{a} = \frac{2 \cdot 81/25}{12/5} = \frac{27}{10}$.
Answer: (D). | ||||
| 15 | 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 The normal of $E$ is perpendicular to both normals: $(2,-2,1)\times(1,-1,2) = (-3,-3,0) \parallel (1,1,0)$.
Through $P(1,-1,1)$: $E: x + y = 0$.
Distance of $Q$: $\frac{|2a|}{\sqrt{2}} = 3\sqrt{2}$, so $|a| = 3$.
$(PQ)^2 = (a-1)^2 + (a+1)^2 + 1 = 2a^2 + 3 = 21$.
Answer: (C). | ||||
| 16 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Line 1: through $(-7, 6, 0)$ along $\vec{b}_1 = (-6, 7, 1)$. Line 2: $\frac{x-7}{-2} = \frac{y-2}{1} = \frac{z-6}{1}$, through $(7, 2, 6)$ along $\vec{b}_2 = (-2, 1, 1)$.
$\vec{b}_1 \times \vec{b}_2 = (7-1,\ -2+6,\ -6+14) = (6, 4, 8)$, $|\vec{b}_1\times\vec{b}_2| = \sqrt{116} = 2\sqrt{29}$.
$\vec{a}_2 - \vec{a}_1 = (14, -4, 6)$; $(14,-4,6)\cdot(6,4,8) = 84 - 16 + 48 = 116$.
Shortest distance $= \frac{116}{2\sqrt{29}} = 2\sqrt{29}$.
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 $\vec{a}\times(\vec{a}\times\vec{b}) = (\vec{a}\cdot\vec{b})\vec{a} - |\vec{a}|^2\vec{b}$, with $|\vec{a}|^2 = 6$.
$\vec{a}\times(2\hat{i} - \hat{k}) = (1,-1,2)\times(2,0,-1) = (1, 5, 2)$.
So $6\vec{b} = 3(1,-1,2) - (1,5,2) = (2,-8,4)$, $\vec{b} = \left(\frac13, -\frac43, \frac23\right)$ (check: $\vec{a}\cdot\vec{b} = 3$).
$\vec{a} - \vec{b} = \left(\frac23, \frac13, \frac43\right)$, $|\vec{a}-\vec{b}| = \frac{\sqrt{21}}{3}$; $\vec{b}\cdot(\vec{a}-\vec{b}) = \frac{2 - 4 + 8}{9} = \frac23$.
Projection $= \frac{2/3}{\sqrt{21}/3} = \frac{2}{\sqrt{21}}$.
Answer: (A). | ||||
| 18 | StatisticsMean deviation | Class 11: Statistics | Single correctMedium | |
A medium single correct question on mean deviation, from Statistics. NCERT: Class 11: Statistics Eight numbers in order, so the median is $M = \frac{2k + 12}{2} = k + 6$, and $7 \le 2k \le 12$ puts $2k \le M \le 12$.
$\sum|x_i - M| = (M-3) + (M-5) + (M-7) + (M-2k) + (12-M) + (16-M) + (21-M) + (24-M) = 58 - 2k$.
Mean deviation $= \frac{58 - 2k}{8} = 6$ gives $k = 5$, so the median is $11$.
Answer: (D). | ||||
| 19 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\sin\frac{(2r-1)\pi}{22} = \cos\left(\frac{\pi}{2} - \frac{(2r-1)\pi}{22}\right) = \cos\frac{(6-r)\pi}{11}$, so the product of sines is $P = \cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}$.
Using $\cos\frac{3\pi}{11} = -\cos\frac{8\pi}{11}$ and $\cos\frac{5\pi}{11} = -\cos\frac{16\pi}{11}$: $P = \cos\theta\cos2\theta\cos4\theta\cos8\theta\cos16\theta$ with $\theta = \frac{\pi}{11}$, which equals $\frac{\sin 32\theta}{32\sin\theta}$. Since $32\theta = 3\pi - \theta$, $\sin 32\theta = \sin\theta$ and $P = \frac{1}{32}$.
The required value is $2P = \frac{1}{16}$.
Answer: (B). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter “Ramu is intelligent and honest” is $a = P \wedge (\sim R)$; “Ramu is not rich” is $b = \sim Q$. The statement is $a \leftrightarrow b$.
