JEE (Main) · 2022
JEE (Main) 2022 — 28 July, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Differential Equations2
- Trigonometry2
- Vector Algebra2
8 easy · 17 medium · 5 hard
Physics
30 Q
- Kinematics2
- Laws of Motion2
- Properties of Bulk Matter2
24 easy · 6 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques3
- p-Block Elements2
- d- and f-Block Elements2
22 easy · 8 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 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations The equation is homogeneous: $\frac{dy}{dx} = \sqrt{1+\frac{y^2}{x^2}} + \frac{y}{x}$.
Put $y = vx$: $v + x\frac{dv}{dx} = \sqrt{1+v^2} + v$, so $\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$.
Integrating: $\ln\left(v + \sqrt{1+v^2}\right) = \ln x + C$.
At $x = 1$, $y = 0$: $v = 0$, so $C = 0$ and $v + \sqrt{1+v^2} = x$.
At $x = 2$: $\sqrt{1+v^2} = 2 - v \Rightarrow 1 + v^2 = 4 - 4v + v^2 \Rightarrow v = \frac{3}{4}$.
Then $\alpha = y = vx = \frac{3}{4}\cdot 2 = \frac{3}{2}$.
Answer: (B). | ||||
| 2 | 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 We need $-1 \le \frac{x^2-4x+2}{x^2+3} \le 1$; the denominator is always positive.
Right inequality: $x^2 - 4x + 2 \le x^2 + 3 \Rightarrow -4x \le 1 \Rightarrow x \ge -\frac{1}{4}$.
Left inequality: $x^2 - 4x + 2 \ge -x^2 - 3 \Rightarrow 2x^2 - 4x + 5 \ge 0$, true for all $x$ (discriminant $16 - 40 < 0$).
Domain: $\left[-\frac{1}{4}, \infty\right)$.
Answer: (B). | ||||
| 3 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra The condition says $\vec{a}, \vec{b}, \vec{c}$ are linearly independent, i.e. $[\vec{a}\ \vec{b}\ \vec{c}] \ne 0$.
$\begin{vmatrix}1+t & 1-t & 1\\ 1-t & 1+t & 2\\ t & -t & 1\end{vmatrix} = (1+t)(1+t+2t) - (1-t)(1-t-2t) + \left(-t(1-t) - t(1+t)\right)$
$= (1+4t+3t^2) - (1-4t+3t^2) - 2t = 6t$.
So the vectors are independent exactly when $t \ne 0$: the set is $\mathbf{R} - \{0\}$.
Answer: (C). | ||||
| 4 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions We need $|x| \le \frac{1}{2}$. Using $\cos^{-1}u = \frac{\pi}{2} - \sin^{-1}u$:
$\frac{\pi}{2} - 3\sin^{-1}x = \frac{\pi}{2} - \sin^{-1}(2x)$, so $\sin^{-1}(2x) = 3\sin^{-1}x$.
Let $\theta = \sin^{-1}x \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$; then $3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and the equation is $\sin 3\theta = 2\sin\theta$:
$3\sin\theta - 4\sin^3\theta = 2\sin\theta \Rightarrow \sin\theta(1 - 4\sin^2\theta) = 0$, so $x = 0, \pm\frac{1}{2}$.
All three satisfy the original equation (e.g. $x = \frac{1}{2}$: $\frac{\pi}{3} - \frac{\pi}{3} = 0 = \cos^{-1}1$).
Sum $= 0 + \frac{1}{2} - \frac{1}{2} = 0$.
Answer: (A). | ||||
| 5 | 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 Test each pair.
$(\vee, \vee)$: $(p \vee q) \vee (p \vee \sim q) = p \vee q \vee \sim q = p \vee T = T$, a tautology.
$(\vee, \wedge)$: $(p \vee q) \wedge (p \wedge \sim q) = p \wedge \sim q$, false when $p$ is false.
$(\wedge, \wedge)$: $(p \wedge q) \wedge (p \wedge \sim q)$ needs $q$ and $\sim q$: a contradiction.
$(\wedge, \vee)$: $(p \wedge q) \vee (p \vee \sim q) = p \vee \sim q$, false when $p$ is false and $q$ true.
Answer: (B). | ||||
| 6 | 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 $(x\vec{a} + y\vec{b})\cdot(6y\vec{a} - 18x\vec{b}) = 6xy|\vec{a}|^2 - 18x^2(\vec{a}\cdot\vec{b}) + 6y^2(\vec{a}\cdot\vec{b}) - 18xy|\vec{b}|^2 = 0$ for all $(x, y)$.
The $x^2$ and $y^2$ coefficients give $\vec{a}\cdot\vec{b} = 0$; the $xy$ coefficient gives $6|\vec{a}|^2 = 18|\vec{b}|^2$, so $|\vec{b}|^2 = \frac{81}{3} = 27$, $|\vec{b}| = 3\sqrt{3}$.
Since $\vec{a} \perp \vec{b}$: $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| = 9 \cdot 3\sqrt{3} = 27\sqrt{3}$.
Answer: (B). | ||||
| 7 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctHard | |
A hard single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines In an equilateral triangle the orthocentre is the centroid:
$H(h, k) = \left(\frac{a + \sin t + \cos t}{3}, \frac{b + \sin t - \cos t}{3}\right)$.
So $3h - a = \sin t + \cos t$ and $3k - b = \sin t - \cos t$. Squaring and adding eliminates $t$:
$(3h - a)^2 + (3k - b)^2 = (\sin t + \cos t)^2 + (\sin t - \cos t)^2 = 2$,
i.e. $\left(h - \frac{a}{3}\right)^2 + \left(k - \frac{b}{3}\right)^2 = \frac{2}{9}$.
The orthocentre therefore lies on the circle with centre $\left(\frac{a}{3}, \frac{b}{3}\right)$. This centre is $\left(1, \frac{1}{3}\right)$, so $a = 3$, $b = 1$.
$a^2 - b^2 = 9 - 1 = 8$.
