JEE (Main) · 2023
JEE (Main) 2023 — 8 April, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Sets, Relations and Functions3
- Integral Calculus3
- Three Dimensional Geometry3
6 easy · 22 medium · 2 hard
Physics
30 Q
- Oscillations and Waves2
- Properties of Bulk Matter2
- Gravitation2
26 easy · 4 medium · 0 hard
Chemistry
30 Q
- s-Block Elements2
- Coordination Compounds2
- Principles Related to Practical Chemistry2
20 easy · 9 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $R = \{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\}$.
Not reflexive: $(1,1) \notin R$ since $1 + 1 \ne 7$.
Symmetric: $x + y = 7 \Rightarrow y + x = 7$.
Not transitive: $(1,6) \in R$ and $(6,1) \in R$ but $(1,1) \notin R$.
Answer: (B). | ||||
| 2 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Write $s = \sin\theta$. $\frac{1 + 2is}{1 - is} = \frac{(1 + 2is)(1 + is)}{1 + s^2} = \frac{(1 - 2s^2) + 3is}{1 + s^2}$.
Purely imaginary: real part $1 - 2s^2 = 0$, so $\sin^2\theta = \frac{1}{2}$ (and then the imaginary part $3s \ne 0$).
In $(0, 2\pi)$: $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$, whose sum is $4\pi$.
Answer: (C). | ||||
| 3 | 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 A homogeneous system has a non-trivial solution exactly when its determinant is zero:
$\begin{vmatrix}1&1&\sqrt{3}\\-1&\tan\theta&\sqrt{7}\\1&1&\tan\theta\end{vmatrix} = 0$.
$R_3 \to R_3 - R_1$ gives the row $(0, 0, \tan\theta - \sqrt{3})$, so the determinant is $(\tan\theta - \sqrt{3})(\tan\theta + 1)$.
$\tan\theta = \sqrt{3}$: $\theta = \frac{\pi}{3}, -\frac{2\pi}{3}$. $\tan\theta = -1$: $\theta = -\frac{\pi}{4}, \frac{3\pi}{4}$.
$\sum\theta = \frac{\pi}{3} - \frac{2\pi}{3} - \frac{\pi}{4} + \frac{3\pi}{4} = \frac{\pi}{6}$, and $\frac{120}{\pi}\cdot\frac{\pi}{6} = 20$.
Answer: (B). | ||||
| 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 The characteristic equation of A is $A^2 - 11A + (10 - 5\lambda)I = 0$ (trace 11, determinant $10 - 5\lambda$).
Multiplying by $A^{-1}$: $A - 11I + (10 - 5\lambda)A^{-1} = 0$, so $A^{-1} = \frac{-1}{10 - 5\lambda}A + \frac{11}{10 - 5\lambda}I$.
$\alpha + \beta = \frac{10}{10 - 5\lambda} = -2$ gives $10 - 5\lambda = -5$, $\lambda = 3$.
Then $\alpha = \frac{1}{5}$, $\beta = -\frac{11}{5}$ and $4\alpha^2 + \beta^2 + \lambda^2 = \frac{4}{25} + \frac{121}{25} + 9 = 5 + 9 = 14$.
Answer: (C). | ||||
| 5 | 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 MATHEMATICS has 11 letters: M, A, T twice each and H, E, I, C, S once.
All words: $\frac{11!}{2!\,2!\,2!} = 4989600$.
C and S together (as CS or SC, one block): $2 \cdot \frac{10!}{2!\,2!\,2!} = 907200$.
Not together: $4989600 - 907200 = 4082400 = 720 \times 5670$, so $k = 5670$.
Answer: (D). | ||||
| 6 | 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 By 17: $(25^{190} - 8^{190}) - (19^{190} - 2^{190})$; each bracket is $a^n - b^n$ with $a - b = 17$, so 17 divides it.
By 2: $25^{190}$ and $19^{190}$ are odd, $8^{190}$ and $2^{190}$ even, so odd − odd − even + even is even. Hence 34 divides the number.
By 7: $25 \equiv 4$, $19 \equiv 5$, $8 \equiv 1$, $2 \equiv 2 \pmod 7$. Since $4^3 \equiv 1$, $5^6 \equiv 1$, $2^3 \equiv 1$ and $190 = 3 \cdot 63 + 1 = 6 \cdot 31 + 4$:
$4^{190} \equiv 4$, $5^{190} \equiv 5^4 \equiv 2$, $1^{190} = 1$, $2^{190} \equiv 2$. The number $\equiv 4 - 2 - 1 + 2 = 3 \pmod 7$, so 7 (and hence 14) does not divide it.
Answer: (C). | ||||
| 7 | 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 $T_{r+1} = {}^{11}C_r (2x^2)^{11-r}\left(\frac{1}{2x}\right)^r = {}^{11}C_r\, 2^{11-2r} x^{22-3r}$.
$x^{10}$: $r = 4$, coefficient ${}^{11}C_4 \cdot 2^3 = 330 \times 8 = 2640$.
$x^7$: $r = 5$, coefficient ${}^{11}C_5 \cdot 2 = 462 \times 2 = 924$.
Difference $= 1716 = 12^3 - 12$.
Answer: (B). | ||||
| 8 | 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 The differences $3, 6, 9, 12, 15, \ldots$ are in AP, so $a_n = 5 + 3(1 + 2 + \ldots + (n-1)) = 5 + \frac{3n(n-1)}{2}$.
$a_{40} = 5 + \frac{3 \cdot 40 \cdot 39}{2} = 2345$.
$S_{30} = 150 + \frac{3}{2}\sum_{n=1}^{30} n(n-1) = 150 + \frac{3}{2}\cdot\frac{29 \cdot 30 \cdot 31}{3} = 150 + 13485 = 13635$.
