JEE (Main) · 2022
JEE (Main) 2022 — 29 July, 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
- Matrices and Determinants3
- Vector Algebra3
- Complex Numbers2
11 easy · 17 medium · 2 hard
Physics
30 Q
- Electrostatics2
- Current Electricity2
- Magnetic Effects of Current and Magnetism2
26 easy · 4 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques3
- Some Basic Concepts of Chemistry2
- p-Block Elements2
24 easy · 4 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 | 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 Let $r = |z| > 0$. By the triangle inequality $|z| - \frac{1}{|z|} \le \left|z - \frac{1}{z}\right| = 2$.
So $r - \frac{1}{r} \le 2$, i.e. $r^2 - 2r - 1 \le 0$, giving $r \le 1 + \sqrt{2}$.
The bound is reached: for the real number $z = 1+\sqrt2$, $\frac1z = \sqrt2 - 1$ and $z - \frac1z = 2$.
Maximum $|z| = \sqrt{2}+1$.
Answer: (D). | ||||
| 2 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants (A) $R_1 \to R_1 + R_2$ gives $(0,\ 1)$ as the first row: possible.
(B) $R_1 \leftrightarrow R_2$: possible.
(D) $R_2 \to R_2 + 2R_1$ gives $(1-2,\ -1+4) = (-1,\ 3)$: possible.
(C) Row 1 is unchanged, so row 2 must come from $R_2 \to cR_2$ or $R_2 \to R_2 + kR_1$. Scaling cannot turn $(1,-1)$ into $(-2,7)$. With $R_2 + kR_1 = (1-k,\ -1+2k)$, the first entry $-2$ needs $k = 3$, which gives $(-2,\ 5) \ne (-2,\ 7)$. Not possible.
Answer: (C). | ||||
| 3 | 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:
$\begin{vmatrix}1&1&1\\2&5&\alpha\\1&2&3\end{vmatrix} = (15-2\alpha) - (6-\alpha) + (4-5) = 8 - \alpha = 0$, so $\alpha = 8$.
The second equation must then be a combination of the other two: $2x+5y+8z = p(x+y+z) + q(x+2y+3z)$ gives $p+q=2$, $p+2q=5$, so $q=3$, $p=-1$ (and $p+3q = 8$ checks).
Hence $\beta = -1\cdot 6 + 3\cdot 14 = 36$.
$\alpha+\beta = 44$.
Answer: (C). | ||||
| 4 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\lim_{x\to0}\frac{\log_e(1+5x)}{x} = 5$ and $\lim_{x\to0}\frac{\log_e(1+\alpha x)}{x} = \alpha$, so the limit is $5 - \alpha$.
Continuity needs $5 - \alpha = 10$, so $\alpha = -5$.
Answer: (D). | ||||
| 5 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $3x^2-5x+2 = (3x-2)(x-1)$ is $\ge 0$ on $[0,\frac23]$ and $\le 0$ on $[\frac23,1]$.
On $[0,\frac23]$: $g(x) = 2x - (3x^2-5x+2) + 1 = -3x^2+7x-1$, increasing from $-1$ to $\frac73$.
On $[\frac23,1]$: $g(x) = 2x + (3x^2-5x+2) + 1 = 3x^2-3x+3$, increasing from $\frac73$ to $3$, so $[g] = 2$ there (except the end point).
On $[0,\frac23]$ the integer part changes where $g = 0, 1, 2$:
$3x^2-7x+1=0 \Rightarrow x_0 = \frac{7-\sqrt{37}}{6}$; $3x^2-7x+2=0 \Rightarrow x_1 = \frac13$; $3x^2-7x+3=0 \Rightarrow x_2 = \frac{7-\sqrt{13}}{6}$.
Integral $= (-1)x_0 + 0\cdot(x_1-x_0) + 1\cdot(x_2 - \frac13) + 2(\frac23 - x_2) + 2\cdot\frac13 = \frac53 - x_0 - x_2$
$= \frac{10}{6} - \frac{14 - \sqrt{37} - \sqrt{13}}{6} = \frac{\sqrt{37}+\sqrt{13}-4}{6}$.
Answer: (A). | ||||
| 6 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Write $b_n = a_{n+1} - 2a_n$. The recurrence gives $a_{n+2} - 2a_{n+1} = (a_{n+1} - 2a_n) + 1$, i.e. $b_{n+1} = b_n + 1$.
With $b_0 = a_1 - 2a_0 = 0$, $b_n = n$.
The expression factorises: $a_{25}(a_{23} - 2a_{22}) - 2a_{24}(a_{23} - 2a_{22}) = (a_{23}-2a_{22})(a_{25}-2a_{24}) = b_{22}\,b_{24}$.
$= 22 \times 24 = 528$.
(Indeed $a_n = 2^n - n - 1$.)
Answer: (B). | ||||
| 7 | 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 Note $r\,(r+1)! - (r-1)\,r! = r!\,[r(r+1) - (r-1)] = (r^2+1)\,r!$.
So the sum telescopes: $\sum_{r=1}^{20}[r\,(r+1)! - (r-1)\,r!] = 20\cdot 21! - 0\cdot 1! = 20\cdot 21!$.
$20\cdot 21! = (22-2)\,21! = 22! - 2(21!)$.
Answer: (B). | ||||
| 8 | Integral CalculusIntegral as an antiderivative | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{d}{dx}\left(\frac{\tan x}{\sin^{2022}x}\right) = \frac{\sec^2 x}{\sin^{2022}x} - 2022\,\frac{\tan x\cos x}{\sin^{2023}x} = \frac{\sec^2 x - 2022}{\sin^{2022}x}$.