Answer: (B). | ||||
| 8 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Reflexive: $(x, x) \in R$ for all $x$ means $7 \mid (3 + \alpha)x$ for all $x$, so $7 \mid 3 + \alpha$, i.e. $\alpha \equiv 4 \pmod 7$.
Conversely, if $\alpha \equiv 4 \equiv -3 \pmod 7$, then $3x + \alpha y \equiv 3(x - y) \pmod 7$, so $(x, y) \in R \iff x \equiv y \pmod 7$, which is reflexive, symmetric and transitive.
Hence R is an equivalence relation if and only if $\alpha$ leaves remainder 4 on division by 7.
Answer: (D). | ||||
| 9 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability Take 100 candidates: 60 female, 40 male, and 60 qualify.
Females qualifying $= 2\times$ males qualifying, so males qualifying $= 20$ and females qualifying $= 40$ (within 60 and 40 respectively).
$P(\text{female} \mid \text{qualified}) = \frac{40}{60} = \frac{2}{3} \approx 0.667$.
None of the options equals $\frac{2}{3}$; the nearest is $\frac{11}{16} = 0.6875$.
Answer: $\frac{2}{3}$ (no option matches; (B) is marked as the nearest).
Read literally (female qualifiers twice the male qualifiers), 40% of candidates are qualified females and 20% qualified males, so P = 40/60 = 2/3, which is not an option. The options fit the reading that the female qualifying RATE is twice the male rate: with r_f = 2r_m, 0.6r_f + 0.4r_m = 0.6 gives r_m = 0.375, r_f = 0.75, so qualified females are 45% of candidates and P = 0.45/0.60 = 3/4.
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 Divide by $\sin^2 2x$; since $\frac{2\sin 4x}{\sin^2 2x} = \frac{4\cos 2x}{\sin 2x} = 4\cot 2x$:
$\frac{dy}{dx} + (8 + 4\cot 2x)y = \frac{2e^{-4x}(2\sin 2x + \cos 2x)}{\sin^2 2x}$.
Integrating factor: $e^{\int(8 + 4\cot 2x)dx} = e^{8x}\sin^2 2x$.
$\frac{d}{dx}\left(y\,e^{8x}\sin^2 2x\right) = 2e^{4x}(2\sin 2x + \cos 2x) = \frac{d}{dx}\left(e^{4x}\sin 2x\right)$.
So $y\,e^{8x}\sin^2 2x = e^{4x}\sin 2x + C$.
At $x = \frac{\pi}{4}$: $e^{-\pi}e^{2\pi}\cdot 1 = e^{\pi} + C \Rightarrow C = 0$, so $y = \frac{e^{-4x}}{\sin 2x}$.
$y\left(\frac{\pi}{6}\right) = \frac{e^{-2\pi/3}}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}e^{-2\pi/3}$.
Answer: (A). | ||||
| 11 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections PQ is the chord of contact from $R(0, 1)$: $y\cdot 1 = (x + 0) - 3$, i.e. $y = x - 3$.
With the parabola: $(x - 3)^2 = 2x - 3 \Rightarrow x^2 - 8x + 12 = 0 \Rightarrow x = 2, 6$. So $P(2, -1)$, $Q(6, 3)$.
Slope $PQ = 1$, slope $PR = \frac{1 - (-1)}{0 - 2} = -1$: $PQ \perp PR$, so the triangle is right-angled at P.
The orthocentre of a right triangle is the right-angle vertex: $(2, -1)$.
Answer: (B). | ||||
| 12 | CirclesRadius and centre | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections Centre $C\left(\frac{1}{2}, -1\right)$, radius $r = \sqrt{\frac{1}{4} + 1 + \frac{11}{4}} = 2$.
QR passes through C, so it is a diameter: $QR = 4$.
The distance of P from QR is $CP\sin\frac{\pi}{4} = 2\cdot\frac{1}{\sqrt{2}} = \sqrt{2}$.
Area $= \frac{1}{2}\cdot 4\cdot\sqrt{2} = 2\sqrt{2}$.
Answer: (B). | ||||
| 13 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $7^{2022} = 49^{1011} = (50 - 1)^{1011} \equiv (-1)^{1011} = -1 \pmod 5$.
$3^{2022} = 9^{1011} = (10 - 1)^{1011} \equiv -1 \pmod 5$.
Sum $\equiv -2 \equiv 3 \pmod 5$.
Answer: (C). | ||||
| 14 | 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 $A^2 = \begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$ and $A^3 = I$. So $A^{49} = A^{48}A = A$ and $A^{98} = A^{96}A^2 = A^2$.
$B_0 = A + 2A^2 = \begin{bmatrix}0&1&2\\2&0&1\\1&2&0\end{bmatrix}$, $\det B_0 = 0 - 1(0 - 1) + 2(4 - 0) = 9$.
For a $3\times 3$ matrix, $\det(\text{Adj}\,M) = (\det M)^2$. So $\det B_n = (\det B_{n-1})^2$ and
$\det B_4 = 9^{2^4} = 9^{16} = 3^{32}$.
Answer: (C). | ||||
| 15 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctHard | |
A hard single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z_2 = x + iy$, $p = |z_2 + 1|$, $q = |z_2 - 1|$ (real, non-negative).
$|z_2 - p|^2 = |z_2 + q|^2 \Rightarrow (x - p)^2 = (x + q)^2$.
Either $x - p = x + q$ (then $p = q = 0$, impossible) or $x - p = -(x + q)$, i.e. $2x = p - q$.
If $x = 0$, then $p = q$ holds automatically: the whole imaginary axis is in $S_2$.
If $x \ne 0$: $p^2 - q^2 = 4x$ and $p - q = 2x$ give $p + q = 2$, the distance between $\pm 1$, so $z_2$ lies on the segment $[-1, 1]$ of the real axis (all of which works).
So $S_2$ = imaginary axis $\cup$ real segment $[-1, 1]$. $S_1$ is the circle centre 3, radius $\frac{1}{2}$.