$\sim(a \leftrightarrow b) \equiv (a \wedge \sim b) \vee (\sim a \wedge b) = ((P \wedge (\sim R)) \wedge Q) \vee ((\sim P \vee R) \wedge (\sim Q))$.
Answer: (D). | ||||
| 21 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalHard | |
A hard numerical question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $A$ has $2^7 = 128$ subsets. $T \notin B$ exactly when $1 \in T$ and $2 \notin T$: $2^5 = 32$ subsets, so $n(B) = 96$.
$n(B \cup C) = 96 + $ (subsets with $1 \in T$, $2 \notin T$ and prime sum). Write $T = \{1\}\cup S$, $S \subseteq \{3,4,5,6,7\}$; the sum $1 + \Sigma S$ is prime for
$S = \{4\}$ (5), $\{6\}$ (7), $\{3,7\}$, $\{4,6\}$ (11), $\{5,7\}$, $\{3,4,5\}$ (13), $\{3,6,7\}$, $\{4,5,7\}$ (17), $\{5,6,7\}$, $\{3,4,5,6\}$ (19), $\{4,5,6,7\}$ (23): 11 subsets; every other choice of $S$ gives a composite sum or 1.
$n(B\cup C) = 96 + 11 = 107$.
Answer: 107. | ||||
| 22 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalHard | |
A hard numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $f(x) = x^2 + bx + p$ with $1 + b + p = \frac13$, so $b = -\frac23 - p$.
If $f(\alpha) = 0$ then $f(f(\alpha)) = f(0) = p$, $f^{(3)}(\alpha) = f(p)$, and $f^{(4)}(\alpha) = f(f(p)) = 0$: $f(p)$ is a root of $f$.
$f(p) = p^2 + bp + p = p^2 + p\left(-\frac23 - p\right) + p = \frac{p}{3}$. So $f\left(\frac p3\right) = 0$: $\frac{p^2}{9} + \frac{bp}{3} + p = 0$, and with $p \ne 0$, $p + 3b + 9 = 0$.
$p + 3\left(-\frac23 - p\right) + 9 = 0$ gives $p = \frac72$, $b = -\frac{25}{6}$. Then $f(x) = x^2 - \frac{25}{6}x + \frac72$ has real roots $3$ and $\frac76$ (and $\frac p3 = \frac76$ is one of them), so the condition holds.
$f(-3) = 9 + \frac{25}{2} + \frac72 = 25$.
Answer: 25. | ||||
| 23 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A = I + N$ with $N = \begin{bmatrix}0&a&a\\0&0&b\\0&0&0\end{bmatrix}$, $N^2 = \begin{bmatrix}0&0&ab\\0&0&0\\0&0&0\end{bmatrix}$, $N^3 = O$.
$A^n = I + nN + \frac{n(n-1)}{2}N^2$, so $na = 48$, $nb = 96$, $na + \frac{n(n-1)}{2}ab = 2160$.
$48 + \frac{n(n-1)}{2}\cdot\frac{48 \cdot 96}{n^2} = 2160$ gives $\frac{2304(n-1)}{n} = 2112$, $n = 12$; then $a = 4$, $b = 8$.
$n + a + b = 24$.
Answer: 24. | ||||
| 24 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives On $\left[\frac54, 2\right]$, $x^2 + 2x$ increases from $4.0625$ to $8$, so $[x^2+2x] \ge 4$, and $|5x - 7| \ge 0$.
Minimum: at $x = \frac75$, $|5x-7| = 0$ and $x^2 + 2x = 4.76$, so $f = 4$; nothing is smaller.
Maximum: at $x = 2$, $f = 3 + 8 = 11$. For $x < 2$, $[x^2 + 2x] \le 7$ and $|5x - 7| < 3$, so $f < 10$.
Sum $= 11 + 4 = 15$.
Answer: 15. | ||||
| 25 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Homogeneous: put $y = vx$. $v + x\frac{dv}{dx} = \frac{4v^3 + 2v}{3v^2 + 1}$, so $x\frac{dv}{dx} = \frac{v^3 + v}{3v^2 + 1}$.
$\int\frac{3v^2+1}{v(v^2+1)}dv = \int\left(\frac1v + \frac{2v}{v^2+1}\right)dv = \ln x + c$, giving $v(v^2 + 1) = Cx$, i.e. $y(x^2 + y^2) = Cx^4$.
