JEE (Main) · 2023
JEE (Main) 2023 — 13 April, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Sequences and Series3
- Integral Calculus3
- Three Dimensional Geometry3
10 easy · 19 medium · 1 hard
Physics
30 Q
- Kinematics2
- Gravitation2
- Properties of Bulk Matter2
25 easy · 5 medium · 0 hard
Chemistry
30 Q
- p-Block Elements3
- Organic Chemistry: Some Basic Principles and Techniques3
- s-Block Elements2
20 easy · 10 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions For real x, $\frac{x^2}{x^2+1} = 1 - \frac{1}{x^2+1}$ takes every value in $[0, 1)$: it is 0 at $x = 0$ and approaches 1 as $|x| \to \infty$ but never reaches it.
$\sin^{-1}$ is increasing, so $\sin^{-1}\left(\frac{x^2}{x^2+1}\right)$ takes every value in $\left[0, \frac{\pi}{2}\right)$.
Multiplying by 4: the range of f is $[0, 2\pi)$.
Answer: (D). | ||||
| 2 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z = x + iy$. Then $Re(\bar{z}) = x$ and $z^2 = x^2 - y^2 + 2ixy$, so
$i(z^2 + x) = -2xy + i(x^2 - y^2 + x)$.
Equating with $\bar{z} = x - iy$:
Real parts: $x = -2xy$, so $x = 0$ or $y = -\frac{1}{2}$.
Imaginary parts: $-y = x^2 - y^2 + x$.
Case $x = 0$: $-y = -y^2$, so $y = 0$ or $y = 1$: $z = 0$ or $z = i$, with $|z|^2 = 0$ and $1$.
Case $y = -\frac{1}{2}$: $\frac{1}{2} = x^2 - \frac{1}{4} + x$, i.e. $x^2 + x - \frac{3}{4} = 0$, so $x = \frac{1}{2}$ or $x = -\frac{3}{2}$. Then $|z|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ and $\frac{9}{4} + \frac{1}{4} = \frac{5}{2}$.
$\sum |z|^2 = 0 + 1 + \frac{1}{2} + \frac{5}{2} = 4$.
Answer: (D). | ||||
| 3 | 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 $x = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm i\sqrt{6}}{2} = \sqrt{2}\left(\frac{1}{2} \pm i\frac{\sqrt{3}}{2}\right) = \sqrt{2}\,e^{\pm i\pi/3}$.
By De Moivre's theorem, $\alpha^{14} + \beta^{14} = (\sqrt{2})^{14}\left(e^{i14\pi/3} + e^{-i14\pi/3}\right) = 2^7 \cdot 2\cos\frac{14\pi}{3}$.
$\frac{14\pi}{3} = 4\pi + \frac{2\pi}{3}$, so $\cos\frac{14\pi}{3} = \cos\frac{2\pi}{3} = -\frac{1}{2}$.
$\alpha^{14} + \beta^{14} = 128 \cdot 2 \cdot \left(-\frac{1}{2}\right) = -128$.
Answer: (C). | ||||
| 4 | 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 For infinitely many solutions the coefficient determinant must vanish:
$\begin{vmatrix}2&1&-1\\2&-5&\lambda\\1&2&-5\end{vmatrix} = 2(25 - 2\lambda) - 1(-10 - \lambda) - 1(4 + 5) = 51 - 3\lambda = 0$, so $\lambda = 17$.
Then the second equation must be a combination of the other two: $2x - 5y + 17z = a(2x + y - z) + b(x + 2y - 5z)$.
x: $2a + b = 2$; y: $a + 2b = -5$. So $a = 3$, $b = -4$; check z: $-3 + 20 = 17$. ✓
Then $\mu = 3(5) - 4(7) = -13$.
$(\lambda + \mu)^2 + (\lambda - \mu)^2 = 2(\lambda^2 + \mu^2) = 2(289 + 169) = 916$.
Answer: (C). | ||||
| 5 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $|A| = 1(6 - 1) - 2(2\alpha - 1) + 3(\alpha - 3) = -2 - \alpha = 2$, so $\alpha = -4$.
For a $3 \times 3$ matrix M: $|kM| = k^3|M|$ and $|\mathrm{adj}\,M| = |M|^2$.
$|2A| = 8 \cdot 2 = 16 = 2^4$.
$|\mathrm{adj}(2A)| = 2^8$, so $|2\,\mathrm{adj}(2A)| = 2^3 \cdot 2^8 = 2^{11}$.
$|\mathrm{adj}(2\,\mathrm{adj}(2A))| = 2^{22}$, so $|2\,\mathrm{adj}(2\,\mathrm{adj}(2A))| = 2^3 \cdot 2^{22} = 2^{25} = 32^5$.
So $n = 5$ and $3n + \alpha = 15 - 4 = 11$.
Answer: (C). | ||||
| 6 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The letters in alphabetical order: A, D, M, N, O, Y (all distinct).
Words before MONDAY:
First letter A or D: $2 \times 5! = 240$.
M, then second letter A, D or N (before O): $3 \times 4! = 72$.
MO, then third letter A or D (before N): $2 \times 3! = 12$.
MON, then fourth letter A (before D): $1 \times 2! = 2$.
MOND, then A Y: this is MONDAY itself.
Words before it: $240 + 72 + 12 + 2 = 326$, so MONDAY is word number 327.
Answer: (C). | ||||
| 7 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom{5}{r}(2x^3)^{5-r}\left(-\frac{1}{3x^2}\right)^r$, with power of x equal to $15 - 3r - 2r = 15 - 5r$.
$15 - 5r = 5$ gives $r = 2$.
Coefficient $= \binom{5}{2} \cdot 2^3 \cdot \frac{1}{9} = 10 \cdot 8 \cdot \frac{1}{9} = \frac{80}{9}$.
Answer: (C). | ||||
| 8 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the common ratio be r (> 1, since the terms are positive and increasing).
$a_3a_5 = a_4^2 = \frac{1}{9}$, so $a_4 = \frac{1}{3}$.
$a_6 + a_8 = a_4(r^2 + r^4) = 2$, so $r^4 + r^2 = 6$, i.e. $(r^2 + 3)(r^2 - 2) = 0$, so $r^2 = 2$.
$a_2 + a_4 = a_4\left(\frac{1}{r^2} + 1\right) = \frac{1}{3} \cdot \frac{3}{2} = \frac{1}{2}$.
$a_4 + a_6 = a_4(1 + r^2) = \frac{1}{3} \cdot 3 = 1$.
