JEE (Main) · 2023
JEE (Main) 2023 — 8 April, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Permutations and Combinations4
- Integral Calculus3
- Three Dimensional Geometry3
13 easy · 17 medium · 0 hard
Physics
30 Q
- Oscillations and Waves2
- Properties of Bulk Matter2
- Gravitation2
26 easy · 3 medium · 1 hard
Chemistry
30 Q
- s-Block Elements2
- p-Block Elements2
- Coordination Compounds2
24 easy · 6 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 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $n(A \times B) = 5 \times 2 = 10$.
A subset with exactly $r$ elements can be chosen in $^{10}C_r$ ways, so the required number is
$^{10}C_3 + {}^{10}C_4 + {}^{10}C_5 + {}^{10}C_6 = 120 + 210 + 252 + 210 = 792$.
Answer: (A). | ||||
| 2 | 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 The left side $|z+2|$ is real, so the right side $(\alpha + 4) + i(\beta + 4)$ must be real: $\beta = -4$.
Then $\sqrt{(\alpha+2)^2 + 16} = \alpha + 4$. Squaring: $\alpha^2 + 4\alpha + 20 = \alpha^2 + 8\alpha + 16$, so $\alpha = 1$ (and $\alpha + 4 = 5 > 0$, as needed).
$\alpha + \beta = -3$ and $\alpha\beta = -4$.
The equation with roots $-3$ and $-4$: $x^2 - (-7)x + 12 = 0$, i.e. $x^2 + 7x + 12 = 0$.
Answer: (B). | ||||
| 3 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha = -1$ and $\beta\gamma = 1$.
Sum of roots: $\alpha + \beta + \gamma = 0$, so $\beta + \gamma = 1$.
Product: $\alpha\beta\gamma = -c$, so $-1 = -c$ and $c = 1$.
$b = \alpha\beta + \beta\gamma + \gamma\alpha = \alpha(\beta + \gamma) + \beta\gamma = -1 + 1 = 0$.
So $\beta, \gamma$ are the roots of $t^2 - t + 1 = 0$. Multiplying by $(t+1)$: $t^3 + 1 = 0$, so $\beta^3 = \gamma^3 = -1$; also $\alpha^3 = -1$.
$b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3 = 0 + 2 + 3 + 6 + 8 = 19$.
Answer: (C). | ||||
| 4 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $|A| = 2(4 - 1) - 1(2 - 0) + 0 = 4$, so $|2A| = 2^3 \cdot 4 = 32 = 2^5$.
For a $3 \times 3$ matrix $M$, $|adj\,M| = |M|^2$, so $|adj(adj(adj\,M))| = |M|^{2^3} = |M|^8$.
$|adj(adj(adj\,2A))| = (2^5)^8 = 2^{40} = 16^{10}$.
$n = 10$.
Answer: (A). | ||||
| 5 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $PP^T = \begin{bmatrix} \frac34 + \frac14 & -\frac{\sqrt3}{4} + \frac{\sqrt3}{4} \\ \cdot & \frac14 + \frac34 \end{bmatrix} = I$, so $P$ is orthogonal: $P^TP = PP^T = I$.
Then $Q^2 = PAP^TPAP^T = PA^2P^T$ and in general $Q^{2007} = PA^{2007}P^T$.
$P^TQ^{2007}P = P^TPA^{2007}P^TP = A^{2007}$.
$A = I + N$ with $N = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, $N^2 = 0$, so $A^{2007} = I + 2007N = \begin{bmatrix} 1 & 2007 \\ 0 & 1 \end{bmatrix}$.
$a = 1, b = 2007, c = 0, d = 1$: $2a + b - 3c - 4d = 2 + 2007 - 0 - 4 = 2005$.
Answer: (B). | ||||
| 6 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations INDEPENDENCE has 12 letters: vowels I, E, E, E, E (5) and consonants N, N, N, D, D, P, C (7).
Treat the five vowels as one block. Arranging the block with the 7 consonants (N three times, D twice): $\frac{8!}{3!\,2!} = \frac{40320}{12} = 3360$.
Inside the block, the vowels I, E, E, E, E can be arranged in $\frac{5!}{4!} = 5$ ways.
Total $= 3360 \times 5 = 16800$.
Answer: (C). | ||||
| 7 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Seat the 7 boys round the table first: $(7-1)! = 6! = 720$ ways.
This leaves 7 gaps between boys; the 5 girls must sit in 5 different gaps: $^7P_5 = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 = 2520$ ways.
Total $= 720 \times 2520 = 1814400$.
$126(5!)^2 = 126 \times 14400 = 1814400$, while $7(360)^2 = 907200$ and $7(720)^2 = 3628800$.
Answer: (A). | ||||
| 8 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Let the coefficients be $^nC_{r-1}, {}^nC_r, {}^nC_{r+1}$.
$\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r} = 5 \Rightarrow n = 6r - 1$.
$\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1} = 4 \Rightarrow n = 5r + 4$.
So $6r - 1 = 5r + 4$, giving $r = 5$ and $n = 29$.
The fourth term is $T_4 = {}^{29}C_3 x^3$, with coefficient $\frac{29 \cdot 28 \cdot 27}{6} = 3654$.
Answer: (B). | ||||
| 9 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S_K = \frac{K(K+1)/2}{K} = \frac{K+1}{2}$, so $S_j^2 = \frac{(j+1)^2}{4}$.
$\sum_{j=1}^{n} (j+1)^2 = \sum_{m=1}^{n+1} m^2 - 1 = \frac{(n+1)(n+2)(2n+3)}{6} - 1 = \frac{2n^3 + 9n^2 + 13n + 6 - 6}{6} = \frac{n(2n^2 + 9n + 13)}{6}$.
Hence $\sum_{j=1}^{n} S_j^2 = \frac{n}{24}(2n^2 + 9n + 13)$.
Since $2, 9, 13$ have no common factor, the least $A$ is 24: $A = 24, B = 2, C = 9, D = 13$.
(A) $A+B+C+D = 48$, not divisible by 5. False.
(B) $A + B = 26 = 2 \times 13$, divisible by $D = 13$. True.
(C) $5(D - C) = 20 \neq 26$. False.
(D) $A + C + D = 46$ is divisible by $B = 2$. False.