Least distance: from the point 1 it is $3 - 1 - \frac{1}{2} = \frac{3}{2}$; from the imaginary axis it is $3 - \frac{1}{2} = \frac{5}{2}$.
Least value $= \frac{3}{2}$.
Answer: (C). | ||||
| 16 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctHard | |
A hard 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 Take the point $(\cos\theta, \sin\theta, 0)$. Its foot on the plane is $(x, y, z) = (\cos\theta + 2t, \sin\theta + 3t, t)$ with $2x + 3y + z = 6$.
So $z = t = 6 - 2x - 3y$, and
$\cos\theta = x - 2t = x - 2(6 - 2x - 3y) = 5x + 6y - 12$,
$\sin\theta = y - 3t = y - 3(6 - 2x - 3y) = 6x + 10y - 18 = 2(3x + 5y - 9)$.
$\cos^2\theta + \sin^2\theta = 1$: $(5x + 6y - 12)^2 + 4(3x + 5y - 9)^2 = 1$, with $z = 6 - 2x - 3y$.
Answer: (B). | ||||
| 17 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 5x - \frac{5\alpha}{x^6} = 0 \Rightarrow x^7 = \alpha$ (a minimum, since $f \to \infty$ at both ends of $(0, \infty)$; $\alpha > 0$ is needed for a minimum).
There $\frac{\alpha}{x^5} = x^2$, so $f_{\min} = \frac{5x^2}{2} + x^2 = \frac{7x^2}{2} = 14 \Rightarrow x^2 = 4$, $x = 2$.
$\alpha = 2^7 = 128$.
Answer: (C). | ||||
| 18 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f'(x) = 5\alpha x^4 + 3\beta x^2 + \gamma > 0$ and $f$ is odd and onto $\mathbf{R}$, so $f$ is a bijection and $g = f^{-1}$; hence $f(g(u)) = u$.
An AP with mean zero is symmetric about 0: $a_i + a_{n+1-i} = 0$. As $f$ is odd, $f(a_i) + f(a_{n+1-i}) = 0$, so $\sum f(a_i) = 0$.
Therefore $f\left(g\left(\frac{1}{n}\sum f(a_i)\right)\right) = \frac{1}{n}\sum f(a_i) = 0$.
Answer: (A). | ||||
| 19 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctHard | |
A hard single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Multiply the recurrence by $a_{n+1}$: $a_{n+1}a_{n+2} = 2 + a_n a_{n+1}$. So $b_n = a_n a_{n+1}$ is an AP with $b_1 = 2$, difference 2: $b_n = 2n$.
Each factor: $\frac{a_n + \frac{1}{a_{n+1}}}{a_{n+2}} = \frac{a_n a_{n+1} + 1}{a_{n+1}a_{n+2}} = \frac{2n + 1}{2n + 2}$.
Product for $n = 1$ to 30: $\frac{3\cdot 5\cdots 61}{4\cdot 6\cdots 62}$.
Numerator $= \frac{61!}{2\cdot 4\cdots 60} = \frac{61!}{2^{30}\,30!}$; denominator $= 2^{30}(2\cdot 3\cdots 31) = 2^{30}\,31!$.
Product $= \frac{61!}{2^{60}\,30!\,31!} = 2^{-60}\,{}^{61}C_{31}$.
So $\alpha = -60$.
Answer: (C). | ||||
| 20 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $e^{-x}f(x) = \int_0^x e^{-t}f'(t)\,dt - (x^2 - x + 1)$. Differentiate:
$e^{-x}f'(x) - e^{-x}f(x) = e^{-x}f'(x) - (2x - 1)$, so $f(x) = (2x - 1)e^x$ (check: $f(0) = -1$ from the original, as here).
$f'(x) = (2x + 1)e^x = 0$ at $x = -\frac{1}{2}$, where $f$ changes from decreasing to increasing.
$f_{\min} = f\left(-\frac{1}{2}\right) = -2e^{-1/2} = -\frac{2}{\sqrt{e}}$.
Answer: (A). | ||||
| 21 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations There are 10 characters. Passwords of length 6 to 8: $10^6 + 10^7 + 10^8 = 111\times 10^6$.
Passwords with letters only: $5^6 + 5^7 + 5^8 = 31\times 5^6$.
At least one number: $111\times 10^6 - 31\times 5^6 = 5^6(111\times 2^6 - 31) = 5^6(7104 - 31) = 7073\times 5^6$.
Answer: 7073. | ||||
| 22 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry In the rhombus PRQS the diagonals are PQ and RS, and they are perpendicular.
$\vec{PQ} = \left(\frac{90}{17}, \frac{60}{17}, \frac{94}{17}\right) \parallel (45, 30, 47)$.
$RS \perp PQ$: $45\alpha - 30 + 47\beta = 0$, i.e. $45\alpha + 47\beta = 30$.
Integer solutions: $\beta = 15 + 45k$, $\alpha = -15 - 47k$ (check $k = 0$: $-675 + 705 = 30$). The smallest absolute values are at $k = 0$: $\alpha = -15$, $\beta = 15$.
$\alpha^2 + \beta^2 = 225 + 225 = 450$.
Answer: 450. | ||||
| 23 | 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 Let $g(x) = f(x) - (2x + 3)$. Then $g(0) = 0$, $g(1) = 0$, and $g$ vanishes at two points $c_1 < c_2$ in $(0, 1)$.
So $g$ has four zeros $0 < c_1 < c_2 < 1$. By Rolle's theorem $g'$ has at least three zeros, one in each gap, and then $g''$ has at least two zeros in $(0, 1)$.
$g'' = f''$, and a function such as a suitable quartic shows two can be attained.
Answer: 2. | ||||
| 24 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $1 + x^2 = s^2$ ($s$ from 1 to 2), $x\,dx = s\,ds$, $x^2 = s^2 - 1$.