$6(a_2 + a_4)(a_4 + a_6) = 6 \cdot \frac{1}{2} \cdot 1 = 3$.
Answer: (D). | ||||
| 9 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Expand near $x = 0$:
$e^{ax} = 1 + ax + \frac{a^2x^2}{2} + \ldots$, $\cos(bx) = 1 - \frac{b^2x^2}{2} + \ldots$, $\frac{cx}{2}e^{-cx} = \frac{c}{2}x - \frac{c^2}{2}x^2 + \ldots$
Numerator $= \left(a - \frac{c}{2}\right)x + \frac{a^2 + b^2 + c^2}{2}x^2 + \ldots$
Denominator $1 - \cos 2x = 2\sin^2 x = 2x^2 + \ldots$
For a finite limit the x-term must vanish: $c = 2a$.
Then the limit is $\frac{a^2 + b^2 + c^2}{4} = 17$, so $a^2 + b^2 + 4a^2 = 68$, i.e. $5a^2 + b^2 = 68$.
Answer: (A). | ||||
| 10 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Denominator: $\tan^{49}x + \tan^{51}x = \tan^{49}x\sec^2 x$, so it is $\int_0^{\pi/4} e^{-x}\tan^{49}x\sec^2 x\,dx$.
Integrate by parts with $\tan^{49}x\sec^2 x = \frac{d}{dx}\left(\frac{\tan^{50}x}{50}\right)$:
$= \left[e^{-x}\frac{\tan^{50}x}{50}\right]_0^{\pi/4} + \frac{1}{50}\int_0^{\pi/4} e^{-x}\tan^{50}x\,dx = \frac{1}{50}\left(e^{-\pi/4} + \int_0^{\pi/4} e^{-x}\tan^{50}x\,dx\right)$.
So the denominator is $\frac{1}{50}$ of the numerator, and the value is 50.
Answer: (C). | ||||
| 11 | 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 region is symmetric about the y-axis; take $x \ge 0$ and double.
$x^2 \le |x^2 - 4|$: for $x^2 \ge 4$ this says $x^2 \le x^2 - 4$, impossible; for $x^2 < 4$ it says $x^2 \le 4 - x^2$, i.e. $x \le \sqrt{2}$.
So for $0 \le x \le \sqrt{2}$ the region runs from $y = \max(x^2, 1)$ up to $y = 4 - x^2$ (which is at least 2 there).
For $0 \le x \le 1$: height $4 - x^2 - 1 = 3 - x^2$; $\int_0^1 (3 - x^2)\,dx = 3 - \frac{1}{3} = \frac{8}{3}$.
For $1 \le x \le \sqrt{2}$: height $4 - 2x^2$; $\int_1^{\sqrt{2}} (4 - 2x^2)\,dx = \left[4x - \frac{2x^3}{3}\right]_1^{\sqrt{2}} = \left(4\sqrt{2} - \frac{4\sqrt{2}}{3}\right) - \frac{10}{3} = \frac{8\sqrt{2} - 10}{3}$.
Half-area $= \frac{8}{3} + \frac{8\sqrt{2} - 10}{3} = \frac{8\sqrt{2} - 2}{3}$.
Area $= \frac{16\sqrt{2} - 4}{3} = \frac{4}{3}(4\sqrt{2} - 1)$.
Answer: (A). | ||||
| 12 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines Vertices:
$15x - y = 82$ and $6x - 5y = -4$: $y = 15x - 82$, $6x - 75x + 410 = -4$, so $x = 6$, $y = 8$: $(6, 8)$.
$15x - y = 82$ and $9x + 4y = 17$: $9x + 60x - 328 = 17$, so $x = 5$, $y = -7$: $(5, -7)$.
$6x - 5y = -4$ and $9x + 4y = 17$: adding $4\times$ the first and $5\times$ the second, $69x = 69$, so $x = 1$, $y = 2$: $(1, 2)$.
Centroid $(\alpha, \beta) = \left(\frac{12}{3}, \frac{3}{3}\right) = (4, 1)$.
$\alpha + 2\beta = 6$ and $2\alpha - \beta = 7$: sum 13, product 42.
The equation is $x^2 - 13x + 42 = 0$.
Answer: (D). | ||||
| 13 | 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 The normal passes through the centre: $4\alpha + 3\beta = 1$.
The centre is at distance r from both tangents: $|3\alpha + 4\beta - 24| = |3\alpha - 4\beta - 32| = 5r$.
Case 1: $3\alpha + 4\beta - 24 = 3\alpha - 4\beta - 32$, so $\beta = -1$ and $\alpha = \frac{1 + 3}{4} = 1$. Then $r = \frac{|3 - 4 - 24|}{5} = 5 < 8$. ✓
Case 2: $3\alpha + 4\beta - 24 = -(3\alpha - 4\beta - 32)$, so $\alpha = \frac{28}{3}$, $\beta = \frac{1 - 112/3}{3} = -\frac{109}{9}$. Then $r = \frac{|28 - 436/9 - 24|}{5} = \frac{400/9}{5} = \frac{80}{9} > 8$: rejected.
So $\alpha = 1$, $\beta = -1$, $r = 5$, and $\alpha - \beta + r = 7$.
Answer: (C). | ||||
| 14 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy 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 Let $A(0, -1, 2)$, $B(-1, 2, 1)$. $\vec{AB} = (-1, 3, -1)$. The line's direction is $(5, 1, -7) - (1, -1, -1) = (4, 2, -6) \parallel (2, 1, -3)$.
Normal to the plane: $(-1, 3, -1) \times (2, 1, -3) = (3(-3) - (-1)(1),\ (-1)(2) - (-1)(-3),\ (-1)(1) - 3(2)) = (-8, -5, -7) \parallel (8, 5, 7)$.
Plane through A: $8x + 5y + 7z = 8(0) + 5(-1) + 7(2) = 9$.
Check the options: $(1, -2, 1)$: $8 - 10 + 7 = 5$; $(-2, 5, 0)$: $-16 + 25 + 0 = 9$ ✓; $(2, 0, 1)$: $16 + 7 = 23$; $(0, 5, -2)$: $25 - 14 = 11$.
Answer: (B). | ||||
| 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 The given line passes through $P(-3, 1, 5)$ with direction $\vec{d} = (-3, 1, 5)$. A line through Q with direction $\vec{e}$ is coplanar with it when $\vec{PQ} \cdot (\vec{d} \times \vec{e}) = 0$.