Answer: (B). | ||||
| 10 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $1 - \cos^2(3x) = \sin^2(3x)$. As $x \to 0$: $\sin^2(3x) \approx 9x^2$, $\cos^3(4x) \to 1$, $\sin^3(4x) \approx 64x^3$, $\log_e(1+2x) \approx 2x$ so $(\log_e(2x+1))^5 \approx 32x^5$.
Limit $= \lim_{x\to 0}\frac{9x^2 \cdot 64x^3}{1 \cdot 32x^5} = \frac{9 \times 64}{32} = 18$.
Answer: (C). | ||||
| 11 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Multiply numerator and denominator by $e^x$: $I = \int \frac{e^x(x+1)}{xe^x(1 + xe^x)^2}dx$.
Put $t = xe^x$, $dt = e^x(x+1)\,dx$: $I = \int \frac{dt}{t(1+t)^2}$.
Partial fractions: $\frac{1}{t(1+t)^2} = \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2}$.
$I = \log_e t - \log_e(1+t) + \frac{1}{1+t} + C = \log_e\frac{t}{1+t} + \frac{1}{1+t} + C$.
As $x \to \infty$, $t \to \infty$, so $I \to 0 + 0 + C$; the condition gives $C = 0$.
At $x = 1$, $t = e$: $I(1) = \log_e\frac{e}{e+1} + \frac{1}{e+1} = 1 + \frac{1}{e+1} - \log_e(e+1) = \frac{e+2}{e+1} - \log_e(e+1)$.
Answer: (A). | ||||
| 12 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The parabolas $y = x^2$ and $y = 8 - x^2$ meet where $x^2 = 4$, $x = \pm 2$ (at $y = 4$).
The upper boundary is $\min(8 - x^2, 7)$: it is $y = 7$ for $|x| \le 1$ and $y = 8 - x^2$ for $1 \le |x| \le 2$.
By symmetry, Area $= 2\left[\int_0^1 (7 - x^2)\,dx + \int_1^2 (8 - 2x^2)\,dx\right]$
$= 2\left[\left(7 - \frac13\right) + \left(8 - \frac{2}{3}(8 - 1)\right)\right] = 2\left[\frac{20}{3} + \frac{10}{3}\right] = 20$.
Answer: (B). | ||||
| 13 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines Slopes: $4x + 3y = 69$ has slope $-\frac43$ and $4y - 3x = 17$ has slope $\frac34$; their product is $-1$, so the triangle is right-angled at their intersection.
The circumcentre is then the midpoint of the hypotenuse, whose ends are where the third line meets the other two.
$4x + 3y = 69$ with $x = 61 - 7y$: $244 - 25y = 69$, $y = 7$, $x = 12$: $(12, 7)$.
$4y - 3x = 17$ with $x = 61 - 7y$: $25y - 183 = 17$, $y = 8$, $x = 5$: $(5, 8)$.
$C = \left(\frac{17}{2}, \frac{15}{2}\right)$, so $\alpha - \beta = 1$ and $\alpha + \beta = 16$.
$(\alpha - \beta)^2 + \alpha + \beta = 1 + 16 = 17$.
Answer: (C). | ||||
| 14 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = 20x$ has $a = 5$, so $R = (5, 0)$.
Let $P(x_1, y_1)$, $Q(x_2, y_2)$. Centroid: $x_1 + x_2 + 5 = 30$ and $y_1 + y_2 + 0 = 30$.
Putting $x = \frac{y - c}{m}$ in $y^2 = 20x$: $my^2 - 20y + 20c = 0$, so $y_1 + y_2 = \frac{20}{m} = 30$, $m = \frac23$, and $c = 6 + m = \frac{20}{3}$.
$y_1y_2 = \frac{20c}{m} = 200$. (Check: $x_1 + x_2 = \frac{(y_1 + y_2) - 2c}{m} = \frac{30 - 40/3}{2/3} = 25$, as required.)
$(y_1 - y_2)^2 = 30^2 - 4(200) = 100$ and $(x_1 - x_2)^2 = \frac{(y_1 - y_2)^2}{m^2} = 100 \cdot \frac94 = 225$.
$(PQ)^2 = 225 + 100 = 325$.
Answer: (B). | ||||
| 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 A plane through the line: $(x + 2y + 3z - 4) + k(2x + y - z + 5) = 0$, with normal $(1 + 2k,\ 2 + k,\ 3 - k)$.
The given plane is spanned by $(1, 1, 1)$ and $(1, -2, 3)$, so its normal is $(1,1,1) \times (1,-2,3) = (3 + 2,\ 1 - 3,\ -2 - 1) = (5, -2, -3)$.
Perpendicular planes have perpendicular normals: $5(1 + 2k) - 2(2 + k) - 3(3 - k) = 0 \Rightarrow 11k - 8 = 0$, $k = \frac{8}{11}$.
Multiplying by 11: $11(x + 2y + 3z - 4) + 8(2x + y - z + 5) = 0$, i.e. $27x + 30y + 25z = 4$.
$a - b + c = 27 - 30 + 25 = 22$.
Answer: (C). | ||||
| 16 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy 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 Points $A(4, -2, -3)$, $B(1, 3, 4)$; directions $\vec{b}_1 = (4, 5, 3)$, $\vec{b}_2 = (3, 4, 2)$.
$\vec{b}_1 \times \vec{b}_2 = (10 - 12,\ 9 - 8,\ 16 - 15) = (-2, 1, 1)$, of magnitude $\sqrt6$.
$\vec{AB} = (-3, 5, 7)$ and $\vec{AB} \cdot (\vec{b}_1 \times \vec{b}_2) = 6 + 5 + 7 = 18$.
Shortest distance $= \frac{18}{\sqrt6} = 3\sqrt6$.
Answer: (B). | ||||
| 17 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra Call the points $A, B, C$. $\vec{AB} = (6 - \alpha,\ 1,\ -2)$ and $\vec{BC} = \left(-\frac32,\ \beta - 11,\ -19\right)$.
Collinear means $\vec{BC} = t\,\vec{AB}$. From the $\hat{k}$ parts, $-19 = -2t$, so $t = \frac{19}{2}$.
$\hat{j}$: $\beta - 11 = \frac{19}{2}$, so $\beta = \frac{41}{2}$ and $6\beta = 123$.
$\hat{i}$: $-\frac32 = \frac{19}{2}(6 - \alpha)$, so $6 - \alpha = -\frac{3}{19}$, $\alpha = \frac{117}{19}$ and $19\alpha = 117$.
