JEE (Main) · 2022
JEE (Main) 2022 — 29 June, 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
- Three Dimensional Geometry3
- Trigonometry3
- Differential Equations2
10 easy · 15 medium · 5 hard
Physics
30 Q
- Current Electricity3
- Kinematics2
- Laws of Motion2
26 easy · 4 medium · 0 hard
Chemistry
30 Q
- d- and f-Block Elements3
- Some Basic Concepts of Chemistry2
- p-Block Elements2
22 easy · 7 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 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each of the 4 entries can be any of the 10 primes: $10^4$ matrices.
$\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is singular when $ad = bc$. As the entries are primes, unique factorisation gives $\{a, d\} = \{b, c\}$ as multisets.
If $a = d$: then $b = c = a$: 10 matrices.
If $a \neq d$: $(a, d)$ can be chosen in $10 \times 9 = 90$ ways and $(b, c)$ is $(a, d)$ or $(d, a)$: $180$ matrices.
Singular matrices: $190$. Probability $= \frac{190}{10^4} = \frac{19}{10^3}$.
Answer: (C). | ||||
| 2 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $y = vx$ (with $x > 0$): $x(v + xv') - vx = x\sqrt{v^2 + 16}$, so $x\frac{dv}{dx} = \sqrt{v^2+16}$.
$\int \frac{dv}{\sqrt{v^2+16}} = \int \frac{dx}{x}$ gives $\ln\left(v + \sqrt{v^2+16}\right) = \ln x + C$, i.e. $v + \sqrt{v^2+16} = Kx$.
At $x = 1$, $v = 3$: $3 + 5 = K$, so $K = 8$.
At $x = 2$: $v + \sqrt{v^2+16} = 16 \Rightarrow v^2 + 16 = (16 - v)^2 \Rightarrow 32v = 240 \Rightarrow v = 7.5$.
$y(2) = 2 \times 7.5 = 15$.
Answer: (A). | ||||
| 3 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry For the image, $\frac{a-2}{3} = \frac{b-4}{-1} = \frac{c-7}{4} = \frac{-2(3\cdot2 - 4 + 4\cdot7 - 2)}{3^2+1^2+4^2} = \frac{-2 \times 28}{26} = -\frac{28}{13}$.
So $a = 2 - \frac{84}{13} = -\frac{58}{13}$, $b = 4 + \frac{28}{13} = \frac{80}{13}$, $c = 7 - \frac{112}{13} = -\frac{21}{13}$.
$2a + b + 2c = \frac{-116 + 80 - 42}{13} = \frac{-78}{13} = -6$.
Answer: (C). | ||||
| 4 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctHard | |
A hard single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $g(t) = t^3 - 3t$: $g' = 3(t^2 - 1)$, local maximum $g(-1) = 2$, local minimum $g(1) = -2$, and $g(2) = 2$.
So for $x \le -1$, $f(x) = x^3 - 3x$ (g is increasing there), and for $-1 \le x \le 2$, $f(x) = 2$.
At $x = -1$: left derivative $g'(-1) = 0$, right derivative 0: differentiable.
At $x = 2$: $f(2) = 2$, $\lim_{x\to2^+}(x^2+2x-6) = 2$: continuous; left derivative 0, right derivative $2x + 2 = 6$: not differentiable.
At $x = 3$: $\lim_{x\to3^-}(x^2+2x-6) = 9 = f(3)$; left derivative 8, right derivative 0: not differentiable.
On $[3, 5]$: $f = 9$ on $[3,4)$, $10$ on $[4,5)$, $f(5) = 11$: jumps at $x = 4$ and $x = 5$, so not differentiable there ($x > 5$ gives $2x + 1 \to 11$, but the left limit at 5 is 10).
Hence $m = 4$ (at $x = 2, 3, 4, 5$).
$I = \int_{-2}^{-1}(x^3 - 3x)\,dx + \int_{-1}^{2} 2\,dx = \left[\frac{x^4}{4} - \frac{3x^2}{2}\right]_{-2}^{-1} + 6 = \left(\frac14 - \frac32\right) - (4 - 6) + 6 = \frac34 + 6 = \frac{27}{4}$.
$(m, I) = \left(4, \frac{27}{4}\right)$.
Answer: (C). | ||||
| 5 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Projection: $\frac{\vec{a}\cdot\vec{c}}{|\vec{c}|} = \frac{\alpha + 6 + 2}{3} = \frac{10}{3}$, so $\alpha = 2$.
$\vec{b}\times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-\beta&4\\1&2&-2\end{vmatrix} = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$.
Comparing with $-6\hat{i} + 10\hat{j} + 7\hat{k}$: $\beta = 1$.
$\alpha + \beta = 3$.
Answer: (A). | ||||
| 6 | 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 curves meet where $2x^2 = 8x$: $x = 0$ and $x = 4$.
Area between them: $\int_0^4 \left(2\sqrt{2}\sqrt{x} - \sqrt{2}x\right)dx = 2\sqrt{2}\cdot\frac{2}{3}\cdot 8 - \sqrt{2}\cdot 8 = \frac{32\sqrt{2}}{3} - 8\sqrt{2} = \frac{8\sqrt{2}}{3}$.
The triangle has vertices $(1, \sqrt{2})$, $(1, 2\sqrt{2})$, $(2, 2\sqrt{2})$; for $1 \le x \le 2$ the parabola is above $y = 2\sqrt{2}$, so the triangle lies inside the region. Its area is $\frac12 \cdot 1 \cdot \sqrt{2} = \frac{\sqrt{2}}{2}$.
Required area $= \frac{8\sqrt{2}}{3} - \frac{\sqrt{2}}{2} = \frac{13\sqrt{2}}{6}$.
Answer: (C). | ||||
| 7 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The first two equations are independent, so for infinitely many solutions the third must be a combination of them: $a(2, 1, -1, 7) + b(1, -3, 2, 1) = (1, 4, \delta, k)$.