(A) $Q(1, 2, 5)$, $\vec{e} = (-1, 2, 5)$: $\vec{e} \times \vec{d} = (2 \cdot 5 - 5 \cdot 1,\ 5(-3) - (-1)5,\ (-1)(1) - 2(-3)) = (5, -10, 5)$; $\vec{PQ} = (4, 1, 0)$; dot product $20 - 10 = 10 \ne 0$.
(B) $Q(-1, 2, 5)$, same $\vec{e}$: $\vec{PQ} = (2, 1, 0)$; $(2, 1, 0) \cdot (5, -10, 5) = 10 - 10 = 0$: coplanar.
(C) $Q(-1, 2, 5)$, $\vec{e} = (-1, 2, 4)$: $\vec{e} \times \vec{d} = (10 - 4, -12 + 5, -1 + 6) = (6, -7, 5)$; $(2, 1, 0) \cdot (6, -7, 5) = 5 \ne 0$.
(D) $Q(-1, 2, 5)$, $\vec{e} = (1, 2, 5)$: $\vec{e} \times \vec{d} = (10 - 5, -15 - 5, 1 + 6) = (5, -20, 7)$; $(2, 1, 0) \cdot (5, -20, 7) = -10 \ne 0$.
Answer: (B). | ||||
| 16 | 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 Direction of the line: $(4, 5, 8) - (1, -7, 5) = (3, 12, 3) \parallel (1, 4, 1)$. A general point is $X = (4 + t, 5 + 4t, 8 + t)$.
$\vec{XP} = (-3 - t, -7 - 4t, -5 - t)$ must be perpendicular to $(1, 4, 1)$:
$-3 - t - 28 - 16t - 5 - t = -36 - 18t = 0$, so $t = -2$ and $N = (2, -3, 6)$.
Distance from the plane: $\frac{|2(2) - 2(-3) + 6 + 5|}{\sqrt{4 + 4 + 1}} = \frac{21}{3} = 7$.
Answer: (B). | ||||
| 17 | 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 $(\vec{a} + 2\vec{b}) \times (2\vec{a} - 3\vec{b}) = -3\,\vec{a} \times \vec{b} + 4\,\vec{b} \times \vec{a} = -7\,\vec{a} \times \vec{b}$.
$|\vec{a} \times \vec{b}| = 2 \cdot 3 \sin\frac{\pi}{4} = 3\sqrt{2}$, so $|\vec{a} \times \vec{b}|^2 = 18$.
Required value $= 49 \times 18 = 882$.
Answer: (B). | ||||
| 18 | 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 $\overrightarrow{CB} = \overrightarrow{CA} + \overrightarrow{AB} = (2, 4, \delta + 3)$.
Area $= \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{CA}|$. $(-2, 1, 3) \times (4, 3, \delta) = (\delta - 9,\ 12 + 2\delta,\ -6 - 4) = (\delta - 9, 2\delta + 12, -10)$.
$|\ldots|^2 = (\delta - 9)^2 + (2\delta + 12)^2 + 100 = 5\delta^2 + 30\delta + 325$, and it must equal $(2 \cdot 5\sqrt{6})^2 = 600$.
$5\delta^2 + 30\delta - 275 = 0$, i.e. $\delta^2 + 6\delta - 55 = 0$, so $\delta = 5$ (as $\delta > 0$).
$\overrightarrow{CB} = (2, 4, 8)$, $\overrightarrow{CA} = (4, 3, 5)$: $\overrightarrow{CB} \cdot \overrightarrow{CA} = 8 + 12 + 40 = 60$.
Answer: (A). | ||||
| 19 | 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 Mean $- $ variance $= np - npq = np^2 = 1$.
$2\binom{n}{2}p^2q^{n-2} = 3np\,q^{n-1}$ gives $(n - 1)p = 3q = 3 - 3p$, i.e. $np + 2p = 3$.
With $np^2 = 1$, $p = \frac{1}{np}$; let $m = np$: $m + \frac{2}{m} = 3$, so $m = 1$ or $m = 2$.
$m = 1$ gives $p = 1$, $n = 1$, for which $P(X = 2) = 0 \ne \frac{3}{2}P(X = 1)$: rejected. So $np = 2$, $p = \frac{1}{2}$, $n = 4$.
$P(X > 1) = 1 - P(0) - P(1) = 1 - \frac{1}{16} - \frac{4}{16} = \frac{11}{16}$.
$n^2P(X > 1) = 16 \cdot \frac{11}{16} = 11$.
Answer: (D). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter The three disjuncts cover the truth-value pairs (T, F), (F, T) and (F, F) of (p, q). The statement is false only when p and q are both true.
So it is $\sim(p \wedge q)$, which by De Morgan's law is $(\sim p) \vee (\sim q)$.
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 Elements of R:
$b = |a|$: $(-4, 4), (-3, 3), (0, 0), (1, 1), (3, 3), (4, 4)$ ($a = -2$ gives $b = 2 \notin A$).
$b^2 = a + 1$: $a = 0$ gives $b = \pm 1$, so $(0, 1)$; $a = 3$ gives $b = \pm 2$, so $(3, -2)$; no other a gives a square of an element of A.
$R = \{(-4, 4), (-3, 3), (0, 0), (1, 1), (3, 3), (4, 4), (0, 1), (3, -2)\}$.
Reflexive: add $(-4, -4), (-3, -3), (-2, -2)$: 3 pairs.
Symmetric: add $(4, -4), (3, -3), (1, 0), (-2, 3)$: 4 pairs.
Minimum number to add $= 3 + 4 = 7$.
Answer: 7. | ||||
| 22 | 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 Divisible by 6 means even (last digit 2 or 4) and digit sum divisible by 3.
Residues mod 3 of the digits: 3 → 0; 1, 4 → 1; 2, 5 → 2. So one digit of residue 0, two of residue 1, two of residue 2.
Last digit 2 (residue 2): the first two digits must have residue sum 1: (0, 1): $1 \cdot 2 = 2$; (1, 0): 2; (2, 2): $2 \cdot 2 = 4$. Total 8.
Last digit 4 (residue 1): the first two must have residue sum 2: (0, 2): 2; (2, 0): 2; (1, 1): 4. Total 8.
Total $= 16$.
Answer: 16. | ||||
| 23 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $7^2 = 49 = 51 - 2 \equiv -2 \pmod{17}$.
$7^8 = (7^2)^4 \equiv (-2)^4 = 16 \equiv -1$, so $7^{16} \equiv 1$ and $7^{96} = (7^{16})^6 \equiv 1$.