$(19\alpha - 6\beta)^2 = (117 - 123)^2 = 36$.
Answer: (C). | ||||
| 18 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(D) = 0.2(0.03) + 0.3(0.04) + 0.5(0.02) = 0.006 + 0.012 + 0.010 = 0.028$.
By Bayes' theorem, $P(C \mid D) = \frac{0.5 \times 0.02}{0.028} = \frac{0.010}{0.028} = \frac{5}{14}$.
Answer: (B). | ||||
| 19 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\sin x + \cos x = \sqrt2\cos\left(x - \frac{\pi}{4}\right)$ and $\sin x - \cos x = \sqrt2\sin\left(x - \frac{\pi}{4}\right)$. With $\theta = x - \frac{\pi}{4}$:
$f = \frac{\cos\theta - 1}{\sin\theta} = \frac{-2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = -\tan\frac{\theta}{2} = -\tan u$, where $u = \frac{x}{2} - \frac{\pi}{8}$.
$f'(x) = -\frac12\sec^2 u$ and $f''(x) = -\frac12 \cdot 2\sec^2 u\tan u \cdot \frac12 = -\frac12\sec^2 u\tan u$.
At $x = \frac{7\pi}{12}$: $u = \frac{7\pi}{24} - \frac{3\pi}{24} = \frac{\pi}{6}$, $\tan u = \frac{1}{\sqrt3}$, $\sec^2 u = \frac43$.
$f = -\frac{1}{\sqrt3}$, $f'' = -\frac12 \cdot \frac43 \cdot \frac{1}{\sqrt3} = -\frac{2}{3\sqrt3}$.
$f f'' = \frac{2}{9}$.
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 $\sim(a \Rightarrow b) \equiv a \wedge \sim b$.
So the negation is $(p \Rightarrow q) \wedge \sim(q \Rightarrow p) \equiv (\sim p \vee q) \wedge (q \wedge \sim p)$.
Since $q \wedge \sim p$ already makes $\sim p \vee q$ true, this is $q \wedge (\sim p)$.
Answer: (C). | ||||
| 21 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Every pair in $R$ has $x - y > 0$, so no reverse pair $(y, x)$ is in $R$. To make $R$ symmetric, the reverse of every pair must be added, so the answer is $n(R)$.
Odd positive difference: $x$ and $y$ of opposite parity with $x > y$. $A$ has 3 odd elements (3, 7, 9) and 5 even ones (0, 4, 6, 8, 10); each odd–even pair gives exactly one such ordered pair: $3 \times 5 = 15$.
Difference 2: $(6, 4), (8, 6), (10, 8), (9, 7)$: 4 pairs.
$n(R) = 15 + 4 = 19$, so 19 pairs must be added.
Answer: 19. | ||||
| 22 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = {}^7C_r (3x^2)^{7-r}\left(-\frac{1}{2x^5}\right)^r$, with power of $x$ equal to $14 - 2r - 5r = 14 - 7r$.
Constant term: $r = 2$. $\alpha = {}^7C_2 \cdot 3^5 \cdot \frac{1}{4} = \frac{21 \times 243}{4} = \frac{5103}{4} = 1275.75$.
$[\alpha] = 1275$.
Answer: 1275. | ||||
| 23 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalEasy | |
A easy numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Take $f(x) = \frac{x^3}{x^4 + 147}$, $x > 0$. $f'(x) = \frac{3x^2(x^4 + 147) - 4x^6}{(x^4 + 147)^2} = \frac{x^2(441 - x^4)}{(x^4 + 147)^2}$.
$f$ increases up to $x^4 = 441$, i.e. $x = \sqrt{21} \approx 4.58$, and decreases after. So the greatest term is $a_4$ or $a_5$.
$a_4 = \frac{64}{403} \approx 0.1588$ and $a_5 = \frac{125}{772} \approx 0.1619$.
The greatest term is $a_5$: $\alpha = 5$.
Answer: 5. | ||||
| 24 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)$, $1 \le \mathrm{cosec}\,x < 2$, so $[\mathrm{cosec}\,x] = 1$ and $\int 8[\mathrm{cosec}\,x]\,dx = 8 \cdot \frac{2\pi}{3} = \frac{16\pi}{3}$.
$\cot x$ falls from $\sqrt3$ to $-\sqrt3$:
$\left(\frac{\pi}{6}, \frac{\pi}{4}\right)$: $[\cot x] = 1$, length $\frac{\pi}{12}$;
$\left(\frac{\pi}{4}, \frac{\pi}{2}\right]$: $[\cot x] = 0$;
$\left(\frac{\pi}{2}, \frac{3\pi}{4}\right]$: $[\cot x] = -1$, length $\frac{\pi}{4}$;
$\left(\frac{3\pi}{4}, \frac{5\pi}{6}\right)$: $[\cot x] = -2$, length $\frac{\pi}{12}$.
$\int [\cot x]\,dx = \frac{\pi}{12} - \frac{\pi}{4} - \frac{2\pi}{12} = -\frac{\pi}{3}$.
Integral $= \frac{16\pi}{3} - 5\left(-\frac{\pi}{3}\right) = 7\pi$, and $\frac{2}{\pi} \cdot 7\pi = 14$.
Answer: 14. | ||||
| 25 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\log_e x^2 = 2\log_e x$, so $\frac{dy}{dx} + \frac{y}{2x\log_e x} = \frac{1}{x}$.
Integrating factor: $e^{\int \frac{dx}{2x\log_e x}} = e^{\frac12\log_e(\log_e x)} = \sqrt{\log_e x}$.
$y\sqrt{\log_e x} = \int \frac{\sqrt{\log_e x}}{x}dx = \frac23(\log_e x)^{3/2} + C$.
At $(e, \frac43)$: $\frac43 = \frac23 + C$, so $C = \frac23$.
At $x = e^4$, $\log_e x = 4$: $2y = \frac23 \cdot 8 + \frac23 = 6$, so $y = 3$.
$\alpha = 3$.
Answer: 3. | ||||
| 26 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The exponent of 3 in $66!$ is $\left[\frac{66}{3}\right] + \left[\frac{66}{9}\right] + \left[\frac{66}{27}\right] + \left[\frac{66}{81}\right] = 22 + 7 + 2 + 0 = 31$.
Answer: 31. | ||||
| 27 | CirclesRadius and centre | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections $C_1$ has centre $(2, 1)$ and radius$^2 = 4 + 1 + \alpha - 5 = \alpha$.