$S_{30} - a_{40} = 13635 - 2345 = 11290$.
Answer: (C). | ||||
| 9 | 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 Dividing $ax^2 + bx + 1 = 0$ by $x^2$ shows that the roots of $x^2 + bx + a = 0$ are $\frac{1}{\alpha}, \frac{1}{\beta}$, so $x^2 + bx + a = \left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)$.
As $x \to \frac{1}{\alpha}$ this tends to 0, and $1 - \cos u \approx \frac{u^2}{2}$. Also $1 - \alpha x = -\alpha\left(x - \frac{1}{\alpha}\right)$.
The expression becomes $\left(\frac{(x - \frac{1}{\alpha})^2(x - \frac{1}{\beta})^2}{4\alpha^2(x - \frac{1}{\alpha})^2}\right)^{1/2} = \frac{|x - \frac{1}{\beta}|}{2\alpha} \to \frac{1}{2\alpha}\left(\frac{1}{\beta} - \frac{1}{\alpha}\right)$ (since $\frac{1}{\beta} > \frac{1}{\alpha}$).
So $k = 2\alpha$.
Answer: (C). | ||||
| 10 | Integral CalculusIntegral as an antiderivative | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Differentiate the options. $\frac{d}{dx}\left(\frac{x}{2}\right)^x = \left(\frac{x}{2}\right)^x\left(\ln\frac{x}{2} + 1\right)$ and $\frac{d}{dx}\left(\frac{2}{x}\right)^x = \left(\frac{2}{x}\right)^x\left(\ln\frac{2}{x} - 1\right) = -\left(\frac{2}{x}\right)^x\left(\ln\frac{x}{2} + 1\right)$.
So $\frac{d}{dx}\left[\left(\frac{x}{2}\right)^x - \left(\frac{2}{x}\right)^x\right] = \left[\left(\frac{x}{2}\right)^x + \left(\frac{2}{x}\right)^x\right]\left(1 + \ln\frac{x}{2}\right)$.
This has the bracket of the integrand, but with the factor $1 + \ln\frac{x}{2}$ where the integrand has $\log_2 x$; options (A), (C), (D) do not even produce the bracket.
$1 + \ln\frac{x}{2} \ne \log_2 x$ (they agree at $x = 2$, but at $x = 4$ they are about $1.69$ and $2$), so none of the options is an antiderivative of the printed integrand.
No option is correct as printed. | ||||
| 11 | CirclesEquation from a diameter | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on equation from a diameter, from Circles. NCERT: Class 11: Conic Sections The centre of the circle is $C(3, -2)$. The radii CP and CQ are perpendicular to the tangents OP and OQ, so $\angle OPC = \angle OQC = 90^\circ$ and O, P, C, Q lie on the circle with diameter OC.
That circle: $x(x - 3) + y(y + 2) = 0$, i.e. $x^2 + y^2 - 3x + 2y = 0$.
Through $\left(\alpha, \frac{1}{2}\right)$: $\alpha^2 + \frac{1}{4} - 3\alpha + 1 = 0$, i.e. $4\alpha^2 - 12\alpha + 5 = 0$, so $\alpha = \frac{1}{2}$ or $\frac{5}{2}$.
Answer: (D). | ||||
| 12 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Each vertex is the sum of the two mid-points next to it minus the opposite one: $(0, 2)$, $(0, 0)$, $(2, 0)$. The triangle is right-angled at the origin with legs 2, 2 and hypotenuse $2\sqrt{2}$.
Inradius $r = \frac{2 + 2 - 2\sqrt{2}}{2} = 2 - \sqrt{2}$, so $D = (2 - \sqrt{2}, 2 - \sqrt{2})$.
$y^2 = 4ax$ through D: $(2 - \sqrt{2})^2 = 4a(2 - \sqrt{2})$, so $a = \frac{2 - \sqrt{2}}{4} = \frac{1}{2} - \frac{1}{4}\sqrt{2}$.
The focus is $(a, 0)$: $\alpha = \frac{1}{2}$, $\beta = -\frac{1}{4}$, and $\frac{\alpha}{\beta^2} = \frac{1/2}{1/16} = 8$.
Answer: (C). | ||||
| 13 | 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 plane contains the point $(1, 2, -5)$, the direction $(1, -3, 7)$ and the vector $(2, 4, -3) - (1, 2, -5) = (1, 2, 2)$.
Normal: $(1, -3, 7) \times (1, 2, 2) = (-20, 5, 5) \parallel (4, -1, -1)$. Plane: $4(x - 1) - (y - 2) - (z + 5) = 0$, i.e. $4x - y - z = 7$.
Image of $(-1, 3, 4)$: $\frac{x + 1}{4} = \frac{y - 3}{-1} = \frac{z - 4}{-1} = \frac{-2(-4 - 3 - 4 - 7)}{18} = 2$, giving $(7, 1, 2)$.
$\alpha + \beta + \gamma = 10$.
Answer: (B). | ||||
| 14 | Three Dimensional GeometryAngle between two lines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on angle between two lines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The line has direction $(1, -1, 1)$; the plane has normal $(a, 1, -1)$.
If the angle between line and plane is $\theta = \cos^{-1}\frac{1}{3}$, then $\sin\theta = \frac{2\sqrt{2}}{3} = \frac{|a - 1 - 1|}{\sqrt{3}\sqrt{a^2 + 2}}$.
$9(a - 2)^2 = 24(a^2 + 2)$ gives $5a^2 + 12a + 4 = 0$, so $a = -2$ or $-\frac{2}{5}$; as $a \in \mathbb{Z}$, $a = -2$.