$7^{103} = 7^{96} \cdot 7^7 \equiv 7^7 = (7^2)^3 \cdot 7 \equiv (-8)(7) = -56 \equiv -56 + 68 = 12 \pmod{17}$.
(With the binomial theorem: $7^{103} = 7 \cdot 49^{51} = 7(51 - 2)^{51} = 17k - 7 \cdot 2^{51}$, and $2^{51} = 8 \cdot 16^{12} = 8(17 - 1)^{12} = 17m + 8$, giving a remainder of $-56 \equiv 12$.)
Answer: 12. | ||||
| 24 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $[\sqrt{k}] = m$ for $m^2 \le k \le (m + 1)^2 - 1$, that is for $2m + 1$ values of k.
For $m = 1$ to $9$ (k from 1 to 99) the blocks are complete: $\sum_{m=1}^{9} m(2m + 1) = 2\sum m^2 + \sum m = 2(285) + 45 = 615$.
For $m = 10$: k from 100 to 120, 21 values, contributing $210$.
Total $= 615 + 210 = 825$.
Answer: 825. | ||||
| 25 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $f(2) = \sum_{k=1}^{10} k\,2^k$ and $f'(2) = \sum_{k=1}^{10} k^2\,2^{k-1}$, so
$2f(2) + f'(2) = \sum_{k=1}^{10} 2^{k-1}(4k + k^2)$.
Term by term ($k = 1$ to $10$): $5, 24, 84, 256, 720, 1920, 4928, 12288, 29952, 71680$.
Sum $= 121857$, so $119 \cdot 2^n = 121856 = 119 \cdot 1024 = 119 \cdot 2^{10}$.
(In closed form, $\sum_{k=1}^{N} k\,2^k = (N - 1)2^{N+1} + 2 = 18434$ and $\sum_{k=1}^{N} k^2\,2^{k-1} = (N^2 - 2N + 3)2^N - 3 = 84989$ for $N = 10$; $2(18434) + 84989 = 121857$.)
$n = 10$.
Answer: 10. | ||||
| 26 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t = \sin x$, $dt = \cos x\,dx$: $f_n = \int_0^1 S(t)T(t)\,dt$ with $S = \sum_{k=1}^{n} t^{k-1}$, $T = \sum_{k=1}^{n} (2k - 1)t^{k-1}$.
Claim: $S\,T = \frac{d}{dt}\left(t\,S^2\right)$. The coefficient of $t^{m}$ in $tS^2$ is the number of pairs $(i, j)$ with $1 \le i, j \le n$ and $i + j = m + 1$, call it $c_m$; so $\frac{d}{dt}(tS^2) = \sum_m m\,c_m t^{m-1}$. In $ST = \sum_{i,j} (2j - 1)t^{i+j-2}$ the coefficient of $t^{m-1}$ is $\sum_{i+j=m+1} (2j - 1)$; pairing $(i, j)$ with $(j, i)$ gives $\frac{1}{2}\sum_{i+j=m+1}(2i + 2j - 2) = \frac{1}{2}c_m(2m) = m\,c_m$. ✓ (For example, $n = 2$: $(1 + t)(1 + 3t) = 1 + 4t + 3t^2 = \frac{d}{dt}\left(t(1 + t)^2\right)$.)
So $f_n = \left[t\,S(t)^2\right]_0^1 = S(1)^2 = n^2$.
$f_{21} - f_{20} = 441 - 400 = 41$.
Answer: 41. | ||||
| 27 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor: $e^{\int \frac{4x}{x^2 - 1}dx} = (x^2 - 1)^2$.
$y(x^2 - 1)^2 = \int \frac{x + 2}{\sqrt{x^2 - 1}}dx = \sqrt{x^2 - 1} + 2\log_e\left(x + \sqrt{x^2 - 1}\right) + C$.
At $x = 2$: $9y(2) = \sqrt{3} + 2\log_e(2 + \sqrt{3}) + C$. Since $9y(2) = 2\log_e(2 + \sqrt{3})$, $C = -\sqrt{3}$.
At $x = \sqrt{2}$: $y(\sqrt{2}) \cdot 1 = 1 + 2\log_e(\sqrt{2} + 1) - \sqrt{3}$.
Comparing with $\alpha\log_e(\sqrt{\alpha} + \beta) + \beta - \sqrt{\gamma}$: $\alpha = 2$, $\beta = 1$, $\gamma = 3$.
$\alpha\beta\gamma = 6$.
Answer: 6. | ||||
| 28 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Let the hyperbola be $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$. $Ae = 2$ with $e = \frac{3}{2}$, so $A = \frac{4}{3}$, $A^2 = \frac{16}{9}$ and $B^2 = A^2(e^2 - 1) = \frac{16}{9} \cdot \frac{5}{4} = \frac{20}{9}$.
The tangent is perpendicular to $2x + 3y = 6$ (slope $-\frac{2}{3}$), so its slope is $m = \frac{3}{2}$.
Tangents of slope m: $y = mx \pm \sqrt{A^2m^2 - B^2} = \frac{3}{2}x \pm \sqrt{4 - \frac{20}{9}} = \frac{3}{2}x \pm \frac{4}{3}$.
The point of contact of $y = mx + c$ is $\left(-\frac{A^2m}{c}, -\frac{B^2}{c}\right)$, which is in the first quadrant only for $c < 0$: so the tangent is $y = \frac{3}{2}x - \frac{4}{3}$ (touching at $\left(2, \frac{5}{3}\right)$).
Intercepts: $a = \frac{8}{9}$, $b = -\frac{4}{3}$.
$|6a| + |5b| = \frac{16}{3} + \frac{20}{3} = 12$.
Answer: 12. | ||||
| 29 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Observed: $\sum x = 10 \times 50 = 500$, $\sum x^2 = 10(12^2 + 50^2) = 26440$.
Corrected: $\sum x = 500 - 45 - 50 + 20 + 25 = 450$, mean $= 45$.
$\sum x^2 = 26440 - 2025 - 2500 + 400 + 625 = 22940$.
Variance $= \frac{22940}{10} - 45^2 = 2294 - 2025 = 269$.
Answer: 269. | ||||
| 30 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta = \tan^{-1}x$; for $x \in (-1, 1]$, $\theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right]$ and $2\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the principal range of $\sin^{-1}$.
So $\sin^{-1}x = 2\theta$ holds exactly when $x = \sin 2\theta = \frac{2x}{1 + x^2}$.