$C_2$: $x^2 + y^2 - 2fx - 2gy + \frac{36}{5} = 0$, centre $(f, g)$, radius$^2 = f^2 + g^2 - \frac{36}{5}$.
Image of $(2, 1)$ in $2x - y + 1 = 0$: $\frac{x - 2}{2} = \frac{y - 1}{-1} = -\frac{2(4 - 1 + 1)}{5} = -\frac85$, so $(f, g) = \left(-\frac65, \frac{13}{5}\right)$.
$r^2 = \frac{36 + 169}{25} - \frac{36}{5} = \frac{205 - 180}{25} = 1$, so $r = 1$.
A mirror image has the same radius: $\alpha = r^2 = 1$.
$\alpha + r = 2$.
Answer: 2. | ||||
| 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 Equal distances from the plane: $|5 + 3 - 6\lambda + 7| = |-4 + 0 - 6 + 7|$, i.e. $|15 - 6\lambda| = 3$.
$\lambda = 2$ or $\lambda = 3$, so $\lambda_1 = 3$, $\lambda_2 = 2$, and the point is $P(1, 2, 3)$.
The line passes through $A(5, 1, -7)$ with direction $(1, 2, 2)$, of length 3.
$\vec{AP} = (-4, 1, 10)$, $|\vec{AP}|^2 = 117$; its projection on the line is $\frac{-4 + 2 + 20}{3} = 6$.
Distance $= \sqrt{117 - 36} = \sqrt{81} = 9$.
Answer: 9. | ||||
| 29 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a} \times (\vec{c} - \vec{b}) = \vec{0}$, so $\vec{c} = \vec{b} + t\vec{a}$ for some scalar $t$.
$\vec{a} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 6 - 18 + 12 = 0$, so $\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = \vec{b} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = \alpha - 22 - 2 = 5$, giving $\alpha = 29$.
$\vec{a} \cdot \vec{c} = \vec{a} \cdot \vec{b} + t|\vec{a}|^2 = (174 + 99 - 24) + 261t = 249 + 261t = -12$, so $t = -1$.
$\vec{c} = \vec{b} - \vec{a} = 23\hat{i} + 2\hat{j} - 14\hat{k}$.
$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 - 14 = 11$.
Answer: 11. | ||||
| 30 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean: $x + y + 52 = 72$, so $x + y = 20$.
Variance: $\frac{\sum x_i^2}{8} - 81 = 9.25$, so $\sum x_i^2 = 722$. The known numbers give $100 + 144 + 36 + 144 + 16 + 64 = 504$, so $x^2 + y^2 = 218$.
$2xy = 400 - 218 = 182$, $xy = 91$: $x, y$ are 13 and 7, and $x > y$ gives $x = 13$, $y = 7$.
$3x - 2y = 39 - 14 = 25$.
Answer: 25. | ||||
| 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 passenger moves with the train, so source (engine) and listener have the same velocity: there is no relative motion between them.
With $v_s = v_o = 10$ ms$^{-1}$ in the same direction, $f' = f\,\frac{v + v_o}{v + v_s} = f$ (or $\frac{v - v_o}{v - v_s}$, depending on which is ahead): either way the factor is 1.
$f' = 400$ Hz.
Answer: (D). | ||||
| 32 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Pressure at the bottom: $P_1 = P_0 + \rho gh = 10^5 + 1000 \times 10 \times 40 = 5 \times 10^5$ Pa. At the surface $P_2 = 10^5$ Pa.
The temperature is the same, so $P_1V_1 = P_2V_2$: $V_2 = \frac{5 \times 10^5 \times 1}{10^5} = 5$ cm$^3$.
Answer: (D). | ||||
| 33 | ThermodynamicsFirst law | Class 11: Thermodynamics | StatementsEasy | |
A easy statements question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Statement I: false. Heat added need not raise the temperature: in an isothermal expansion of an ideal gas all the heat goes into work ($\Delta U = 0$), and during melting or boiling the heat goes into the change of phase.
Statement II: true. Work done by a system is $W = \int P\,dV$; with a positive pressure, $W > 0$ requires $dV > 0$, so the volume increases.
Answer: (D). | ||||
| 34 | Properties of Bulk MatterElasticity | 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 elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Strain $= 0.04\% = 4 \times 10^{-4}$.
Energy per unit volume $u = \frac12 Y(\text{strain})^2 = \frac12 \times 7 \times 10^{10} \times 16 \times 10^{-8} = 5600$ J/m$^3$.
Answer: (B). | ||||
| 35 | GravitationOrbital motion | Class 11: Gravitation | StatementsEasy | |
A easy statements question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation For a circular orbit of radius $r$: $K = \frac{GMm}{2r}$, $U = -\frac{GMm}{r}$, $E = K + U = -\frac{GMm}{2r}$.
So $U = 2E$, not $\frac{E}{2}$: Statement I is incorrect.
$K = -E = |E|$, the whole magnitude of $E$, not half of it: Statement II is incorrect.
Answer: (B). | ||||
| 36 | 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 $\vec{a} = \frac{d\vec{v}}{dt} = 2\hat{i} + 6t\hat{j}$; at $t = 1$ s, $\vec{a} = 2\hat{i} + 6\hat{j}$ m s$^{-2}$.
$\vec{F} = m\vec{a} = 0.5(2\hat{i} + 6\hat{j}) = \hat{i} + 3\hat{j}$ N.
$x = 3$.
Answer: (C). | ||||
| 37 | 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 For perpendicular vectors, $R = \sqrt{A^2 + \left(\frac{A}{2}\right)^2} = \sqrt{\frac{5A^2}{4}} = \frac{\sqrt5 A}{2}$.
Answer: (B). | ||||
| 38 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At depth $d$, $g_d = g\left(1 - \frac{d}{R}\right)$. With $d = \frac{R}{2}$, $g_d = \frac{g}{2}$.
Weight $= \frac{400}{2} = 200$ N.
Answer: (C). | ||||
| 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 $R = \frac{u^2\sin 2\theta}{g}$. Here $\sin 60^\circ = \sin 120^\circ$, so $\frac{R_A}{R_B} = \frac{40^2}{60^2} = \frac{1600}{3600} = \frac49$.
Answer: (C). | ||||
| 40 | 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 $\rho = \frac{m}{\pi r^2 L}$, so $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L}$.