So $I(x) = \frac{\tan x}{\sin^{2022}x} + C$. At $\frac{\pi}{4}$: $\frac{1}{(1/\sqrt2)^{2022}} = 2^{1011}$, so $C = 0$.
$I\left(\frac{\pi}{3}\right) = \frac{\sqrt3}{(\sqrt3/2)^{2022}} = \frac{\sqrt3\cdot 2^{2022}}{3^{1011}}$, $I\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt3}\cdot 2^{2022}$.
$\frac{I(\pi/3)}{I(\pi/6)} = \frac{3}{3^{1011}} = 3^{-1010}$, so $3^{1010}I\left(\frac{\pi}{3}\right) - I\left(\frac{\pi}{6}\right) = 0$.
Answer: (A). | ||||
| 9 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations The lines $x+y-2=0$ and $x-y=0$ meet at $(1,1)$. Put $X = x-1$, $Y = y-1$: $\frac{dY}{dX} = \frac{X+Y}{X-Y}$.
With $Y = vX$: $X\frac{dv}{dX} = \frac{1+v}{1-v} - v = \frac{1+v^2}{1-v}$, so $\int\frac{1-v}{1+v^2}dv = \int\frac{dX}{X}$:
$\tan^{-1}v - \frac12\log_e(1+v^2) = \log_e|X| + c$, i.e. $\tan^{-1}\frac{Y}{X} = \frac12\log_e(X^2+Y^2) + c$.
At $(2,1)$: $X=1, Y=0$, so $c = 0$.
At $(k+1,2)$: $X=k, Y=1$: $\tan^{-1}\frac1k = \frac12\log_e(k^2+1)$, i.e. $2\tan^{-1}\left(\frac1k\right) = \log_e(k^2+1)$.
Answer: (A). | ||||
| 10 | 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 $x^3+6x^2+11x+6 = (x+1)(x+2)(x+3)$ and $\frac{2x^2+11x+13}{(x+1)(x+2)(x+3)} = \frac{2}{x+1} + \frac{1}{x+2} - \frac{1}{x+3}$.
Integrating factor: $e^{\int P\,dx} = \frac{(x+1)^2(x+2)}{x+3}$.
$y\cdot\frac{(x+1)^2(x+2)}{x+3} = \int \frac{x+3}{x+1}\cdot\frac{(x+1)^2(x+2)}{x+3}\,dx = \int (x+1)(x+2)\,dx = \frac{x^3}{3} + \frac{3x^2}{2} + 2x + C$.
At $(0,1)$: $\frac{2}{3} = C$.
At $x = 1$: $y\cdot\frac{4\cdot3}{4} = \frac13 + \frac32 + 2 + \frac23 = \frac92$, so $3y = \frac92$ and $y(1) = \frac32$.
Answer: (B). | ||||
| 11 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctHard | |
A hard single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines Substituting the vertex in the diagonal: $10[(\cos\alpha-\sin\alpha)^2 + (\sin\alpha+\cos\alpha)^2] = 20 \ne 10$, so the vertex is not on this diagonal. Its distance from the diagonal is half the other diagonal, $\frac{a}{\sqrt2}$:
$\frac{|20-10|}{\sqrt{(\cos\alpha-\sin\alpha)^2+(\sin\alpha+\cos\alpha)^2}} = \frac{10}{\sqrt2}$, so $a = 10$.
Then $100 + 110 + 3(m_1^2+m_2^2) = 220$, so $m_1^2 + m_2^2 = \frac{10}{3}$; with $m_1m_2 = -1$, $m_1^2 = 3$ or $\frac13$: the sides make $60^\circ$ and $150^\circ$, or $30^\circ$ and $120^\circ$, with the $x$-axis.
The diagonal bisects them, so it makes $15^\circ$ or $105^\circ$, or $75^\circ$ or $165^\circ$. Its slope is $-\frac{\cos\alpha-\sin\alpha}{\sin\alpha+\cos\alpha} = \tan\left(\alpha - \frac{\pi}{4}\right)$ with $\alpha - \frac{\pi}{4} \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$, so $\alpha - \frac{\pi}{4} = \pm 15^\circ$: $\alpha = 60^\circ$ or $30^\circ$.
Either way $\sin^4\alpha + \cos^4\alpha = 1 - 2\sin^2\alpha\cos^2\alpha = 1 - \frac{2\cdot3}{16} = \frac58$.
$72\cdot\frac58 + 100 - 30 + 13 = 45 + 83 = 128$.
Answer: (B). | ||||
| 12 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions The left side is at most 2; by AM–GM $4^x + 4^{-x} \ge 2$, with equality only at $x = 0$.
So both sides must equal 2, which forces $x = 0$; there $2\cos 0 = 2$ holds.
$S = \{0\}$ has one element.
Answer: (A). | ||||
| 13 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines $AB$ is vertical and $AC$ horizontal, so the angle at $A$ is right and the circumcentre is the midpoint of $BC$: $\left(\frac{5\alpha}{8}, 2\right)$.
Equating with $\left(5, \frac{\alpha}{4}\right)$: $\alpha = 8$ (and $\frac{\alpha}{4} = 2$ checks).
$A(8,-2)$, $B(8,6)$, $C(2,-2)$: $AB = 8$, $AC = 6$, $BC = 10$.
Area $= \frac12\cdot8\cdot6 = 24$; perimeter $= 24$; circumradius $= 5$; inradius $= \frac{24}{12} = 2$.
The statement "perimeter is 25" is not correct.
Answer: (B). | ||||
| 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 $PQ$ is the distance of $P$ from the plane: $\frac{|1+4+3-14|}{\sqrt{1+4+1}} = \frac{6}{\sqrt6} = \sqrt6$.