$\sqrt{1 + x^2 + (1 + x^2)^{3/2}} = \sqrt{s^2 + s^3} = s\sqrt{1 + s}$.
Integral $= 15\int_1^2 \frac{(s^2 - 1)s}{s\sqrt{1 + s}}ds = 15\int_1^2 (s - 1)\sqrt{1 + s}\,ds$.
Put $w = 1 + s$ (2 to 3): $15\int_2^3 (w - 2)\sqrt{w}\,dw = 15\left[\frac{2}{5}w^{5/2} - \frac{4}{3}w^{3/2}\right]_2^3$.
At 3: $\frac{18\sqrt{3}}{5} - 4\sqrt{3} = -\frac{2\sqrt{3}}{5}$. At 2: $\frac{8\sqrt{2}}{5} - \frac{8\sqrt{2}}{3} = -\frac{16\sqrt{2}}{15}$.
Integral $= 15\left(-\frac{2\sqrt{3}}{5} + \frac{16\sqrt{2}}{15}\right) = 16\sqrt{2} - 6\sqrt{3}$.
$\alpha = 16$, $\beta = -6$, $\alpha + \beta = 10$.
Answer: 10. | ||||
| 25 | 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 $(A + B)^2 = A^2 + AB + BA + B^2$, with
$AB = \begin{bmatrix}\beta - 1&1\\2\beta + \alpha&2\end{bmatrix}$, $BA = \begin{bmatrix}\beta + 2&\alpha - \beta\\1&-1\end{bmatrix}$, $B^2 = \begin{bmatrix}\beta^2 + 1&\beta\\\beta&1\end{bmatrix}$, $A^2 = \begin{bmatrix}-1&-1-\alpha\\2+2\alpha&\alpha^2-2\end{bmatrix}$.
First condition: $AB + BA + B^2 = \begin{bmatrix}\beta^2 + 2\beta + 2&1 + \alpha\\3\beta + \alpha + 1&2\end{bmatrix} = \begin{bmatrix}2&2\\2&2\end{bmatrix}$, giving $\alpha = 1$, $\beta = 0$. So $\alpha_1 = 1$.
Second condition: $A^2 + AB + BA = \begin{bmatrix}2\beta&-\beta\\3\alpha + 3\beta + 3&\alpha^2 - 1\end{bmatrix} = O$, giving $\beta = 0$, $\alpha = -1$. So $\alpha_2 = -1$.
$|\alpha_1 - \alpha_2| = 2$.
Answer: 2. | ||||
| 26 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations With common difference $d > 0$ and mean $p$: $a_i - p = \pm\frac{3d}{2}, \pm\frac{d}{2}$.
So $\left|\frac{d^2}{4} - q\right| = 500$ and $\left|\frac{9d^2}{4} - q\right| = 500$. The two inner values differ, so they are $-500$ and $+500$:
$\frac{d^2}{4} - q = -500$, $\frac{9d^2}{4} - q = 500$. Subtracting: $2d^2 = 1000$, $d^2 = 500$, $q = 125 + 500 = 625$.
Roots of $f(x) = 0$: $p \pm \sqrt{q} = p \pm 25$; difference $= 50$.
Answer: 50. | ||||
| 27 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections H has $e = \sqrt{2}$, so E has $e = \frac{1}{\sqrt{2}}$: $b^2 = a^2(1 - \frac{1}{2}) = \frac{a^2}{2}$.
Tangent $y = mx + K$ with $m^2 = \frac{5}{2}$.
To H: $K^2 = m^2 - 1 = \frac{3}{2}$. To E: $K^2 = a^2m^2 + b^2 = \frac{5a^2}{2} + \frac{a^2}{2} = 3a^2$.
So $3a^2 = \frac{3}{2}$, $a^2 = \frac{1}{2}$, $b^2 = \frac{1}{4}$.
$4(a^2 + b^2) = 4\cdot\frac{3}{4} = 3$.
Answer: 3. | ||||
| 28 | StatisticsMean, median and mode | Class 11: Statistics | NumericalHard | |
A hard numerical question on mean, median and mode, from Statistics. NCERT: Class 11: Statistics $\bar{x} = \frac{1}{20}\left[\sum x_i^2 - 2\sum i\,x_i + \sum i^2\right]$ with $x_i = 3\left(\frac{1}{2}\right)^{i-1}$.
$\sum x_i^2 = 9\cdot\frac{1 - 4^{-20}}{1 - \frac{1}{4}} = 12(1 - 4^{-20})$.
$\sum_{i=1}^{20} i\,r^{i-1} = \frac{1 - 21r^{20} + 20r^{21}}{(1 - r)^2}$ with $r = \frac{1}{2}$: $= 4\left(1 - \frac{11}{2^{20}}\right)$, so $\sum i\,x_i = 12\left(1 - \frac{11}{2^{20}}\right)$.
$\sum i^2 = \frac{20\cdot 21\cdot 41}{6} = 2870$.
Sum $= 12 - \frac{12}{4^{20}} - 24 + \frac{264}{2^{20}} + 2870 = 2858 + \varepsilon$, where $0 < \varepsilon < 0.001$.
$\bar{x} = 142.9 + \frac{\varepsilon}{20}$, so $[\bar{x}] = 142$.
Answer: 142. | ||||
| 29 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $g(u) = u^3 + 2u^2 + 3\sin u$. The base is $\frac{g(x + 2\cos x)}{g(x + 2)} \to \frac{g(2)}{g(2)} = 1$, so the limit is of the form $1^\infty$ and equals $e^L$ with
$L = \lim_{x\to 0}\frac{100}{x}\left(\frac{g(x + 2\cos x) - g(x + 2)}{g(x + 2)}\right)$.
By the mean value theorem, $g(x + 2\cos x) - g(x + 2) \approx g'(2)\,(2\cos x - 2) = g'(2)\cdot O(x^2)$, since $2\cos x - 2 = -x^2 + \ldots$
So the bracket is $O(x^2)$, $L = \lim \frac{100\cdot O(x^2)}{x} = 0$ ($g(2) = 16 + 3\sin 2 \ne 0$).