$y(1) = 1$: $C = 2$. At $x = 2$: $y(4 + y^2) = 32$, i.e. $y^3 + 4y - 32 = 0$. At $y = 2$ the left side is $-16$, at $y = 3$ it is $7$, so $y(2) \in (2, 3)$ (about 2.7).
$y(2) \in [2, 3)$, so $n = 3$.
Answer: 3. | ||||
| 26 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\int_0^x (x-t)f'(t)dt = x\int_0^x f'(t)dt - \int_0^x tf'(t)dt$; its derivative is $\int_0^x f'(t)dt + xf'(x) - xf'(x) = f(x) - f(0)$.
Differentiating the equation: $f'(x) + f(x) - f(0) = (2e^{2x} - 2e^{-2x})\cos 2x - 2(e^{2x} + e^{-2x})\sin 2x + \frac2a$.
At $x = 0$: $f'(0) + 0 = 0 - 0 + \frac2a$, so $4 = \frac2a$, $a = \frac12$.
$(2a + 1)^5a^2 = 2^5\cdot\frac14 = 8$.
Answer: 8. | ||||
| 27 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $a_n = \sum_{k=1}^{n}\left[\frac{x^k}{k^2}\right]_{-1}^{n} = \sum_{k=1}^{n}\frac{n^k - (-1)^k}{k^2}$.
$a_1 = 2$ (not in the open interval). $a_2 = 3 + \frac34 = 3.75$. $a_3 = 4 + 2 + \frac{28}{9} \approx 9.11$. $a_4 = 5 + \frac{15}{4} + \frac{65}{9} + \frac{255}{16} \approx 31.9 > 30$, and $a_n$ keeps increasing.
The set is $\{2, 3\}$; the sum is 5.
Answer: 5. | ||||
| 28 | CirclesStandard and general form | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on standard and general form, from Circles. NCERT: Class 11: Conic Sections $C_1 = (-3, -4)$, $r_1 = \sqrt{9 + 16 - 16} = 3$.
$C_2 = (\sqrt3 - 3, \sqrt6 - 4)$, $r_2^2 = (3-\sqrt3)^2 + (4-\sqrt6)^2 + k + 6\sqrt3 + 8\sqrt6 = 34 + k$.
$C_1C_2 = |(\sqrt3, \sqrt6)| = 3 = |r_2 - r_1|$, so $r_2 = 6$ ($k = 2 > 0$).
The point of contact lies on line $C_2C_1$ at distance $r_2 = 6$ from $C_2$, beyond $C_1$: $P = C_2 + 2(C_1 - C_2) = (\sqrt3 - 3 - 2\sqrt3, \sqrt6 - 4 - 2\sqrt6) = (-3 - \sqrt3, -4 - \sqrt6)$.
$(\alpha + \sqrt3)^2 + (\beta + \sqrt6)^2 = 9 + 16 = 25$.
Answer: 25. | ||||
| 29 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $F_x = 12x^2 - 3y^2 + 12x - 5y + 9 = 48 - 27 - 24 - 15 + 9 = -9$; $F_y = -6xy - 5x - 16y = 36 + 10 - 48 = -2$ at $(-2, 3)$.
Slope of tangent $= -\frac{F_x}{F_y} = -\frac92$; slope of normal $= \frac29$.
Tangent meets $y = 0$ at $x = -2 + \frac{3}{9/2} = -\frac43$; normal meets it at $x = -2 - \frac{3}{2/9} = -\frac{31}{2}$.
Base $= -\frac43 + \frac{31}{2} = \frac{85}{6}$, height 3: $A = \frac12\cdot\frac{85}{6}\cdot 3 = \frac{85}{4}$, $8A = 170$.
Answer: 170. | ||||
| 30 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $x = \frac{2\alpha}{1 + \alpha^2}$. With $\theta = \tan^{-1}\frac43$, $\cos\theta = \frac35$, so $y^2 = \sin^2\frac\theta2 = \frac{1 - \cos\theta}{2} = \frac15$.
$1 - x = \frac15$ gives $\frac{2\alpha}{1+\alpha^2} = \frac45$, $2\alpha^2 - 5\alpha + 2 = 0$, $\alpha = 2$ or $\frac12$.
$\sum 16\alpha^3 = 16\left(8 + \frac18\right) = 130$.
Answer: 130. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter $\mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}} = \frac{6 - 2}{6 + 2} = 0.5 = 50\%$.