$x(1 + x^2) = 2x$ gives $x = 0$ or $x^2 = 1$. In $(-1, 1]$: $x = 0$ and $x = 1$ (check: $\sin^{-1}1 = \frac{\pi}{2} = 2 \cdot \frac{\pi}{4}$).
Number of solutions $= 2$.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 can be added or subtracted.
$\frac{a}{Y^2}$ has the dimensions of pressure, so $a = [\text{pressure}][\text{volume}]^2$.
b has the dimensions of volume.
$\frac{a}{b} = [\text{pressure}][\text{volume}] = [ML^{-1}T^{-2}][L^3] = [ML^2T^{-2}]$, the dimensions of energy.
Answer: (A). | ||||
| 32 | 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 $v = \frac{ds}{dt} = 2(2.5)t = 5t$.
At $t = 5$ s: $v = 25\ ms^{-1}$.
Answer: (C). | ||||
| 33 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The trains move in opposite directions, so the speed of B relative to the passenger is $90 + 54 = 144$ km/h $= 144 \times \frac{5}{18} = 40$ m/s.
B passes the passenger completely in 8 s, so its length is $40 \times 8 = 320$ m.
Answer: (D). | ||||
| 34 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion $F = m\omega^2r = 200 \times (0.2)^2 \times 70 = 200 \times 0.04 \times 70 = 560$ N.
Answer: (C). | ||||
| 35 | GravitationEscape velocity | Class 11: Gravitation | StatementsEasy | |
A easy statements question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}}$, so $v_e$ grows with $\frac{M}{R}$: Statement I is correct.
$v_e$ depends on R (for a given M it varies as $\frac{1}{\sqrt{R}}$): Statement II is incorrect.
Answer: (C). | ||||
| 36 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Assertion and reasonMedium | |
A medium assertion and reason question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter By Stokes' law the terminal velocity is $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$, so $v_t \propto r^2$.
Percentage error: $\frac{\Delta v_t}{v_t} = 2\frac{\Delta r}{r} = 2 \times \frac{0.1}{5} \times 100\% = 4\%$. Assertion A is true.
$v_t$ is proportional to $r^2$, not inversely proportional to r: Reason R is false.
Answer: (C). | ||||
| 37 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g = \frac{GM}{R^2} = \frac{G \cdot \frac{4}{3}\pi R^3\rho}{R^2} = \frac{4}{3}\pi G\rho R$, so $g \propto \rho R$.
$\frac{g_B}{g_A} = \frac{(\rho/2)(1.5R)}{\rho R} = 0.75 = \frac{3}{4}$.
Answer: (C). | ||||
| 38 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics A sudden compression allows no time for heat exchange, so the process is adiabatic: $PV^{\gamma} = $ constant.
$P_oV_o^{\gamma} = P\left(\frac{V_o}{4}\right)^{\gamma}$, so $P = P_o(4)^{\gamma}$.
Answer: (C). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $\lambda = \frac{k_BT}{\sqrt{2}\pi d^2P}$, so at constant pressure $\lambda \propto T$.
$\lambda' = 1500d \times \frac{373}{273} \approx 2049d$.
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 $KE = \frac{1}{2}k(A^2 - x^2)$ and $PE = \frac{1}{2}kx^2$.
Equal when $A^2 - x^2 = x^2$, i.e. $x = \frac{A}{\sqrt{2}}$.
Answer: (C). | ||||
| 41 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With parts q and $Q - q$: $F = \frac{kq(Q - q)}{r^2}$.
$\frac{dF}{dq} = \frac{k(Q - 2q)}{r^2} = 0$ gives $q = \frac{Q}{2}$ (a maximum, since $\frac{d^2F}{dq^2} < 0$).
So the parts are 5 $\mu$C and 5 $\mu$C.
Answer: (C). | ||||
| 42 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity From the figure: the 3 V cell and the 4 Ω resistor are in series along the top wire, between the left and right vertical wires. Across those two wires are connected (i) the 4 $\mu$F capacitor in series with a 6 Ω resistor and (ii) a 6 Ω resistor.
In steady state no current flows through the capacitor branch. The current is $I = \frac{3}{4 + 6} = 0.3$ A, through the 4 Ω and the bottom 6 Ω.
The potential difference across the bottom 6 Ω is $0.3 \times 6 = 1.8$ V. The capacitor branch is across the same two wires and its 6 Ω carries no current, so the capacitor's voltage is 1.8 V.
$Q = CV = 4\ \mu F \times 1.8\ V = 7.2\ \mu C$.
Answer: (C). | ||||
| 43 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsEasy | |
A easy statements 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 $\vec{F} = q(\vec{v} \times \vec{B})$ with $q = -e$, $\vec{v} = v\hat{i}$, $\vec{B} = -B\hat{k}$.
$\vec{v} \times \vec{B} = -vB(\hat{i} \times \hat{k}) = -vB(-\hat{j}) = vB\hat{j}$, so $\vec{F} = -evB\hat{j}$: along the negative y-axis. B is true, A and C are false.
The force is perpendicular to the velocity, which is perpendicular to the uniform field, so the electron moves in a circle (in the x-y plane): E is true, D is false.
Answer: (D). | ||||
| 44 | 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 Resonance ($X_L = X_C$) needs both an inductor and a capacitor in the circuit; with only one of them there is no resonance. Statement I is false.
In a purely capacitive or purely inductive circuit the phase difference is $90^\circ$, so the power factor $\cos\phi = 0$ and the average power consumed is zero. Statement II is false.
Answer: (B). | ||||
| 45 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves The wave travels along $\vec{E} \times \vec{B}$.
$(-\hat{k}) \times \hat{i} = -(\hat{k} \times \hat{i}) = -\hat{j}$.
The wave travels along the negative y-axis.
Answer: (B). | ||||
| 46 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \frac{(2 + 1)^2}{(2 - 1)^2} = 9 : 1$.
Answer: (C). | ||||
| 47 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | StatementsEasy | |
A easy statements question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Ultraviolet rays have the highest frequency, hence the most energetic photons, of the three; they are the most effective in ejecting photoelectrons. Statement I is true.
Einstein's equation $K_{max} = h\nu - \phi_0$: $K_{max}$ increases linearly with $\nu$; it is not inversely proportional to it. Statement II is false.
Answer: (C). | ||||
| 48 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Assertion and reasonEasy | |
A easy assertion and reason question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The binding energy curve is nearly flat (about 8 MeV per nucleon) for mass numbers 30 to 170: A is true.