Distance of $(6, -6, 4)$ from $-2x + y - z = b$: $\frac{|-12 - 6 - 4 - b|}{\sqrt{6}} = 3\sqrt{6}$, so $|22 + b| = 18$, $b = -4$ or $-40$. $|a - b| \le 10$ leaves $b = -4$.
$a^4 + b^2 = 16 + 16 = 32$.
Answer: (B). | ||||
| 15 | 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 The four points are coplanar: with $\vec{AB} = (-1, 1, 4)$, $\vec{AC} = (-4, -4, 4)$, $\vec{AD} = (-1, -7, -8)$, $\vec{AC} \times \vec{AD} = (60, -36, 24)$ and $\vec{AB}\cdot(60, -36, 24) = -60 - 36 + 96 = 0$.
Area of a plane quadrilateral $= \frac{1}{2}|\vec{AC} \times \vec{BD}|$ with $\vec{BD} = (0, -8, -12)$:
$\vec{AC} \times \vec{BD} = (48 + 32, 0 - 48, 32 - 0) = (80, -48, 32) = 16(5, -3, 2)$, of length $16\sqrt{38}$.
Area $= 8\sqrt{38}$.
Answer: (B). | ||||
| 16 | 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 $\vec{u}$'s coplanar: $\begin{vmatrix}1&1&a\\1&b&1\\c&1&1\end{vmatrix} = (b - 1) - (1 - c) + a(1 - bc) = a + b + c - abc - 2 = 0$.
$\vec{v}$'s coplanar: $\begin{vmatrix}a+b&c&c\\a&b+c&a\\b&b&c+a\end{vmatrix} = (a + b)\,c(a + b + c) - c\,a(a + c - b) + c\,b(a - b - c) = 4abc = 0$.
So $abc = 0$ and then $a + b + c = 2$: $6(a + b + c) = 12$.
Answer: (D). | ||||
| 17 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $\sum_{x \ge 0}(x + 1)t^x = \frac{1}{(1 - t)^2}$; with $t = \frac{1}{3}$ the total is $k \cdot \frac{9}{4} = 1$, so $k = \frac{4}{9}$.
$P(0) = \frac{4}{9}$, $P(1) = \frac{4}{9}\cdot\frac{2}{3} = \frac{8}{27}$.
$P(X \ge 2) = 1 - \frac{12}{27} - \frac{8}{27} = \frac{7}{27}$.
Answer: (C). | ||||
| 18 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $\sum x = 12 \cdot \frac{9}{2} = 54$; $\sum x^2 = 12\left(4 + \frac{81}{4}\right) = 291$.
Corrected: $\sum x = 54 - 19 + 21 = 56$, $\sum x^2 = 291 - (81 + 100) + (49 + 196) = 355$.
Variance $= \frac{355}{12} - \left(\frac{56}{12}\right)^2 = \frac{355}{12} - \frac{196}{9} = \frac{1065 - 784}{36} = \frac{281}{36}$.
$m + n = 281 + 36 = 317$.
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 $\sin 3\theta = \sin\theta(3 - 4\sin^2\theta) = \sin\theta(4\cos^2\theta - 1)$, so $4\cos^2\theta - 1 = \frac{\sin 3\theta}{\sin\theta}$.
The product telescopes: $\frac{\sin 27^\circ}{\sin 9^\circ}\cdot\frac{\sin 81^\circ}{\sin 27^\circ}\cdot\frac{\sin 243^\circ}{\sin 81^\circ}\cdot\frac{\sin 729^\circ}{\sin 243^\circ} = \frac{\sin 729^\circ}{\sin 9^\circ} = 1$, since $729^\circ = 720^\circ + 9^\circ$.
The value is $36 \times 1 = 36$.
Answer: (C). | ||||
| 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 $\sim[(p \wedge \sim q) \vee \sim p] = \sim(p \wedge \sim q) \wedge p = (\sim p \vee q) \wedge p = (\sim p \wedge p) \vee (q \wedge p) = p \wedge q$.
Answer: (B). | ||||
| 21 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Logarithm: $\frac{(2x + 1)(3x + 1)}{2x - 1} > 0$. Sign scheme with critical points $-\frac{1}{2}, -\frac{1}{3}, \frac{1}{2}$: $x \in \left(-\frac{1}{2}, -\frac{1}{3}\right) \cup \left(\frac{1}{2}, \infty\right)$.
Inverse cosine: $-1 \le \frac{2x^2 - 3x + 4}{3x - 5} \le 1$. The numerator is always positive (discriminant $9 - 32 < 0$).
If $3x - 5 > 0$: $\le 1$ needs $2x^2 - 6x + 9 \le 0$, impossible (discriminant $< 0$).
If $3x - 5 < 0$: the ratio is negative and $\ge -1$ needs $2x^2 - 3x + 4 \le 5 - 3x$, i.e. $x^2 \le \frac{1}{2}$, $|x| \le \frac{1}{\sqrt{2}}$.
Domain: $\left(-\frac{1}{2}, -\frac{1}{3}\right) \cup \left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right]$.
$18\left(\frac{1}{4} + \frac{1}{9} + \frac{1}{4} + \frac{1}{2}\right) = 18 \cdot \frac{10}{9} = 20$.
Answer: 20. | ||||
| 22 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations First equation: put $[x] = k$ (an integer). Then $x^2 - 12x + 31 + k = 0$ gives $x = 6 \pm \sqrt{5 - k}$, which needs $k \le 5$.
$x = 6 + \sqrt{5 - k} \ge 6 > k$, so $[x] \ne k$.
$x = 6 - \sqrt{5 - k}$ with $t = 5 - k \ge 0$: $[x] = k$ needs $k \le 6 - \sqrt{t} < k + 1$, i.e. $\sqrt{t} \le t + 1$ and $\sqrt{t} > t$. The second needs $0 < t < 1$, which no integer satisfies. So $m = 0$.