Limit $= e^0 = 1$.
Answer: 1. | ||||
| 30 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Both denominators are always positive (discriminants $9 - 40 < 0$, $25 - 144 < 0$).
Subtract 1 from each side: $\frac{2x^2 - 12x + 7}{x^2 + 3x + 10} = \frac{2x^2 - 12x + 7}{3x^2 + 5x + 12}$.
So either $2x^2 - 12x + 7 = 0$, or $x^2 + 3x + 10 = 3x^2 + 5x + 12$, i.e. $x^2 + x + 1 = 0$, which has no real root.
$2x^2 - 12x + 7 = 0$ has discriminant $144 - 56 = 88 > 0$: two real roots with sum $\frac{12}{2} = 6$.
Answer: 6. | ||||
| 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 The energy density of a magnetic field is $u = \frac{B^2}{2\mu_0}$, so $\frac{B^2}{\mu_0}$ has the dimensions of energy per volume:
$\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
Answer: (C). | ||||
| 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 Time of fall: $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times 19.6}{9.8}} = 2$ s.
Speed of the parade: $9$ km/h $= 2.5$ m/s. In 2 s a cadet covers $2.5\times 2 = 5$ m.
So the cadet who catches it was 5 m from the tree when the mango was dropped.
Answer: (A). | ||||
| 33 | 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 Impulse $= \Delta p = 5\times 25 = 125$ N s in both cases (same initial momentum, both end at rest).
Average force $= \frac{\Delta p}{\Delta t}$: $\frac{125}{3} \approx 41.7$ N and $\frac{125}{5} = 25$ N, different.
Answer: (B). | ||||
| 34 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion The escaping air exerts a thrust $F = v\frac{dm}{dt}$.
Rate of loss of mass: $\frac{10\text{ g}}{5\text{ s}} = 2$ g/s; relative speed 4.5 cm/s.
$F = 4.5\times 2 = 9$ g cm s$^{-2}$ = 9 dyne.
Answer: (B). | ||||
| 35 | 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 $g = \frac{GM}{R^2}$, so $\frac{\Delta g}{g} = -2\frac{\Delta R}{R} = -2(-2\%) = +4\%$.
Answer: (D). | ||||
| 36 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Doubling the length means strain $\frac{\Delta L}{L} = 1$.
$F = YA\frac{\Delta L}{L} = 2\times 10^{11}\times 10^{-4}\times 1 = 2\times 10^7$ N (taking Hooke's law to hold, as the question intends).
Answer: (C). | ||||
| 37 | ThermodynamicsSecond law | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\eta = 1 - \frac{T_2}{T_1} = 0.5 \Rightarrow T_2 = 0.5T_1$.
'Increases by 30%' of its value: $\eta' = 0.5\times 1.3 = 0.65$, so $T_2 - 40 = 0.35T_1$.
$0.5T_1 - 40 = 0.35T_1 \Rightarrow 0.15T_1 = 40 \Rightarrow T_1 = 266.7$ K.
Answer: (C). | ||||
| 38 | 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 Statement I: molecular velocities are random in direction, so the average momentum (a vector) is zero at every temperature. False.
Statement II: $v_{rms} = \sqrt{\frac{3RT}{M}}$. Doubling $T$ and halving $M$ (O$_2$ → O) multiplies $\frac{T}{M}$ by 4, so $v_{rms}$ becomes $2v$. True.
Answer: (D). | ||||
| 39 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The form $y = A\sin\frac{2\pi}{\lambda}(vt - x)$ is a wave moving along $+x$ with speed $v$.
Here $v = 400$ m/s.
Answer: (C). | ||||
| 40 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $\frac{1}{24} = \frac{1}{40} + \frac{1}{40K} \Rightarrow \frac{1}{40K} = \frac{1}{24} - \frac{1}{40} = \frac{5 - 3}{120} = \frac{1}{60}$.
$40K = 60 \Rightarrow K = 1.5$.
Answer: (A). | ||||
| 41 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Increased by twice its length: $L' = L + 2L = 3L$. The volume is unchanged, so $A' = \frac{A}{3}$.
$R = \frac{\rho L}{A} \propto L^2$, so $\frac{R'}{R_1} = 3^2 = 9$.
Answer: (A). | ||||
| 42 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Current sensitivity $\frac{\theta}{I} = \frac{NAB}{k}$.
It increases with more turns N, larger area A, stronger field B, and a smaller torsional constant k. So (B) and (D) increase it; (A) and (C) decrease it.
Answer: (D). | ||||
| 43 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium 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 rod is horizontal and B is vertical, so the magnetic force $F = IlB$ is horizontal (pushing into the incline).
Along the smooth incline: $IlB\cos 45^\circ = mg\sin 45^\circ \Rightarrow I = \frac{mg}{lB} = \frac{\lambda g}{B}$.
$I = \frac{0.45\times 10}{0.15} = 30$ A.
Answer: (A). | ||||
| 44 | 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 $X_L = \omega L = 49\times\frac{22}{7}\times 0.030 = 154\times 0.03 = 4.62\ \Omega$.
$V_0 = I_0X_L = 5\times 4.62 = 23.1$ V, and in a pure inductor the voltage leads the current by 90°.
$V = 23.1\sin(49\pi t - 30^\circ + 90^\circ) = 23.1\sin(49\pi t + 60^\circ)$.
Answer: (D). | ||||
| 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 speed in a medium depends only on that medium: $v = \frac{c}{\sqrt{\mu_r\epsilon_r}}$.
Medium 2: $v_2 = \frac{3\times 10^8}{\sqrt{1\times 9}} = 1.0\times 10^8$ ms$^{-1}$.
Answer: (A). | ||||
| 46 | OpticsOptical instruments | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on optical instruments, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Normal adjustment: $f_o + f_e = 30$ cm and $m = \frac{f_o}{f_e} = 2$.
