JEE (Main) · 2022
JEE (Main) 2022 — 27 June, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Matrices and Determinants2
- Sequences and Series2
9 easy · 18 medium · 3 hard
Physics
30 Q
- Kinematics2
- Magnetic Effects of Current and Magnetism2
- Work, Energy and Power2
24 easy · 6 medium · 0 hard
Chemistry
30 Q
- Solutions2
- Hydrogen2
- p-Block Elements2
20 easy · 10 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | 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 Taking moduli, $|z| = |z|^2$, so $|z| = 0$ or $|z| = 1$. The root $z = 0$ is real, so the non-real roots have $|z| = 1$; write $z = e^{i\theta}$.
Then $e^{-i\theta} = i e^{2i\theta}$, i.e. $e^{-3i\theta} = i = e^{i\pi/2}$, so $\theta = -\frac{\pi}{6} - \frac{2k\pi}{3}$: $\theta = -\frac{\pi}{6}, \frac{\pi}{2}, -\frac{5\pi}{6}$.
The roots $\frac{\sqrt{3}}{2} - \frac{i}{2}$, $i$, $-\frac{\sqrt{3}}{2} - \frac{i}{2}$ are all non-real and form an equilateral triangle inscribed in the unit circle.
Area $= \frac{3\sqrt{3}}{4}r^2 = \frac{3\sqrt{3}}{4}$.
Answer: (A). | ||||
| 2 | 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 $\Delta = \begin{vmatrix}1&2&1\\ \alpha&3&-1\\ -\alpha&1&2\end{vmatrix} = 1(6+1) - 2(2\alpha - \alpha) + 1(\alpha + 3\alpha) = 7 + 2\alpha$.
Inconsistency needs $\Delta = 0$, so $\alpha = -\frac{7}{2}$.
Check: adding the second and third equations gives $4y + z = 0$, so $z = -4y$ and the first gives $x = 2 + 2y$. The second then reads $-\frac{7}{2}(2 + 2y) + 3y + 4y = -7$, but its right side is $-\frac{7}{2}$: no solution, so the system is indeed inconsistent.
Answer: (D). | ||||
| 3 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Each sum is an infinite G.P.: $x = \frac{1}{1-a}$, $y = \frac{1}{1-b}$, $z = \frac{1}{1-c}$.
So $\frac{1}{x} = 1 - a$, $\frac{1}{y} = 1 - b$, $\frac{1}{z} = 1 - c$. Since a, b, c are in A.P., so are $1-a, 1-b, 1-c$; hence $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.
(Option D fails: the sum is $3 - (a+b+c)$.)
Answer: (C). | ||||
| 4 | CirclesRadius and centre | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections A circle $x^2 + y^2 + 2gx + 2fy + k = 0$ satisfies $\frac{dy}{dx} = -\frac{x + g}{y + f}$: the numerator has no y and the denominator no x. So $b = 0$, and $\frac{dy}{dx} = \frac{a(x+1)}{cy + a}$.
Matching $-\frac{x+g}{y+f}$ needs $c = -a$, giving $\frac{dy}{dx} = -\frac{x+1}{y-1}$, i.e. $(x+1)dx + (y-1)dy = 0$.
Integrating: $(x+1)^2 + (y-1)^2 = r^2$. Through (2, 5): $r^2 = 9 + 16 = 25$, so the centre is $(-1, 1)$ and $r = 5$.
Distance of (11, 6) from the centre $= \sqrt{144 + 25} = 13$, so the shortest distance to the circle is $13 - 5 = 8$.
Answer: (B). | ||||
| 5 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Since a is an integer, $[x - 3a] = [x] - 3a$.
As $x \to 7^+$: $[1 - x] = -7$ and $[x] = 7$, so the expression is $\frac{25}{7 - 3a}$.
As $x \to 7^-$: $[1 - x] = -6$ and $[x] = 6$, so it is $\frac{24}{6 - 3a}$.
For the limit to exist: $25(6 - 3a) = 24(7 - 3a)$, so $150 - 75a = 168 - 72a$ and $a = -6$ (both one-sided limits are then $\frac{25}{25} = \frac{24}{24} = 1$).
Answer: (A). | ||||
| 6 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $f(x) = x^4 - 4x + 1$. $f'(x) = 4x^3 - 4 = 0$ only at $x = 1$; f decreases on $(-\infty, 1)$ and increases on $(1, \infty)$.