Second equation: for $x \ge -2$, $x^2 - 5x - 14 = 0$, $x = 7$ or $-2$ (both allowed). For $x < -2$, $x^2 + 5x + 6 = 0$, $x = -3$ (and $-2$, not $< -2$). So $n = 3$ (roots $7, -2, -3$).
$m^2 + mn + n^2 = 0 + 0 + 9 = 9$.
Answer: 9. | ||||
| 23 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Onto functions from a 5-set to a 4-set: one value is taken twice. Choose the pair of elements sharing a value, ${}^5C_2 = 10$, and assign the 4 blocks to 4 values, $4! = 24$: 240 in all.
By symmetry among the values, $f(a)$ equals each of 1, 2, 3, 4 in a quarter of them, so $f(a) = 1$ in 60.
Required number $= 240 - 60 = 180$.
Answer: 180. | ||||
| 24 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Dividing the last condition by $xyz$: $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{\sqrt{2}}$. In AP, $\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$, so $\frac{3}{y} = \frac{3}{\sqrt{2}}$ and $y = \sqrt{2}$.
GP: $xz = (\sqrt{2}y)^2 = 4$. AP: $\frac{x + z}{xz} = \frac{2}{y} = \sqrt{2}$, so $x + z = 4\sqrt{2}$.
(Then $x, z = 2\sqrt{2} \pm 2$, and indeed $z < y < x$.)
$x + y + z = 5\sqrt{2}$ and $3(x + y + z)^2 = 3 \times 50 = 150$.
Answer: 150. | ||||
| 25 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Continuity at 1: $3 + k\sqrt{2} = m + k^2$.
Equal derivatives at 1: $6 + \frac{k}{2\sqrt{2}} = 2m$, so $m = 3 + \frac{k}{4\sqrt{2}}$.
Substituting: $k\sqrt{2} - \frac{k}{4\sqrt{2}} = k^2$, so ($k > 0$) $k = \frac{7}{4\sqrt{2}}$ and $m = 3 + \frac{7}{32} = \frac{103}{32}$.
$f'(8) = 16m$. $f'\left(\frac{1}{8}\right) = \frac{6}{8} + \frac{k}{2\sqrt{9/8}} = \frac{3}{4} + \frac{k\sqrt{2}}{3} = \frac{3}{4} + \frac{7}{12} = \frac{4}{3}$.
$\frac{8 \cdot 16m}{4/3} = 96m = 96 \cdot \frac{103}{32} = 309$.
Answer: 309. | ||||
| 26 | 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 $[x^2]$ changes at $x = 1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}$ (and $\sqrt{6} > 2.4$):
$\int_0^{2.4}[x^2]dx = 0 + 1(\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3}) + 4(\sqrt{5} - 2) + 5(2.4 - \sqrt{5})$
$= (-1 + 6 - 8 + 12) + (1 - 2)\sqrt{2} + (2 - 3)\sqrt{3} + (4 - 5)\sqrt{5} = 9 - \sqrt{2} - \sqrt{3} - \sqrt{5}$.
$\alpha + \beta + \gamma + \delta = 9 - 1 - 1 - 1 = 6$.
Answer: 6. | ||||
| 27 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals On $[0, 1)$: $1 + [x] = 1$, and $x^2 + \frac{3}{4} \le 1$ for $x \le \frac{1}{2}$, so $f = x^2 + \frac{3}{4}$ on $[0, \frac{1}{2}]$ and $f = 1$ on $[\frac{1}{2}, 1)$. Here the line $y = 2 - x \ge 1$ lies above f.
On $[1, 2]$: $f \ge \frac{7}{4}$ while $2 - x \le 1$, so the line bounds the region.
$A = \int_0^{1/2}\left(x^2 + \frac{3}{4}\right)dx + \frac{1}{2} \cdot 1 + \int_1^2(2 - x)dx = \left(\frac{1}{24} + \frac{3}{8}\right) + \frac{1}{2} + \frac{1}{2} = \frac{17}{12}$.
$12A = 17$.
Answer: 17. | ||||
| 28 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $t = \log_e(\cos y)$, so $dt = -\tan y\,dy$. The equation $t^2\cos y\,dx = (1 + 3xt)\sin y\,dy$ becomes $t^2\,dx = -(1 + 3xt)\,dt$, i.e. $\frac{dx}{dt} + \frac{3}{t}x = -\frac{1}{t^2}$.
Integrating factor $t^3$: $\frac{d}{dt}(xt^3) = -t$, so $xt^3 = -\frac{t^2}{2} + C$ and $x = -\frac{1}{2t} + \frac{C}{t^3}$.
At $y = \frac{\pi}{3}$, $t = -\log_e 2$ and $x = \frac{1}{2\log_e 2} + \frac{C}{t^3}$; the given value forces $C = 0$. So $x = -\frac{1}{2\log_e\cos y}$.
At $y = \frac{\pi}{6}$: $2\log_e\frac{\sqrt{3}}{2} = \log_e\frac{3}{4}$, so $x = \frac{1}{\log_e 4 - \log_e 3}$: $m = 4$, $n = 3$, $mn = 12$.
Answer: 12. | ||||
| 29 | Conic SectionsParabola | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The parabola is $y^2 = 12x$ ($a = 3$); points $(3t^2, 6t)$. Ordinates in ratio 3 : 1 means $t_1 = 3t_2 = 3t$.
Tangents at $t_1, t_2$ meet at $(at_1t_2, a(t_1 + t_2)) = (9t^2, 12t)$.
$\frac{\beta^2}{\alpha} = \frac{144t^2}{9t^2} = 16$.