$f_e = 10$ cm, $f_o = 20$ cm.
Answer: (A). | ||||
| 47 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter For an electron accelerated through $V$ volts, $\lambda = \frac{h}{\sqrt{2meV}}$.
Putting in the constants: $\lambda = \frac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.6\times 10^{-19}V}}$ m $= \frac{1.227}{\sqrt{V}}$ nm.
So $x = \sqrt{V}$.
Answer: (D). | ||||
| 48 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei When $\frac{7}{8}$ has decayed, $\frac{1}{8} = \left(\frac{1}{2}\right)^3$ remains: 3 half-lives.
$t = 3\times 60 = 180$ days.
Answer: (C). | ||||
| 49 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A solar cell works with no external bias: it delivers current against the load, so its characteristic is drawn in the fourth quadrant, from the short-circuit current on the I-axis (negative) to the open-circuit voltage on the V-axis.
(A) is a forward-biased diode, (C) an ohmic resistor, (D) a Zener diode.
Answer: (B). | ||||
| 50 | Communication SystemsAmplitude modulation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on amplitude modulation, from Communication Systems. NCERT: No matching NCERT chapter $\mu = \frac{A_m}{A_c}$. For $\mu > 1$ the envelope over-modulates and the signal is distorted; so $\mu \le 1$ (and $\mu = 0$ means no modulation at all).
Answer: (A). | ||||
| 51 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Zero projection means the dot product is zero: $2(1) + 4(2) - 2\alpha = 0 \Rightarrow \alpha = 5$.
Answer: 5. | ||||
| 52 | 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 Activity must fall by $64 = 2^6$: six half-lives.
$t = 6\times 2.5 = 15$ hours.
Answer: 15. | ||||
| 53 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Fringe width $\beta = \frac{\lambda D}{d} \propto \lambda$ (same apparatus).
$\lambda_2 = 560\times\frac{8.1}{7.2} = 630$ nm.
Answer: 630. | ||||
| 54 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current The half-power frequencies satisfy $\omega_2 - \omega_1 = \frac{R}{L}$.
$L = \frac{R}{\omega_2 - \omega_1} = \frac{5}{232 - 212} = 0.25$ H $= 250$ mH.
Answer: 250. | ||||
| 55 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity Potential gradient: $\frac{20\text{ mV}}{60\text{ cm}}$, so across the full 300 cm wire the drop is $100$ mV $= 0.1$ V.
Current in the wire: $I = \frac{0.1}{20} = 0.005$ A.
For the main circuit (driving cell taken to have no internal resistance): $4 = I(20 + R) \Rightarrow 20 + R = 800 \Rightarrow R = 780\ \Omega$.
Answer: 780. | ||||
| 56 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $\tau_{max} = pE$.
$\frac{\tau_1}{\tau_2} = \frac{1.2\times 10^{-30}\times 5\times 10^4}{2.4\times 10^{-30}\times 15\times 10^4} = \frac{6}{36} = \frac{1}{6}$.
So $x = 6$.
Answer: 6. | ||||
| 57 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Train speed $u = 36$ km/h $= 10$ m/s. The hill receives $f_1 = f\frac{v}{v - u}$ and reflects it; the driver, moving towards the hill, hears
$f' = f_1\frac{v + u}{v} = f\frac{v + u}{v - u} = 320\times\frac{340}{320} = 340$ Hz.
Answer: 340. | ||||
| 58 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium 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, buoyancy minus weight (air weight negligible) balances Stokes' drag:
$6\pi\eta rv = \frac{4}{3}\pi r^3\rho g \Rightarrow \eta = \frac{2r^2\rho g}{9v}$.
$r = 1$ mm $= 10^{-3}$ m, $v = 0.35$ cm/s $= 3.5\times 10^{-3}$ m/s:
$\eta = \frac{2\times 10^{-6}\times 1750\times 9.8}{9\times 3.5\times 10^{-3}} = \frac{0.0343}{0.0315} = 1.089$ Pa s $= 10.89$ poise (with $g = 10$: 11.1 poise).
Nearest integer: 11.
Answer: 11. | ||||
| 59 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power When the speed is halved the kinetic energy is $\frac{E}{4}$, so the spring has stored $\frac{3E}{4}$ (smooth surface).
$\frac{1}{2}kx^2 = \frac{3E}{4}$ with $x = 0.25$ m: $k = \frac{3E}{2\times 0.0625} = 24E$ N m$^{-1}$.
Answer: 24. | ||||
| 60 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Each disc has radius $\frac{a}{2}$. Its moment of inertia about a diameter is $\frac{M(a/2)^2}{4} = \frac{Ma^2}{16}$.
Top and bottom discs: OO' is a diameter, $\frac{Ma^2}{16}$ each.
Side discs: their centres are $\frac{a}{2}$ from OO' (they touch the axis), so by the parallel-axis theorem $\frac{Ma^2}{16} + M\left(\frac{a}{2}\right)^2 = \frac{5Ma^2}{16}$ each.
Total $= 2\cdot\frac{Ma^2}{16} + 2\cdot\frac{5Ma^2}{16} = \frac{12Ma^2}{16} = \frac{3}{4}Ma^2$.
So $x = 3$.
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom (A), (B) and (C) are correct statements.
(D) is incorrect: an atomic orbital is specified by three quantum numbers, $n$, $l$ and $m_l$ ($n$ and $l$ fix only the subshell).
Answer: (D). | ||||
| 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 $H = U + PV$, so at constant pressure $\Delta H = \Delta U + P\Delta V$; (A) has the wrong sign.
(B) is the first law (W = work done on the system), (C) the second law, (D) the definition of Gibbs energy change at constant T.
Answer: (A). | ||||
| 63 | ElectrochemistryBatteries and fuel cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on batteries and fuel cells, from Electrochemistry. NCERT: Class 12: Electrochemistry (A) The nickel–cadmium cell giving CdO and Ni(OH)$_2$ is a rechargeable cell delivering current: discharging of a secondary battery (II).