$= \frac{0.01}{0.4} + 2 \times \frac{0.03}{6} + \frac{0.04}{8} = 0.025 + 0.010 + 0.005 = 0.04$.
Maximum error $= 4\%$.
Answer: (C). | ||||
| 41 | Communication SystemsRange of a TV tower | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on range of a tv tower, from Communication Systems. NCERT: No matching NCERT chapter Range $d = \sqrt{2Rh}$, so the area covered is $\pi d^2 = 2\pi Rh$.
$2\pi Rh = 2\pi \times 6400 \times 0.098$ km$^2 = 2\pi \times 627.2 \approx 3941$ km$^2$, i.e. about 3942 km$^2$.
Answer: (C). | ||||
| 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 Each input passes through a NAND gate with its two inputs joined, which is a NOT gate, giving $\bar{A}$ and $\bar{B}$. These feed a NAND gate:
$Y = \overline{\bar{A}\cdot\bar{B}} = A + B$ (De Morgan), an OR gate.
From the waveforms: A is 0 for $0$–$2$ and 1 for $2$–$4$; B is 0 for $0$–$1$, 1 for $1$–$2$, 0 for $2$–$3$, 1 for $3$–$4$.
$Y = A + B$: 0 for $0$–$1$, and 1 from $t = 1$ to $t = 4$ (B is 1 on $1$–$2$, A is 1 on $2$–$4$).
This is the first waveform.
Answer: (A). | ||||
| 43 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctHard | |
A hard single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei In the liquid-drop picture the binding energy is $B = a_vA - a_sA^{2/3} - a_c\frac{Z(Z-1)}{A^{1/3}} - \ldots$
The statements are read literally, each as written.
A. The total surface term is $-a_sA^{2/3}$; per nucleon, as A says, it is $-a_sA^{-1/3}$. A is wrong.
B. The Coulomb contribution to the binding energy is $-a_c\frac{Z(Z-1)}{A^{1/3}}$; $\frac{Z(Z-1)}{A^{4/3}}$ is that contribution per nucleon, which B does not say. B is wrong.
C. The volume energy is proportional to the number of nucleons: $b_v = a_3A$. C is right.
D. Surface nucleons have fewer neighbours, so the binding energy is reduced by an amount proportional to the surface area ($\propto R^2 \propto A^{2/3}$). D is right.
E. The count of 12 nearest neighbours (close packing) is the assumption behind the volume energy; for the surface energy the point is that surface nucleons have fewer than 12. E is wrong.
Correct: C and D only.
Answer: (C). | ||||
| 44 | 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 $\lambda = \frac{h}{p}$ depends only on the momentum. Equal wavelengths need equal momenta, whatever the masses.
$p_p : p_e = 1 : 1$.
Answer: (D). | ||||
| 45 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics In a reflecting telescope the primary mirror brings the light to a focus inside the tube, where the observer would block the light. A small secondary mirror (as in the Cassegrain design) reflects the converging beam through a hole in the primary or to the side, so the eyepiece is outside the tube.
Spherical aberration is removed by a paraboloidal primary, a mirror has no chromatic aberration anyway, and mechanical support concerns the size of the primary.
Answer: (B). | ||||
| 46 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$, so $\frac{1}{\mu_0\epsilon_0} = c^2$, with dimensions $\left[\frac{L}{T}\right]^2 = $ L$^2$/T$^2$.
Answer: (C). | ||||
| 47 | Electromagnetic Induction and Alternating CurrentsLenz's law and eddy currents | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lenz's law and eddy currents, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current When the coil moves, the metallic core cuts magnetic flux and eddy currents are set up in it. By Lenz's law they oppose the motion, damping the oscillations (electromagnetic damping), so the coil comes to rest quickly — a dead-beat galvanometer.
A non-magnetic metal does not concentrate the field, so it neither makes the field radial nor increases the torque.
Answer: (D). | ||||
| 48 | 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 force $q\vec{v} \times \vec{B}$ acts only on the perpendicular component, turning it into uniform circular motion about the field direction; the parallel component is unaffected and carries the particle steadily along $\vec{B}$.
Together: a helix whose axis is along $\vec{B}$.
Answer: (C). | ||||
| 49 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity In the steady state no current flows through the capacitors. The only conducting path from A to D is A → $6\,\Omega$ → B → G ($2\,\Omega$) → C → $8\,\Omega$ → D.
$I = \frac{4}{6 + 2 + 8} = 0.25$ A.
$C_1$ is across A and C: $V_{AC} = I(6 + 2) = 2$ V.
$C_2$ is across B and D: $V_{BD} = I(2 + 8) = 2.5$ V.
Ratio $= \frac{2}{2.5} = \frac45$.
Answer: (A). | ||||
| 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 By Gauss's law, inside a uniformly charged solid sphere $E = \frac{kQr}{R^3}$, rising linearly from zero at the centre; outside $E = \frac{kQ}{r^2}$, falling as $\frac{1}{r^2}$. The field is greatest, $\frac{kQ}{R^2}$, at $r = R$.
That is the first graph: a straight rise to $r = R$, then a $\frac{1}{r^2}$ fall.
Answer: (A). | ||||
| 51 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves For a pipe open at both ends, $f_n = \frac{nv}{2L}$. Second harmonic: $f_2 = \frac{2 \times 360}{2 \times 0.4} = 900$ Hz.
Answer: 900. | ||||
| 52 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical 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 At the steady (terminal) speed, buoyancy balances viscous drag (the bubble's weight is neglected):
$\frac43\pi r^3\rho g = 6\pi\eta rv \Rightarrow \eta = \frac{2r^2\rho g}{9v}$.
$r = 3 \times 10^{-3}$ m, $v = 3.5 \times 10^{-3}$ m/s:
$\eta = \frac{2 \times 9 \times 10^{-6} \times 1750 \times 10}{9 \times 3.5 \times 10^{-3}} = \frac{0.315}{0.0315} = 10$ Pa s.
Answer: 10. | ||||
| 53 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Every element of the semicircular ring is at distance R from the axis through the centre perpendicular to its plane, so $I = \sum m_i R^2 = MR^2$.
$\frac{1}{x} = 1$, so $x = 1$.
Answer: 1. | ||||
| 54 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $K = \frac{p^2}{2m}$. With $p' = 1.5p$, $K' = 2.25K$.
Increase $= (2.25 - 1) \times 100\% = 125\%$.