$2a + b = 1$ and $a - 3b = 4$ give $a = 1$, $b = -1$.
Then $\delta = -a + 2b = -3$ and $k = 7a + b = 6$.
$\delta + k = 3$.
Answer: (B). | ||||
| 8 | 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 $\alpha^2 = \beta^2 = 1 - 2i$ (the roots are $\pm$ a square root of $1 - 2i$).
$\alpha^8 + \beta^8 = 2(1 - 2i)^4$, so $|\alpha^8 + \beta^8| = 2|1 - 2i|^4 = 2(\sqrt{5})^4 = 50$.
Answer: (A). | ||||
| 9 | 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 $(p \vee q) \Rightarrow q \equiv \sim(p \vee q) \vee q \equiv (\sim p \wedge \sim q) \vee q \equiv \sim p \vee q$.
$\Rightarrow$: when $p \wedge q$ is true, q is true, so $\sim p \vee q$ is true; the implication is never false: a tautology.
$\wedge$ and $\Leftrightarrow$: with p, q both false, $p \wedge q$ is F and $\sim p \vee q$ is T, so both are false there.
$\vee$: with p T, q F, both sides are F.
Answer: (C). | ||||
| 10 | 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 $a_{ij} = 2^{-i}\cdot 2^{j}$, so $A = uv^T$ with $u = (2^{-1}, 2^{-2}, 2^{-3})^T$ and $v = (2, 4, 8)^T$.
$v^Tu = 1 + 1 + 1 = 3$, hence $A^2 = u(v^Tu)v^T = 3A$ and $A^n = 3^{n-1}A$.
$A^2 + \ldots + A^{10} = (3 + 3^2 + \ldots + 3^9)A = \frac{3(3^9 - 1)}{2}A = \left(\frac{3^{10} - 3}{2}\right)A$.
Answer: (A). | ||||
| 11 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $(x, y) \in R$ exactly when x and y lie in the same block $A_i$ of the partition.
Reflexive: x is in the same block as itself.
Symmetric: if x, y are in the same block, so are y, x.
Transitive: if x, y share a block and y, z share a block, the blocks are the same (they are disjoint), so x, z share it.
R is an equivalence relation.
Answer: (D). | ||||
| 12 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctHard | |
A hard single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $a_{n+2} - a_{n+1} = (a_{n+1} - a_n) + 1$: the differences $0, 1, 2, \ldots$, so $a_n = 0 + 1 + \ldots + (n-1) = \frac{n(n-1)}{2}$ ($a_2 = 1$, $a_3 = 3$, …).
With $S = \sum_{n\ge2}\binom{n}{2}x^n = \frac{x^2}{(1-x)^3}$ (from differentiating $\sum x^n$ twice) and $x = \frac17$:
$S = \frac{1/49}{(6/7)^3} = \frac{1}{49}\cdot\frac{343}{216} = \frac{7}{216}$.
Answer: (B). | ||||
| 13 | Straight LinesSlope, parallel and perpendicular lines | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines OB has slope $\frac32$. A line through O making $\frac{\pi}{4}$ with OB has slope $m$ with $\tan$ of its inclination $= \tan(\theta \pm \frac{\pi}{4})$, $\tan\theta = \frac32$:
$\frac{\frac32 + 1}{1 - \frac32} = -5$ and $\frac{\frac32 - 1}{1 + \frac32} = \frac15$.
Line $y = -5x$ meets $y = 2$ at $\left(-\frac25, 2\right)$; line $y = \frac{x}{5}$ meets it at $(10, 2)$.
$AA' = 10 + \frac25 = \frac{52}{5}$.
Answer: (C). | ||||
| 14 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let the square have side x and the triangle side y: $4x + 3y = 22$.
Area $A = x^2 + \frac{\sqrt{3}}{4}y^2$ with $x = \frac{22 - 3y}{4}$.
$\frac{dA}{dy} = 2x\left(-\frac34\right) + \frac{\sqrt{3}}{2}y = 0 \Rightarrow x = \frac{y}{\sqrt{3}}$ (a minimum, since $A$ is a convex quadratic in y).
Then $\frac{4y}{\sqrt{3}} + 3y = 22 \Rightarrow y = \frac{22\sqrt{3}}{4 + 3\sqrt{3}} = \frac{66}{4\sqrt{3} + 9}$.
Answer: (B). | ||||
| 15 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions For $u \in [-1, 1]$, $\sin^{-1}u \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\frac{2\sin^{-1}u}{\pi} \in [-1, 1]$ and the outer $\cos^{-1}$ is always defined. The only condition is $-1 \le \frac{1}{4x^2 - 1} \le 1$, i.e. $|4x^2 - 1| \ge 1$ (with $4x^2 \ne 1$).
$4x^2 - 1 \ge 1 \Rightarrow x^2 \ge \frac12 \Rightarrow |x| \ge \frac{1}{\sqrt{2}}$.
$4x^2 - 1 \le -1 \Rightarrow x = 0$.
Domain: $\left(-\infty, -\frac{1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, \infty\right) \cup \{0\}$.
Answer: (D). | ||||
| 16 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem A general term is $\frac{10!}{a!\,b!\,c!}(3x^3)^a(-2x^2)^b\left(\frac{5}{x^5}\right)^c$ with $a + b + c = 10$.
Constant term: $3a + 2b - 5c = 0$. With $c = 10 - a - b$: $8a + 7b = 50$, whose only solution in non-negative integers is $a = 1$, $b = 6$, $c = 3$.
Term $= \frac{10!}{1!\,6!\,3!}\cdot 3 \cdot (-2)^6 \cdot 5^3 = 840 \cdot 3 \cdot 64 \cdot 125$.
$840 = 2^3 \cdot 105$, so the term is $2^{3+6}\cdot(105 \cdot 3 \cdot 125)$, with the bracket odd.
