JEE (Main) · 2023
JEE (Main) 2023 — 12 April, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Three Dimensional Geometry3
- Sets, Relations and Functions2
7 easy · 20 medium · 3 hard
Physics
30 Q
- Gravitation2
- Work, Energy and Power2
- Electromagnetic Induction and Alternating Currents2
25 easy · 5 medium · 0 hard
Chemistry
30 Q
- Aldehydes, Ketones and Carboxylic Acids3
- Amines3
- Some Basic Concepts of Chemistry2
15 easy · 13 medium · 2 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctHard | |
A hard single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Write $t = \log_3 x$ and $A = \frac{6 + 2t}{-5x}$.
The logarithm needs $x > 0$, $3x \neq 1$ and $A > 0$. Since $-5x < 0$, $A > 0$ needs $6 + 2t < 0$, i.e. $t < -3$, so $x < \frac{1}{27}$. The base $3x$ is then less than 1.
$\sin^{-1}$ needs $-1 \le \log_{3x} A \le 1$. With a base below 1 the logarithm is decreasing, so this is $3x \le A \le \frac{1}{3x}$.
Upper condition: $\frac{-(6 + 2t)}{5x} \le \frac{1}{3x}$ gives $-(6 + 2t) \le \frac{5}{3}$, so $t \ge -\frac{23}{6}$, i.e. $x \ge 3^{-23/6}$.
Lower condition: $-(6 + 2t) \ge 15x^2$. At $x = \frac{1}{27}$ the left side is 0 and the right side about 0.0206, so the condition fails near $\frac{1}{27}$; it holds up to $x_0 \approx 0.036630$, the root of $-(6 + 2\log_3 x) = 15x^2$.
So $D = [3^{-23/6}, x_0]$. On $D \subset (0, 1)$, $[x] = 0$ and $g(x) = x$, so the range of $g$ is the closed interval $[3^{-23/6}, x_0]$, not an open interval $(\alpha, \beta)$.
With its ends, $\alpha^2 + \frac{5}{\beta} = 3^{-23/3} + \frac{5}{x_0} \approx 136.50$, which is none of the options; 135 needs the upper end taken as $\frac{1}{27}$, which ignores the condition $\log_{3x} A \le 1$.
No option is correct as printed. | ||||
| 2 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $x = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2} = \sqrt{3}\cdot\frac{-1 \pm i}{\sqrt{2}} = \sqrt{3}\,e^{\pm i\,3\pi/4}$.
So $\alpha^n + \beta^n = 2\cdot 3^{n/2}\cos\frac{3n\pi}{4}$.
$n = 14$: $\cos\frac{42\pi}{4} = \cos\frac{21\pi}{2} = 0$; $n = 10$: $\cos\frac{30\pi}{4} = \cos\frac{15\pi}{2} = 0$.
$n = 23$: $\cos\frac{69\pi}{4} = \cos\left(17\pi + \frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$, so $\alpha^{23} + \beta^{23} = -\sqrt{2}\cdot 3^{23/2}$.
$n = 15$: $\cos\frac{45\pi}{4} = \cos\left(11\pi + \frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$, so $\alpha^{15} + \beta^{15} = -\sqrt{2}\cdot 3^{15/2}$.
Ratio $= \frac{3^{23/2}}{3^{15/2}} = 3^4 = 81$.
Answer: (C). | ||||
| 3 | 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 Let $P = \begin{bmatrix}1 & 2\\-1 & -1\end{bmatrix}$. Then $P^2 = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix} = -I$, so $P^{-1} = -P = \begin{bmatrix}-1 & -2\\1 & 1\end{bmatrix}$.
Hence $B = PAP^{-1}$ and $B^n = PA^nP^{-1}$, with $A^n = \begin{bmatrix}1 & \frac{n}{51}\\0 & 1\end{bmatrix}$.
$\sum_{n=1}^{50} A^n = \begin{bmatrix}50 & \frac{1275}{51}\\0 & 50\end{bmatrix} = \begin{bmatrix}50 & 25\\0 & 50\end{bmatrix} = M$.
$PM = \begin{bmatrix}50 & 125\\-50 & -75\end{bmatrix}$, and $PMP^{-1} = \begin{bmatrix}75 & 25\\-25 & 25\end{bmatrix}$.
Sum of the elements $= 75 + 25 - 25 + 25 = 100$.
Answer: (C). | ||||
| 4 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The first digit must be 5, 7 or 9 (to exceed 40000); the last digit must be 0 or 5.
Last digit 0: first digit 3 ways, middle three digits from the remaining 4: $4\cdot3\cdot2 = 24$; total 72.
Last digit 5: first digit 7 or 9 (2 ways), middle three from the remaining 4 (including 0): 24; total 48.
Total $= 72 + 48 = 120$.
Answer: (C). | ||||
| 5 | 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 The sum is $\sum_{r=0}^{49} (-1)^r\,{}^{100}C_r$.
Using ${}^{100}C_r = {}^{99}C_r + {}^{99}C_{r-1}$ the sum telescopes: $\sum_{r=0}^{k} (-1)^r\,{}^{n}C_r = (-1)^k\,{}^{n-1}C_k$.
With $n = 100$, $k = 49$: the sum is $(-1)^{49}\,{}^{99}C_{49} = -{}^{99}C_{49}$.
Answer: (B). | ||||
| 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 Since ${}^{n}C_{n-k} = {}^{n}C_k$, the sum is $\sum_{k=0}^{n} \frac{{}^{n}C_k}{k+1}$.
$\frac{{}^{n}C_k}{k+1} = \frac{{}^{n+1}C_{k+1}}{n+1}$, so the sum is $\frac{2^{n+1} - 1}{n+1}$ (equivalently $\int_0^1 (1+x)^n\,dx$).
$\frac{2^{n+1} - 1}{n+1} = \frac{1023}{10}$ gives $n + 1 = 10$, $n = 9$.
Answer: (D). | ||||
| 7 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z_1 - z_0 = \frac{1}{2} - \frac{i}{2}$, so $|z_1 - z_0| = \frac{1}{\sqrt{2}}$ and $|z_2 - z_0| = \sqrt{2}$ (outside the circle, as required).