Answer: (D). | ||||
| 32 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The wire length is fixed: $2(2\pi r_1) = 5(2\pi r_2)$, so $r_2 = \frac{2r_1}{5}$.
At the centre $B = \frac{\mu_0 NI}{2r}$, so $\frac{B_2}{B_1} = \frac{N_2}{N_1}\cdot\frac{r_1}{r_2} = \frac52\cdot\frac52 = \frac{25}{4}$.
Answer: (B). | ||||
| 33 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Weight = buoyancy on the immersed half + surface tension acting upward along the circle of contact (circumference $2\pi r$):
$\frac43\pi r^3\rho g = \frac23\pi r^3\sigma g + 2\pi rT$, so $\frac23\pi r^3 g(2\rho - \sigma) = 2\pi rT$ and $r^2 = \frac{3T}{g(2\rho - \sigma)}$.
With $T = 7.5\times10^{-4}\ N\,cm^{-1} = 7.5\times10^{-2}\ N\,m^{-1}$ and densities in SI units: $r^2 = \frac{0.225}{10(2\rho-\sigma)} = \frac{2.25\times10^{-2}}{2\rho - \sigma}\ m^2$, so $r = \frac{0.15}{\sqrt{2\rho-\sigma}}\ m = \frac{15}{\sqrt{2\rho - \sigma}}$ cm.
Answer: (A). | ||||
| 34 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Change in momentum of one ball $= m(v - (-v)) = 0.05 \times 20 = 1\ kg\,m\,s^{-1}$.
$F = \frac{\Delta p}{\Delta t} = \frac{1}{0.005} = 200$ N.
Answer: (B). | ||||
| 35 | Laws of MotionEquilibrium of concurrent forces | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on equilibrium of concurrent forces, from Laws of Motion. NCERT: Class 11: Laws of Motion Net force: $F_x = 5 - 6 = -1$ N, $F_y = 7 - 8 = -1$ N.
The additional force must be $(+1, +1)$ N: magnitude $\sqrt{2}$ N at $45^\circ$ to the positive x-axis.
Answer: (A). | ||||
| 36 | 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 Isolated: $C_0 = 4\pi\varepsilon_0R_1$. With the earthed outer sphere: $C = \frac{4\pi\varepsilon_0R_1R_2}{R_2 - R_1}$.
$\frac{C}{C_0} = \frac{R_2}{R_2 - R_1} = n$, so $R_2(n - 1) = nR_1$ and $\frac{R_2}{R_1} = \frac{n}{n-1}$.
Answer: (A). | ||||
| 37 | 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 $\lambda = \frac{h}{\sqrt{2mqV}}$; both carry charge $e$ and $m_d = 2m_p$.
$\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{m_dV_d}{m_pV_p}} = \sqrt{\frac{2V_d}{V_p}} = \frac{1}{\sqrt2}$, so $\frac{2V_d}{V_p} = \frac12$ and $V_p : V_d = 4 : 1$.
Answer: (D). | ||||
| 38 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\frac1f = \frac1v - \frac1u = \frac1{12} + \frac1{240} = \frac{21}{240}$ (cm).
The plate shifts the image away from the lens by $t\left(1 - \frac1\mu\right) = 1\left(1 - \frac{1}{1.5}\right) = \frac13$ cm. For the final image to fall on the screen, the lens must form its image at $v' = 12 - \frac13 = \frac{35}{3}$ cm.
$\frac1{u'} = \frac{3}{35} - \frac{21}{240} = \frac{144 - 147}{1680} = -\frac{1}{560}$, so $u' = -560$ cm $= 5.6$ m.
The object must be moved from 2.4 m to 5.6 m, i.e. shifted by 3.2 m (away from the lens).
Answer: (B). | ||||
| 39 | 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 $B_0 = \frac{E_0}{c} = \frac{540}{3\times10^8} = 1.8\times10^{-6} = 18\times10^{-7}$ T.
Answer: (A). | ||||
| 40 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current A metal detector contains a coil that is part of a circuit tuned to resonance. A metal object passing through changes the coil's impedance, the circuit goes off resonance, the current changes markedly, and this change sets off the alarm. The detector is the standard textbook example of resonance in ac circuits.
Answer: (B). | ||||
| 41 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy 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 frequency does not depend on the energy: $f = \frac{eB}{2\pi m} = \frac{1.6\times10^{-19}\times10^{-4}}{2\pi\times9\times10^{-31}} \approx 2.8\times10^6$ Hz.