$k = 9$.
Answer: (D). | ||||
| 17 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\cos\left(\pi x - \pi\left[\frac{x}{2}\right]\right) = (-1)^{[x/2]}\cos\pi x$.
On $[0, 2)$: $\int_0^2 \cos\pi x\,dx = 0$. On $[2, 4)$: $-\int_2^4 \cos\pi x\,dx = 0$. On $[4, 5]$: $\int_4^5 \cos\pi x\,dx = \left[\frac{\sin\pi x}{\pi}\right]_4^5 = 0$.
The integral is 0.
Answer: (D). | ||||
| 18 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctHard | |
A hard single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Let $P(t_1^2, 2t_1)$, $Q(t_2^2, 2t_2)$; focal chord: $t_1t_2 = -1$.
Right angle at (3, 0): $(t_1^2 - 3)(t_2^2 - 3) + 4t_1t_2 = 0 \Rightarrow 1 - 3(t_1^2 + t_2^2) + 9 - 4 = 0 \Rightarrow t_1^2 + t_2^2 = 2$.
With $t_1t_2 = -1$: $(t_1 + t_2)^2 = 0$, so $t_1 = 1$, $t_2 = -1$: $P(1, 2)$, $Q(1, -2)$, the chord $x = 1$.
For the ellipse, $x = 1$ is a focal chord perpendicular to the axis, so $ae = 1$ and it is a latus rectum: $\frac{b^2}{a} = 2$.
$b^2 = a^2 - a^2e^2 = a^2 - 1$, so $a^2 - 1 = 2a \Rightarrow a = 1 + \sqrt{2}$.
$\frac{1}{e^2} = a^2 = 3 + 2\sqrt{2}$.
Answer: (B). | ||||
| 19 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Tangent to $C_1$ at $(-1, 1)$: $-x + y = 2$.
Centre of $C_2$ is $(3, 2)$, radius $\sqrt{5}$. Its distance from the line: $d = \frac{|-3 + 2 - 2|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.
Half-chord $= \sqrt{5 - \frac92} = \frac{1}{\sqrt{2}}$, so $AB = \sqrt{2}$.
N is the pole of AB: its distance from the centre is $\frac{r^2}{d} = \frac{5\sqrt{2}}{3}$, so its distance from AB is $\frac{5\sqrt{2}}{3} - \frac{3}{\sqrt{2}} = \frac{1}{3\sqrt{2}}$.
Area $= \frac12 \cdot \sqrt{2} \cdot \frac{1}{3\sqrt{2}} = \frac16$.
Answer: (C). | ||||
| 20 | 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 $\sum_{1}^{5} x_i = 24$ and $\frac{\sum x_i^2}{5} - \left(\frac{24}{5}\right)^2 = \frac{194}{25}$, so $\sum x_i^2 = \frac{194 + 576}{5} = 154$.
First four: sum $= 14$, so $x_5 = 10$; their squares sum to $154 - 100 = 54$.
$a = \frac{54}{4} - \left(\frac72\right)^2 = \frac{54 - 49}{4} = \frac54$.
$4a + x_5 = 5 + 10 = 15$.
Answer: (B). | ||||
| 21 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | NumericalHard | |
A hard numerical question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x + iy$: $z(1+i) + \bar{z}(1-i) = 2(x - y)$, so the second condition is $x - y \le 1$.
S is the part of the disc with centre (2, 0), radius 1, on the side $x - y \le 1$. The line $x - y = 1$ cuts the circle at (1, 0) and (2, 1), so S is bounded by the arc $(2 + \cos\theta, \sin\theta)$, $\theta \in \left[\frac{\pi}{2}, \pi\right]$, and the chord from (1, 0) to (2, 1).
Minimum: the point of the disc nearest (0, 4) lies on the line from (2, 0) to (0, 4): $z_1 = \left(2 - \frac{2}{\sqrt5}, \frac{4}{\sqrt5}\right)$; here $x - y = 2 - \frac{6}{\sqrt5} < 1$, so $z_1 \in S$. $|z_1|^2 = 4 - \frac{8}{\sqrt5} + \frac45 + \frac{16}{5} = 8 - \frac{8}{\sqrt5}$.
Maximum: on the arc, $|z - 4i|^2 = 21 + 4\cos\theta - 8\sin\theta$, which is 13 at $\theta = \frac{\pi}{2}$ and 17 at $\theta = \pi$ and has only a minimum in between; on the chord the largest value is at an end. So $z_2 = 1$, $|z_2|^2 = 1$.
$5(|z_1|^2 + |z_2|^2) = 5\left(9 - \frac{8}{\sqrt5}\right) = 45 - 8\sqrt5$, so $\alpha = 45$, $\beta = -8$.
$\alpha + \beta = 37$.
Answer: 37. | ||||
| 22 | 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 With $c = \cos 2x$: $2\cos^4 x - \cos 2x = \frac{(1 + c)^2}{2} - c = \frac{1 + c^2}{2}$, so the coefficient of y is $P = \frac{2\sqrt2}{1 + \cos^2 2x}$.
$\frac{d}{dx}\tan^{-1}(\sqrt2\cot 2x) = \frac{-2\sqrt2\csc^2 2x}{1 + 2\cot^2 2x} = \frac{-2\sqrt2}{\sin^2 2x + 2\cos^2 2x} = -\frac{2\sqrt2}{1 + \cos^2 2x}$.
So $\int P\,dx = -\tan^{-1}(\sqrt2\cot 2x)$ and the integrating factor is $e^{-\tan^{-1}(\sqrt2\cot 2x)}$.
$\frac{d}{dx}\left(y\,e^{-\tan^{-1}(\sqrt2\cot 2x)}\right) = x$, so $y\,e^{-\tan^{-1}(\sqrt2\cot 2x)} = \frac{x^2}{2} + C$.