Collinear: $z_2 = z_0 \pm \sqrt{2}\cdot\frac{1 - i}{\sqrt{2}} = z_0 \pm (1 - i)$.
$z_2 = \frac{3}{2} + \frac{1}{2}i$ gives $|z_2|^2 = \frac{9}{4} + \frac{1}{4} = \frac{5}{2}$; $z_2 = -\frac{1}{2} + \frac{5}{2}i$ gives $\frac{1}{4} + \frac{25}{4} = \frac{13}{2}$.
The smaller value is $\frac{5}{2}$.
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 $S_n = \frac{n^2 + 3n}{(n+1)(n+2)} = 1 - \frac{2}{(n+1)(n+2)}$.
$a_n = S_n - S_{n-1} = \frac{2}{n(n+1)} - \frac{2}{(n+1)(n+2)} = \frac{4}{n(n+1)(n+2)}$ (also true for $n = 1$: $a_1 = \frac{2}{3}$).
$\sum_{k=1}^{10} \frac{1}{a_k} = \frac{1}{4}\sum_{k=1}^{10} k(k+1)(k+2) = \frac{1}{4}\cdot\frac{10\cdot11\cdot12\cdot13}{4} = \frac{4290}{4}$.
$28 \times \frac{4290}{4} = 30030 = 2\cdot3\cdot5\cdot7\cdot11\cdot13$, the first 6 primes. So $m = 6$.
Answer: (B). | ||||
| 9 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctHard | |
A hard single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $s = \sin^2 x \in (0, 1)$. Then $\ln f = s\ln\frac{\sqrt{3e}}{2\sqrt{s}} = \frac{s}{2}\left(\ln\frac{3e}{4} - \ln s\right)$.
$\frac{d}{ds}\ln f = \frac{1}{2}\left(\ln\frac{3e}{4} - \ln s\right) - \frac{1}{2} = \frac{1}{2}\ln\frac{3}{4s}$, which is zero at $s = \frac{3}{4}$, positive before and negative after: a local maximum (at $x = \frac{\pi}{3}$).
There $\ln f = \frac{3}{8}\left(\ln\frac{3e}{4} - \ln\frac{3}{4}\right) = \frac{3}{8}$, so the maximum value is $e^{3/8} = \frac{k}{e}$ and $k = e^{11/8}$, $k^8 = e^{11}$.
$\left(\frac{k}{e}\right)^8 + \frac{k^8}{e^5} + k^8 = e^3 + e^6 + e^{11}$.
Answer: (D). | ||||
| 10 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Slope at $x = -1$ is $3x^2 = 3$: tangent $y = 3x + 2$.
$x^3 = 3x + 2 \Rightarrow (x + 1)^2(x - 2) = 0$, so the tangent meets the curve again at $x = 2$.
On $[-1, 2]$ the tangent lies above the curve: area $= \int_{-1}^{2} (3x + 2 - x^3)\,dx = \left[\frac{3x^2}{2} + 2x - \frac{x^4}{4}\right]_{-1}^{2} = 6 - \left(-\frac{3}{4}\right) = \frac{27}{4}$.
Answer: (C). | ||||
| 11 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dy}{dx} - \frac{x}{1 + x^2}y = -\frac{y^2}{1 + x^2}$ (Bernoulli). Put $v = \frac{1}{y}$: $\frac{dv}{dx} + \frac{x}{1 + x^2}v = \frac{1}{1 + x^2}$.
Integrating factor $\sqrt{1 + x^2}$: $\frac{d}{dx}\left(v\sqrt{1 + x^2}\right) = \frac{1}{\sqrt{1 + x^2}}$, so $\frac{\sqrt{1 + x^2}}{y} = \ln\left(x + \sqrt{1 + x^2}\right) + C$.
$y(0) = 1$ gives $C = 1$.
At $x = 2\sqrt{2}$: $\sqrt{1 + 8} = 3$, so $\frac{3}{\beta} = \ln(3 + 2\sqrt{2}) + 1$, i.e. $e^{3\beta^{-1}} = e(3 + 2\sqrt{2})$.
Answer: (D). | ||||
| 12 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$ (P lies on it: $9\cdot\frac{12}{7} + 4\cdot\frac{36}{7} = 36$).
PQ and RS are diameters. A point at distance $\rho$ from the centre in direction $\theta$ satisfies $\frac{1}{\rho^2} = \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{9}$, so for perpendicular directions $\frac{1}{OP^2} + \frac{1}{OR^2} = \frac{1}{4} + \frac{1}{9} = \frac{13}{36}$.
$PQ = 2\,OP$ and $RS = 2\,OR$, so $\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4}\cdot\frac{13}{36} = \frac{13}{144}$.
$p + q = 13 + 144 = 157$.
Answer: (D). | ||||
| 13 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines The line meets the axes at $\left(\frac{7}{\cos\theta}, 0\right)$ and $\left(0, \frac{7}{\sin\theta}\right)$; the mid-point is $h = \frac{7}{2\cos\theta}$, $k = \frac{7}{2\sin\theta}$, so the locus is $\frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{49}$, with $h, k > 0$ for $\theta \in \left(0, \frac{\pi}{2}\right)$.
$k = \frac{7\sqrt{3}}{3} = \frac{7}{\sqrt{3}}$: $\frac{1}{h^2} = \frac{4}{49} - \frac{3}{49} = \frac{1}{49}$, so $h = 7$ (positive). ($\theta = \frac{\pi}{3}$.)
$\alpha = 7$.
Answer: (C). | ||||
| 14 | 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 new plane contains the line of intersection: $4x - y + z - 10 + \lambda(x + y - z - 4) = 0$, normal $(4 + \lambda, \lambda - 1, 1 - \lambda)$.
It is perpendicular to P: $4(4 + \lambda) - (\lambda - 1) + (1 - \lambda) = 18 + 2\lambda = 0$, $\lambda = -9$.
New plane: $-5x - 10y + 10z + 26 = 0$, i.e. $5x + 10y - 10z - 26 = 0$.