$f(1) = 1 - 4 + 1 = -2 < 0$ while $f(x) \to +\infty$ on both sides, so the graph crosses the x-axis exactly twice.
Answer: (B). | ||||
| 7 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The triangle is isosceles with equal sides $a = 10 + x^2$ and base $b = 20 - 2x^2$. Its area is $\frac{b}{4}\sqrt{4a^2 - b^2}$.
$4a^2 - b^2 = (2a - b)(2a + b) = (4x^2)(40) = 160x^2$, so Area $= \frac{\sqrt{160}}{4}(20 - 2x^2)|x|$.
With $t = x^2$, maximise $g(t) = (20 - 2t)\sqrt{t}$: $g'(t) = -2\sqrt{t} + \frac{20 - 2t}{2\sqrt{t}} = 0 \Rightarrow 20 - 6t = 0 \Rightarrow t = \frac{10}{3}$.
So $k^2 = \frac{10}{3}$ and $3k^2 = 10$.
Answer: (C). | ||||
| 8 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $y = 2\cos\left(5\log_e\frac{x}{5}\right)$. Then $y' = -2\sin\left(5\log_e\frac{x}{5}\right)\cdot\frac{5}{x}$, so $xy' = -10\sin\left(5\log_e\frac{x}{5}\right)$.
Differentiating again: $y' + xy'' = -10\cos\left(5\log_e\frac{x}{5}\right)\cdot\frac{5}{x}$, so $xy' + x^2y'' = -50\cos\left(5\log_e\frac{x}{5}\right) = -25y$.
Hence $x^2y'' + xy' + 25y = 0$.
Answer: (D). | ||||
| 9 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Use $\int e^x(f + f')\,dx = e^x f + C$. Try $f(x) = \frac{x - 1}{x + 1} = 1 - \frac{2}{x+1}$: $f'(x) = \frac{2}{(x+1)^2}$ and $f + f' = \frac{(x^2 - 1) + 2}{(x+1)^2} = \frac{x^2 + 1}{(x+1)^2}$. So $f(x) = 1 - \frac{2}{x+1}$.
$f' = 2(x+1)^{-2}$, $f'' = -4(x+1)^{-3}$, $f''' = 12(x+1)^{-4}$. At $x = 1$: $f''' = \frac{12}{16} = \frac{3}{4}$.
Answer: (B). | ||||
| 10 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Using $\int_{-a}^{a} g(x)\,dx = \int_0^a [g(x) + g(-x)]\,dx$: the numerator $|x^3 + x|$ is even, and $\frac{1}{e^{x|x|} + 1} + \frac{1}{e^{-x|x|} + 1} = \frac{1}{e^{x|x|} + 1} + \frac{e^{x|x|}}{1 + e^{x|x|}} = 1$.
So $I = \int_0^2 (x^3 + x)\,dx = \frac{16}{4} + \frac{4}{2} = 6$.
Answer: (D). | ||||
| 11 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $\frac{2^y}{2^y - 1}\,dy = -\frac{2^x}{2^x - 1}\,dx$.
Integrating, $\frac{1}{\ln 2}\ln(2^y - 1) = -\frac{1}{\ln 2}\ln(2^x - 1) + c$, i.e. $(2^y - 1)(2^x - 1) = K$.
At $x = 1$, $y = 1$: $K = 1 \cdot 1 = 1$. At $x = 2$: $(2^y - 1)\cdot 3 = 1$, so $2^y = \frac{4}{3}$ and $y = \log_2\frac{4}{3} = 2 - \log_2 3$.
Answer: (D). | ||||
| 12 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines B is on both $2x + y = 4$ and $x + 3y = 7$: $B = (1, 2)$.
With A the vertex (AB = AC), the foot M of the perpendicular from A to BC is the midpoint of BC. $2(6) + 1 - 4 = 9$, so $M = (6, 1) - \frac{9}{5}(2, 1) = \left(\frac{12}{5}, -\frac{4}{5}\right)$.
$C = 2M - B = \left(\frac{19}{5}, -\frac{18}{5}\right)$.
Centroid: $\alpha = \frac{6 + 1 + \frac{19}{5}}{3} = \frac{18}{5}$, $\beta = \frac{1 + 2 - \frac{18}{5}}{3} = -\frac{1}{5}$. So $15(\alpha + \beta) = 15 \cdot \frac{17}{5} = 51$.