Answer: 16. | ||||
| 30 | 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 $P_2$: vectors $(0, 1, -1)$ and $(3, 2, 1)$ from $(2, -1, 0)$; normal $(0, 1, -1) \times (3, 2, 1) = (3, -3, -3) \parallel (1, -1, -1)$. $P_2$: $x - y - z = 3$.
Line of intersection: direction $(3, -1, -7) \times (1, -1, -1) = (-6, -4, -2) \parallel (3, 2, 1)$; subtracting the planes, $2x - 6z = 8$, and $z = 0$ gives the point $(4, 1, 0)$.
A point of the line is $(4 + 3s, 1 + 2s, s)$. Perpendicular from $(7, 4, -1)$: $(3s - 3, 2s - 3, s + 1)\cdot(3, 2, 1) = 14s - 14 = 0$, $s = 1$.
Foot $(7, 3, 1)$: $\alpha + \beta + \gamma = 11$.
Answer: 11. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Matching listEasy | |
A easy matching list question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement A. Torque = force × distance: $MLT^{-2} \cdot L = ML^2T^{-2}$ (II).
B. Stress = force / area: $MLT^{-2}/L^2 = ML^{-1}T^{-2}$ (IV).
C. Pressure gradient = pressure / length: $ML^{-1}T^{-2}/L = ML^{-2}T^{-2}$ (I).
D. Viscosity, from $F = \eta A\frac{dv}{dx}$: $\eta = \frac{MLT^{-2}}{L^2 \cdot T^{-1}} = ML^{-1}T^{-1}$ (III).
A-II, B-IV, C-I, D-III.
Answer: (C). | ||||
| 32 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At $t = 0$, OP makes $30^\circ = \frac{\pi}{6}$ with the x-axis, and P moves anticlockwise (towards the y-axis, as the arrow shows).
So at time t the angle of OP with the x-axis is $\omega t + \frac{\pi}{6}$, and its projection on the x-axis is $x(t) = r\cos\left(\omega t + \frac{\pi}{6}\right)$.
Answer: (C). | ||||
| 33 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The average kinetic energy of a molecule is proportional to the absolute temperature.
Doubling it needs $T = 2 \times 300 = 600$ K $= 327^\circ$C.
Answer: (A). | ||||
| 34 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{400} = \frac{1}{4}$.
Heat taken from the source $Q_1 = \frac{W}{\eta} = \frac{2}{1/4} = 8$ kJ.
Answer: (C). | ||||
| 35 | Properties of Bulk MatterFluid mechanics | 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 fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter By Pascal's law the pressure is the same on both pistons.
$P = \frac{mg}{A} = \frac{5000 \times 10}{250 \times 10^{-4}} = \frac{5 \times 10^4}{2.5 \times 10^{-2}} = 2 \times 10^6$ Pa.
Answer: (D). | ||||
| 36 | 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 $g' = \frac{GM}{(R + h)^2} = g\left(1 + \frac{h}{R}\right)^{-2} \approx g\left(1 - \frac{2h}{R}\right)$ for $h \ll R$.
Answer: (A). | ||||
| 37 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation In a circular orbit of radius r, $v = \sqrt{\frac{GM}{r}}$, so $L = mvr = m\sqrt{GMr} \propto \sqrt{r}$.
Increased by eight times its initial value: $r' = r + 8r = 9r$, so $L' = \sqrt{9}\,L = 3L$.
Answer: (A). | ||||
| 38 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | StatementsEasy | |
A easy statements question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Statement I: $\int v\,dt$ is the displacement, not the distance (the two differ when the velocity changes sign). So it is incorrect.
Statement II: $\int a\,dt = \int dv = \Delta v$, the change in velocity. True.
Answer: (D). | ||||
| 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 $\frac{dy}{dx} = 1 - \frac{x}{10} = 0$ at $x = 10$ m.
$y_{max} = 10 - \frac{100}{20} = 5$ m.
Answer: (A). | ||||
| 40 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Momentum is conserved in the impact: $0.1 \times 400 = 4v$, $v = 10$ m/s.
Friction then stops the block: $v^2 = 2\mu g s$, so $100 = 2\mu \times 10 \times 20$ and $\mu = 0.25$.
Answer: (A). | ||||
| 41 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $V = \frac{kq}{r}$, so $r = \frac{9 \times 10^9 \times 5 \times 10^{-9}}{50} = \frac{45}{50} = 0.9$ m $= 90$ cm.
Answer: (A). | ||||
| 42 | Communication SystemsAntenna size and radiated power | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on antenna size and radiated power, from Communication Systems. NCERT: No matching NCERT chapter The power radiated by a linear antenna of length l is proportional to $\left(\frac{l}{\lambda}\right)^2$; this is why an antenna must be comparable in size to the wavelength, and why high-frequency (short-wavelength) carriers are used.
Answer: (C). | ||||
| 43 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits In the collector loop, taking $V_{CE} \approx 0$: $I_C = \frac{V_{CC}}{R_C} = \frac{1}{1000} = 1$ mA.
$I_b = \frac{I_C}{\beta} = \frac{1\text{ mA}}{100} = 10\ \mu$A.
Answer: (B). | ||||
| 44 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $\frac{1}{8} = \left(\frac{1}{2}\right)^3$ in 3 days, so the half-life is 1 day.
After 5 days, $N = \frac{N_0}{2^5} = \frac{N_0}{32} = 8$ g, so $N_0 = 256$ g.
Answer: (C). | ||||
| 45 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | StatementsEasy | |
A easy statements question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter A: true; the number of photoelectrons, and so the saturation current, is proportional to the intensity.
B: false; $K_{max} = h\nu - \phi$ does not depend on intensity.