Distance of $(2, 3, -4)$: $\alpha = \frac{|10 + 30 + 40 - 26|}{\sqrt{25 + 100 + 100}} = \frac{54}{15} = \frac{18}{5}$.
$35\alpha = 126$.
Answer: (D). | ||||
| 15 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Direction of $l_2$: $(3, 2, 1)\times(1, -3, 2) = (7, -5, -11)$; a point on $l_2$ is $(1, -2, 3)$ (check: $3 - 4 + 3 - 2 = 0$, $1 + 6 + 6 - 13 = 0$).
Coplanarity with $l_1$ (through $(-5, -4, \alpha)$, direction $(3, 1, -2)$): $(3, 1, -2)\times(7, -5, -11) = (-21, 19, -22)$ and
$(6, 2, 3 - \alpha)\cdot(-21, 19, -22) = -126 + 38 - 66 + 22\alpha = 0$, so $\alpha = 7$.
A point of $l_1$: $(-5 + 3t, -4 + t, 7 - 2t)$. For the nearest point, $Q - P = (1 - 3t, 1 - t, -5 + 2t)$ is perpendicular to $(3, 1, -2)$: $14 - 14t = 0$, $t = 1$.
$P = (-2, -3, 5)$ and $|a| + |b| + |c| = 10$.
Answer: (B). | ||||
| 16 | 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 Coplanar: $\begin{vmatrix}a & 1 & 1\\1 & b & 1\\1 & 1 & c\end{vmatrix} = 0$.
Subtract row 1 from rows 2 and 3: $\begin{vmatrix}a & 1 & 1\\1 - a & b - 1 & 0\\1 - a & 0 & c - 1\end{vmatrix} = 0$.
Expanding and dividing by $(1 - a)(1 - b)(1 - c)$: $\frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 0$.
Since $\frac{a}{1 - a} = \frac{1}{1 - a} - 1$, this gives $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 1$.
Answer: (B). | ||||
| 17 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $(\vec{a} + \vec{b})\times\vec{c} = \vec{0}$, so $\vec{c} = \mu(\vec{a} + \vec{b}) = \mu(\lambda + 3, 0, 1)$.
$\vec{a}\cdot\vec{c} = \mu(\lambda^2 + 3\lambda - 1) = -17$ and $\vec{b}\cdot\vec{c} = \mu(3\lambda + 11) = -20$.
Dividing: $20(\lambda^2 + 3\lambda - 1) = 17(3\lambda + 11)$, i.e. $20\lambda^2 + 9\lambda - 207 = 0$, so $\lambda = 3$ or $-\frac{69}{20}$; $\lambda \in \mathbb{Z}$ gives $\lambda = 3$ and then $\mu = -1$.
$\vec{c} = (-6, 0, -1)$; $\vec{c}\times(3, 1, 1) = (0\cdot1 - (-1)\cdot1,\ (-1)\cdot3 - (-6)\cdot1,\ (-6)\cdot1 - 0\cdot3) = (1, 3, -6)$.
$|\ldots|^2 = 1 + 9 + 36 = 46$.
Answer: (A). | ||||
| 18 | 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 For one fair die, $E(X) = \frac{7}{2}$, $E(X^2) = \frac{91}{6}$, so $\mathrm{Var}(X) = \frac{91}{6} - \frac{49}{4} = \frac{35}{12}$.
The dice are independent, so $\mathrm{Var}(\alpha - \beta) = \frac{35}{12} + \frac{35}{12} = \frac{35}{6}$.
$p = 35$; its divisors $1 + 5 + 7 + 35 = 48$.
Answer: (C). | ||||
| 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 $\cos A + \cos C = 2(1 - \cos B) = 4\sin^2\frac{B}{2}$. Also $\cos A + \cos C = 2\sin\frac{B}{2}\cos\frac{A - C}{2}$, so $\cos\frac{A - C}{2} = 2\sin\frac{B}{2}$.
Then $\sin A + \sin C = 2\cos\frac{B}{2}\cos\frac{A - C}{2} = 4\sin\frac{B}{2}\cos\frac{B}{2} = 2\sin B$, so $a + c = 2b$ and $b = 5$.
$\cos A = \frac{25 + 49 - 9}{70} = \frac{13}{14}$, $\cos C = \frac{9 + 25 - 49}{30} = -\frac{1}{2}$.
$\cos A - \cos C = \frac{13}{14} + \frac{7}{14} = \frac{10}{7}$.
Answer: (B). | ||||
| 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 (S1): $p \wedge (\sim q)$ is true only when $p$ is true and $q$ false, and exactly then $p \Rightarrow q$ is false. The conjunction is never true: a contradiction. (S1) is true.
(S2): the four terms cover all four truth combinations of $p$ and $q$, so for every assignment one of them is true: a tautology. (S2) is true.
Answer: (A). | ||||
| 21 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Reflexive: $(1,1), (2,2), (3,3)$ are in. Transitive with $(1,2), (2,3)$: $(1,3)$ is in. So every such relation contains $R_0 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$, and may add some of $(2,1), (3,2), (3,1)$.
$R_0$: transitive, not symmetric. ✓
$R_0 \cup \{(2,1)\}$: transitive (e.g. $(2,1),(1,3) \to (2,3)$ present), not symmetric. ✓
$R_0 \cup \{(3,2)\}$: transitive (e.g. $(1,3),(3,2) \to (1,2)$ present), not symmetric. ✓
$R_0 \cup \{(3,1)\}$: $(3,1),(1,2)$ needs $(3,2)$ — not transitive.
$\{(2,1),(3,2)\}$: needs $(3,1)$; $\{(2,1),(3,1)\}$: needs $(3,2)$; $\{(3,2),(3,1)\}$: $(2,3),(3,1)$ needs $(2,1)$. All three added gives the universal relation, which is symmetric.
So there are 3 such relations.
Answer: 3. | ||||
| 22 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Only row 1 depends on $k$: $\sum_{k=1}^{n} (1, 2k, 2k - 1) = (n, n(n+1), n^2)$.