At $x = \frac{\pi}{4}$: $\cot\frac{\pi}{2} = 0$, so $\frac{\pi^2}{32} = \frac{\pi^2}{32} + C$, $C = 0$.
At $x = \frac{\pi}{3}$: $\sqrt2\cot\frac{2\pi}{3} = -\sqrt{\frac23}$, so $y\,e^{\tan^{-1}\sqrt{2/3}} = \frac{\pi^2}{18}$, i.e. $y = \frac{\pi^2}{18}e^{-\tan^{-1}\sqrt{2/3}}$.
$\alpha = \sqrt{\frac23}$, $3\alpha^2 = 2$.
Answer: 2. | ||||
| 23 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Normal $\vec{n} = (-1, 1, 1)$. $\vec{PQ} = (1, -3, 4)$ and $\vec{PQ}\cdot\vec{n} = -1 - 3 + 4 = 0$, so PQ is parallel to the plane (both points are at the same signed distance: $-1 + 2 - 1 - 1 = -1$ and $-2 - 1 + 3 - 1 = -1$).
The feet of the perpendiculars are then PQ translated along the normal, so $d = |PQ|$ and $d^2 = 1 + 9 + 16 = 26$.
Answer: 26. | ||||
| 24 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $10\cos^2\theta = 5 + 5\cos 2\theta$, so the equation becomes $3\cos^2 2\theta + \cos 2\theta = 0$: $\cos 2\theta = 0$ or $\cos 2\theta = -\frac13$.
For $\theta \in [-4\pi, 4\pi]$, $2\theta$ runs over $[-8\pi, 8\pi]$, eight full periods of cosine. Each of $\cos\phi = 0$ and $\cos\phi = -\frac13$ has 2 solutions per period and none at the end points (where $\cos\phi = 1$).
Number of solutions $= 8 \times 2 + 8 \times 2 = 32$.
Answer: 32. | ||||
| 25 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | NumericalEasy | |
A easy numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $f(\theta) = 2\theta - \cos^2\theta + \sqrt2$. Then $f'(\theta) = 2 + 2\sin\theta\cos\theta = 2 + \sin 2\theta \ge 1 > 0$, so f is strictly increasing.
$f \to -\infty$ as $\theta \to -\infty$ and $f \to \infty$ as $\theta \to \infty$, and f is continuous, so it has exactly one zero.
Answer: 1. | ||||
| 26 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $A = \tan^{-1}\frac12$. Then $\cos^{-1}\frac{1}{\sqrt5} = \tan^{-1}2 = \frac{\pi}{2} - A$.
So $3A + 2\left(\frac{\pi}{2} - A\right) = \pi + A$ and $\tan(\pi + A) = \tan A = \frac12$. First term: $50 \times \frac12 = 25$.
Let $\phi = \tan^{-1}(2\sqrt2)$: $\cos\phi = \frac13$, $\sin\phi = \frac{2\sqrt2}{3}$, so $\tan\frac{\phi}{2} = \frac{\sin\phi}{1 + \cos\phi} = \frac{2\sqrt2/3}{4/3} = \frac{1}{\sqrt2}$. Second term: $4\sqrt2 \times \frac{1}{\sqrt2} = 4$.
Total $= 25 + 4 = 29$.
Answer: 29. | ||||
| 27 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $x = y = 0$: $f(0) = 2f(0)$, so $f(0) = 2k = 0$.
Then $(c + 1)(x + y)^2 + (1 - c^2)(x + y) = (c + 1)(x^2 + y^2) + (1 - c^2)(x + y) - xy$, so $2(c + 1)xy = -xy$ and $c = -\frac32$.
$1 - c^2 = -\frac54$, so $f(x) = -\frac{x^2}{2} - \frac{5x}{4}$.
$\sum_{1}^{20} f(x) = -\frac{2870}{2} - \frac{5 \times 210}{4} = -1435 - 262.5 = -1697.5$.
$|2 \times (-1697.5)| = 3395$.
Answer: 3395. | ||||
| 28 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $e^2 = 1 + \frac{b^2}{a^2} = \frac{11}{4}$, so $b = \frac{\sqrt7}{2}a$.
$2a + 2b = 4(2\sqrt2 + \sqrt{14})$ gives $a\left(1 + \frac{\sqrt7}{2}\right) = 2\sqrt2(2 + \sqrt7)$, i.e. $\frac{a(2 + \sqrt7)}{2} = 2\sqrt2(2 + \sqrt7)$, so $a = 4\sqrt2$.
$a^2 = 32$, $b^2 = \frac74 \times 32 = 56$; $a^2 + b^2 = 88$.
Answer: 88. | ||||
| 29 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry In $P_2$: $\vec{u} = (2, -2, -3) - (2, -3, 2) = (0, 1, -5)$, $\vec{v} = (1, -4, 2) - (2, -3, 2) = (-1, -1, 0)$.
Normal $\vec{n}_2 = \vec{u}\times\vec{v} = (0 - 5,\ 5 - 0,\ 0 + 1) = (-5, 5, 1)$.
The line of intersection is along $\vec{n}_1 \times \vec{n}_2 = (2, 1, -3)\times(-5, 5, 1) = (1 + 15,\ 15 - 2,\ 10 + 5) = (16, 13, 15)$.
So $\alpha = 13$, $\beta = 15$ and $\alpha + \beta = 28$.
Answer: 28. | ||||
| 30 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Read \"consecutive integers\" in the order written: $b_2 = b_1 + 1$, $b_3 = b_1 + 2$.
A: $b_1, b_2, b_3$ consecutive. $b_1 \in \{1, \ldots, 98\}$: 98 ways; $b_4$ any of the other 97 numbers: $98 \times 97 = 9506$.
B: $b_2, b_3, b_4$ consecutive: likewise 9506.