Triangle $PQR$ is right-angled at $Q$, with $\tan 60^\circ = \frac{PQ}{QR}$, so $QR = \frac{\sqrt6}{\sqrt3} = \sqrt2$.
Area $= \frac12\cdot\sqrt6\cdot\sqrt2 = \sqrt3$.
Answer: (B). | ||||
| 15 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Taking $(2,3,9)$ as origin, the other points give $(3,-1,-8)$, $(-1,\lambda-3,-1)$, $(\lambda-2,-1,-6)$. Coplanarity needs their scalar triple product to vanish:
$\begin{vmatrix}3&-1&-8\\-1&\lambda-3&-1\\\lambda-2&-1&-6\end{vmatrix} = 3(-6\lambda+17) + (\lambda+4) - 8(-\lambda^2+5\lambda-5) = 8\lambda^2 - 57\lambda + 95 = 0$.
Product of the roots $= \frac{95}{8}$.
Answer: (D). | ||||
| 16 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability After a transfer Bag II has 10 balls. $P(\text{black}\mid\text{red moved}) = \frac{5}{10}$, $P(\text{black}\mid\text{black moved}) = \frac{6}{10}$, $P(\text{black}\mid\text{white moved}) = \frac{5}{10}$.
By Bayes' theorem: $P(\text{red moved}\mid\text{black}) = \frac{\frac{3}{10}\cdot\frac{5}{10}}{\frac{3}{10}\cdot\frac{5}{10} + \frac{4}{10}\cdot\frac{6}{10} + \frac{3}{10}\cdot\frac{5}{10}} = \frac{15}{15+24+15} = \frac{15}{54} = \frac{5}{18}$.
Answer: (B). | ||||
| 17 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations For $z = X + iY$:
$|z+i| = |z-1| \Rightarrow X^2 + (Y+1)^2 = (X-1)^2 + Y^2 \Rightarrow Y = -X$.
$|z-1+i| \ge |z| \Rightarrow (X-1)^2 + (Y+1)^2 \ge X^2 + Y^2 \Rightarrow Y \ge X - 1$; with $Y = -X$ this is $X \le \frac12$.
$|z| < 2 \Rightarrow 2X^2 < 4 \Rightarrow -\sqrt2 < X < \sqrt2$.
So $S$ is the segment $Y = -X$, $X \in \left(-\sqrt2, \frac12\right]$.
For $w = 2x + iy$ the real part is $2x$ (and $y = -2x$ can be chosen): $-\sqrt2 < 2x \le \frac12$, i.e. $x \in \left(-\frac{1}{\sqrt2}, \frac14\right]$.
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 Three coplanar vectors from one point with equal angles between each pair make $120^\circ$ with one another (equal angle $0^\circ$ would make every cross product zero).
$(\vec a\times\vec b)\cdot(\vec b\times\vec c) = (\vec a\cdot\vec b)(\vec b\cdot\vec c) - (\vec a\cdot\vec c)(\vec b\cdot\vec b) = \frac{ab\cdot bc}{4} + \frac{ac\,b^2}{2} = \frac34\,ab^2c$, where $a, b, c$ are the magnitudes.
Similarly the other two terms are $\frac34\,abc^2$ and $\frac34\,a^2bc$.
Sum $= \frac34\,abc\,(a+b+c) = \frac34\cdot14\,(a+b+c) = 168$, so $a+b+c = 16$.
Answer: (C). | ||||
| 19 | 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 $x^2+2x+7 = (x+1)^2 + 6 > 0$, so we need $-(x^2+2x+7) \le x^2-3x+2 \le x^2+2x+7$.
Right inequality: $-3x+2 \le 2x+7 \Rightarrow x \ge -1$.
Left inequality: $2x^2 - x + 9 \ge 0$, true for all $x$ (discriminant $1 - 72 < 0$).
Domain $= [-1, \infty)$.
Answer: (C). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $(p\Rightarrow q)\vee(p\Rightarrow r) \equiv \sim p \vee q \vee r$.
(A) $\sim(p\wedge\sim r)\vee q \equiv \sim p \vee r \vee q$: equivalent.
(B) $q \vee \sim r \vee p$: not equivalent (e.g. $p$ true, $q, r$ false makes the given statement false but this true).
(C) $\sim p \vee q \vee r$: equivalent.
(D) $\sim p \vee q \vee r$: equivalent.
Answer: (B). | ||||
| 21 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalEasy | |
A easy numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability Mean $np$ and variance $npq$ are the roots of $t^2 - 82.5t + 1350 = 0$: $t = \frac{82.5 \pm 37.5}{2} = 60$ or $22.5$.
The variance is the smaller: $np = 60$, $npq = 22.5$, so $q = \frac{3}{8}$, $p = \frac58$, and $n = \frac{60}{5/8} = 96$.
Answer: 96. | ||||
| 22 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Since $\alpha^2 = \alpha + 4$ and $\beta^2 = \beta + 4$, $P_{n+2} = P_{n+1} + 4P_n$, i.e. $P_{n+2} - P_{n+1} = 4P_n$.
Numerator $= P_{16}(P_{15} - P_{14}) - P_{15}(P_{15} - P_{14}) = (P_{15}-P_{14})(P_{16}-P_{15}) = 4P_{13}\cdot4P_{14}$.
The ratio is $16$.
Answer: 16. | ||||
| 23 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $X'A^kX$ is the sum of all entries of $A^k$.
$A^2 = \begin{bmatrix}1&0&6\\0&1&0\\0&0&1\end{bmatrix}$, and by induction $A^{2m} = \begin{bmatrix}1&0&6m\\0&1&0\\0&0&1\end{bmatrix}$, with entry sum $3 + 6m$.