$\sum D_k = \begin{vmatrix}n & n^2 + n & n^2\\ n & n^2 + n + 2 & n^2\\ n & n^2 + n & n^2 + n + 2\end{vmatrix}$. Subtract row 1 from rows 2 and 3: rows become $(0, 2, 0)$ and $(0, 0, n + 2)$.
$\sum D_k = n\cdot2\cdot(n + 2) = 96 \Rightarrow n^2 + 2n - 48 = 0 \Rightarrow n = 6$.
Answer: 6. | ||||
| 23 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalHard | |
A hard numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Take $a, b, c$ distinct with $b$ the middle term. A run of three consecutive digits in A.P. (with non-zero common difference) is $abc$ or $cba$.
Total arrangements of $a^3b^3c^3$: $\frac{9!}{3!\,3!\,3!} = 1680$.
Count those with no window $abc$ or $cba$ and subtract. Exact inclusion–exclusion over the windows (or listing all 1680 arrangements) gives 736 arrangements with no such window, so $1680 - 736 = 944$ arrangements contain $abc$ or $cba$ at least once.
(Treating $abc$ as a block among the other six digits gives $2\times\frac{7!}{2!\,2!\,2!} = 1260$, but that counts an arrangement with two such blocks, e.g. $abc\,abc\,\ldots$, more than once.)
Answer: 944. | ||||
| 24 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Sum $S = \frac{31}{2}$ and sum of reciprocals $= \frac{31}{8}$. For a G.P., $\sum\frac{1}{a_k} = \frac{S}{a_1a_5}$, so $a_1a_5 = a_3^2 = 4$ and $a_3 = 2$.
$a_3 + a_4 + a_5 = 2(1 + r + r^2) = 14 \Rightarrow r^2 + r - 6 = 0 \Rightarrow r = 2$ (positive terms).
Terms: $\frac{1}{2}, 1, 2, 4, 8$ (sum 15.5 ✓).
Mean of squares $= \frac{1}{5}\left(\frac{1}{4} + 1 + 4 + 16 + 64\right) = \frac{341}{20}$; variance $= \frac{341}{20} - \left(\frac{31}{10}\right)^2 = \frac{1705 - 961}{100} = \frac{744}{100} = \frac{186}{25}$.
$m + n = 186 + 25 = 211$.
Answer: 211. | ||||
| 25 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Between integers $f$ is continuous; check $x = -1$ and $x = 0$.
At $x = -1$: from the left $[x] = -2$, $f \to 2 + \sqrt{1} = 3$; at and right of $-1$, $f \to 1 + 0 = 1$. Discontinuous.
At $x = 0$: from the left $[x] = -1$, $f \to 1 + 1 = 2$; at and right of 0, $f \to 0$. Discontinuous.
Number of points $= 2$.
Answer: 2. | ||||
| 26 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalEasy | |
A easy numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The integrand is even and changes sign at $x = 0.1$.
$\int_0^{0.1} (1 - 100x^2)\,dx = 0.1 - \frac{100}{3}(0.001) = \frac{1}{15}$.
$\int_{0.1}^{0.15} (100x^2 - 1)\,dx = \frac{100}{3}(0.003375 - 0.001) - 0.05 = \frac{7}{240}$.
Total $= 2\left(\frac{16}{240} + \frac{7}{240}\right) = \frac{23}{120} = \frac{575}{3000}$, so $k = 575$.
Answer: 575. | ||||
| 27 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $I(x) = \sqrt{x(x + 7)} + 7\ln\left(\sqrt{x} + \sqrt{x + 7}\right) + C$ (check: the derivative is $\frac{2x + 7}{2\sqrt{x(x+7)}} + \frac{7}{2\sqrt{x(x+7)}} = \sqrt{\frac{x + 7}{x}}$).
$I(9) = 12 + 7\ln 7 + C$, so $C = 0$.
$I(1) = \sqrt{8} + 7\ln(1 + 2\sqrt{2})$, so $\alpha = 2\sqrt{2}$ and $\alpha^4 = 64$.
Answer: 64. | ||||
| 28 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections A circle touching both axes in the first quadrant has centre $(r, r)$ and radius $r$.
Distance of the centre from $x + y = 2$: $\frac{|2r - 2|}{\sqrt{2}}$. Half the chord is 1: $r^2 - \frac{(2r - 2)^2}{2} = 1 \Rightarrow r^2 - 4r + 3 = 0$, so $r = 1, 3$.
$r_1^2 + r_2^2 - r_1r_2 = 1 + 9 - 3 = 7$.
Answer: 7. | ||||
| 29 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium 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 $A(-6, 0, 0)$, $B(0, -2, 0)$, $C(0, 0, 3)$. With O the origin, $OA, OB, OC$ are mutually perpendicular, so the orthocentre of ABC is the foot of the perpendicular from O to the plane. (For example $OH \perp$ plane and $OA \perp BC$, $OH \perp BC$ give $AH \perp BC$.)
Foot: $t(1, 3, -2)$ with $t + 9t + 4t + 6 = 0$, $t = -\frac{3}{7}$: $H = \left(-\frac{3}{7}, -\frac{9}{7}, \frac{6}{7}\right)$ (the $z$-coordinate matches).
$\alpha + \beta = -\frac{12}{7}$; $98\cdot\frac{144}{49} = 288$.
Answer: 288. | ||||
| 30 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each toss succeeds (shows less than $n$) with probability $p = \frac{n - 1}{n}$. The number of tosses is geometric with mean $\frac{1}{p} = \frac{n}{n - 1}$.
$\frac{n}{n - 1} = \frac{n}{9} \Rightarrow n = 10$.
Answer: 10. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Orbital speed $v = \sqrt{\frac{GM}{r}}$ does not depend on the satellite's mass, so it is the same for both.
Kinetic, potential and total energy are all proportional to the satellite's mass, so they differ by a factor of 2.
Answer: (A). | ||||
| 32 | 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 Spring constant $k = \frac{F}{x}$: $[MLT^{-2}]/[L] = [MT^{-2}]$ — II.
Angular speed $\omega = \frac{\theta}{t}$: $[T^{-1}]$ — I.
Angular momentum $L = mvr$: $[ML^2T^{-1}]$ — IV.
Moment of inertia $I = mr^2$: $[ML^2]$ — III.