$A \cap B$: $b_1, b_2, b_3, b_4$ consecutive: $b_1 \in \{1, \ldots, 97\}$: 97.
$|A \cup B| = 9506 + 9506 - 97 = 18915$.
Answer: 18915. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Both balls fall $180 - 100 = 80$ m to the meeting point.
Ball A: $80 = \frac12 \cdot 10\, t^2$, so $t = 4$ s.
Ball B starts at $t = 2$ s, so it takes 2 s: $80 = u \cdot 2 + \frac12 \cdot 10 \cdot 2^2 = 2u + 20$, giving $u = 30\ ms^{-1}$.
Answer: (D). | ||||
| 32 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion Let the masses be m, m, 2m. The two small pieces have perpendicular momenta 30m and 40m, so their resultant is $\sqrt{(30m)^2 + (40m)^2} = 50m$.
The total momentum stays zero, so the third piece has momentum 50m opposite to it: $2m\,v = 50m$, $v = 25\ ms^{-1}$.
Answer: (B). | ||||
| 33 | 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 $\frac{A_0}{A} = \frac{2.56 \times 10^{-3}}{2 \times 10^{-5}} = 128 = 2^7$, so 7 half-lives pass.
Time $= 7 \times 5 = 35$ days.
Answer: (B). | ||||
| 34 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctMedium | |
A medium single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Angular momentum about O = spin about the centre + that of the centre of mass moving about O.
Spin: $I_{cm}\omega = \frac23 mR^2\omega = \frac23 R^2\omega$ (m = 1 kg), clockwise.
The centre moves along x with $v = \omega R$ (rolling) at height R above the line through O: $mvR = R^2\omega$, also clockwise for the shell rolling to the right as shown.
Total $L = \frac23 R^2\omega + R^2\omega = \frac53 R^2\omega$, so $a = 5$.
Answer: (C). | ||||
| 35 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory At STP, 44.8 L is 2 mol of gas. The capacity is fixed, so the heating is at constant volume; helium is monatomic, $C_V = \frac32 R$.
$Q = nC_V\Delta T = 2 \times \frac32 \times 8.3 \times 20 = 498$ J.
Answer: (C). | ||||
| 36 | 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 Extension is proportional to load: $L_1 = L + k$ and $L_2 = L + 2k$ (k the extension per kg).
So $k = L_2 - L_1$ and $L = L_1 - k = 2L_1 - L_2$.
Answer: (C). | ||||
| 37 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Assertion and reasonEasy | |
A easy assertion and reason question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter A: a photon with energy below the work function cannot free an electron, so no emission occurs. Correct.
R: by $K_{max} = h\nu - \phi$, when $h\nu = \phi$ the kinetic energy is zero. Correct.
R describes the threshold case $h\nu = \phi$; it does not explain why there is no emission for $h\nu < \phi$ (that follows from the work function being the minimum energy needed to free an electron). So R is not the explanation of A.
Answer: (B). | ||||
| 38 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power At $x = 0$, $v = 0$; at $x = 4$ m, $v = 0.25 \times 4^{5/2} = 0.25 \times 32 = 8\ ms^{-1}$.
Work-energy theorem: $W = \frac12 \times 0.5 \times 8^2 = 16$ J.
Answer: (D). | ||||
| 39 | 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 $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, so $r \propto \sqrt{K}$.
K becomes 4 times, so r becomes 2 times: ratio 2 : 1.
Answer: (C). | ||||
| 40 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | StatementsEasy | |
A easy statements question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Below resonance ($\omega < \omega_r$), $X_C = \frac{1}{\omega C} > X_L = \omega L$: the circuit is mainly capacitive. (a) true, (b) false.
At resonance $X_L = X_C$, so $Z = \sqrt{R^2 + (X_L - X_C)^2} = R$, not zero. (c) true, (d) false.
Answer: (C). | ||||
| 41 | 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 The jet delivers momentum at the rate $F = \frac{dm}{dt}v = 1 \times 10 = 10$ N (the water is stopped by the block).
Initial acceleration $a = \frac{F}{M} = \frac{10}{2} = 5\ ms^{-2}$.
Answer: (C). | ||||
| 42 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Only like quantities add: $\frac{a}{V^2}$ has the dimensions of P, so $[a] = [PV^2]$; b has the dimensions of V.
$\left[\frac{a}{b}\right] = \frac{PV^2}{V} = [PV]$.
Answer: (C). | ||||
| 43 | 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 With $|\vec{A}| = |\vec{B}| = A$: $|\vec{A} + \vec{B}|^2 = 2A^2(1 + \cos\theta)$ and $|\vec{A} - \vec{B}|^2 = 2A^2(1 - \cos\theta)$.
$1 + \cos\theta = 4(1 - \cos\theta) \Rightarrow \cos\theta = \frac35$.
$\theta = \cos^{-1}\left(\frac35\right)$.
Answer: (C). | ||||
| 44 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{8\pi G\rho R^2}{3}} \propto R\sqrt{\rho}$.
For B: $\frac{v_B}{v_A} = \frac12 \times \sqrt{4} = 1$, so $v_B = 12\ kms^{-1}$.
Answer: (A). | ||||
| 45 | Magnetic Effects of Current and MagnetismEarth magnetism | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on earth magnetism, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B_H = B\cos\delta$, so $B = \frac{0.5}{\cos 30^\circ} = \frac{0.5 \times 2}{\sqrt3} = \frac{1}{\sqrt3}$ G.
Answer: (A). | ||||
| 46 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Maximum particle velocity $= A\omega = 10 \times 2\pi n = 20\pi n$; wave velocity $= n\lambda$.
$20\pi n = 4n\lambda \Rightarrow \lambda = 5\pi$ (cm).
Answer: (B). | ||||
| 47 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The battery stays connected, so V is fixed and $U = \frac12 CV^2 \propto C \propto K$.
$\frac{U'}{U} = \frac{15}{10} = 1.5$: the energy increases by 50%.