Answer: (C). | ||||
| 13 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $b^2 = a^2(1 - e^2) = \frac{15}{16}a^2$.
The point gives $\frac{32/5}{a^2} + \frac{9}{\frac{15}{16}a^2} = 1$, i.e. $\frac{32}{5a^2} + \frac{48}{5a^2} = \frac{16}{a^2} = 1$.
So $a^2 = 16$, $b^2 = 15$ and $a^2 + b^2 = 31$.
Answer: (B). | ||||
| 14 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Put $n = l + m$ in the second relation: $3l^2 + m^2 + cl(l + m) = 0$, i.e. $(3 + c)l^2 + c\,lm + m^2 = 0$.
The two lines are parallel when this quadratic in $\frac{l}{m}$ has equal roots: $c^2 - 4(3 + c) = 0$, so $c^2 - 4c - 12 = 0$ and $c = 6$ or $c = -2$.
The positive value is $c = 6$.
Answer: (A). | ||||
| 15 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{b} \times \vec{c}$ is always perpendicular to $\vec{c}$, so a solution needs $\vec{a} \cdot \vec{c} = 0$.
But $\vec{a} \cdot \vec{c} = 2 - 3 - 2 = -3 \ne 0$, so no vector $\vec{b}$ satisfies $\vec{b} \times \vec{c} = \vec{a}$.
Answer: (A). | ||||
| 16 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Total selections: $\binom{18}{5} = 8568$.
Favourable: $x_1$ from $\{1, \ldots, 6\}$ (6 ways), $x_3$ from $\{8, 9, 10\}$ (3 ways), $x_5$ from $\{12, \ldots, 18\}$ (7 ways): $6 \cdot 3 \cdot 7 = 126$.
Probability $= \frac{126}{8568} = \frac{1}{68}$.
Answer: (C). | ||||
| 17 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $\binom{7}{3}p^3q^4 = 5\binom{7}{4}p^4q^3$ and $\binom{7}{3} = \binom{7}{4}$, so $q = 5p$. With $p + q = 1$: $p = \frac{1}{6}$, $q = \frac{5}{6}$.
Mean $= np = \frac{7}{6}$, variance $= npq = \frac{35}{36}$. Sum $= \frac{42 + 35}{36} = \frac{77}{36}$.
Answer: (C). | ||||
| 18 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\theta = \frac{\pi}{7}$ and $S = \cos 2\theta + \cos 4\theta + \cos 6\theta$. Multiply by $2\sin\theta$:
$2\sin\theta\,S = (\sin 3\theta - \sin\theta) + (\sin 5\theta - \sin 3\theta) + (\sin 7\theta - \sin 5\theta) = \sin 7\theta - \sin\theta = -\sin\theta$, since $\sin\pi = 0$.
So $S = -\frac{1}{2}$.
Answer: (B). | ||||
| 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 $\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}$ (principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$).
$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = 2\pi - \frac{7\pi}{6} = \frac{5\pi}{6}$ (range $[0, \pi]$).
$\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \frac{3\pi}{4} - \pi = -\frac{\pi}{4}$ (range $(-\frac{\pi}{2}, \frac{\pi}{2})$).
Sum $= \frac{4\pi + 10\pi - 3\pi}{12} = \frac{11\pi}{12}$.
Answer: (A). | ||||
| 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 $\sim(p \wedge q) \vee q = (\sim p \vee \sim q) \vee q = \sim p \vee (\sim q \vee q)$, which is a tautology (always T).
Check the options with $a \to b \equiv \sim a \vee b$:
(A) $\sim q \vee (p \wedge q) = \sim q \vee p$: false when $p$ is F and $q$ is T.
(B) $\sim p \vee q$: false when $p$ is T and $q$ is F.
(C) $\sim p \vee (\sim p \vee q) = \sim p \vee q$: not a tautology.
(D) $\sim p \vee (p \vee q) = (\sim p \vee p) \vee q$: a tautology.
Answer: (D). | ||||
| 21 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f(1 - x) = \frac{2e^{2 - 2x}}{e^{2-2x} + e}$; multiplying above and below by $e^{2x - 1}$ gives $f(1 - x) = \frac{2e}{e + e^{2x}}$.
So $f(x) + f(1 - x) = \frac{2e^{2x} + 2e}{e^{2x} + e} = 2$.