A-II, B-I, C-IV, D-III.
Answer: (B). | ||||
| 33 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | StatementsEasy | |
A easy statements question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Statement I: by the work–energy theorem $F\,s = K$; equal K and equal F give equal stopping distance s. Correct.
Statement II: the direction of the velocity changes, so the velocity changes and the acceleration is not zero. Incorrect.
Answer: (C). | ||||
| 34 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter Carrier: $\omega_c = 1000\pi$, $f_c = 500$ Hz. Message: $\omega_m = 4\pi$, $f_m = 2$ Hz.
An AM wave contains the carrier $f_c$ and the side bands $f_c \pm f_m$: 500 Hz, 498 Hz and 502 Hz.
So A, D and E.
Answer: (A). | ||||
| 35 | 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 Statement I: $I = \frac{V}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}}$. As $\omega$ rises from low values the reactance falls to zero at resonance and then grows, so the current rises to a maximum $V/R$ and then falls. True.
Statement II: at resonance $Z = R$, so $\cos\phi = R/Z = 1$. True.
Answer: (A). | ||||
| 36 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}}$, so $\frac{v_p}{v_E} = \sqrt{\frac{16}{4}} = 2$. Ratio 2 : 1.
Answer: (A). | ||||
| 37 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium single correct question on 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 Newton's law of cooling in its average-temperature form: $\frac{\Delta T}{t} = k\left(\bar{T} - T_0\right)$.
First interval: $\frac{20}{5} = k(70 - 20)$, so $k = 0.08$ per minute.
Second interval: $\frac{20}{t} = 0.08(50 - 20) = 2.4$, so $t = \frac{25}{3}$ min $= 500$ s.
Answer: (B). | ||||
| 38 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics The Carnot efficiency between 373 K and 273 K is $1 - \frac{273}{373} = \frac{100}{373} \approx 26.8\%$, the greatest possible.
A practical (irreversible) engine has a lower efficiency than the Carnot engine (B true), and so less than 26.8%, i.e. less than 27% (D true). A and C are false.
So B and D only.
Answer: (B). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms} = \sqrt{\frac{3RT}{M}}$, so at the same temperature $v_{Ar} = v_{Cl_2}\sqrt{\frac{M_{Cl_2}}{M_{Ar}}} = 490\sqrt{\frac{70.9}{39.9}} = 490 \times 1.333 \approx 653$ m/s.
The nearest option is 651.7 m/s.
Answer: (C). | ||||
| 40 | 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 $U = \frac{1}{2}kx^2$ and $K = \frac{1}{2}k(A^2 - x^2)$. At $x = \frac{A}{2}$: $U \propto \frac{A^2}{4}$, $K \propto \frac{3A^2}{4}$.
$U : K = 1 : 3$.
Answer: (B). | ||||
| 41 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Assertion and reasonEasy | |
A easy assertion and reason question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance By Gauss's law the net flux is $\frac{q_{enc}}{\varepsilon_0}$. A dipole is two equal and opposite charges, so $q_{enc} = 0$ and the flux is zero.
A is true, R is true, and R is exactly why A holds.
Answer: (A). | ||||
| 42 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity The volume is unchanged, so $A \propto \frac{1}{l}$ and $R = \rho\frac{l}{A} \propto l^2$.
$l \to \frac{l}{4}$: $R \to \frac{160}{16} = 10$ Ω.
Answer: (A). | ||||
| 43 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsEasy | |
A easy statements 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 Statement I: diamagnetism arises from induced orbital moments and is (practically) independent of temperature; it is paramagnetism that follows Curie's law. Incorrect.
Statement II: the induced moment in a diamagnetic substance opposes the applied field (negative susceptibility). True.
Answer: (D). | ||||
| 44 | 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 $v = u - gt$: $v_3 = 150 - 30 = 120$ m/s, $v_5 = 150 - 50 = 100$ m/s.
$\frac{120}{100} = \frac{6}{5} = \frac{x + 1}{x}$, so $x = 5$.
Answer: (A). | ||||
| 45 | Laws of MotionEquilibrium of concurrent forces | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on equilibrium of concurrent forces, from Laws of Motion. NCERT: Class 11: Laws of Motion In equilibrium, $F_1$ balances the resultant of $F_2$ and $F_3$: $\sqrt{8^2 + 6^2} = 10$ N.
Removing $F_1$ leaves a net force of 10 N, so $a = \frac{10}{5} = 2$ m s$^{-2}$.
Answer: (A). | ||||
| 46 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Real depths from the two faces are $12\mu$ and $4\mu$ (apparent depth = real depth / $\mu$), and they add up to the side:
$12\mu + 4\mu = 24$, so $\mu = \frac{3}{2}$.
Answer: (D). | ||||
| 47 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{\sqrt{2mqV}}$. $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha V_\alpha}{m_p q_p V_p}} = \sqrt{\frac{4 \times 2 \times 4}{1 \times 1 \times 2}} = \sqrt{16} = 4$.
Ratio 4 : 1.
Answer: (B). | ||||
| 48 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei At room temperature the atoms are in $n = 1$. Excitation energies: $n = 3$ needs $13.6\left(1 - \frac{1}{9}\right) = 12.09$ eV; $n = 4$ needs 12.75 eV.
A 12.5 eV electron can raise the atom to $n = 3$ but not $n = 4$.
From $n = 3$ the lines are $3\to2$, $3\to1$, $2\to1$: $\frac{3 \times 2}{2} = 3$ lines.
Answer: (C). | ||||
| 49 | Electronic DevicesSemiconductors | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on semiconductors, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Current gain $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{16 - 5\ \text{mA}}{200 - 100\ \mu\text{A}} = \frac{11 \times 10^{-3}}{100 \times 10^{-6}} = 110$.
Answer: (C). | ||||
| 50 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Assertion and reasonEasy | |
A easy assertion and reason question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Optical communication uses visible/infrared light (wavelength about 1 µm), far shorter than radar microwaves (centimetres). A is false.
Infrared has a higher frequency than microwaves, so its photons are more energetic. R is true.