C: true; $K_{max}$ rises linearly with frequency.
D: false; there is a threshold frequency, not a threshold intensity.
E: false (the opposite of C).
A and C only.
Answer: (C). | ||||
| 46 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Fringe width $\beta = \frac{\lambda D}{d} \propto \lambda$: $\beta' = 2 \times \frac{600}{400} = 3$ mm.
Answer: (B). | ||||
| 47 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Fast electrons stopped by a metal target produce X-rays (the X-ray tube: continuous bremsstrahlung plus characteristic lines).
Answer: (D). | ||||
| 48 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Motional emf $e = Bvl$: $v = \frac{e}{Bl} = \frac{0.08}{0.4 \times 0.1} = 2$ m s$^{-1}$.
Answer: (B). | ||||
| 49 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Assertion and reasonEasy | |
A easy assertion and reason question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter An electromagnet must become strongly magnetised when current flows (high permeability) and lose its magnetism when the current is switched off (low retentivity, low coercivity). Soft iron has exactly these properties, which is why electromagnets are made of it.
Both A and R are true and R explains A.
Answer: (A). | ||||
| 50 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity The upper wire joins A directly to D, and the lower wire joins C directly to B. So D is at the potential of A and C at that of B.
The first 20 k$\Omega$ (A to C) lies between A and B; the second 20 k$\Omega$ (C to D) lies between B and A; the 10 k$\Omega$ (D to B) lies between A and B.
All three are in parallel: $\frac{1}{R} = \frac{1}{20} + \frac{1}{20} + \frac{1}{10} = \frac{1}{5}$, $R = 5$ k$\Omega$.
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 Fundamental $f = \frac{v}{2L}$; with the same tension and string, $f \propto \frac{1}{L}$.
$L' = 90 \times \frac{120}{180} = 60$ cm.
Answer: 60. | ||||
| 52 | 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 The prevented strain is $\alpha\Delta T$, so the force is $F = YA\alpha\Delta T = 2 \times 10^{11} \times 10^{-4} \times 10^{-5} \times 200 = 4 \times 10^4$ N.
Answer: 4. | ||||
| 53 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For a hollow sphere $I = \frac{2}{3}mR^2$. Rolling without slipping, its kinetic energy is $\frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{2}{3}mR^2\cdot\frac{v^2}{R^2} = \frac{5}{6}mv^2$.
Static friction does no work in pure rolling, so at the highest point all of it has become potential energy: $mgh = \frac{5}{6}mv^2$.
$h = \frac{5v^2}{6g} = \frac{5 \times 9}{60} = 0.75$ m $= 75$ cm.
Answer: 75. | ||||
| 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 Impulse $= 2 \times 5 = 10$ N s, so the momentum becomes $10 + 10 = 20$ kg m s$^{-1}$.
$K = \frac{p^2}{2m}$: $\Delta K = \frac{400 - 100}{10} = 30$ J.
Answer: 30. | ||||
| 55 | 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 Initial energy $\frac{1}{2}CV^2 = \frac{1}{2} \times 600 \times 10^{-12} \times 4 \times 10^4 = 12\ \mu$J.
The charge shares equally between the two equal capacitors: common potential 100 V, final energy $2 \times \frac{1}{2} \times 600 \times 10^{-12} \times 10^4 = 6\ \mu$J.
Energy lost $= 12 - 6 = 6\ \mu$J.
Answer: 6. | ||||
| 56 | 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 $\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right)$. $H_\alpha$ ($n = 3$): $\frac{5}{36}R$; $H_\beta$ ($n = 4$): $\frac{3}{16}R$.
$\frac{\lambda_\alpha}{\lambda_\beta} = \frac{3/16}{5/36} = \frac{108}{80} = \frac{27}{20}$, so $x = 27$.
Answer: 27. | ||||
| 57 | 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 $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$ with light going from the rarer medium into the denser: $\mu_1 = 1$, $\mu_2 = 1.5$, $u = -15$ cm, $R = +30$ cm (centre in the denser medium).
$\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30} = \frac{1}{60}$, so $\frac{1.5}{v} = -\frac{3}{60}$ and $v = -30$ cm.
The image is virtual, 30 cm from the pole on the side of the object.
Answer: 30. | ||||
| 58 | 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 $Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{100}\sqrt{\frac{1}{6.25 \times 10^{-6}}} = \frac{1}{100} \times 400 = 4$.
Answer: 4. | ||||
| 59 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B_{centre} = \frac{\mu_0 NI}{2r}$ and on the axis $B = \frac{\mu_0 NIr^2}{2(r^2 + d^2)^{3/2}}$.
At $d = r$: $\frac{B_{centre}}{B_{axis}} = \frac{(2r^2)^{3/2}}{r^3} = 2^{3/2} = \sqrt{8}$, so $x = 8$.
Answer: 8. | ||||
| 60 | 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 $v_d = \frac{I}{neA} = \frac{3.2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}} = \frac{3.2}{2.56 \times 10^4} = 1.25 \times 10^{-4}$ m s$^{-1}$ $= 125 \times 10^{-6}$ m s$^{-1}$.
Answer: 125. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | StatementsEasy | |
A easy statements question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry A. 0.00253: leading zeros are not significant, so 3.
B. 1.0003: zeros between non-zero digits count, so 5.
C. 15.0: a trailing zero after the decimal point counts, so 3.
D. 163: 3.
A, C and D each have three significant figures.
Answer: (D). | ||||
| 62 | Chemical Bonding and Molecular StructureHydrogen bonding | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on hydrogen bonding, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure The constant a measures the intermolecular attraction. Approximate values (L$^2$ bar mol$^{-2}$): Ar 1.34, $CH_4$ 2.25, $H_2O$ 5.46 (hydrogen bonding), $C_6H_6$ 18.2 (large, highly polarisable molecule).