The reason is that the nuclear force is short ranged: a nucleon interacts only with its few nearest neighbours, so in a nucleus large enough the binding energy per nucleon saturates and does not depend on the size of the nucleus. R is true and explains A.
Answer: (A). | ||||
| 49 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Reading the inputs from the figure, before the first dashed line and then in the six intervals between the seven dashed lines:
A: 1 | 0, 0, 1, 1, 0, 0
B: 1 | 0, 1, 0, 1, 0, 1
NAND gives 0 only when both inputs are 1:
Y: 0 | 1, 1, 1, 0, 1, 1
So the output is low at the start, high for the first three intervals, low in the fourth and high in the last two: option (B).
Answer: (B). | ||||
| 50 | Communication SystemsAntenna size | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on antenna size, from Communication Systems. NCERT: No matching NCERT chapter For efficient radiation the antenna should be comparable to the wavelength; the minimum size required is $\frac{\lambda}{4}$ (a quarter-wave antenna).
Answer: (D). | ||||
| 51 | 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 By the work-energy theorem, $E = \frac{1}{2}mu^2$.
From u to 2u: $\frac{1}{2}m(2u)^2 - \frac{1}{2}mu^2 = 3 \cdot \frac{1}{2}mu^2 = 3E$.
$n = 3$.
Answer: 3. | ||||
| 52 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion The cylinder turns about its fixed axis. Moment of inertia of a hollow cylinder: $I = MR^2 = 5 \times (0.7)^2 = 2.45\ kg\,m^2$.
Torque: $\tau = FR = 52.5 \times 0.7 = 36.75$ N m.
$\alpha = \frac{\tau}{I} = \frac{36.75}{2.45} = 15\ rad\ s^{-2}$ (equivalently $\alpha = \frac{F}{MR} = \frac{52.5}{3.5}$).
Answer: 15. | ||||
| 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 steady state the same heat current flows through both plates (same area A and thickness l):
$\frac{84A(100 - T)}{l} = \frac{126A(T - 0)}{l}$
$8400 - 84T = 126T$, so $T = \frac{8400}{210} = 40$ ℃.
Answer: 40. | ||||
| 54 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Fundamental frequency $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, so with the same string and length $f \propto \sqrt{T} \propto \sqrt{m}$.
$\frac{m}{180} = \left(\frac{50}{30}\right)^2 = \frac{25}{9}$, so $m = 500$ g.
Answer: 500. | ||||
| 55 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Force on $-2q$ (at $x = \frac{3}{4}R$):
From q (distance $\frac{3R}{4}$), attraction towards the origin: $F_1 = \frac{k \cdot q \cdot 2q}{(3R/4)^2} = \frac{32}{9}\frac{kq^2}{R^2}$, along $-x$.
From 2q (distance $\frac{R}{4}$), attraction towards $x = R$: $F_2 = \frac{k \cdot 2q \cdot 2q}{(R/4)^2} = 64\frac{kq^2}{R^2}$, along $+x$.
Net: $\left(64 - \frac{32}{9}\right)\frac{kq^2}{R^2} = \frac{544}{9}\frac{kq^2}{R^2}$, along $+x$.
$\frac{kq^2}{R^2} = \frac{9 \times 10^9 \times 4 \times 10^{-12}}{4 \times 10^{-4}} = 90$ N.
$F = \frac{544}{9} \times 90 = 5440$ N.
Answer: 5440. | ||||
| 56 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity From the figure: the 12 V cell is connected across the top and bottom corners of a bridge. Top to left corner 3 Ω, top to right corner 4 Ω, left to bottom 9 Ω, right to bottom 2 Ω; the 6 $\mu$F capacitor joins the left and right corners.
In steady state no current flows through the capacitor. Take the bottom corner at 0 V and the top at 12 V.
Left arm: 3 Ω + 9 Ω = 12 Ω, current 1 A; left corner at $12 - 1 \times 3 = 9$ V.
Right arm: 4 Ω + 2 Ω = 6 Ω, current 2 A; right corner at $12 - 2 \times 4 = 4$ V.
Capacitor voltage $= 9 - 4 = 5$ V.
$U = \frac{1}{2}CV^2 = \frac{1}{2} \times 6 \times 25 = 75\ \mu J$.
Answer: 75. | ||||
| 57 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The field is into the page, perpendicular to the wire. The tension vanishes when the magnetic force balances the weight: $BIL = mg$.
$I = \frac{mg}{BL} = \frac{0.040 \times 10}{0.40 \times 0.50} = 2$ A (flowing from A to B, so that $I\vec{L} \times \vec{B}$ points upwards).
Answer: 2. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Charge driven through the circuit: $q = \frac{\Delta\Phi_{total}}{R} = \frac{NA\,\Delta B}{R}$.
$\Delta B = 1.5 - (-1.5) = 3$ T; $A = 24 \times 10^{-4}\ m^2$; $N = 100$.
$q = \frac{100 \times 24 \times 10^{-4} \times 3}{12} = \frac{0.72}{12} = 0.06$ C $= 60$ mC.
Answer: 60. | ||||
| 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 Cutting along a plane perpendicular to the principal axis gives two identical plano-convex lenses, each with one of the curved faces.
For the biconvex lens (radius R each side): $\frac{1}{10} = (\mu - 1)\frac{2}{R}$. For each half: $\frac{1}{f'} = (\mu - 1)\frac{1}{R} = \frac{1}{20}$, so $f' = 20$ cm.
Power $= \frac{100}{20} = 5$ D.
Answer: 5. | ||||
| 60 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $hc = 6.6 \times 10^{-34} \times 3 \times 10^8 = 1.98 \times 10^{-25}$ J m; with $1\ eV = 1.6 \times 10^{-19}$ J, $hc = 1237.5$ eV nm.
Energy absorbed: $\frac{1237.5}{500} = 2.475$ eV. Energy emitted: $\frac{1237.5}{600} = 2.0625$ eV.
Net energy absorbed $= 0.4125$ eV $= 4125 \times 10^{-4}$ eV.
Answer: 4125. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | StatementsEasy | |
A easy statements question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $SO_2$: S has two bonding domains and one lone pair; $H_2O$: O has two bond pairs and two lone pairs. Both molecules are bent (V-shaped): Statement I is correct.
The bond angle of $SO_2$ is about $119^\circ$ (one lone pair, $sp^2$ S), that of $H_2O$ about $104.5^\circ$ (two lone pairs). So the bond angle of $SO_2$ is greater: Statement II is incorrect.
Answer: (C). | ||||
| 62 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Combustion of methane is exothermic.