Answer: (C). | ||||
| 42 | 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 From A the current passes the 5 k$\Omega$, then the parallel pair 10 k$\Omega$ and 5 k$\Omega$ ($\frac{10\times5}{15} = \frac{10}{3}$ k$\Omega$), then the 10 k$\Omega$ to B.
$R = 5 + \frac{10}{3} + 10 = \frac{55}{3}$ k$\Omega$; $V_{AB} = 15\ mA \times \frac{55}{3}\ k\Omega = 275$ V.
Answer: (D). | ||||
| 43 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g' = g\left(\frac{R}{3R}\right)^2 = \frac{\pi^2}{9}$.
Seconds pendulum, $T = 2$ s: $L = \frac{g'T^2}{4\pi^2} = \frac{\pi^2}{9}\cdot\frac{4}{4\pi^2} = \frac19$ m.
Answer: (D). | ||||
| 44 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$, $v_s = \sqrt{\frac{\gamma RT}{M}}$, so $\frac{v_{rms}}{v_s} = \sqrt{\frac3\gamma} = \sqrt2$ and $\gamma_{mix} = \frac32$.
$C_V$: He $\frac32R$, $H_2$ $\frac52R$. $\gamma_{mix} = \frac{2\cdot\frac52 + n\cdot\frac72}{2\cdot\frac32 + n\cdot\frac52} = \frac{5 + 3.5n}{3 + 2.5n} = \frac32$.
$10 + 7n = 9 + 7.5n$, so $n = 2$.
Answer: (B). | ||||
| 45 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics Carnot efficiencies: $\eta_1 = 1 - \frac{420}{720} = \frac{5}{12}$; $\eta_2 = 1 - \frac{320}{1220} = \frac{45}{61}$.
$\frac{\eta_1}{\eta_2} = \frac{5}{12}\cdot\frac{61}{45} = \frac{61}{108} \approx 0.56$.
Answer: (B). | ||||
| 46 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $\frac{g'}{g} = \left(\frac{R}{5R/4}\right)^2 = \frac{16}{25}$, so the weight falls by $\frac{9}{25} = 36\%$.
Answer: (A). | ||||
| 47 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Momentum: $0.2\times10 = 10v$, $v = 0.2\ m\,s^{-1}$.
KE before $= \frac12(0.2)(100) = 10$ J; after $= \frac12(10)(0.04) = 0.2$ J. Loss $= 9.8$ J.
Answer: (B). | ||||
| 48 | 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 $\frac{u^2\sin2\theta}{g} = \frac{u^2\sin^2\theta}{2g}$ gives $2\sin\theta\cos\theta = \frac{\sin^2\theta}{2}$, so $\tan\theta = 4$.
Answer: (D). | ||||
| 49 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $H = I^2Rt$: $\frac{\Delta H}{H} = 2\frac{\Delta I}{I} + \frac{\Delta R}{R} + \frac{\Delta t}{t} = 4\% + 1\% + 3\% = 8\%$.
Answer: (D). | ||||
| 50 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac1\lambda = R\left(1 - \frac1{n^2}\right)$, so $\frac1{n^2} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}$ and $n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$.
Answer: (B). | ||||
| 51 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\frac{dv}{dx} = 5\ s^{-1}$; $a = v\frac{dv}{dx} = 20\times5 = 100\ m\,s^{-2}$.
Answer: 100. | ||||
| 52 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Masses at $(0,0)$, $(3,0)$, $(0,3)$: centre of mass $(1, 1)$, at distance $\sqrt2$ m from the origin. $x = 2$.
Answer: 2. | ||||
| 53 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Heat given by the water cooling to 0°C: $0.3\times4200\times25 = 31500$ J.
Ice melted $= \frac{31500}{3.5\times10^5} = 0.09$ kg $= 90$ g (less than 120 g, so consistent).
Answer: 90. | ||||
| 54 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $E_{3\to2} \propto \frac14 - \frac19 = \frac{5}{36}$; $E_{\infty\to2} \propto \frac14 = \frac{9}{36}$. Ratio $= \frac59 = \frac{x}{x+4}$ gives $x = 5$.
Answer: 5. | ||||
| 55 | 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 Balancing length $\propto$ emf: $l_2 = 36\times\frac{1.80}{1.20} = 54$ cm. Difference $= 54 - 36 = 18$ cm.