(B) The mercury cell (Zn amalgam, HgO) is a primary battery (I).
(C) PbSO$_4$ turning back into Pb, PbO$_2$ and H$_2$SO$_4$ is the lead storage battery being charged (IV).
(D) H$_2$ + O$_2$ → H$_2$O is the fuel cell reaction (III).
Answer: (C). | ||||
| 64 | Surface ChemistryPhysisorption and chemisorption | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on physisorption and chemisorption, from Surface Chemistry. NCERT: No matching NCERT chapter (A) Ostwald process, oxidation of NH$_3$: Pt gauze (III).
(B) Haber process: iron catalyst (IV).
(C) Inversion of cane sugar: acid (H$_2$SO$_4$) catalyst (II).
(D) Lead chamber process: NO catalyses the oxidation of SO$_2$ (I).
Answer: (C). | ||||
| 65 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Standard values ($\Delta_{eg}H$, kJ mol$^{-1}$): Na −53, K −48, Rb −47, Cs −46, Ar +96, Kr +96, I −296, At −270.
Rb/Cs (−47, −46) and Ar/Kr (+96, +96) are nearly the same; Na/K differ by 5 and I/At by about 26.
Answer: (C). | ||||
| 66 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Leaching dissolves the ore in a suitable reagent, leaving the impurities: bauxite dissolves in NaOH as sodium aluminate (Bayer's process).
(A) is roasting, (B) reduction in the blast furnace, (D) a reduction of the oxide.
Answer: (C). | ||||
| 67 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Clark's method adds slaked lime to temporary hard water:
$\text{Ca(HCO}_3)_2 + \text{Ca(OH)}_2 \rightarrow 2\text{CaCO}_3\downarrow + 2\text{H}_2\text{O}$
$\text{Mg(HCO}_3)_2 + 2\text{Ca(OH)}_2 \rightarrow 2\text{CaCO}_3\downarrow + \text{Mg(OH)}_2\downarrow + 2\text{H}_2\text{O}$
Magnesium is removed as the less soluble Mg(OH)$_2$. Products: CaCO$_3$ and Mg(OH)$_2$.
Answer: (B). | ||||
| 68 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter (A) is incorrect: Li$^+$ has a very large hydration enthalpy; LiF is sparingly soluble because of its very high lattice enthalpy.
(B) KO$_2$ contains the superoxide ion O$_2^-$ with one unpaired electron: paramagnetic.
(C) Sodium in liquid ammonia gives ammoniated cations and electrons: conducting.
(D) Na (0.97 g cm$^{-3}$) is denser than K (0.86 g cm$^{-3}$).
Answer: (A). | ||||
| 69 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) $(\text{NH}_4)_2\text{Cr}_2\text{O}_7 \rightarrow \text{N}_2 + \text{Cr}_2\text{O}_3 + 4\text{H}_2\text{O}$: N$_2$ (II).
(B) KMnO$_4$ oxidises HCl to Cl$_2$ (IV).
(C) Al dissolves in aqueous NaOH giving Na[Al(OH)$_4$] and H$_2$ (I).
(D) $2\text{NaNO}_3 \rightarrow 2\text{NaNO}_2 + \text{O}_2$ (III).
Answer: (C). | ||||
| 70 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Standard reduction potentials $E^\circ(\text{M}^{2+}/\text{M})$: Mn −1.18 V, Zn −0.76 V, Ni −0.25 V, Cu +0.34 V.
Only Cu has a positive value, so it cannot displace H$_2$ from non-oxidising mineral acids.
Answer: (A). | ||||
| 71 | 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 Statement I: polluted water has low dissolved oxygen but a high BOD (much organic matter to oxidise). False.
Statement II: eutrophication causes algal blooms whose decay uses up dissolved oxygen. True.
Answer: (D). | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) Tetrahydrofuran: a saturated five-membered ring containing O, puckered, so a non-planar heterocyclic compound (III).
(B) Two three-membered rings sharing a C–C bond (a bicyclo[1.1.0]butane skeleton): bicyclo compound (IV).
(C) Two three-membered rings sharing one carbon atom (spiropentane skeleton): spiro compound (I).
(D) Furan: planar ring with 6 π electrons (two C=C plus a lone pair of O): aromatic (II).
Answer: (C). | ||||
| 73 | 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 Oxymercuration–demercuration adds H and OH in the Markovnikov sense without rearrangement: A = 3,3-dimethylbutan-2-ol.
Hydroboration–oxidation adds them anti-Markovnikov: B = 3,3-dimethylbutan-1-ol.
Answer: (B). | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The marked hydrogens are all α to a carbonyl group; the most acidic is the one whose carbanion is best stabilised.
An α-H of a ketone (pK$_a$ about 19–20) is more acidic than an α-H of a carboxylic acid: in base the acid is first deprotonated at O, and the carboxylate stabilises an adjacent negative charge poorly.
A phenyl group on the α-carbon adds benzylic resonance stabilisation.
So the benzylic α-H of phenylacetone (C), between a phenyl and a ketone C=O, has the lowest pK$_a$.
Answer: (C). | ||||
| 75 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Friedel–Crafts alkylation gives cumene; air oxidation gives cumene hydroperoxide; acid hydrolysis gives phenol (P) and acetone.
A: phenol is oxidised by acidified dichromate to p-benzoquinone.
B: bromine in a non-polar solvent (CS$_2$, low temperature) gives monobromination, mainly 4-bromophenol (bromine water would give 2,4,6-tribromophenol).
Answer: (B). | ||||
| 76 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Elimination from a quaternary ammonium ion follows the Hofmann rule: the bulky, poor leaving group –N(CH$_3$)$_3^+$ makes the base remove the more accessible, more acidic hydrogen of the CH$_3$ group.
So the less substituted alkene, pent-1-ene, is major (A), and pent-2-ene is minor (B): Hofmann products.