System A (adiabatic): no heat can leave, so the heat released stays in the system and its temperature rises.
System B (diathermic walls): heat passes freely to the surroundings, so the system stays at the temperature of the surroundings: its temperature remains the same.
Answer: (A). | ||||
| 63 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter The Tyndall effect is seen only when the diameter of the dispersed particles is not much smaller than the wavelength of the light used: A is correct.
Particles that large scatter the light in all directions, which makes the path of the beam visible: this is the cause of the Tyndall effect, so R is correct and explains A.
Answer: (A). | ||||
| 64 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter In $[BF_4]^-$ boron forms four B–F bonds (three normal and one coordinate bond from $F^-$), so its covalency is 4.
Oxidation state: $x + 4(-1) = -1$, so $x = +3$.
Answer: (B). | ||||
| 65 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | StatementsEasy | |
A easy statements question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Ellingham diagrams can be drawn for the formation of oxides, sulfides and halides of metals: Statement I is correct.
An Ellingham diagram plots $\Delta_fG^\circ$ (Gibbs energy of formation), not $\Delta_fH^\circ$, against T: Statement II is incorrect.
Answer: (C). | ||||
| 66 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Protium, deuterium and tritium have the same electronic configuration, so their chemical properties are almost the same: A is correct.
They differ in mass, so bonds to them have different bond dissociation enthalpies ($H_2$ 435.9, $D_2$ 443.4 kJ mol$^{-1}$); this is the main reason their reaction rates differ. R is correct and explains A.
Answer: (A). | ||||
| 67 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Anhydrous $BeF_2$ is best prepared by heating ammonium tetrafluoroberyllate: $(NH_4)_2BeF_4 \xrightarrow{\Delta} BeF_2 + 2NH_3 + 2HF$. The product is pure and free of oxide.
Direct fluorination of Be, or of $BeH_2$, with $F_2$ is violent and hard to control, and is not used; the carbon reduction route ($BeO + C + Cl_2$) is the method for $BeCl_2$, not for the fluoride.
Answer: (B). | ||||
| 68 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter For a given alkali metal, $\Delta_fH^\circ$ of the halide becomes less negative from fluoride to iodide (lattice enthalpy falls as the anion grows).
Approximate values (kJ mol$^{-1}$): NaF $-573$, NaCl $-411$, NaBr $-361$, NaI $-288$.
As signed numbers: $NaF < NaCl < NaBr < NaI$.
Answer: (A). | ||||
| 69 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter In acidic medium $O_2 + 4H^+ + 4e^- \to 2H_2O$, $E^\circ = 1.23$ V.
A halide ion is oxidised by $O_2$ when $E^\circ(X_2/X^-)$ is below 1.23 V:
$Cl_2/Cl^-$: 1.36 V (not oxidised); $Br_2/Br^-$: 1.09 V (oxidised); $I_2/I^-$: 0.54 V (oxidised).
So $Br^-$ and $I^-$ only.
Answer: (A). | ||||
| 70 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The attraction grows with the number of unpaired electrons.
$[Co(en)_3]^{3+}$: $Co^{3+}$, $d^6$ with the strong-field ligand en: $t_{2g}^6$, 0 unpaired.
$[Co(H_2O)_6]^{2+}$: $Co^{2+}$, $d^7$ with weak-field $H_2O$: $t_{2g}^5e_g^2$, 3 unpaired.
$[Zn(H_2O)_6]^{2+}$: $d^{10}$, 0 unpaired.
$[Ni(H_2O)_6]^{2+}$: $d^8$, $t_{2g}^6e_g^2$, 2 unpaired.
Maximum: $[Co(H_2O)_6]^{2+}$.
Answer: (B). | ||||
| 71 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | StatementsEasy | |
A easy statements question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Greenhouse gases absorb infrared radiation from the earth: carbon dioxide, methane, water vapour, nitrous oxide, CFCs and ozone.
Water vapour (A) and ozone (B) are greenhouse gases. $I_2$ is not an atmospheric greenhouse gas, and molecular hydrogen, a homonuclear diatomic molecule, does not absorb infrared radiation.
Answer: (A). | ||||
| 72 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds The complex is of the type $[M(AA)_2ab]$ (octahedral).
Trans isomer (Cl and Br opposite): it has a mirror plane through Cl, Cr, Br, so it is optically inactive: 1 isomer.
Cis isomer: no plane of symmetry, so it exists as a pair of enantiomers: 2 isomers.
Total stereoisomers $= 3$.
Answer: (B). | ||||
| 73 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter In group analysis $Ba^{2+}$ belongs to group V: it is precipitated as $BaCO_3$ (white) by ammonium carbonate in the presence of $NH_4Cl$ and $NH_4OH$. (The precipitate is then dissolved in acetic acid and confirmed as yellow $BaCrO_4$.)
$Ba(OAc)_2$ is soluble and is not a precipitate.
Answer: (C). | ||||
| 74 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | StatementsMedium | |
A medium statements question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Tropolone (2-hydroxycyclohepta-2,4,6-trien-1-one) has three ring C=C bonds and one C=O bond: 8 $\pi$ electrons in all.
The C=O bond is strongly polarised towards oxygen ($C^+–O^-$): the carbonyl $\pi$ electrons sit on oxygen, leaving a tropylium-like ring with 6 $\pi$ electrons (4n + 2, n = 1). The compound is aromatic: Statement I is true.
The $\pi$ electrons of the C=O group are therefore not part of the aromatic sextet: Statement II is false.
Answer: (C). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Matching listEasy | |
A easy matching list question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes A. Alcoholic KOH brings about dehydrohalogenation: $CH_3CH_2CH_2Br \to CH_3CH=CH_2$, an alkene (III).
B. KCN is ionic; $CN^-$ attacks through carbon: $CH_3CH_2CH_2CN$, a nitrile (I).
C. $AgNO_2$ is covalent; the nitrite attacks through nitrogen: $CH_3CH_2CH_2NO_2$, a nitroalkane (IV).
D. Silver acetate gives the ester $CH_3COOCH_2CH_2CH_3$ (II).
A-III, B-I, C-IV, D-II.
Answer: (B). | ||||
| 76 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Assertion and reasonMedium | |
A medium assertion and reason question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The compounds are cyclohexanols: A is 2-chlorocyclohexanol, B is 4-fluorocyclohexanol, C is 3-methylcyclohexanol. There is no resonance, so acidity is set by inductive effects, which weaken quickly with distance.