Answer: 125. | ||||
| 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 Initial binding energy: $242 \times 7.6 = 1839.2$ MeV. Final: $2 \times 121 \times 8.1 = 1960.2$ MeV.
Gain $= 1960.2 - 1839.2 = 121$ MeV (equivalently $242 \times 0.5$).
Answer: 121. | ||||
| 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 Put mirror A at $x = 0$, mirror B at $x = 10$ cm and O at $x = 2$ cm.
Behind A, the images are: O reflected in A, at $x = -2$ (2 cm behind A); O reflected first in B (at $x = 18$) and then in A, at $x = -18$ (18 cm behind A); the next ones are farther still (O → A → B → A gives $x = -22$).
The second nearest image behind A is 18 cm from A.
Answer: 18. | ||||
| 57 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{75 \times 10^{-3} \times 1.2 \times 10^{-6}}} = \frac{1}{\sqrt{9 \times 10^{-8}}} = \frac{10^4}{3}$ rad/s.
$I_{max} = \omega Q_{max} = \frac{10^4}{3} \times 2.7 \times 10^{-6} = 9 \times 10^{-3}$ A $= 9$ mA.
Answer: 9. | ||||
| 58 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Inside a long solenoid $H = nI$, with $n = 8$ per cm $= 800$ m$^{-1}$.
$I = \frac{1.6 \times 10^3}{800} = 2$ A.
Answer: 2. | ||||
| 59 | Current ElectricityDrift velocity and mobility | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity $v_d = \frac{I}{neA} = \frac{2}{2 \times 10^{28} \times 1.6 \times 10^{-19} \times 25 \times 10^{-6}} = \frac{2}{8 \times 10^4} = 2.5 \times 10^{-5}$ m s$^{-1}$ $= 25 \times 10^{-6}$ m s$^{-1}$.
Answer: 25. | ||||
| 60 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $W = pE(\cos\theta_1 - \cos\theta_2) = pE(\cos 0^\circ - \cos 180^\circ) = 2pE$.
$W = 2 \times 6 \times 10^{-6} \times 1.5 \times 10^3 = 1.8 \times 10^{-2}$ J $= 18$ mJ.
Answer: 18. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Iodine atoms: left $2 + x$, right $6 \times 2 = 12$, so $x = 10$.
Check by electrons: each $\mathrm{IO_3^-}$ (I at +5) gains 5 electrons, 2 of them gain 10; each $\mathrm{I^-}$ loses 1, so 10 $\mathrm{I^-}$ are needed. Charge: $-2 - 10 + 12 = 0$ on both sides.
Answer: (B). | ||||
| 62 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Anode (oxidation): $\frac12\mathrm{H_2} \rightarrow \mathrm{H^+} + e^-$, a hydrogen electrode, $\mathrm{Pt\,|\,H_2(g)}$.
Cathode (reduction): $\mathrm{AgCl(s)} + e^- \rightarrow \mathrm{Ag(s)} + \mathrm{Cl^-}$, a silver–silver chloride electrode, $\mathrm{AgCl(s)\,|\,Ag}$.
Both $\mathrm{H^+}$ and $\mathrm{Cl^-}$ are formed in one solution, which is HCl: $\mathrm{Pt\,|\,H_2(g)\,|\,HCl(sol^n)\,|\,AgCl(s)\,|\,Ag}$.
(A) has no hydrogen electrode; (B) reduces $\mathrm{Ag^+}$, not AgCl; (D) uses KCl, where the $\mathrm{H^+}$ of the reaction would not be the electrolyte.
Answer: (C). | ||||
| 63 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich: $\frac{x}{m} = kp^{1/n}$ ($n \ge 1$).
A. $\frac{x}{m}$ against $p$ rises steeply and then levels off: the Freundlich curve. Represents it.
B. $\log\frac{x}{m} = \log k + \frac1n\log p$: a straight line of slope $\frac1n$ with intercept $\log k$ (the constant). Represents it.
C. $\frac{x}{m}$ against $c$ as a straight line with a positive intercept is not of the form $kc^{1/n}$, which passes through the origin. Does not.
D. $\frac{x}{m}$ against $p^{1/n}$ is a straight line through the origin of slope $k$. Represents it.
A, B, D only.
Answer: (C). | ||||
| 64 | Classification of Elements and PeriodicityElectronegativity | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on electronegativity, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Pauling electronegativities: Br 2.96, C 2.55, At about 2.2, P 2.19.
Br > C > At > P.
Answer: (B). | ||||
| 65 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter In the cyanide process, gold is leached from its ore with dilute NaCN (an alkaline solution) in the presence of air: $4\mathrm{Au} + 8\mathrm{CN^-} + 2\mathrm{H_2O} + \mathrm{O_2} \rightarrow 4[\mathrm{Au(CN)_2}]^- + 4\mathrm{OH^-}$, and the gold is then displaced by zinc.
Cu, Pb and Sn are obtained by roasting/smelting or reduction of their sulphide or oxide ores, not by alkali leaching.
Answer: (A). | ||||
| 66 | 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 Water gas (syngas) is a mixture of CO and $\mathrm{H_2}$. With a cobalt catalyst it is used to make methanol:
$\mathrm{CO(g)} + 2\mathrm{H_2(g)} \xrightarrow{\text{Co catalyst}} \mathrm{CH_3OH(l)}$.
Answer: (B). | ||||
| 67 | 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 About 2–3% gypsum ($\mathrm{CaSO_4\cdot 2H_2O}$) is added to cement clinker to slow down the setting of cement, so that it hardens properly and can be worked.
Answer: (C). | ||||
| 68 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | StatementsEasy | |
A easy statements question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Statement I: correct. Owing to their small, highly polarising cations, Li and Mg burn in air to give the normal oxides $\mathrm{Li_2O}$ and MgO (and nitrides); they do not form superoxides.
Statement II: correct. The ionic radius of $\mathrm{Li^+}$ is about 76 pm and that of $\mathrm{Mg^{2+}}$ about 72 pm, so $\mathrm{Li^+}$ is slightly larger (the two are close, which underlies the Li–Mg diagonal relationship).
Answer: (A). | ||||
| 69 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter The reaction needs $\mathrm{X^-}$ to reduce $\mathrm{Cu^{2+}}$ to $\mathrm{Cu^+}$. Only iodide is a strong enough reducing agent (and CuI is very insoluble):
$2\mathrm{Cu^{2+}} + 4\mathrm{I^-} \rightarrow \mathrm{Cu_2I_2(s)} + \mathrm{I_2}$. $\mathrm{Cl^-}$ and $\mathrm{Br^-}$ do not reduce $\mathrm{Cu^{2+}}$.