Pair $\frac{k}{100}$ with $\frac{100 - k}{100}$ for $k = 1, \ldots, 49$: $49 \times 2 = 98$. The middle term is $f\left(\frac{1}{2}\right) = \frac{2e}{e + e} = 1$.
Sum $= 98 + 1 = 99$.
Answer: 99. | ||||
| 22 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalHard | |
A hard numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $t = e^x > 0$ and multiply by $2t$: $2t^3 - 22t^2 + 81t - 90 = 0$.
$t = 2$ is a root ($16 - 88 + 162 - 90 = 0$), so $2t^3 - 22t^2 + 81t - 90 = (t - 2)(2t^2 - 18t + 45)$.
The quadratic factor has discriminant $324 - 360 = -36 < 0$: no real $t$. So the only real root is $t = 2$, i.e. $x = \log_e 2$.
The sum of all the (real) roots is $\log_e 2$, so $p = 2$.
Answer: 2. | ||||
| 23 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3 \times 3$ matrix, $\text{adj}(\text{adj}\,A) = |A|^{3-2}A = |A|\,A$, so $\det(\text{adj}(\text{adj}\,A)) = |A|^3 \cdot |A| = |A|^4$.
$\det\begin{pmatrix}14&28&-14\\-14&14&28\\28&-14&14\end{pmatrix} = 14^3\begin{vmatrix}1&2&-1\\-1&1&2\\2&-1&1\end{vmatrix} = 14^3[1(1+2) - 2(-1-4) - 1(1-2)] = 14^3 \cdot 14 = 14^4$.
So $|A|^4 = 14^4$ and the positive value is $|A| = 14$.
Answer: 14. | ||||
| 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 There are 5 red and 11 blue cubes. The 5 reds make 6 gaps (two ends and 4 between). Each of the 4 inner gaps needs at least 2 blues: that uses 8, leaving 3 blues to place freely in the 6 gaps.
Number of ways $= \binom{3 + 6 - 1}{6 - 1} = \binom{8}{5} = 56$.
Answer: 56. | ||||
| 25 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom{60}{r}\left(x^{1/2}5^{-1/4}\right)^{60-r}\left(5^{1/2}x^{-1/3}\right)^r$. The power of x is $\frac{60 - r}{2} - \frac{r}{3} = 10$, so $180 - 5r = 60$ and $r = 24$.
Coefficient $= \binom{60}{24}5^{-36/4}5^{24/2} = \binom{60}{24}\cdot 5^{3}$.
Power of 5 in $\binom{60}{24} = \frac{60!}{24!\,36!}$: $(12 + 2) - 4 - (7 + 1) = 2$.
So the coefficient is $5^{2+3}\cdot l = 5^5 \cdot l$ with l co-prime to 5, and $k = 5$.
Answer: 5. | ||||
| 26 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $A_1$ is taken as the bounded region above the x-axis ($y \ge 0$) between the parabola arms $|x| = y^2$ and the lines $|x| + 2y = 8$; it is symmetric about the y-axis.
For $x \ge 0$ the curves $x = y^2$ and $x = 8 - 2y$ meet where $y^2 + 2y - 8 = 0$, i.e. $y = 2$. So, integrating in y, the half-area is $\int_0^2 y^2\,dy + \int_2^4 (8 - 2y)\,dy = \frac{8}{3} + 4 = \frac{20}{3}$.
Area $A_1 = \frac{40}{3}$. $A_2$ is a square with diagonals $2k$: area $2k^2$.
$27 \cdot \frac{40}{3} = 5 \cdot 2k^2$, so $360 = 10k^2$, $k^2 = 36$ and $k = 6$.
Answer: 6. | ||||
| 27 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The r-th term is $\frac{r}{4r^4 + 1}$ (check: $4 + 1 = 5$, $64 + 1 = 65$, $324 + 1 = 325$).
$4r^4 + 1 = (2r^2 + 2r + 1)(2r^2 - 2r + 1)$, and the difference of the factors is $4r$, so $\frac{r}{4r^4 + 1} = \frac{1}{4}\left[\frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1}\right]$.
Since $2(r+1)^2 - 2(r+1) + 1 = 2r^2 + 2r + 1$, the sum telescopes: $S_{10} = \frac{1}{4}\left[1 - \frac{1}{2(100) + 20 + 1}\right] = \frac{1}{4}\cdot\frac{220}{221} = \frac{55}{221}$.