Answer: 18. | ||||
| 56 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits With A positive, the upper-left diode (from the A junction to the top junction) is forward biased: a short, so the top junction is at A's potential. The upper-right diode points from the B junction to the top junction, so it is reverse biased: open.
The vertical 20 $\Omega$ (top to bottom junction) is then in parallel with the lower-left 20 $\Omega$ (A to bottom junction): 10 $\Omega$. This is in series with the 15 $\Omega$ from the bottom junction to B.
$R_{AB} = 10 + 15 = 25\ \Omega$.
Answer: 25. | ||||
| 57 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $A_R = 2A\cos\frac\phi2 = \sqrt3A$, so $\cos\frac\phi2 = \frac{\sqrt3}{2}$, $\phi = 60^\circ$.
Answer: 60. | ||||
| 58 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charge $Q = (C + 3C)\times18 = 72C$, conserved after the battery is removed.
New capacitance $= 9C + 3C = 12C$, so $V = \frac{72C}{12C} = 6$ V.
Answer: 6. | ||||
| 59 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens: $\frac1v = \frac1{20} - \frac1{60} = \frac1{30}$, so the lens alone would image the object 30 cm beyond it, i.e. 20 cm behind the mirror.
For the final image to fall on the object the rays must retrace their path, so they must strike the mirror normally: the converging rays head for the mirror's centre of curvature. Hence $R = 20$ cm and $f = \frac R2 = 10$ cm.
Answer: 10. | ||||
| 60 | 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 $|e| = \left|\frac{d\phi}{dt}\right| = |16t - 9| = |4 - 9| = 5$ V at $t = 0.25$ s.
$I = \frac{5}{20} = 0.25$ A $= 250$ mA.
Answer: 250. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $XeO_3$: 3 bond pairs + 1 lone pair, $sp^3$, pyramidal (II).
$XeF_2$: 2 bond pairs + 3 lone pairs, $sp^3d$, linear (I).
$XeOF_4$: 5 bond pairs + 1 lone pair, $sp^3d^2$, square pyramidal (IV).
$XeF_6$: 6 bond pairs + 1 lone pair, $sp^3d^3$, distorted octahedral (III).
Answer: (A). | ||||
| 62 | SolutionsColligative properties | Class 12: Solutions | Single correctEasy | |
A easy single correct question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = K_f\cdot\frac{w/M}{1\ kg}$ with the same $w$, so $\Delta T_f \propto \frac1M$.
$\frac{\Delta T_A}{\Delta T_B} = \frac{M_Y}{M_X} = \frac14$, so $M_X : M_Y = 4 : 1 = 1 : 0.25$.
Answer: (B). | ||||
| 63 | EquilibriumIonic equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Reaction (c) is (a) + (b): $H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}$. When reactions are added their equilibrium constants multiply:
$K_{a_3} = \frac{[H^+]^2[C_2O_4^{2-}]}{[H_2C_2O_4]} = \frac{[H^+][HC_2O_4^-]}{[H_2C_2O_4]}\cdot\frac{[H^+][C_2O_4^{2-}]}{[HC_2O_4^-]} = K_{a_1}\times K_{a_2}$.
Answer: (D). | ||||
| 64 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry $\Lambda_m = \frac{\kappa}{c}$ with the same $\kappa$ for both cells.
$c_1 = \frac{10}{20}$ mol mL$^{-1}$ $= 0.5$; $c_2 = \frac{20}{80} = 0.25$ mol mL$^{-1}$. So $\frac{\Lambda_{m2}}{\Lambda_{m1}} = \frac{c_1}{c_2} = 2$.
Answer: (A). | ||||
| 65 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | StatementsMedium | |
A medium statements question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter Micelles form above the Kraft temperature and the critical micelle concentration. The process absorbs a little heat ($\Delta H$ slightly positive: endothermic) and is driven by entropy: the ordered water around the separate hydrocarbon tails is released when the tails cluster, so $\Delta S > 0$ and $\Delta G = \Delta H - T\Delta S < 0$.
So B and C are correct.
Answer: (C). | ||||
| 66 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across period 2 ionization enthalpy rises, with two exceptions: B ($2p^1$) is lower than Be ($2s^2$, the $2p$ electron is easier to remove), and O ($2p^4$, paired electron) is lower than N (stable half-filled $2p^3$).