Answer: (A). | ||||
| 48 | 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 Force $F = qE = 40 \times 10^{-6} \times 10^5 = 4$ N, opposing the motion. Mass $= 100\ mg = 10^{-4}$ kg, so the deceleration is $a = 4 \times 10^4\ ms^{-2}$.
$s = \frac{v^2}{2a} = \frac{200^2}{8 \times 10^4} = 0.5$ m.
Answer: (D). | ||||
| 49 | 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 $\beta = \frac{\lambda D}{d}$, so $\beta' = 0.5 \times \frac{6000}{5000} \times \frac12 = 0.3$ mm.
Answer: (D). | ||||
| 50 | Communication SystemsBandwidth of signals and channels | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on bandwidth of signals and channels, from Communication Systems. NCERT: No matching NCERT chapter Source frequency $f = \frac{c}{\lambda} = \frac{3 \times 10^8}{10^{-6}} = 3 \times 10^{14}$ Hz.
Available bandwidth $= 0.02 \times 3 \times 10^{14} = 6 \times 10^{12}$ Hz.
Number of channels $= \frac{6 \times 10^{12}}{8 \times 10^3} = 7.5 \times 10^8 = 75 \times 10^7$.
Answer: (B). | ||||
| 51 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity With the same voltage, $H = \frac{V^2t}{R}$, so $R \propto t$: $R_1 \propto 20$, $R_2 \propto 60$.
In parallel the powers add: $\frac{1}{t} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60}$, so $t = 15$ min.
Answer: 15. | ||||
| 52 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalMedium | |
A medium numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves $I = \frac12\epsilon_0 c E_0^2$ with $E_0 = cB_0$: $I = \frac12\epsilon_0 c^3 B_0^2$.
$B_0^2 = \frac{2I}{\epsilon_0 c^3} = \frac{0.44}{8.85 \times 10^{-12} \times 27 \times 10^{24}} = \frac{0.44}{2.3895 \times 10^{14}} = 1.841 \times 10^{-15}$.
$B_0 = 4.29 \times 10^{-8}$ T $= 42.9 \times 10^{-9}$ T, which rounds to 43.
Answer: 43. | ||||
| 53 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter In series the thermal resistances add: $\frac{L_1 + L_2}{K_{eq}A} = \frac{L_1}{KA} + \frac{L_2}{2KA}$.
$\frac{6.5}{K_{eq}} = \frac{4}{K} + \frac{2.5}{2K} = \frac{5.25}{K}$, so $K_{eq} = \frac{6.5}{5.25}K = \frac{26}{21}K = \left(1 + \frac{5}{21}\right)K$.
$\alpha = 21$.
Answer: 21. | ||||
| 54 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At $y = 5$ cm: $v = \omega\sqrt{A^2 - y^2} = \omega\sqrt{100 - 25} = \omega\sqrt{75}$.
Tripled: $v' = 3\omega\sqrt{75}$, with $\omega$ unchanged (it depends only on the system).
New amplitude: $A'^2 = y^2 + \frac{v'^2}{\omega^2} = 25 + 9 \times 75 = 700$, so $A' = \sqrt{700}$ cm.
$x = 700$.
Answer: 700. | ||||
| 55 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity The V–I line makes 45° with the I-axis, so $R = \tan 45^\circ = 1\ \Omega$.
$\rho = \frac{RA}{L} = \frac{1 \times \pi (1.2)^2}{31.4} = \frac{3.14 \times 1.44}{31.4} = 0.144\ \Omega$ cm $= 144 \times 10^{-3}\ \Omega$ cm.
$x = 144$.
Answer: 144. | ||||
| 56 | ThermodynamicsSecond law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics For a reversible (Carnot) engine $\frac{Q_2}{Q_1} = \frac{T_2}{T_1}$.
$T_1 = 227 + 273 = 500$ K, so $T_2 = 500 \times \frac{225}{300} = 375$ K $= 102$°C.
Answer: 102. | ||||
| 57 | Atoms and NucleiRutherford model and alpha scattering | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on rutherford model and alpha scattering, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Impact parameter $b \propto \cot\frac{\theta}{2}$.
$\frac{\sqrt{d_1}}{\sqrt{d_2}} = \frac{\cot 30^\circ}{\cot 45^\circ} = \sqrt3$, so $\frac{d_1}{d_2} = 3$.
$x = 3$.
Answer: 3. | ||||
| 58 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Current gain $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{10\ mA}{100\ \mu A} = 100$.
Voltage gain $A_V = \beta\frac{R_L}{R_i} = 100 \times 2 = 200$.
Power gain $= \beta \times A_V = 100 \times 200 = 2 \times 10^4$.
$x = 2$.
Answer: 2. | ||||
| 59 | 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 Radius R = 15 cm. First surface (air to glass, $u = \infty$): $\frac{1.5}{v} = \frac{1.5 - 1}{15}$, so $v = 45$ cm from the first surface.
This is $45 - 30 = 15$ cm beyond the second surface: a virtual object at $u = +15$ cm for the second surface (glass to air, $R = -15$ cm).
$\frac{1}{v} - \frac{1.5}{15} = \frac{1 - 1.5}{-15} \Rightarrow \frac1v = \frac{1}{10} + \frac{1}{30} = \frac{4}{30}$, so $v = 7.5$ cm beyond the second surface.
Distance from the centre $= 15 + 7.5 = 22.5$ cm $= 225$ mm.
Answer: 225. | ||||
| 60 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity The single loop holds the 15 V cell, 1 Ω and 2 Ω in series: $I = \frac{15}{1 + 2} = 5$ A.
The cell's longer (positive) plate faces B, so the current leaves it towards the 1 Ω resistor and B, then flows through the 2 Ω resistor from B to A.