Answer: (D). | ||||
| 51 | Magnetic Effects of Current and MagnetismEarth magnetism | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on earth magnetism, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter A compass needle oscillates in the horizontal plane under the horizontal component $B_H = B\cos\delta$: $f = \frac{1}{2\pi}\sqrt{\frac{mB_H}{I}}$, so $f^2 \propto B\cos\delta$.
$\frac{f_1^2}{f_2^2} = \frac{B_1\cos30^\circ}{B_2\cos60^\circ}$: $\frac{400}{900} = \frac{B_1}{B_2}\cdot\sqrt{3}$.
$\frac{B_1}{B_2} = \frac{4}{9\sqrt{3}} = \frac{4}{\sqrt{243}}$, so $x = 243$.
Answer: 243. | ||||
| 52 | 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 At constant speed the engine's force equals the friction force $\mu mg = 0.04 \times 500 \times 10 = 200$ N.
Work $= 200 \times 4000 = 8 \times 10^5$ J $= 800$ kJ.
Answer: 800. | ||||
| 53 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For a thin spherical shell $I = \frac{2}{3}mR^2$. Rolling: $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{3}mv^2$, $K_{total} = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2$.
Ratio $= \frac{1/3}{5/6} = \frac{2}{5}$, so $x = 2$.
Answer: 2. | ||||
| 54 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical 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 Continuity: $A_1v_1 = A_2v_2 = Q$. Bernoulli (horizontal pipe): $\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}\rho Q^2\left(\frac{1}{A_2^2} - \frac{1}{A_1^2}\right)$.
$Q = A_1A_2\sqrt{\frac{2\Delta P}{\rho(A_1^2 - A_2^2)}} = (10 \times 10^{-4})(5 \times 10^{-4})\sqrt{\frac{2 \times 3}{1250 \times 0.75 \times 10^{-6}}}$
$= 5 \times 10^{-7} \times \sqrt{6400} = 5 \times 10^{-7} \times 80 = 4 \times 10^{-5}$ m$^3$ s$^{-1}$.
$x = 4$.
Answer: 4. | ||||
| 55 | 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 Resonances in the ratio 1 : 3 : 5 mean a pipe closed at one end, with $f_n = \frac{nv}{4L}$ for odd $n$.
Fifth harmonic: $\frac{5v}{4L} = 405$, so $\frac{v}{4L} = 81$ Hz and $L = \frac{324}{4 \times 81} = 1$ m.
Answer: 1. | ||||
| 56 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Volume conserved: $R = n^{1/3}r = 4r$. Charge $Q = 64q$.
$V = \frac{kQ}{R} = \frac{64}{4}\cdot\frac{kq}{r} = 16 \times 10 = 160$ mV.
Answer: 160. | ||||
| 57 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $R_T = R_0(1 + \alpha T)$ and $I \propto \frac{1}{R}$ for a fixed source voltage.
$\frac{R_{100}}{R_0} = \frac{2}{1.2} = \frac{5}{3}$, so $100\alpha = \frac{2}{3}$ and $R_{50} = R_0\left(1 + \frac{1}{3}\right) = \frac{4}{3}R_0$.
$I_{50} = \frac{3}{4} \times 2 = 1.5$ A $= 1500$ mA $= 15 \times 10^2$ mA.
Answer: 15. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\Phi = B\pi r^2$, $\varepsilon = B \cdot 2\pi r\frac{dr}{dt} = 0.4 \times 2\pi \times 0.02 \times 10^{-3} = 1.6\pi \times 10^{-5}$ V $\approx 50.3$ µV.
Nearest integer: 50.
Answer: 50. | ||||
| 59 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The first lens forms the image of the distant object at its focus, 20 cm behind it, i.e. 40 cm in front of the second lens.
Second lens: $u = -40$ cm, $f = 20$ cm: $\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40}$, $v = 40$ cm beyond the second lens.
Distance from the first lens $= 60 + 40 = 100$ cm.
Answer: 100. | ||||
| 60 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Mass defect $\Delta m = 238.05060 - 234.04360 - 4.00260 = 0.00440$ u.
$Q = 0.00440 \times 931.5 = 4.0986$ MeV $\approx 4$ MeV.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Molar mass $= 0.57 \times \frac{22400}{100} = 127.7$ g mol$^{-1}$ (22.4 L mol$^{-1}$ at STP).
Mass of chlorine per mole $= 0.55 \times 127.7 = 70.2$ g, i.e. $\frac{70.2}{35.5} \approx 2$ Cl atoms.
The formula is $MCl_2$.
Answer: (B). | ||||
| 62 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Acetylide ion $C_2^{2-}$ has $12 + 2 = 14$ electrons, like $N_2$: bond order 3, all electrons paired (diamagnetic).
$NO^+$: $15 - 1 = 14$ electrons, bond order 3, diamagnetic — the same.
$O_2^+$ (15 electrons, bond order 2.5), $N_2^+$ (13, 2.5) and $O_2^-$ (17, 1.5) are all paramagnetic.
Answer: (C). | ||||
| 63 | ElectrochemistryBatteries and fuel cells | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on batteries and fuel cells, from Electrochemistry. NCERT: Class 12: Electrochemistry In the lead storage battery the anode (during use) is lead and the cathode is a grid of lead packed with $PbO_2$; the electrolyte is about 38% $H_2SO_4$.
On discharge both electrodes become coated with $PbSO_4$. On charging the reactions reverse: $PbSO_4$ on the cathode (the $PbO_2$ electrode) is converted back into $PbO_2$, and $PbSO_4$ on the anode back into Pb.
A is false (the anode's $PbSO_4$ becomes Pb), B is true, C is false ($PbO_2$ is the cathode), D is true.
So B, D only.
Answer: (B). | ||||
| 64 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter Gases with higher critical temperatures liquefy more easily, have stronger van der Waals forces, and are adsorbed to a greater extent by charcoal.
Critical temperatures: D (154.3 K) > C (126.0 K) > B (33.2 K) > A (5.3 K), so the order of adsorption is D > C > B > A.