$A^{2m+1} = A^{2m}A = \begin{bmatrix}-1&2&3-6m\\0&1&6\\0&0&-1\end{bmatrix}$, with entry sum $10 - 6m$, which is never 33 for $m \ge 0$.
$3 + 6m = 33 \Rightarrow m = 5$, so $k = 10$.
Answer: 10. | ||||
| 24 | 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 55 means divisible by 5 and 11. The last digit must be 5 (no 0 is available).
A five-digit number below 23421 would have to begin $2\,3$, and then its third digit is 4, 5 or 6, giving at least $23456 > 23421$: none.
Four-digit numbers $abc5$ from $\{2,3,4,6\}$: divisibility by 11 needs $(a + c) - (b + 5) \equiv 0 \pmod{11}$, i.e. $a + c = b + 5$ (the difference lies between $-6$ and $7$, so it must be 0).
$b = 2$: $a+c = 7$ from $\{3,4,6\}$: $\{3,4\}$ → 2 numbers (3245, 4235).
$b = 3$: $a+c = 8$ from $\{2,4,6\}$: $\{2,6\}$ → 2 numbers (2365, 6325).
$b = 4$: $a+c = 9$ from $\{2,3,6\}$: $\{3,6\}$ → 2 numbers (3465, 6435).
$b = 6$: $a+c = 11$ from $\{2,3,4\}$: none.
Total $6$ (all exceed 1012).
Answer: 6. | ||||
| 25 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $K\,{}^{10}C_K = 10\,{}^{9}C_{K-1}$, so $\sum K^2({}^{10}C_K)^2 = 100\sum_{K=1}^{10}{}^9C_{K-1}\,{}^9C_{K-1} = 100\sum_{j=0}^{9}{}^9C_j\,{}^9C_{9-j} = 100\cdot{}^{18}C_9$.
${}^{18}C_9 = 48620$, so the sum is $4862000 = 22000 \times 221$.
$L = 221$.
Answer: 221. | ||||
| 26 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $[x+20] = [x] + 20$, so $f(x) = 4|2x+3| + 9\left[x+\frac12\right] - 12[x] - 240$.
$9\left[x+\frac12\right]$ jumps at $x = n - \frac12$, where $[x]$ does not jump: $f$ is discontinuous there. In $(-20,20)$ these are $-19.5, -18.5, \dots, 19.5$: 40 points (this includes $x = -\frac32$, where $|2x+3|$ also has its corner).
$-12[x]$ jumps at every integer, where $\left[x+\frac12\right]$ does not: $-19, \dots, 19$: 39 points.
Elsewhere $f$ is linear, hence differentiable.
Total $40 + 39 = 79$.
Answer: 79. | ||||
| 27 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives For $y = 5x^2+2x-25$ at $(2,-1)$ the slope is $10\cdot2 + 2 = 22$: the tangent is $y = 22x - 45$.
On $y = x^3 - x^2 + x$: $3a^2 - 2a + 1 = 22 \Rightarrow 3a^2 - 2a - 21 = 0 \Rightarrow a = 3$ or $a = -\frac73$.
The point must lie on $y = 22x - 45$: for $a = 3$, $b = 27 - 9 + 3 = 21 = 66 - 45$ ✓. For $a = -\frac73$, $b = -\frac{553}{27}$ but the line gives $-\frac{289}{3}$ ✗.
So $(a,b) = (3, 21)$ and $|2a + 9b| = |6 + 189| = 195$.
Answer: 195. | ||||
| 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 Radius $r = \frac{13}{2}$, half-chord $6$, so the centre $O$ is at distance $d = \sqrt{\frac{169}{4} - 36} = \frac52$ from $AB$.
$P$ lies on the line through $O$ perpendicular to $AB$, with $OP = \frac{r^2}{d} = \frac{169/4}{5/2} = \frac{169}{10}$.
Distance of $P$ from $AB$ $= \frac{169}{10} - \frac52 = \frac{144}{10} = 14.4$; five times this is $72$.
Answer: 72. | ||||
| 29 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2 + 2\vec a\cdot\vec b = |\vec a|^2 + 2|\vec b|^2$ gives $|\vec b|^2 = 2\vec a\cdot\vec b = 6$.
$|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2$: $75 = 6|\vec a|^2 - 9$, so $|\vec a|^2 = 14$.
Answer: 14. | ||||
| 30 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $S$: lattice points with $x, y \ge 1$ inside the ellipse $\frac{(x-3)^2}{16} + \frac{(y-4)^2}{9} \le 1$; $T$: the disc of radius 6 about $(7,4)$.
Count column by column ($y$ values allowed by both):
$x=1$: ellipse needs $16(y-4)^2 \le 108$, $|y-4| \le 2$; disc needs $(y-4)^2 \le 0$: $y = 4$ → 1.
$x=2$: ellipse $|y-4| \le 2$ ($16(y-4)^2 \le 135$); disc $(y-4)^2 \le 11$: $y = 2..6$ → 5.
$x=3$: ellipse $|y-4| \le 3$; disc $(y-4)^2 \le 20$: $y = 1..7$ → 7.
$x=4$: ellipse $16(y-4)^2 \le 135$, $|y-4| \le 2$; disc $(y-4)^2 \le 27$: $y = 2..6$ → 5.
$x=5$: ellipse $16(y-4)^2 \le 108$, $|y-4| \le 2$; disc $\le 32$: → 5.
$x=6$: ellipse $16(y-4)^2 \le 63$, $|y-4| \le 1$; → 3.
$x=7$: ellipse $16(y-4)^2 \le 0$, $y = 4$; → 1.