Across the 2 Ω resistor: $V_B - V_A = IR = 5 \times 2 = 10$ V. (Check along the top: $V_A + 15 - 5 \times 1 = V_A + 10$.)
Answer: 10. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $M(Fe_3O_4) = 232$, so $\frac{4640}{232} = 20$ mol; $M(CO) = 28$, so $\frac{2520}{28} = 90$ mol.
20 mol of $Fe_3O_4$ needs only 80 mol CO, so $Fe_3O_4$ is the limiting reagent.
Iron formed $= 3 \times 20 = 60$ mol $= 60 \times 56 = 3360$ g.
Answer: (C). | ||||
| 62 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | StatementsEasy | |
A easy statements question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom (A) True: half-filled 3d gives Cr $[Ar]3d^54s^1$.
(B) True: $m_l$ runs from $-l$ to $+l$.
(C) True: this is the aufbau principle.
(D) False: the total number of nodes is $n - 1$ (radial $n - l - 1$ plus angular $l$).
Correct: (A), (B) and (C).
Answer: (D). | ||||
| 63 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure By Fajans' rules, with a common anion the smaller cation polarises the chloride more and gives more covalent character. Cation size: $Li^+ < Na^+ < K^+ < Cs^+$.
Covalent character: LiCl > NaCl > KCl > CsCl.
Answer: (C). | ||||
| 64 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium KCl and HCl supply $Cl^-$ and $AgNO_3$ supplies $Ag^+$: each common ion lowers the solubility of AgCl ($K_{sp} = [Ag^+][Cl^-]$ fixed).
In deionised water there is no common ion, so the solubility is greatest there.
Answer: (D). | ||||
| 65 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter (A) False: Brownian motion keeps particles from settling and so stabilises a sol.
(B) False: an emulsion takes only a limited amount of dispersed phase; beyond that it breaks.
(C) False: mixing oppositely charged sols neutralises the charges and coagulates them.
(D) True: like charges make the particles repel each other, which prevents them from coming together.
Answer: (D). | ||||
| 66 | 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 Pt (Z = 78) has the irregular ground-state configuration $[Xe]4f^{14}5d^96s^1$ (Xe = 54; 54 + 14 + 9 + 1 = 78).
Answer: (A). | ||||
| 67 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Gold and silver are leached from their ores with dilute NaCN in the presence of air, forming $[Au(CN)_2]^-$ and $[Ag(CN)_2]^-$.
Zinc: NaCN is used as a depressant in froth flotation to keep ZnS back when separating it from PbS in a mixed sulphide ore.
Copper: its sulphide ores are concentrated by froth flotation, roasted, smelted and refined electrolytically; cyanide is not part of the usual process.
Answer: (D). | ||||
| 68 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions As a reducing agent, $H_2O_2$ is itself oxidised to $O_2$.
(A) $H_2O_2$ is oxidised to $O_2$, but the medium is acidic ($H_3O^+$ formed).
(B) and (D): $H_2O_2$ is reduced to $H_2O$/$OH^-$; it acts as an oxidant.
(C) $MnO_4^-$ (Mn +7) is reduced to $MnO_2$ (Mn +4) while $H_2O_2$ is oxidised to $O_2$, and $OH^-$ shows the medium is basic.
Answer: (C). | ||||
| 69 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter The characteristic flame emission lines: Li crimson red 670.8 nm, Na yellow 589.2 nm, Rb red-violet 780.0 nm, Cs blue 455.5 nm.
(A)-(I), (B)-(II), (C)-(III), (D)-(IV).
Answer: (A). | ||||
| 70 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Cs: very low ionisation enthalpy, used in photoelectric cells → (III).
Ga: liquid over a very wide range (m.p. 303 K, b.p. about 2500 K), used in high-temperature thermometers → (I).
B: boron carbide fibres are used in bullet-proof vests → (IV).
Si: silicones are water-repellent and used in water-repellent sprays → (II).
Answer: (A). | ||||
| 71 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$: the alkali gives sodium hypophosphite, the salt of hypophosphorous acid $H_3PO_2$.
$H_3PO_2$, $HP(O)(H)(OH)$ in effect, has two P–H bonds and one P–OH; it is phosphinic acid.
(Phosphonic acid $H_3PO_3$ has one P–H; hypophosphoric acid $H_4P_2O_6$ has none.)
Answer: (B). | ||||
| 72 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter $SO_2$ (and $NO_x$) from industries form acid rain; sulphuric acid in it reacts with the marble: $CaCO_3 + H_2SO_4 \rightarrow CaSO_4 + H_2O + CO_2$, eating away the monument.
Answer: (A). | ||||
| 73 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Acidic $KMnO_4$ cleaves C=C: a $=CH_2$ end becomes $CO_2$ (effervescence), a $=CR_2$ end becomes a ketone.
2-Methylpropene $(CH_3)_2C=CH_2 \rightarrow (CH_3)_2C=O$ (acetone) $+ CO_2$.
But-1-ene gives propanoic acid + $CO_2$ (no ketone); the but-2-enes give acetic acid only.
Answer: (D). | ||||
| 74 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctHard | |
A hard single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $(CH_3)_3CLi$ is a very strong base. It first removes the acidic phenolic proton, giving $ArO^-Li^+$; a second equivalent then exchanges Br for Li (lithium–halogen exchange), giving the dilithio compound with Li on the ring and OLi ortho to it. That is A.
$CO_2$ adds to the C–Li carbon to give the carboxylate, and $H_3O^+$ protonates both oxygens: salicylic acid (2-hydroxybenzoic acid), as shown.
Answer: (B). | ||||
| 75 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | StatementsMedium | |
A medium statements question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Statement I: the alcohol (nucleophile) attacks the protonated carbonyl carbon of the acid and, through a tetrahedral intermediate, $-OH$ is replaced by $-OR$: a nucleophilic acyl substitution. Correct.