Increasing order: Ar < $CH_4$ < $H_2O$ < $C_6H_6$, i.e. A, B, C, D.
Answer: (A). | ||||
| 63 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics A positive catalyst gives an alternative path of lower activation energy. It does not change the energies of the reactants or the products (so ΔH is unchanged).
Only diagram (C) shows a lower peak with the same reactant and product levels. (A) and (D) change the product energy; (B) raises the barrier.
Answer: (C). | ||||
| 64 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Moseley's law: $\sqrt{\nu} = a(Z - b)$, so a plot of $\sqrt{\nu}$ against Z is a straight line.
Answer: (A). | ||||
| 65 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter In the Hall–Héroult process purified $Al_2O_3$ is dissolved in molten cryolite ($Na_3AlF_6$, with $CaF_2$) to lower the melting point and raise conductivity, and electrolysed. Graphite anodes take part in the reduction: $2Al_2O_3 + 3C \rightarrow 4Al + 3CO_2$.
So graphite is the reducing agent; cryolite and fluorspar are only fluxes/solvents.
Answer: (B). | ||||
| 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 $H_2O_2$ decomposes slowly in light; the decomposition is catalysed by traces of alkali, dust and metal surfaces (including the alkali in ordinary glass). It is therefore kept in wax-lined glass or plastic vessels in the dark, and urea is added as a stabiliser.
Answer: (B). | ||||
| 67 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Assertion and reasonMedium | |
A medium assertion and reason question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter A: true. Sea water holds about 10.8 g of $Na^+$ but only about 0.4 g of $K^+$ per kg, roughly 30 : 1 (although the two are similarly abundant in the earth's crust).
R: true; K is larger than Na (atomic radius 227 pm against 186 pm).
But R does not explain A: the lower amount of potassium in the sea comes from the way potassium is held back on land (retained by clay minerals of the soil and taken up by plants), not simply from the size of the atom.
Answer: (B). | ||||
| 68 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter For good Portland cement, the ratio of lime (CaO) to the total of silica ($SiO_2$), alumina ($Al_2O_3$) and iron oxide ($Fe_2O_3$) should be as close as possible to 2; the ratio of silica to alumina plus iron oxide should be 2.5 to 4.
Answer: (B). | ||||
| 69 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Matching listEasy | |
A easy matching list question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A: $Cr^{3+}$, $d^3$ ($t_{2g}^3$) whatever the ligand: 3 unpaired (II).
B: $Fe^{2+}$, $d^6$ with weak-field water, high spin $t_{2g}^4e_g^2$: 4 unpaired (IV).
C: $Co^{3+}$, $d^6$ with ammonia, low spin $t_{2g}^6$: 0 unpaired (I).
D: $Ni^{2+}$, $d^8$ ($t_{2g}^6e_g^2$): 2 unpaired (III).
A-II, B-IV, C-I, D-III.
Answer: (A). | ||||
| 70 | Environmental ChemistryOzone depletion | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ozone depletion, from Environmental Chemistry. NCERT: No matching NCERT chapter The chain is: CFCs are split by UV light into chlorine radicals (A); $\dot{C}l$ destroys ozone giving $Cl\dot{O}$ (B); $Cl\dot{O}$ reacts with atomic oxygen to regenerate $\dot{C}l$ (C), so one Cl atom destroys many $O_3$ molecules.
Reaction (D), $2Cl\dot{O} \rightarrow ClO_2 + \dot{C}l$, is not one of these steps.
Answer: (D). | ||||
| 71 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $NaHCO_3$ turns pent-4-enoic acid into its carboxylate. $Br_2$ adds to the C=C to form a cyclic bromonium ion; the carboxylate oxygen, held close by the chain, opens it intramolecularly at the inner carbon (5-exo closure), instead of a bromide ion doing so.
The product is the γ-lactone 5-(bromomethyl)dihydrofuran-2-one (bromolactonisation).
Answer: (D). | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The compound is $CH_3COCH_2CH_2CH(CH_3)COOH$. The principal group, COOH, is C-1; the chain of six carbons runs C-1 (COOH), C-2 (CH bearing $CH_3$), C-3, C-4, C-5 (C=O), C-6 ($CH_3$).
Name: 2-methyl-5-oxohexanoic acid. (The ketone is 'oxo', not 'formyl', which would mean CHO.)
Answer: (D). | ||||
| 73 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Nucleophilic aromatic substitution goes through a carbanion (Meisenheimer) intermediate, which electron-withdrawing groups ortho/para to Cl stabilise.
D (2,4,6-trinitrochlorobenzene, three –NO2 groups) > B (o-nitrochlorobenzene, one) > A (chlorobenzene) > C (o-chloroanisole: –OMe donates electrons by resonance and makes the ring even less open to a nucleophile).
Answer: (D). | ||||
| 74 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Electron-withdrawing halogens stabilise the carboxylate ion by the –I effect, more strongly the more electronegative and the more numerous they are.
$CF_3COOH$ (three F) > $FCH_2COOH$ > $ClCH_2COOH$ > $BrCH_2COOH$ > $CH_3COOH$: B > D > C > E > A.
Answer: (D). | ||||
| 75 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Two molecules of phenol (carbolic acid) condense with phthalic anhydride in the presence of conc. $H_2SO_4$ (at their para positions) to give phenolphthalein, an acid–base indicator.
Answer: (D). | ||||
| 76 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Starting material: 4-nitrotoluene.
(i) $Br_2$: the activating $-CH_3$ directs ortho/para and the $-NO_2$ group directs meta; both point to the carbon ortho to $CH_3$, giving 2-bromo-4-nitrotoluene.