Values (kJ mol$^{-1}$): B 801 < Be 899 < O 1314 < N 1402.
Answer: (D). | ||||
| 67 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | StatementsEasy | |
A easy statements question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Pig iron comes from the blast furnace (about 4% carbon). Cast iron is made from pig iron by melting it with scrap iron and coke, and has slightly less carbon (about 3%).
Statement I reverses the relation (cast iron is made from pig iron, not the other way), so it is wrong; Statement II is wrong because pig iron has more carbon than cast iron.
Answer: (B). | ||||
| 68 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter High-purity (>99.95%) dihydrogen is prepared by electrolysing warm aqueous barium hydroxide solution between nickel electrodes. The other methods give hydrogen with impurities (or are laboratory methods not aimed at high purity).
Answer: (C). | ||||
| 69 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Densities (g cm$^{-3}$): Be 1.85, Mg 1.74, Ca 1.55, Sr 2.63. In group 2 the density falls from Be to Ca and then rises.
Order: $Sr > Be > Mg > Ca$.
Answer: (C). | ||||
| 70 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $NO$, $N_2O$ and $CO$ are neutral oxides.
$B_2O_3$, $N_2O_5$, $SO_3$ and $P_4O_{10}$ are acidic (they give boric, nitric, sulphuric and phosphoric acids with water): 4.
Answer: (B). | ||||
| 71 | 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 All three are octahedral $Ni^{2+}$ complexes; the energy absorbed equals $\Delta_o$, which follows the spectrochemical series $H_2O < NH_3 < en$.
So $C < B < A$.
Answer: (A). | ||||
| 72 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Excess sulphate in drinking water has a laxative effect (A-III). Excess fluoride hardens and deforms bones and teeth: bending of bones (B-II). Nicotine is used as a pesticide (insecticide) (C-I). Sodium arsinite is a herbicide (weed killer) (D-IV).
Answer: (C). | ||||
| 73 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The substrate is allyl acrylate, $CH_2=CH-CO-O-CH_2-CH=CH_2$, with two different C=C bonds.
Acrylate part: the C=C is conjugated with C=O. Protonation occurs on the carbonyl oxygen and $Br^-$ adds to the $\beta$-carbon (conjugate addition), giving $Br-CH_2-CH_2-CO-$.
Allyl part: an isolated C=C; Markovnikov addition gives the secondary bromide $-O-CH_2-CH(Br)-CH_3$.
Product: $BrCH_2CH_2COOCH_2CH(Br)CH_3$.
Answer: (D). | ||||
| 74 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The reactant is 2-methylpropanal, $(CH_3)_2CH-CHO$, which has one $\alpha$-hydrogen, so dilute aqueous base brings about aldol addition.
The enolate $(CH_3)_2C^--CHO$ attacks the carbonyl carbon of a second molecule: $(CH_3)_2CH-CH(OH)-C(CH_3)_2-CHO$ (3-hydroxy-2,2,4-trimethylpentanal).
The carbon next to CHO is now quaternary (no $\alpha$-H), so the aldol cannot dehydrate to an enal.
Answer: (B). | ||||
| 75 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) 4-nitrophenol: $-NO_2$ withdraws by $-I$ and $-R$ (conjugation with the phenoxide) - most acidic (pKa 7.1).
(B) 3-nitrophenol: only $-I$ acts from the meta position (pKa 8.4).
(C) 3-methoxyphenol: from meta, $-OCH_3$ acts only by $-I$, so it is slightly more acidic than phenol (pKa 9.6).
(D) 4-methoxyphenol: the $+R$ effect of $-OCH_3$ destabilises the phenoxide - least acidic (pKa 10.2).
Order: $A > B > C > D$.
Answer: (A). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $CH_3MgBr$ adds to the nitrile: A is the imine salt $CH_3CH_2C(CH_3)=NMgBr$.
Acid hydrolysis gives the ketone B, butan-2-one $CH_3CH_2COCH_3$.
Clemmensen reduction ($Zn-Hg$/HCl) turns $C=O$ into $CH_2$: C is butane, $CH_3CH_2CH_2CH_3$.
Answer: (A). | ||||
| 77 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Nylon 6,6: fibres, bristles for brushes (A-III). Low density polythene: squeeze bottles, toys, flexible pipes (B-IV). High density polythene: buckets, dustbins, bottles (C-I). Teflon: non-stick coating on utensils (D-II).