Statement II: an electron-withdrawing group makes the carbonyl carbon more electrophilic, which speeds up the nucleophilic attack, the slow step. Correct.
Answer: (A). | ||||
| 76 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines A: the strongly activating $-NH_2$ makes aniline react with excess bromine water at all ortho and para positions: 2,4,6-tribromoaniline.
B: acetylation ($(CH_3CO)_2O$) gives acetanilide, whose $-NHCOCH_3$ is only moderately activating and is bulky, so $Br_2$/$CH_3COOH$ gives mainly the para-bromo product; acid hydrolysis (HCl) removes the acetyl group: 4-bromoaniline.
Answer: (C). | ||||
| 77 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Elastomers have weak intermolecular forces and a few cross-links that pull the chains back when the force is removed. Buna-N (acrylonitrile–butadiene rubber) is an elastomer.
Bakelite is a thermosetting polymer; nylon 6,6 and terylene are fibres.
Answer: (C). | ||||
| 78 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules DNA contains β-D-2-deoxyribose; RNA contains β-D-ribose.
Answer: (B). | ||||
| 79 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Cimetidine (a thioether and a cyano-guanidine) and ranitidine (a thioether and a nitro-ethenediamine) contain sulfur; saccharin is a cyclic sulphonimide ($-SO_2-NH-$).
Histamine is 2-(imidazol-4-yl)ethanamine, $C_5H_9N_3$: no sulfur.
Answer: (C). | ||||
| 80 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | StatementsEasy | |
A easy statements question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Statement I: phenol ($pK_a \approx 10$) is a weak acid. Correct.
Statement II: phenols do dissolve in NaOH (forming phenoxide), but they are stronger acids than alcohols and water, because the phenoxide ion is resonance-stabilised. Incorrect.
Answer: (C). | ||||
| 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 $M = \frac{dRT}{P}$ with $T = 257 + 273 = 530$ K and $P = \frac{100}{760}$ atm.
$M = \frac{0.46 \times 0.082 \times 530 \times 760}{100} = \frac{19.99 \times 760}{100} = 151.9\ g\ mol^{-1}$.
To the nearest integer: 152.
Answer: 152. | ||||
| 82 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics 85 g of $NH_3$ is $\frac{85}{17} = 5$ mol.
$\Delta H = 5 \times 23.4 = 117$ kJ.
Answer: 117. | ||||
| 83 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Mass of acid $= 1.2 \times 1.02 = 1.224$ g $= 0.0204$ mol. In 2.0 L of this dilute solution (about 2 kg of water), molality $m \approx 0.0102\ mol\ kg^{-1}$.
$i = \frac{\Delta T_f}{K_f m} = \frac{0.0198}{1.85 \times 0.0102} = 1.049$.
For $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$: $i = 1 + \alpha$, so $\alpha = 0.049$, about 4.9%.
To the nearest integer: 5.
Answer: 5. | ||||
| 84 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Charge $= 0.10 \times 2 \times 3600 = 720$ C, i.e. $\frac{720}{96500} = 7.461 \times 10^{-3}$ mol of electrons.
Cathode: $2H^+ + 2e^- \rightarrow H_2$: $3.731 \times 10^{-3}$ mol. Anode: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$: $1.865 \times 10^{-3}$ mol.
Total $= 5.596 \times 10^{-3}$ mol; volume $= 5.596 \times 10^{-3} \times 22700 = 127.0\ cm^3$.
Answer: 127. | ||||
| 85 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln\frac{k_{300}}{k_{310}} = -\frac{E_a}{R}\left(\frac{1}{300} - \frac{1}{310}\right) = -\frac{532611}{8.3} \times \frac{10}{93000} = -64170 \times 1.0753 \times 10^{-4} = -6.9$.
$6.9 = 3 \times 2.3 = 3\ln 10$, so $\frac{k_{300}}{k_{310}} = 10^{-3}$.
$x = 1$.
Answer: 1. | ||||
| 86 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Fusing chromite with sodium carbonate in air gives sodium chromate: A = $Na_2CrO_4$.
Acidification: $2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$: B = $Na_2Cr_2O_7$.
In $Cr_2O_7^{2-}$ two $CrO_4$ tetrahedra share one bridging oxygen; each Cr carries three terminal O.
Terminal oxygen atoms: 6.
Answer: 6. | ||||
| 87 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$ (Mn +6 to +7 and +4).
The higher oxidation state product is $MnO_4^-$, with Mn(+7): $3d^0$, no unpaired electrons.
Spin-only moment $= \sqrt{n(n+2)} = 0$ B.M.
Answer: 0. | ||||
| 88 | 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 Moles of $H_2SO_4 = 12.5 \times 10^{-3} \times 1 = 0.0125$; $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$, so $NH_3 = 0.025$ mol.
Mass of N $= 0.025 \times 14 = 0.35$ g.
% N $= \frac{0.35}{0.55} \times 100 = 63.6$.
To the nearest integer: 64.
Answer: 64. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques 2-Chlorobutane: C2 carries H, Cl, $CH_3$, $C_2H_5$: asymmetric.
Glycine ($H_2NCH_2COOH$): no carbon with four different groups.
Alanine: the α-carbon carries H, $NH_2$, $CH_3$, COOH: asymmetric.
Styrene: none.
The substituted cyclohexanol (2-isopropyl-5-methylcyclohexan-1-ol, menthol): the carbons bearing OH, isopropyl and methyl are all asymmetric.
Compounds with asymmetric carbon: 3.
Answer: 3. | ||||
| 90 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | NumericalEasy | |
A easy numerical question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Zymase ferments glucose: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2$, so A = ethanol.
Iodoform reaction: $C_2H_5OH + 4I_2 + 6NaOH \rightarrow CHI_3 + HCOONa + 5NaI + 5H_2O$, so B = sodium formate, HCOONa.
Carbon atoms in B: 1.
Answer: 1. | ||||
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