Total $1+5+7+5+5+3+1 = 27$.
Answer: 27. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Let each of A and B carry $q$, a distance $r$ apart: $F = \frac{kq^2}{r^2}$.
C touches A: both share, A has $\frac q2$, C has $\frac q2$. C then touches B: $\frac{q/2 + q}{2} = \frac{3q}{4}$ each.
At the midpoint ($\frac r2$ from each), A pushes C towards B with $\frac{k(q/2)(3q/4)}{(r/2)^2} = \frac32F$, and B pushes C towards A with $\frac{k(3q/4)^2}{(r/2)^2} = \frac94F$.
Net force $= \frac94F - \frac32F = \frac34F$ (towards A).
Answer: (B). | ||||
| 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 Torque = force × arm: N m (III). Stress = force per area: N m⁻² (IV). Latent heat = heat per mass: J kg⁻¹ (II). Power = work per time = N m s⁻¹ (I).
A→III, B→IV, C→II, D→I.
Answer: (B). | ||||
| 33 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charges distribute so that the outer faces carry $\frac{q_1+q_2}{2}$ each and the inner faces $\pm\frac{q_1-q_2}{2}$.
Only the inner charges produce the field between the plates, so $V = \frac{(q_1-q_2)/2}{C} = \frac{q_1-q_2}{2C}$.
Answer: (C). | ||||
| 34 | Current ElectricityResistivity | Class 12: Current Electricity | Assertion and reasonEasy | |
A easy assertion and reason question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity A standard resistance must keep its value as the temperature changes. Constantan and manganin have very small temperature coefficients of resistance (and high resistivity), which is exactly why they are used for standard coils.
Both are true and R explains A.
Answer: (A). | ||||
| 35 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Let the lengths be $x$ and $y$, with area $A$ and resistivity $\rho$. Stretching X to twice its length at constant volume halves its area, so $R_W = \rho\frac{2x}{A/2} = 4\rho\frac{x}{A}$.
$R_Y = \rho\frac{y}{A}$. Given $R_W = 2R_Y$: $4x = 2y$, so $x : y = 1 : 2$.
Answer: (B). | ||||
| 36 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The long wire Y sets up $B = \frac{\mu_0 I_Y}{2\pi d}$ at X; the force on the 0.5 m length of X is $F = \frac{\mu_0 I_X I_Y \ell}{2\pi d} = \frac{2\times10^{-7}\times2\times3\times0.5}{0.05} = 1.2\times10^{-5}$ N.
By Newton's third law Y feels the same force. Parallel currents in the same direction attract, so it is directed towards X.
Answer: (A). | ||||
| 37 | 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 The time between throws is $\frac1n$, and this is the time to reach the top: $\frac ug = \frac1n$, so $u = \frac gn$.
Maximum height $H = \frac{u^2}{2g} = \frac{g}{2n^2}$.
Answer: (D). | ||||
| 38 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current X is a resistor, $R = \frac{100}{5} = 20\ \Omega$; Y is an inductor, $X_L = 20\ \Omega$.
In series $Z = \sqrt{20^2 + 20^2} = 20\sqrt2\ \Omega$, so $I_0 = \frac{100}{20\sqrt2} = \frac{5}{\sqrt2}$ A.
$I_{rms} = \frac{I_0}{\sqrt2} = \frac52$ A.
Answer: (D). | ||||
| 39 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics After P: $\frac{2I_0}{2} = I_0$. After Q (Malus's law): $I_0\cos^2 30^\circ = \frac{3I_0}{4}$.
Answer: (C). | ||||
| 40 | 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 Kinetic energy gained is $qV$, so $p = \sqrt{2mqV}$.
$\frac{p_\alpha}{p_p} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4\times2} = 2\sqrt2$.
Answer: (B). | ||||
| 41 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | StatementsEasy | |
A easy statements question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $R = R_0A^{1/3}$, so the volume $\frac43\pi R^3 \propto A$: (A) true, (B) false.
Density $= \frac{\text{mass}}{\text{volume}} \propto \frac{A}{A}$ is independent of $A$: (E) true, (C) and (D) false.
Answer: (B). | ||||
| 42 | GravitationGravitational potential energy | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation $\Delta U = \frac{GMm}{R} - \frac{GMm}{4R} = \frac34\,mgR$ (using $GM = gR^2$).
$= \frac34\times1\times10\times6.4\times10^6 = 4.8\times10^7$ J $= 48$ MJ.
Answer: (A). | ||||
| 43 | 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 From rest, $t_1 = \sqrt{\frac{2(h/2)}{g}} = \sqrt{\frac hg}$ and the whole fall takes $\sqrt{\frac{2h}{g}} = \sqrt2\,t_1$.
So $t_2 = \sqrt2\,t_1 - t_1 = (\sqrt2 - 1)t_1$.
Answer: (D). | ||||
| 44 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion At rest, the tension is $T = m_2g = 30$ N. Along the smooth incline $T = m_1g\sin\theta$: $\sin\theta = \frac{30}{50} = \frac35$, so $\cos\theta = \frac45$.
The smooth plane exerts only the normal force: $N = m_1g\cos\theta = 50\times\frac45 = 40$ N.
Answer: (B). | ||||
| 45 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $K = \frac{p^2}{2m}$, so $\frac{K'}{K} = (1.2)^2 = 1.44$: an increase of 44%.
Answer: (C). | ||||
| 46 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\vec\tau = \vec r\times\vec F = \begin{vmatrix}\hat i&\hat j&\hat k\\2&2&1\\5&3&-7\end{vmatrix} = \hat i(-14-3) - \hat j(-14-5) + \hat k(6-10) = -17\hat i + 19\hat j - 4\hat k$.