Answer: (B). | ||||
| 78 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Sucrose is $\alpha$-D-glucopyranose and $\beta$-D-fructofuranose joined C1-C2 (both anomeric carbons used, so it is non-reducing). Maltose and amylose have $\alpha$(1→4) links between glucose units; lactose is galactose $\beta$(1→4) glucose.
Answer: (B). | ||||
| 79 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Some inhibitors bind not to the active site but to another site of the enzyme, the allosteric site, changing the shape of the active site so the substrate cannot bind.
Answer: (B). | ||||
| 80 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Titrating a base with an acid, the end point is reached when the solution just turns acidic. Methyl orange is yellow in its basic (benzenoid, azo) form and red in acid, where it is protonated to the quinonoid form.
At the end point it is present as the quinonoid form.
Answer: (A). | ||||
| 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 $N_2 + 3H_2 \to 2NH_3$: by volume, 20 L of $NH_3$ uses 10 L of $N_2$.
Unused $N_2 = 56 - 10 = 46$ L.
Answer: 46. | ||||
| 82 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $n = \frac{11}{44} = 0.25$ mol.
$T = \frac{PV}{nR} = \frac{2\times10^6 \times 2\times10^{-3}}{0.25\times8.3} = \frac{4000}{2.075} = 1927.7$ K.
In °C: $1927.7 - 273 = 1654.7 \approx 1655$ (with 273.15 it is 1654.6, still 1655).
Answer: 1655. | ||||
| 83 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom From $n = 5$ to $n = 1$, the possible transitions number $\frac{n(n-1)}{2} = \frac{5\times4}{2} = 10$ (counting all lines a sample of such atoms can emit).
Answer: 10. | ||||
| 84 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Neutralising the weak acid releases less heat because part of the heat is used to ionise it:
$\Delta H_{ion} = -55.3 - (-57.3) = 2\ kJ\ mol^{-1}$.
Answer: 2. | ||||
| 85 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics For a first-order reaction $\ln\frac{p}{p_0} = -kt$, so the slope of $\ln\frac{p}{p_0}$ against $t$ is $-k$: $k = 3.465\times10^4\ s^{-1}$.
$t_{1/2} = \frac{0.693}{3.465\times10^4} = 2\times10^{-5}$ s.
Answer: 2. | ||||
| 86 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Xe has 8 valence electrons.
$XeO_3$: 3 double bonds use 6, leaving 1 lone pair. $XeOF_4$: one double and four single bonds use 6, leaving 1 lone pair. $XeF_6$: six bonds use 6, leaving 1 lone pair.
Sum $= 1 + 1 + 1 = 3$.
Answer: 3. | ||||
| 87 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Standard potentials $E^\circ(M^{3+}/M^{2+})$: Cr $-0.41$ V, Mn $+1.57$ V, Fe $+0.77$ V, Co $+1.97$ V. Only $Cr^{3+}/Cr^{2+}$ is negative.
$Cr^{3+}$ is $3d^3$ with 3 unpaired electrons: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ B.M. $\approx 4$.
Answer: 4. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $ROH + CH_3MgI \to CH_4\uparrow + Mg(OR)I$: one mole of methane per mole of alcohol.
Moles of $CH_4$ at STP $= \frac{3.1}{22400} = 1.384\times10^{-4}$ mol.
$M = \frac{4.5\times10^{-3}}{1.384\times10^{-4}} = 32.5$ g mol$^{-1}$, which rounds to 33 (with 22 700 mL mol$^{-1}$, 32.95, also 33).
Answer: 33. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $R_f = \frac{\text{distance moved by substance}}{\text{distance moved by solvent front}}$; $R_f(A) = \frac{2.08}{3.25} = 0.64$, $R_f(B) = \frac{1.05}{3.25} = 0.323$.
Ratio $= \frac{2.08}{1.05} = 1.98 \approx 2$.
Answer: 2. | ||||
| 90 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques n-Pentane: 1-, 2- and 3-bromopentane (3).
Isopentane (2-methylbutane): 1-bromo-2-methylbutane, 2-bromo-2-methylbutane, 2-bromo-3-methylbutane, 1-bromo-3-methylbutane (4).
Neopentane: 1-bromo-2,2-dimethylpropane (1).
Total $= 3 + 4 + 1 = 8$.
Answer: 8. | ||||
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