Answer: (A). | ||||
| 65 | 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 Densities (g cm$^{-3}$): Na 0.97, K 0.86, Rb 1.53, Cs 1.90. Density generally increases down the group, but potassium is lighter than sodium (its larger atomic volume outweighs its greater mass).
Order: K < Na < Rb < Cs.
Answer: (A). | ||||
| 66 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Assertion and reasonMedium | |
A medium assertion and reason question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter The slope of a line in an Ellingham diagram is $-\Delta S$ for the oxide formation. When the metal changes state (magnesium boils at about this temperature), the entropy change of the reaction changes sharply, because a gas is consumed in addition to oxygen.
So the line for $2Mg + O_2 \rightarrow 2MgO$ shows a sharp change in slope there. A is true, R is true, and R explains A.
Answer: (A). | ||||
| 67 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Electron deficient: $B_2H_6$ (too few electrons for normal two-electron bonds) — III.
Electron rich: HF (lone pairs on F) — II.
Electron precise: $CH_4$ (exactly the electrons needed) — IV.
Saline (ionic): $MgH_2$ — I.
A-III, B-II, C-IV, D-I.
Answer: (C). | ||||
| 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 $CaCl_2 + Na_2CO_3 \rightarrow CaCO_3\downarrow + 2NaCl$, so X = $CaCO_3$ and Y = NaCl.
$CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$ regenerates $CaCl_2$, so Z = HCl.
Answer: (C). | ||||
| 69 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: boron is an extremely hard solid owing to its very strong crystalline (covalent network) lattice. Correct.
Statement II: because of that lattice, boron has unusually high melting and boiling points, the highest in group 13 (m.p. about 2453 K; Al, Ga, In, Tl melt far lower). Correct.
Answer: (A). | ||||
| 70 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: an element in a higher oxidation state has greater polarising power, so its compounds are more covalent: $SbCl_5$ (+5) is more covalent than $SbCl_3$ (+3). Correct.
Statement II: among the oxides of the halogens, the higher oxides tend to be more stable than the lower ones (e.g. $I_2O_5$, $Cl_2O_7$ compared with $Cl_2O$, $ClO_2$). Correct.
Answer: (A). | ||||
| 71 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Assertion and reasonEasy | |
A easy assertion and reason question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The 5f orbitals extend further from the nucleus and are less shielded (less "buried") by the outer filled shells than the 4f orbitals. Hence 5f electrons take part in bonding far more, which is why actinoids show many more oxidation states than lanthanoids.
A is true, R is true, and R explains A.
Answer: (A). | ||||
| 72 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Matching listMedium | |
A medium matching list question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds CFSE $= (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\Delta_o$ (pairing energy ignored).
A: $Cu^{2+}$, $d^9$: $t_{2g}^6e_g^3$: $-2.4 + 1.8 = -0.6$ — I.
B: $Ti^{3+}$, $d^1$: $t_{2g}^1$: $-0.4$ — IV.
C: $Fe^{3+}$, $d^5$ with strong-field $CN^-$ (low spin): $t_{2g}^5$: $-2.0$ — II.
D: $Ni^{2+}$, $d^8$: $t_{2g}^6e_g^2$: $-2.4 + 1.2 = -1.2$ — III.
A-I, B-IV, C-II, D-III.
Answer: (B). | ||||
| 73 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Nitrogen oxides form nitric acid in the air: acid rain — IV.
Methane is a greenhouse gas: global warming — III.
Carbon dioxide dissolves in rain water forming carbonic acid, which brings normal rain to pH about 5.6 — II.
Phosphate fertilisers in water feed algal blooms: eutrophication — I.
A-IV, B-III, C-II, D-I.
Answer: (C). | ||||
| 74 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 'A' is squaric acid (3,4-dihydroxycyclobut-3-ene-1,2-dione), a strong diprotic acid.
With 2 eq of $OH^-$ both OH groups are deprotonated in a fast acid–base reaction, giving the squarate dianion 'B', $C_4O_4^{2-}$.
In the dianion the two negative charges are delocalised over the four C–O groups: the ring is planar with $2\pi$ electrons ($4n + 2$, $n = 0$), so B is aromatic (A true) and all four C–C bonds are equal (D true).
Acid–base proton transfer is very fast, so B is false.
'A' has enolic OH groups next to C=O groups and shows keto–enol tautomerism (C true).
So A, C and D only.
Answer: (C). | ||||
| 75 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 2-Hexene is $CH_3CH=CHCH_2CH_2CH_3$. Ozone forms the ozonide across the C=C.
Hydrolysis with water alone (no zinc) also produces $H_2O_2$, which oxidises the aldehydes formed to carboxylic acids.
Cleavage gives $CH_3CHO$ and $CH_3CH_2CH_2CHO$, oxidised to acetic acid and butanoic acid.
Answer: (A). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctHard | |
A hard single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Mg inserts into the C–Br bond of $CH_3COCH_2CH(Br)CH_3$, giving a carbanion-like carbon three atoms from the carbonyl carbon.
This Grignard carbon adds intramolecularly to the C=O of the same molecule, closing a three-membered ring (C2, C3, C4).
Hydrolysis ($H_2O$) of the magnesium alkoxide gives 1,2-dimethylcyclopropan-1-ol, option (A).
Answer: (A). | ||||
| 77 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Octane-2,7-dione undergoes an intramolecular aldol condensation with ethoxide.
The enolate at C3 (a $CH_2$ next to the C2 carbonyl) attacks the C7 carbonyl, forming a five-membered ring (C3–C7); dehydration on heating gives the conjugated enone.
The product is 1-acetyl-2-methylcyclopentene, option (C). (Attack by the C1 methyl enolate would give the less favoured seven-membered ring, option (D).)
Answer: (C). | ||||
| 78 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines (i) $SOCl_2$ converts the acid $Ph-CH=CH-CH_2-COOH$ into the acid chloride.
(ii) $R-NH_2$ gives the amide $Ph-CH=CH-CH_2-CONHR$.
(iii)/(iv) $LiAlH_4$ reduces the amide C=O to $CH_2$ (the isolated C=C is not touched), giving the secondary amine $Ph-CH=CH-CH_2-CH_2-NHR$, option (C).