Answer: (C). | ||||
| 47 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics D (2.0 m³, 600 N m⁻²) → E (5.0 m³, 300 N m⁻²): work = area under the line $= \frac{600+300}{2}\times3 = 1350$ J.
E → F at 300 N m⁻² from 5.0 to 2.0 m³: work $= 300\times(-3) = -900$ J.
Total $= 450$ J.
Answer: (B). | ||||
| 48 | Magnetic Effects of Current and MagnetismEarth magnetism | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on earth magnetism, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B_V = B\sin\delta$, and $\tan 37^\circ = \frac34$ gives $\sin 37^\circ = \frac35$.
$B = \frac{6\times10^{-5}}{3/5} = 1\times10^{-4}$ T.
Answer: (D). | ||||
| 49 | 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 $v_{rms} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3\times1.38\times10^{-23}\times273}{5\times10^{-17}}} = \sqrt{2.26\times10^{-4}} \approx 1.5\times10^{-2}$ m s⁻¹ $= 15$ mm s⁻¹.
Answer: (C). | ||||
| 50 | 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 At $45^\circ$ to the interface the angle of incidence is also $45^\circ$. A deviation of $15^\circ$ (bending towards the normal) gives $r = 30^\circ$.
$\mu = \frac{\sin45^\circ}{\sin30^\circ} = \sqrt2 \approx 1.414$.
Answer: (C). | ||||
| 51 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | NumericalMedium | |
A medium numerical question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion The far end must supply the centripetal force for all the liquid: $F = \int_0^L \frac mL\,dr\,\omega^2 r = \frac{m\omega^2L}{2}$.
$F = \frac{0.25\times\omega^2\times0.5}{2} = \frac{\omega^2}{16}$, so $\omega = 4\sqrt F$.
$x = 4$.
Answer: 4. | ||||
| 52 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | NumericalEasy | |
A easy numerical question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Visible power $P = 11$ W, spread over a sphere: $I = \frac{P}{4\pi r^2}$.
$I_1 = \frac{11}{4\pi} = 0.8754$ W m⁻², $I_5 = \frac{11}{100\pi} = 0.0350$ W m⁻².
Change $= 0.8403$ W m⁻² $= 84.03\times10^{-2}$ W m⁻², so $a = 84$ to the nearest integer.
Answer: 84. | ||||
| 53 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Maximum tension $T = 5\times10^8\times10^{-4} = 5\times10^4$ N.
$\frac{mv^2}{r} = T$: $v^2 = \frac{5\times10^4\times0.5}{10} = 2500$, so $v = 50$ m s⁻¹.
Answer: 50. | ||||
| 54 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal velocity, viscous force = weight − buoyancy $= mg\left(1 - \frac{\sigma}{\rho}\right) = 0.3\times10^{-3}\times10\times\left(1 - \frac{1.3}{8}\right)$.
$= 3\times10^{-3}\times0.8375 = 2.51\times10^{-3}$ N $= 25.1\times10^{-4}$ N.
$x = 25$ to the nearest integer.
Answer: 25. | ||||
| 55 | Communication SystemsAmplitude modulation | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter The modulating frequency is $f_m = \frac{6.28\times10^6}{2\pi} = 1$ MHz.
The band-pass filter keeps the AM wave, with sidebands at $f_c \pm f_m$: bandwidth $= 2f_m = 2$ MHz.
Answer: 2. | ||||
| 56 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Linear density $\mu = \frac{0.01}{0.5} = 0.02$ kg m⁻¹. Tension $T = \mu v^2 = 0.02\times3600 = 72$ N.
Extension $\Delta L = \frac{TL}{AY} = \frac{72\times0.5}{2\times10^{-6}\times1.2\times10^{11}} = 1.5\times10^{-4}$ m $= 15\times10^{-5}$ m.
Answer: 15. | ||||
| 57 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves In water the effective weight is reduced by the buoyancy: $g_{eff} = g\left(1 - \frac15\right) = \frac45g$.
$T' = T\sqrt{\frac{g}{g_{eff}}} = 10\sqrt{\frac54} = 5\sqrt5$ s, so $x = 5$.
Answer: 5. | ||||
| 58 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalEasy | |
A easy numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The Zener current is greatest when the load draws nothing, so the whole current through R flows in the Zener: $I = \frac{20 - 8}{R} \le 25$ mA.
$R_{min} = \frac{12}{0.025} = 480\ \Omega$.
Answer: 480. | ||||
| 59 | 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 $\frac{N_B}{N_A} = \frac{N_0e^{-16\lambda t}}{N_0e^{-25\lambda t}} = e^{9\lambda t} = e$, so $t = \frac{1}{9\lambda}$.
$a = 9$.
Answer: 9. | ||||
| 60 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Energy conservation: $\frac12CV^2 = \frac12LI_{max}^2$, so $I_{max} = V\sqrt{\frac CL} = 100\sqrt{\frac{500\times10^{-6}}{50\times10^{-3}}} = 100\times0.1 = 10$ A.
Answer: 10. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $M(KNO_3) = 39+14+48 = 101$ g mol⁻¹, so 110.0 g is $\frac{110}{101} = 1.089$ mol.
$HNO_3$ needed $= \frac43\times1.089 = 1.452$ mol; $M(HNO_3) = 63$ g mol⁻¹, so mass $= 1.452\times63 = 91.5$ g.
Answer: (C). | ||||
| 62 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom The orbitals are A: 3d ($n+l = 5$), B: 4p ($5$), C: 4d ($6$), D: 3p ($4$).
By the $(n+l)$ rule, lower $n+l$ is lower in energy; for equal $n+l$, lower $n$ is lower: 3p < 3d < 4p < 4d.