Answer: (B). | ||||
| 70 | 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 (A) $\mathrm{Co^{3+}}$, $3d^6$, with the strong-field ligand $\mathrm{CN^-}$: low spin $t_{2g}^6e_g^0$, no unpaired electron, diamagnetic; large crystal-field stabilisation and a +3 metal make it very stable.
(B) $\mathrm{Co^{2+}}$, $3d^7$, weak-field water: high spin, 3 unpaired electrons.
(C) $\mathrm{Ni^{2+}}$, $3d^8$: $t_{2g}^6e_g^2$, 2 unpaired electrons.
(D) $\mathrm{Co^{3+}}$ with weak-field $\mathrm{Cl^-}$: high spin, 4 unpaired electrons.
Answer: (A). | ||||
| 71 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Drinking-water limits: fluoride about 1 ppm (above 2 ppm it mottles teeth), so F$^-$ → III (< 2 ppm); sulphate < 500 ppm → IV; nitrate 50 ppm → I; zinc 5 ppm → II.
The correct matching is A-III, B-IV, C-I, D-II, which is not among the options.
Of the options, (D) A-III, B-II, C-I, D-IV is the only one with any correct pair (A and C); it interchanges sulphate and zinc.
Answer: (D) (closest; see note). | ||||
| 72 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[\mathrm{Fe(CN)_6}]^{3-}$: $\mathrm{Fe^{3+}}$ $d^5$, strong field, low spin $t_{2g}^5$: 1 unpaired, $\mu = \sqrt3 = 1.73$ BM.
$[\mathrm{Mn(CN)_6}]^{3-}$: $\mathrm{Mn^{3+}}$ $d^4$, low spin $t_{2g}^4$: 2 unpaired, $\mu = \sqrt8 = 2.83$ BM.
$[\mathrm{CoF_6}]^{3-}$: $\mathrm{Co^{3+}}$ $d^6$, weak field, high spin: 4 unpaired, $\mu = \sqrt{24} = 4.90$ BM.
$[\mathrm{MnBr_4}]^{2-}$: $\mathrm{Mn^{2+}}$ $d^5$, tetrahedral, high spin: 5 unpaired, $\mu = \sqrt{35} = 5.92$ BM.
Order: $[\mathrm{Fe(CN)_6}]^{3-} < [\mathrm{Mn(CN)_6}]^{3-} < [\mathrm{CoF_6}]^{3-} < [\mathrm{MnBr_4}]^{2-}$.
Answer: (B). | ||||
| 73 | 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 $\mathrm{S_N1}$ reactivity follows the stability of the carbocation left when the halide departs.
A. $\mathrm{Br_{(a)}}$ is on the carbon next to the benzene ring: its loss gives a resonance-stabilised benzylic cation; $\mathrm{Br_{(b)}}$ gives an ordinary secondary cation. → $\mathrm{Br_{(a)}}$.
B. $\mathrm{I_{(a)}}$ is on a $\mathrm{CH_2}$ next to the C=C (allylic): an allylic cation, resonance-stabilised; $\mathrm{I_{(b)}}$ is a plain primary iodide. → $\mathrm{I_{(a)}}$.
C. $\mathrm{Br_{(a)}}$ is at a bridgehead of the bicyclic ring: the cation there cannot become planar, so it hardly forms. $\mathrm{Br_{(b)}}$ is on a tertiary carbon (bearing Me) and gives a tertiary cation. → $\mathrm{Br_{(b)}}$.
D. $\mathrm{Br_{(a)}}$ is tertiary, $\mathrm{Br_{(b)}}$ primary. → $\mathrm{Br_{(a)}}$.
A – $\mathrm{Br_{(a)}}$; B – $\mathrm{I_{(a)}}$; C – $\mathrm{Br_{(b)}}$; D – $\mathrm{Br_{(a)}}$.
Answer: (A). | ||||
| 74 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $\mathrm{LiBH_4}$ is a milder reducing agent than $\mathrm{LiAlH_4}$: it reduces esters to primary alcohols but does not reduce free carboxylic acids (the acid is only deprotonated to its carboxylate, which $\mathrm{H_3O^+}$ restores).
So the $-\mathrm{CO_2Et}$ chain becomes $-\mathrm{CH_2OH}$ and the $-\mathrm{CO_2H}$ chain is unchanged.
The product keeps $\mathrm{CO_2H}$ on the upper chain and has OH on the lower chain: the third structure.
Answer: (C). | ||||
| 75 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Assertion and reasonEasy | |
A easy assertion and reason question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Butan-1-ol (b.p. about 391 K) and ethoxyethane (b.p. about 308 K) are isomers, $\mathrm{C_4H_{10}O}$, of the same molar mass. The alcohol has an O–H group and its molecules are associated through intermolecular hydrogen bonds, which must be broken on boiling; the ether has no O–H and no such association.
A is true, R is true, and R explains A.
Answer: (A). | ||||
| 76 | AminesDiazonium salts | Class 12: Amines | Matching listEasy | |
A easy matching list question on diazonium salts, from Amines. NCERT: Class 12: Amines List II: I is fluorobenzene, II benzonitrile, III p-aminoazobenzene ($\mathrm{C_6H_5{-}N{=}N{-}C_6H_4{-}NH_2}$), IV chlorobenzene.
A. Aniline couples with the diazonium ion at its para position: p-aminoazobenzene → III.
B. $\mathrm{HBF_4}$ gives the diazonium fluoroborate, which on heating gives fluorobenzene (Balz–Schiemann) → I.
C. Cu/HCl replaces $\mathrm{N_2^+}$ by Cl (Gattermann) → IV.
D. CuCN/KCN replaces $\mathrm{N_2^+}$ by CN (Sandmeyer) → II.
A-III, B-I, C-IV, D-II.
Answer: (B). | ||||
| 77 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Cysteine ($\mathrm{HS{-}CH_2{-}CH(NH_2)COOH}$) has a thiol group and methionine ($\mathrm{CH_3S{-}CH_2CH_2{-}CH(NH_2)COOH}$) a thioether group. Isoleucine, lysine and glutamic acid contain no sulphur.
b, d.