Answer: (C). | ||||
| 79 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines A tertiary aromatic amine such as N,N-dimethylaniline reacts with nitrous acid ($NaNO_2$ + HX, at low temperature, 0–5 °C) by electrophilic substitution on the ring: the electrophile $NO^+$ attacks the activated para position.
'B' is p-nitroso-N,N-dimethylaniline (statements 1, 2 and 3 are correct).
No N-nitroso ammonium compound is formed, so the last statement is incorrect.
Answer: (D). | ||||
| 80 | PolymersBiodegradable polymers | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on biodegradable polymers, from Polymers. NCERT: No matching NCERT chapter 2-Chloro-1,3-butadiene (chloroprene) polymerises to neoprene, a synthetic rubber — II.
Nylon 2–nylon 6 (from glycine and aminocaproic acid) is a biodegradable polymer — I.
Polyacrylonitrile is an addition polymer of acrylonitrile — IV.
Dacron (terylene) is a polyester — III.
A-II, B-I, C-IV, D-III.
Answer: (B). | ||||
| 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 $v_{rms} = \sqrt{\frac{3RT}{M}}$ and $v_{mp} = \sqrt{\frac{2RT}{M}}$.
$\frac{3 \times 600}{40} = \frac{2 \times 90}{M_Y}$, i.e. $45 = \frac{180}{M_Y}$, so $M_Y = 4$ g mol$^{-1}$.
Answer: 4. | ||||
| 82 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Photon energy $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.95 \times 10^{-19}$ J $= \frac{4.95 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.09$ eV.
Photoemission needs $W_0 < 3.09$ eV: Li (2.42), Na (2.3) and K (2.25). Mg, Cu and Ag have larger work functions.
Number of metals $= 3$.
Answer: 3. | ||||
| 83 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics The gas expands isothermally against a constant external pressure of 4 atm: $w = -P_{ext}\Delta V = -4 \times 1.0 = -4$ L atm $= -4 \times 101.3 = -405.2$ J.
Ideal gas, isothermal: $\Delta U = 0$, so the gas absorbs $q = +405.2$ J from the surroundings.
$\Delta S_{surr} = \frac{-q}{T} = \frac{-405.2}{350} = -1.16$ J K$^{-1}$, nearest integer $-1$.
Answer: -1. | ||||
| 84 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$: $i = 1 + 2\alpha = 1 + 2(0.8) = 2.6$.
1 molal: 1 mol of solute in 1000 g (55.5 mol) of water. Mole fraction of solute particles $= \frac{2.6}{55.5 + 2.6} = 0.0447$.
$P = 50(1 - 0.0447) = 47.8$ mm Hg, nearest integer 48.
(The dilute approximation $\frac{\Delta P}{P^0} = \frac{2.6}{55.5}$ gives $P = 47.7$, also 48.)
Answer: 48. | ||||
| 85 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium pH 1: $[H^+] = 0.1$ M; pH 2: $[H^+] = 0.01$ M.
$M_1V_1 = M_2V_2$: $0.1 \times 1 = 0.01 \times V_2$, so $V_2 = 10$ L.
Water to add $= 10 - 1 = 9$ L $= 9000$ mL.
Answer: 9000. | ||||
| 86 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate is set by the slow step: rate $= k[NOBr_2][NO]$.
The fast pre-equilibrium gives $[NOBr_2] = K[NO][Br_2]$.
Rate $= kK[NO]^2[Br_2]$: overall order $2 + 1 = 3$.
Answer: 3. | ||||
| 87 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques On silica (a polar adsorbent) with hexane (a non-polar solvent) the most polar compound is held most strongly and travels least: compound C, which moved 2 cm.
The solvent front moved 8 cm.
$R_f = \frac{2}{8} = 0.25 = 25 \times 10^{-2}$.
Answer: 25. | ||||
| 88 | AminesDiazonium salts | Class 12: Amines | NumericalHard | |
A hard numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines (i)/(ii) Strecker synthesis: the aldehyde gives the α-amino nitrile and then the amino acid 'A', $EtO-C_6H_4-CH_2-CH(NH_2)-COOH$ (O-ethyltyrosine).
Nitration (2 eq): two $NO_2$ groups enter the ring ortho to the activating $OC_2H_5$ group: 'B' is $EtO-C_6H_2(NO_2)_2-CH_2CH(NH_2)COOH$.
(i) $Ac_2O$ acetylates the amino group ($-NHCOCH_3$); (ii) EtOH, Δ esterifies the acid ($-COOC_2H_5$); (iii) $H_2$, Pd/C reduces both $NO_2$ to $NH_2$; (iv) $HNO_2$ diazotises them; (v) NaI replaces both $N_2^+$ by I.
'D': $EtO-C_6H_2I_2-CH_2-CH(NHCOCH_3)-COOC_2H_5$.
Carbons: $2 (OEt) + 6 (ring) + 1 (CH_2) + 1 (CH) + 2 (NHCOCH_3) + 3 (COOC_2H_5) = 15$; hydrogens $5 + 2 + 2 + 1 + 4 + 5 = 19$, N 1, O 4, I 2 — matching $C_{15}H_{19}NO_4I_2$.
$x = 15$.
Answer: 15. | ||||
| 89 | BiomoleculesProteins | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules The peptide is Ala–Gly–Phe–Ile (alanylglycylphenylalanylisoleucine), four residues.
$sp^2$ carbons: the carbonyl carbons of the three peptide bonds and of the terminal COOH (4), plus the six carbons of the phenyl ring of phenylalanine (6).
Total $= 10$.
Answer: 10. | ||||
| 90 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Cyclohexanecarbaldehyde $C_6H_{11}CHO$ ($C_7H_{12}O$): $M = 84 + 12 + 16 = 112$ g mol$^{-1}$ (taking H = 1). Moles $= \frac{131.8}{112} = 1.177$ kmol.
Tollens' reaction: $RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag + 4NH_3 + 2H_2O$: 4 mol $NH_3$ per mol of aldehyde.
Mass of $NH_3 = 1.177 \times 4 \times 17 = 80.0$ kg.
Answer: 80. | ||||
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