So D < A < B < C.
Answer: (B). | ||||
| 63 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics Carbon in 0.6 kg of coal $= 0.6\times600 = 360$ g $= 30$ mol.
60% to CO: 18 mol × 100 kJ = 1800 kJ. 40% to $CO_2$: 12 mol × 400 kJ = 4800 kJ.
Total $= 6600$ kJ.
Answer: (D). | ||||
| 64 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium $H^+$ from HCl: $200\times0.01 = 2$ mmol; from $H_2SO_4$: $400\times0.01\times2 = 8$ mmol. Total 10 mmol in 600 mL.
$[H^+] = \frac{10}{600} = \frac{1}{60}$ M, so pH $= \log 60 = \log 2 + \log 3 + 1 = 1.78$.
Answer: (B). | ||||
| 65 | 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 In physisorption, a gas with a higher critical temperature is more easily liquefied and is adsorbed more. He has the lowest critical temperature (5.2 K), so it is adsorbed least.
Answer: (A). | ||||
| 66 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Liquation refines a low-melting metal such as tin: the impure metal is melted on a sloping hearth and the molten metal flows down, leaving the higher-melting impurities behind.
Answer: (C). | ||||
| 67 | HydrogenHydrogen and its compounds | No matching NCERT chapter | StatementsEasy | |
A easy statements question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Stannane, $SnH_4$, is a covalent (molecular, electron-precise) hydride of a group 14 element: Statement I is true.
Sn is $sp^3$ hybridised with four bond pairs, so $SnH_4$ is tetrahedral, not planar: Statement II is false.
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 Gypsum, $CaSO_4\cdot2H_2O$, is added to cement clinker to slow down the setting so that the cement gets sufficiently hardened ("enhance the setting time" means lengthen it).
Answer: (B). | ||||
| 69 | 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 On heating, borax gives a glassy mixture of $NaBO_2$ and $B_2O_3$; the $B_2O_3$ combines with CoO to form cobalt metaborate: $CoO + B_2O_3 \rightarrow Co(BO_2)_2$, which is blue.
Answer: (B). | ||||
| 70 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Hydration enthalpies of the $M^{2+}$ ions grow in magnitude across the series, but crystal-field stabilisation adds to them. $Mn^{2+}$ is $d^5$ (high spin in water) with zero CFSE, and it also lies early in the series, so its hydration enthalpy has the smallest magnitude of the four (it sits in the dip of the double-humped curve).
Answer: (B). | ||||
| 71 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctHard | |
A hard single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All four are $d^9$ Cu(II), so each suffers a Jahn–Teller distortion. The overall departure from a regular octahedron also depends on how unlike the ligands are and how they are placed.
In $[Cu(H_2O)_6]^{2+}$ all six ligands are alike; in $[Cu(en)(H_2O)_4]^{2+}$ and trans-$[Cu(en)_2Cl_2]$ the different ligands keep a symmetrical arrangement (in the trans isomer the two weak $Cl^-$ lie on the elongated axis, in line with the Jahn–Teller elongation).
In cis-$[Cu(en)_2Cl_2]$ the two $Cl^-$ ligands are mutually cis, so the elongated axis cannot hold both; the bond lengths and angles are disturbed in all directions, giving the least symmetrical, most distorted structure.
Answer: (C). | ||||
| 72 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter High concentrations of $NO_2$ damage the leaves of plants and retard the rate of photosynthesis. (NO is converted to $NO_2$; the other two are ions, not oxides formed in air.)
Answer: (C). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In the common name, Greek letters count from the carbon carrying the –CHO group: that ring carbon is α, its neighbours β, the next ones γ.
So the methyl group is on the ring carbon 1,3 to the one carrying –CHO directly (3-methylcyclohexane-1-carbaldehyde): structure (A). (B) and (D) are acetaldehydes, and (C) is the β (2-methyl) compound.
Answer: (A). | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A is 2-bromo-3-methylbutane, $(CH_3)_2CH{-}CHBr{-}CH_3$. With Mg in dry ether it gives the Grignard reagent $(CH_3)_2CH{-}CH(MgBr){-}CH_3$, which $D_2O$ deuterates at that carbon: B $= (CH_3)_2CH{-}CHD{-}CH_3$.
That carbon now carries four different groups (H, D, $CH_3$, isopropyl), so B is chiral.
Answer: (C). | ||||
| 75 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsHard | |
A hard statements question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques In A the central carbon carries $NO_2$, H, a (Z)-prop-1-enyl group (upper branch: C* and CH₃ lie on the same side of the double bond) and an (E)-prop-1-enyl group (lower branch). The Z and E groups are different, so the carbon is a stereocentre and A is chiral: Statement I is correct.
Priorities: $NO_2$ (1) > (Z)-propenyl (2) > (E)-propenyl (3; Z ranks above E) > H (4). In A, H is hashed; $NO_2$ points up-right, Z up-left, E down: 1 → 2 → 3 runs anticlockwise, so A is S.
In the second structure H is again hashed; $NO_2$ points up-left, the E group (top) up-right and the Z group down: 1 → 2 → 3 again runs anticlockwise, so it is also S.
It is the same compound redrawn (a mirror image with the Z and E branches then exchanged), not the mirror image: Statement II is incorrect.
Answer: (C). | ||||
| 76 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Ethanol with hot conc. $H_2SO_4$ dehydrates to ethene, $CH_2{=}CH_2$. Cold dilute alkaline $KMnO_4$ (Baeyer's reagent) hydroxylates it to ethane-1,2-diol, $HOCH_2CH_2OH$ (glycol).