Answer: (B). | ||||
| 78 | Chemistry in Everyday LifeChemicals in food | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on chemicals in food, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Saccharin was the first artificial sweetening agent → II.
Aspartame is unstable at cooking temperature (used only in cold foods and drinks) → IV.
Alitame is a high-potency sweetener (about 2000 times as sweet as sucrose) → I.
Sucralose is stable at cooking temperature → III.
A-II, B-IV, C-I, D-III.
Answer: (A). | ||||
| 79 | Practical Organic ChemistryDetection of functional groups | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on detection of functional groups, from Practical Organic Chemistry. NCERT: No matching NCERT chapter List II: I is 1-phenylethanol $\mathrm{C_6H_5CH(OH)CH_3}$; II isopropylamine $\mathrm{(CH_3)_2CHNH_2}$; III phenylacetaldehyde $\mathrm{C_6H_5CH_2CHO}$; IV 3-ethylphenol.
A. Alkaline copper sulphate with sodium citrate is Benedict's reagent, which detects an aliphatic aldehyde → III.
B. Neutral $\mathrm{FeCl_3}$ gives a violet colour with phenols → IV.
C. Chloroform with alkali (carbylamine test) detects a primary amine → II.
D. KI with NaOCl is the iodoform test, given by the $\mathrm{CH_3CH(OH)-}$ group → I.
A-III, B-IV, C-II, D-I.
Answer: (D). | ||||
| 80 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In chromyl chloride, $\mathrm{CrO_2Cl_2}$: $x + 2(-2) + 2(-1) = 0$, so Cr is +6, $3d^0$.
Mn(VII): $25 - 7 = 18$ electrons, $3d^0$. Fe(III) is $3d^5$, V(IV) $3d^1$, Ti(III) $3d^1$.
Answer: (A). | ||||
| 81 | 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 At constant temperature the amount of each gas is proportional to $PV$: $\mathrm{CH_4}$ $2 \times 2 = 4$, $\mathrm{CO_2}$ $4 \times 3 = 12$, Ne $3 \times 4 = 12$ (atm L).
After mixing the gases fill $2 + 3 + 4 = 9$ L, so $P = \frac{4 + 12 + 12}{9} = \frac{28}{9} = 3.11$ atm (Dalton's law: the sum of the partial pressures $\frac49 + \frac{12}{9} + \frac{12}{9}$).
Nearest integer: 3.
Answer: 3. | ||||
| 82 | 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 (A) Correct: each element's line spectrum is characteristic of its electronic energy levels.
(B) Incorrect: atoms in the gas phase emit only certain wavelengths, giving a line spectrum, not a continuous spread from red to violet.
(C) Correct: an absorption spectrum shows dark lines where the emission spectrum shows bright ones.
(D) Correct: helium was discovered through its lines in the solar spectrum.
One statement is incorrect.
Answer: 1. | ||||
| 83 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure By Fajans' rules, covalent character in an ionic bond arises from the cation distorting the electron cloud of the anion. It depends on the polarising power of the cation (A), the polarisability of the anion (C), and so on the extent to which the anion is distorted (B).
The polarising power of the anion (D) is not a factor: the small cation is hardly polarised.
Three factors: A, B, C.
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 Heat supplied $q = 60 \times 100 = 6000$ J, all of it to the gas (adiabatic walls).
Heat capacity $C = \frac{q}{\Delta T} = \frac{6000}{5} = 1200$ J K$^{-1}$ (at constant volume).
Answer: 1200. | ||||
| 85 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions A liquid boils when its vapour pressure reaches 1 atm. A non-volatile solute lowers the vapour pressure, so the solution's curve lies below the solvent's and reaches 1 atm at a higher temperature.
The upper curve (solvent) reaches 1 atm at 82 $^\circ$C; the lower one (solution) at 83 $^\circ$C.
Boiling point of the solvent: 82 $^\circ$C.
Answer: 82. | ||||
| 86 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium $\mathrm{p}K_{In} = -\log(4 \times 10^{-10}) = 10 - 2\log 2 = 9.4$. The colour changes over $\mathrm{p}K_{In} \pm 1$, i.e. pH 8.4 to 10.4.
A. Incorrect: a weak acid–weak base titration has no sharp pH jump through this range, so no indicator (phenolphthalein included) is suitable.
B. Correct: the colour change begins at pH $9.4 - 1 = 8.4$.
C. Incorrect: phenolphthalein is a weak organic acid.
D. Correct: it is colourless in acid (and pink in base).
Two statements are correct.
Answer: 2. | ||||
| 87 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Arrhenius: $\frac{d\ln k}{dT} = \frac{E_a}{RT^2}$.
(A) Correct: a larger $E_a$ means a steeper dependence of $k$ on $T$.
(B) Correct: with $E_a = 0$, $k = A$, independent of temperature (taking A as constant).
(C) Incorrect: the opposite of (A).
(D) Incorrect: no dependence of $k$ on $T$ means $E_a = 0$, not negative.
Two statements are correct.
Answer: 2. | ||||
| 88 | 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 $\mathrm{SbF_5}$ is a fluoride-ion acceptor: $\mathrm{XeF_4} + \mathrm{SbF_5} \rightarrow [\mathrm{XeF_3}]^+[\mathrm{SbF_6}]^-$.
$m = 3$, $n = 1$, $y = 6$, $z = 1$: $m + n + y + z = 11$.
Answer: 11. | ||||
| 89 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons One mole of $\mathrm{O_3}$ means one C=C. Joining the carbonyl carbons of ethanal ($\mathrm{CH_3CHO}$) and propanone ($\mathrm{(CH_3)_2CO}$) by a double bond gives X: $\mathrm{CH_3CH{=}C(CH_3)_2}$, 2-methylbut-2-ene, $\mathrm{C_5H_{10}}$.
Molar mass $= 5 \times 12 + 10 \times 1 = 70$ g mol$^{-1}$.
Answer: 70. | ||||
| 90 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy 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 Mass of carbon $= 0.6 \times 0.5 = 0.3$ g. Each 12 g of C gives 44 g of $\mathrm{CO_2}$:
mass of $\mathrm{CO_2} = 0.3 \times \frac{44}{12} = 1.1$ g $= 11 \times 10^{-1}$ g.
Answer: 11. | ||||
Sit this paper free.
Timed like the exam or chapter by chapter, with a solution for each question. Run by a non-profit trust.