(ii) $H_2$/Pd reduces $-NO_2$ to $-NH_2$.
(iii) $NaNO_2$/HCl at 0 °C gives the diazonium salt.
(iv) $H_3PO_2$ replaces $-N_2^+$ by H.
Product: 2-bromotoluene (Br ortho to CH3). The amino group served only as a temporary director.
Answer: (A). | ||||
| 77 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | StatementsEasy | |
A easy statements question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Antagonists bind to the receptor site (A) and block it, inhibiting the body's natural communication (C). They act at the receptor on the cell surface, not by entering the cell (B false). Drugs that mimic the natural messenger are agonists (D false).
Answer: (D). | ||||
| 78 | BiomoleculesProteins | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Glutamic acid E, glutamine Q, tyrosine Y, tryptophan W.
A-III, B-I, C-IV, D-II.
Answer: (A). | ||||
| 79 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | StatementsMedium | |
A medium statements question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter In Ostwald's theory methyl orange is treated as a weak base (phenolphthalein is the weak acid), so I is incorrect.
In the quinonoid theory, the quinonoid form is the deeply coloured one and the benzenoid form the lightly coloured one, so II is incorrect.
Answer: (B). | ||||
| 80 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | StatementsEasy | |
A easy statements question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter The two statements are swapped: redox indicators respond to a change in oxidation (electrode) potential, and acid–base indicators respond to a change in pH. Both are incorrect.
Answer: (B). | ||||
| 81 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Radial nodes $= n - l - 1$.
7s: 6; 7p: 5; 6s: 5; 8p: 6; 8d: 5.
Three of them (7p, 6s, 8d) have 5 radial nodes.
Answer: 3. | ||||
| 82 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Lone pairs on Xe = (8 − charge correction − electrons used in bonds)/2, counting each Xe–F as using 1 electron and each Xe=O as 2:
$XeF_5^+$: $(7 - 5)/2 = 1$. $XeO_3$: $(8 - 6)/2 = 1$. $XeO_2F_2$: $(8 - 4 - 2)/2 = 1$. $XeF_5^-$: $(9 - 5)/2 = 2$. $XeO_3F_2$: $(8 - 6 - 2)/2 = 0$. $XeOF_4$: $(8 - 2 - 4)/2 = 1$. $XeF_4$: $(8 - 4)/2 = 2$.
One lone pair: $XeF_5^+$, $XeO_3$, $XeO_2F_2$, $XeOF_4$, i.e. 4.
Answer: 4. | ||||
| 83 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics A bomb calorimeter works at constant volume: $\Delta U = -1406$ kJ mol$^{-1}$.
$\Delta n_g = 2 - 4 = -2$ (water is liquid), so $\Delta H = \Delta U + \Delta n_g RT = -1406 - 2 \times 8.3 \times 300 \times 10^{-3} = -1406 - 4.98 = -1410.98$ kJ mol$^{-1}$.
At equilibrium $\Delta G = \Delta H - T\Delta S = 0$, so $T\Delta S = \Delta H = -1410.98 \approx -1411$ kJ.
Answer: 1411. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $K_b = \frac{RT_b^2 M}{1000\,\Delta_{vap}H}$; with equal M, $K_b \propto \frac{T_b^2}{\Delta_{vap}H}$.
$m = \frac{K_b(X)}{K_b(Y)} = \frac{2^2}{1/2} = 8$.
Answer: 8. | ||||
| 85 | EquilibriumSolubility product | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium With the common ion, $[SO_4^{2-}] \approx 0.1$ M, so $s = \frac{K_{sp}}{0.1} = 10^{-9}$ mol L$^{-1}$.
In grams: $10^{-9} \times 233 = 233 \times 10^{-9}$ g L$^{-1}$.
Answer: 233. | ||||
| 86 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry A: correct; the maximum (reversible) electrical work at constant T and P equals the Gibbs energy of reaction ($\Delta_rG = -nFE$).
B: incorrect; $E^\circ$ refers to the fixed standard state (1 bar), so it does not depend on the pressure; only E does, through the Nernst equation.
C: correct; from $\Delta_rG^\circ = -nFE^\circ$ and $\left(\frac{\partial\Delta G}{\partial T}\right)_P = -\Delta S$, $\frac{dE^\circ}{dT} = \frac{\Delta_rS^\circ}{nF}$.
D: correct; that is the definition of reversible operation.
One incorrect statement.
Answer: 1. | ||||
| 87 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Coagulating power is inversely proportional to the coagulating value (the least concentration that causes coagulation).
$x = \frac{50.04}{0.09} = 556$.
Answer: 556. | ||||
| 88 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Pyrophosphoric acid, $H_4P_2O_7$: $(HO)_2P(=O)-O-P(=O)(OH)_2$.
σ bonds: 4 O–H + 4 P–OH + 2 P–O (bridge) + 2 (σ part of P=O) = 12. π bonds: 2 (one in each P=O).
Ratio $= \frac{12}{2} = 6$.
Answer: 6. | ||||
| 89 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions $Fe(CO)_5$: CO is neutral, Fe is 0. $VO^{2+}$: $x - 2 = +2$, V is +4. $WO_3$: W is +6.
Sum $= 0 + 4 + 6 = 10$.
Answer: 10. | ||||
| 90 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mu = \sqrt{n(n+2)}$: 6.06 BM is close to $\sqrt{35} = 5.92$, so n = 5 unpaired electrons, which is $Mn^{2+}$ ($d^5$, high spin with weak-field N-bonded thiocyanate).
Charge: $+2 + 6(-1) = -4$, so $x = 4$.
Answer: 4. | ||||
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