Answer: (C). | ||||
| 77 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Terylene (Dacron, PET) is the polyester of ethylene glycol (ethane-1,2-diol) and terephthalic acid (benzene-1,4-dicarboxylic acid).
Answer: (C). | ||||
| 78 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Convert the Fischer projection X to a Haworth ring: a group on the right goes down, a group on the left goes up; C5 has the ring O on the right (a D-sugar), so CH$_2$OH is up.
In X: C1 OH left (up), C2 OH left (up), C3 OH right (down), C4 OH left (up).
Option (D) has exactly: C1 OH up, C2 OH up, C3 OH down, C4 OH up, CH$_2$OH up.
(A) and (C) differ at C2 and C3; (B) differs at C4.
Answer: (D). | ||||
| 79 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter (A) True. (B) True: inhibitors bind either at the active site or at an allosteric site.
(C) False: competitive inhibitors compete with the substrate for the active site.
(D) False: non-competitive inhibitors bind at an allosteric site.
Answer: (C). | ||||
| 80 | Principles Related to Practical ChemistryPhysical chemistry experiments | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on physical chemistry experiments, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter In this clock reaction, $\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{I}_2 + 2\text{H}_2\text{O}$, and the small fixed amount of thiosulphate turns the I$_2$ back to I$^-$ until it is used up; then free I$_2$ turns starch blue.
(A) True: starch solution decomposes on standing, so it must be fresh.
(B) True, (E) false: thiosulphate must be the limiting reagent (less than KI), or no blue colour would appear in a reasonable time.
(C) True, (D) false: the time is noted the moment the blue colour appears.
Answer: (A). | ||||
| 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 Z is the limiting reagent: 0.05 mol Z gives $\frac{0.05}{3}$ mol XYZ$_3$.
Molar mass of XYZ$_3$ $= 10 + 20 + 3(30) = 120$ g mol$^{-1}$.
Mass $= \frac{0.05}{3}\times 120 = 2$ g.
Answer: 2. | ||||
| 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 Volume of the unit cell: $(300\times 10^{-10}\text{ cm})^3 = 2.7\times 10^{-23}$ cm$^3$; mass of a cell $= 6.0\times 2.7\times 10^{-23} = 1.62\times 10^{-22}$ g, holding 2 atoms (bcc).
Number of atoms in 180 g $= \frac{180}{1.62\times 10^{-22}}\times 2 = 2.22\times 10^{24} = 22.2\times 10^{23}$.
Nearest integer: 22.
Answer: 22. | ||||
| 83 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure By molecular orbital theory:
B$_2$ (10 e$^-$): $\pi2p_x^1\pi2p_y^1$, 2 unpaired — paramagnetic.
Li$_2$ (6 e$^-$): all paired — diamagnetic.
C$_2$ (12 e$^-$): $\pi2p_x^2\pi2p_y^2$ — diamagnetic.
C$_2^-$ (13 e$^-$): one electron in $\sigma2p_z$ — paramagnetic.
O$_2^{2-}$ (18 e$^-$): $\pi^*$ full — diamagnetic.
O$_2^+$ (15 e$^-$): one $\pi^*$ electron — paramagnetic.
He$_2^+$ (3 e$^-$): $\sigma^*1s^1$ — paramagnetic.
Paramagnetic: B$_2$, C$_2^-$, O$_2^+$, He$_2^+$ = 4.
Answer: 4. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions Molality $= \frac{10.2/176}{0.150} = \frac{0.05795}{0.150} = 0.3864$ mol kg$^{-1}$.
$\Delta T_f = K_f m = 3.9\times 0.3864 = 1.507$ °C $= 15.07\times 10^{-1}$ °C.
Nearest integer: 15.
Answer: 15. | ||||
| 85 | EquilibriumIonic equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium For a weak acid, $[\text{H}^+] = \sqrt{K_aC} = \sqrt{2\times 10^{-5}\times 0.2} = \sqrt{4\times 10^{-6}} = 2\times 10^{-3}$ M.
pH $= 3 - \log 2 = 3 - 0.30 = 2.70 = 27\times 10^{-1}$.
Answer: 27. | ||||
| 86 | 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 $k = \frac{0.693}{t_{1/2}} = \frac{2.303\times 0.3010}{0.3010} = 2.303$ min$^{-1}$.
$\log\frac{[A]_0}{[A]} = \frac{kt}{2.303} = \frac{2.303\times 2}{2.303} = 2$, so $\frac{[A]_0}{[A]} = 10^2 = 100$.
Answer: 100. | ||||
| 87 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter XF$_5$ interhalogens have 5 bond pairs and 1 lone pair: square pyramidal. These are BrF$_5$, IF$_5$, ClF$_5$.
ClF$_3$ and BrF$_3$ are T-shaped, IF$_7$ pentagonal bipyramidal, I$_2$Cl$_6$ planar (bridged dimer), ClF linear.
Answer: 3. | ||||
| 88 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalEasy | |
A easy numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}$.
A = MnO$_4^-$ (Mn +7), B = MnO$_2$ (Mn +4, $d^3$, 3 unpaired electrons).
$\mu = \sqrt{n(n + 2)} = \sqrt{15} = 3.87$ BM ≈ 4.
Answer: 4. | ||||
| 89 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[\text{M(AA)}_2\text{b}_2]$ has a trans isomer (with a plane of symmetry, optically inactive) and a cis isomer that is chiral, existing as a pair of enantiomers.
Stereoisomers: trans + cis-d + cis-l = 3.
Linkage isomers (–SCN against –NCS binding) are not counted: for each complex the question asks only for the more stable bonding mode.
Answer: 3. | ||||
| 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 C $= 0.7938\times\frac{12}{44} = 0.2165$ g; mass of H $= 0.4428\times\frac{2}{18} = 0.0492$ g.
Mass of O $= 0.492 - 0.2165 - 0.0492 = 0.2263$ g.
% O $= \frac{0.2263}{0.492}\times 100 = 46.0\%$.
Answer: 46. | ||||
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