A: Cl on the carbon next to the C–OH carbon: strong –I, the most acidic.
B: F three carbons away: its –I effect reaches the O–H only weakly, so B is less acidic than A.
C: $CH_3$ has a +I effect, which lowers acidity: least acidic.
So A > B > C: Assertion A is correct.
F is more electronegative than Cl, so it is the stronger electron-withdrawing (–I) group: R is correct.
But A > B holds in spite of F being the stronger –I group, because of the positions; R does not explain A.
Answer: (B). | ||||
| 77 | Organic Chemistry: Some Basic Principles and TechniquesTypes of organic reactions | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on types of organic reactions, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Acrylonitrile, $CH_2=CH–CN$, is an electron-poor alkene: the $\beta$-carbon is electrophilic, and nucleophiles add to it in a conjugate (Michael) addition.
2-Mercaptoethanol has two nucleophilic sites, SH and OH. Sulfur is the softer and much stronger nucleophile (and the thiol is the more acidic), so the thiol adds: $HOCH_2CH_2SH + CH_2=CHCN \to HOCH_2CH_2–S–CH_2CH_2CN$.
Addition to the nitrile carbon (options C and D) does not take place under these conditions, and option A would need the O–H to add instead of the S–H.
Answer: (B). | ||||
| 78 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Work backwards. The last step, $H_3PO_2$, replaces a diazonium group by H. So C is 2,4,6-tribromobenzenediazonium chloride, and B, which was diazotised with $NaNO_2/HCl$, is 2,4,6-tribromoaniline.
B is formed from A by bromination; the $–NH_2$ group activates the ring so strongly that bromination goes to all three ortho/para positions even in $CS_2$ at 0–5 °C. A is aniline.
(Phenol would give a phenol, not an amine to diazotise; benzoic and salicylic acids do not give B.)
Answer: (D). | ||||
| 79 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Lysine has a second amino group in the side chain, arginine a guanidino group and histidine an imidazole ring: each has two basic groups.
Asparagine's side chain is an amide ($–CONH_2$), which is not basic; its only basic group is the $\alpha$-amino group.
Answer: (A). | ||||
| 80 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter A. Weakest intermolecular forces: elastomers, such as neoprene from 2-chloro-1,3-butadiene (III).
B. Hydrogen bonding between chains: fibres such as nylon 6,6 from hexamethylenediamine + adipic acid (I).
C. Heavily branched (cross-linked) polymer: bakelite from phenol + formaldehyde (IV).
D. High density polythene is made with the Ziegler–Natta catalyst $AlEt_3 + TiCl_4$ (II).
A-III, B-I, C-IV, D-II.
Answer: (C). | ||||
| 81 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $M_2CO_3 + 2HCl \to 2MCl + H_2O + CO_2$: one mole of carbonate gives one mole of $CO_2$.
Moles of carbonate $= 0.01$, so molar mass $= \frac{1}{0.01} = 100\ g\ mol^{-1}$.
Answer: 100. | ||||
| 82 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter In a bcc lattice the atoms touch along the body diagonal: $4r = \sqrt{3}a$.
$r = \frac{\sqrt{3} \times 4}{4} = \sqrt{3} = 1.732$ Å $= 17.32 \times 10^{-1}$ Å.
Nearest integer: 17.
Answer: 17. | ||||
| 83 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom Orbital angular momentum $= \sqrt{l(l + 1)}\frac{h}{2\pi}$.
For an s orbital $l = 0$, so the orbital angular momentum is 0, and $x = 0$.
Answer: 0. | ||||
| 84 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Take 100 g of sea water: NaCl 29.25 g $= 0.5$ mol; $MgCl_2$ 19 g $= 0.2$ mol; water $= 100 - 48.25 = 51.75$ g.
With full ionisation, NaCl gives 2 ions and $MgCl_2$ 3: moles of particles $= 0.5 \times 2 + 0.2 \times 3 = 1.6$.
Effective molality $= \frac{1.6}{0.05175} = 30.92$ mol kg$^{-1}$.
$\Delta T_b = 0.52 \times 30.92 = 16.08$ K.
Boiling point $= 100 + 16.08 = 116.08$ °C, nearest integer 116.
Answer: 116. | ||||
| 85 | EquilibriumpH and buffers | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium NaOH: $20 \times 0.1 = 2$ mmol; acetic acid: $50 \times 0.1 = 5$ mmol.
After neutralisation: 3 mmol $CH_3COOH$ and 2 mmol $CH_3COO^-$ — a buffer.
$pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]} = 4.76 + \log\frac{2}{3} = 4.76 + 0.30 - 0.48 = 4.58 = 458 \times 10^{-2}$.
Answer: 458. | ||||
| 86 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry At pH 14, $[OH^-] = 1$ M, so $[Cu^{2+}] = \frac{K_{sp}}{[OH^-]^2} = 10^{-20}$ M.
$Cu^{2+} + 2e^- \to Cu$: $E = E^\circ - \frac{0.059}{2}\log\frac{1}{[Cu^{2+}]} = 0.34 - 0.0295 \times 20 = 0.34 - 0.59 = -0.25$ V.
$-0.25\ V = (-)25 \times 10^{-2}$ V, so $x = 25$.
Answer: 25. | ||||
| 87 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics If the pressure of A falls by p, the total pressure is $(800 - p) + 2p + p = 800 + 2p$.
After 10 min: $800 + 2p = 1600$, so $p = 400$: half of A has reacted, and $t_{1/2} = 10$ min.
After 30 min (three half-lives): $p_A = \frac{800}{8} = 100$, so $p = 700$.
Total pressure $= 800 + 2(700) = 2200$ mm Hg.
Answer: 2200. | ||||
| 88 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Cr balance: $Y = 2$. O balance: $Z = 7$. H balance: $X = 2Z = 14$.
Charge check: left $-2 + 14 + 12 = 24$; right $2(3) + 6(3) = 24$. ✓
$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \to 2Cr^{3+} + 6Fe^{3+} + 7H_2O$.
$X + Y + Z = 14 + 2 + 7 = 23$.
Answer: 23. | ||||
| 89 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$ (often written $Na_2B_4O_7 \cdot 10H_2O$).
$x = 5$, $y = 4$, $z = 8$, so $x + y + z = 17$.
Answer: 17. | ||||
| 90 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Mass of Br in 0.376 g of AgBr $= \frac{80}{188} \times 0.376 = 0.16$ g.
% Br $= \frac{0.16}{0.400} \times 100 = 40$.
Answer: 40. | ||||
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