Answer: (C). | ||||
| 77 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hinsberg's reagent is benzenesulfonyl chloride, $C_6H_5SO_2Cl$, used to distinguish primary, secondary and tertiary amines.
Answer: (A). | ||||
| 78 | PolymersNatural rubber and cellulose | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on natural rubber and cellulose, from Polymers. NCERT: No matching NCERT chapter Proteins, starch and (natural) rubber occur in nature. Rayon is regenerated cellulose made chemically, a semi-synthetic polymer.
Answer: (D). | ||||
| 79 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Assertion and reasonEasy | |
A easy assertion and reason question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Amylose is the water-soluble component of starch (about 15–20%); amylopectin is the insoluble one. So A is not correct.
Amylose is a long unbranched chain of 200–1000 α-D-glucose units: R is correct.
Answer: (D). | ||||
| 80 | Principles Related to Practical ChemistryTitrimetric analysis | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on titrimetric analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter A weak acid–strong base titration has its equivalence point above 7 (about 8–9). Phenolphthalein, a weak acid, changes colour in the range 8.2–10 and is pink in its ionised form in basic medium.
Answer: (C). | ||||
| 81 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Ne: $\frac{200}{20} = 10$ mol at 20 bar. $p_{O_2} = 25 - 20 = 5$ bar.
At the same T and V, partial pressure ∝ moles: $n_{O_2} = 10\times\frac{5}{20} = 2.5$ mol, so $x = 2.5\times32 = 80$ g.
Answer: 80. | ||||
| 82 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Count steric number (bond pairs + lone pairs) on the central atom:
$PF_5$: 5 → $sp^3d$. $BrF_5$: 5 + 1 = 6 → $sp^3d^2$. $PCl_3$: 3 + 1 = 4 → $sp^3$. $SF_6$: 6 → $sp^3d^2$. $[ICl_4]^-$: 4 + 2 = 6 → $sp^3d^2$. $ClF_3$: 3 + 2 = 5 → $sp^3d$. $IF_5$: 5 + 1 = 6 → $sp^3d^2$.
Four: $BrF_5$, $SF_6$, $[ICl_4]^-$, $IF_5$.
Answer: 4. | ||||
| 83 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Mass of ethanol $= 62.5\times0.80 = 50$ g $= 0.050$ kg. $\Delta T_f = 156.0 - 155.1 = 0.9$ K.
$\Delta T_f = K_f\,\frac{w/M}{W}$: $0.9 = 2.00\times\frac{1.80/M}{0.050}$, so $M = \frac{2\times1.80}{0.9\times0.050} = 80$ g mol⁻¹.
Answer: 80. | ||||
| 84 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag$, $n = 2$.
$E = E^\ominus - \frac{0.06}{2}\log\frac{[Cu^{2+}]}{[Ag^+]^2} = E^\ominus - 0.03\log\frac{10^{-3}}{10^{-4}} = E^\ominus - 0.03$.
$0.43 = E^\ominus - 0.03$, so $E^\ominus_{cell} = 0.46$ V $= 0.80 - E^\ominus_{Cu^{2+}/Cu}$, giving $E^\ominus_{Cu^{2+}/Cu} = 0.34$ V $= 34\times10^{-2}$ V.
Answer: 34. | ||||
| 85 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Number of half-lives $= \frac{100}{30} = \frac{10}{3}$. Remaining $= 1\times2^{-10/3}$ μg.
$\log(2^{-10/3}) = -\frac{10}{3}\times0.30 = -1$, so the amount is $10^{-1}$ μg $= 1\times10^{-1}$ μg.
Answer: 1. | ||||
| 86 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds bpy (2,2'-bipyridine) is neutral and bidentate. Oxidation state: $x + 0 + 4(-1) = -1$, so $x = +3$.
Coordination number $= 2$ (bpy) $+ 4$ (Cl) $= 6$.
Sum $= 3 + 6 = 9$.
Answer: 9. | ||||
| 87 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Only peroxodisulphuric acid, $H_2S_2O_8$ (HO₃S–O–O–SO₃H), has an O–O bond. $H_2SO_3$ and $H_2SO_4$ have none, and $H_2S_2O_7$ has an S–O–S bridge.
Answer: 1. | ||||
| 88 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | NumericalMedium | |
A medium numerical question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Moles of X $= \frac{1.84\times10^{-3}}{92} = 2\times10^{-5}$ mol. Moles of $H_2 = \frac{1.344}{22400} = 6\times10^{-5}$ mol, i.e. 3 mol of $H_2$ per mole of X.
Each alcoholic hydrogen gives half a mole of $H_2$ ($2R{-}OH + 2Na \rightarrow 2R{-}ONa + H_2$), so the data give $2 \times 3 = 6$ alcoholic hydrogens per molecule.
The molar mass of 92, however, fits glycerol, $C_3H_5(OH)_3$, with 3 alcoholic hydrogens; 3 is also the mole ratio of $H_2$ to X if each hydrogen is (wrongly) taken to give a whole $H_2$.
Answer: 3 or 6. | ||||
| 89 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids HCN adds to the C=O: the product is $Ph{-}C(OH)(CN){-}C(OH)(CN){-}Ph$.
It has two stereocentres with identical sets of substituents, so the possible stereoisomers are the (R,R) and (S,S) enantiomers and the (R,S) meso form.
Number of stereoisomers $= 3$.
Answer: 3. | ||||
| 90 | Chemistry in Everyday LifeAntibiotics and antiseptics | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on antibiotics and antiseptics, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Bithionol is 2,2'-thiobis(4,6-dichlorophenol): two dichlorophenol rings joined by S, so it has 4 chlorine atoms.
Answer: 4. | ||||
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