$221 = 13 \cdot 17$ is co-prime to 55, so $m + n = 55 + 221 = 276$.
Answer: 276. | ||||
| 28 | CirclesRadius and centre | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on radius and centre, from Circles. NCERT: Class 11: Conic Sections The centre is on the perpendicular bisector of the chord from (1, 2) to (3, 6): through (2, 4) with normal (2, 4), i.e. $x + 2y = 10$.
With $2x - y + 4 = 0$: $x = 10 - 2y$, $20 - 5y = -4$, so $y = \frac{24}{5}$, $x = \frac{2}{5}$. Centre $O = \left(\frac{2}{5}, \frac{24}{5}\right)$.
The centre of the circle is the centre of the rectangle. The side through (1, 2), (3, 6) has length $\sqrt{4 + 16} = 2\sqrt{5}$; the other side is twice the distance from O to this side's midpoint (2, 4): $2\sqrt{\left(\frac{8}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = 2\cdot\frac{4\sqrt{5}}{5} = \frac{8}{\sqrt{5}}$.
Area $= 2\sqrt{5}\cdot\frac{8}{\sqrt{5}} = 16$.
Answer: 16. | ||||
| 29 | Conic SectionsParabola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2 = 2x$ has vertex (0, 0) and focus $\left(\frac{1}{2}, 0\right)$. The centre lies on the perpendicular bisector $x = \frac{1}{4}$: centre $\left(\frac{1}{4}, h\right)$ with $\frac{1}{16} + h^2 = 4$, so $h = \pm\frac{\sqrt{63}}{4}$.
The parabola $y = \left(x - \frac{1}{4}\right)^2 + \alpha$ opens upward with vertex $\left(\frac{1}{4}, \alpha\right)$ on the same vertical line as the centre. Its radius of curvature at the vertex is $\frac{1}{2} < 2$, so it touches the circle at its vertex, which must be the top point of the circle: $\alpha = h + 2$.
Then $4\alpha - 8 = 4h = \pm\sqrt{63}$ (both signs give $\alpha > 0$), and $(4\alpha - 8)^2 = 63$.
Answer: 63. | ||||
| 30 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The image is $(a, b, c) - 2t(3, -4, 12)$ with $t = \frac{3a - 4b + 12c + 19}{169}$.
From the x-coordinate, $-6t = -6$, so $t = 1$: $\beta = b + 8$, $\gamma = c - 24$, and $3a - 4b + 12c + 19 = 169$, i.e. $3a - 4b + 12c = 150$.
With $a = 5 - b - c$: $15 - 7b + 9c = 150$, so $7b - 9c = -135$.
$7\beta - 9\gamma = 7b + 56 - 9c + 216 = -135 + 272 = 137$.
Answer: 137. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $v_x = 20\cos\alpha$ (constant), $v_y = 20\sin\alpha - 10 \times 10 = 20\sin\alpha - 100$.
$\tan\beta = \frac{v_y}{v_x} = \frac{20\sin\alpha - 100}{20\cos\alpha} = \tan\alpha - 5\sec\alpha$.
Answer: (B). | ||||
| 32 | KinematicsRelative velocity | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on relative velocity, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Standing still, she tilts the umbrella at 45°, so the rain's horizontal and vertical components are equal: $v_h = v_v$.
Running at $15\sqrt{2}$ km/h, the rain appears vertical, so the rain's horizontal component equals her speed: $v_h = 15\sqrt{2}$ km/h, hence $v_v = 15\sqrt{2}$ km/h.
Relative to her, only the vertical component remains: $15\sqrt{2} = \frac{30}{\sqrt{2}}\ kmh^{-1}$.
Answer: (C). | ||||
| 33 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\rho = \frac{m}{\pi r^2 L}$, so $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L} = 1\% + 2(1\%) + 1\% = 4\%$.
Answer: (A). | ||||
| 34 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Only friction accelerates m, so the largest common acceleration is $a_{max} = \mu g = 0.5 \times 9.8 = 4.9\ m/s^2$.
For the whole system on the smooth table: $F_{max} = (m + M)a_{max} = 10 \times 4.9 = 49$ N.
Answer: (C). | ||||
| 35 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion For the centre of mass to stay fixed: $m_1\Delta x_1 + m_2\Delta x_2 = 0$, so $10(+6) + 30\,\Delta x_2 = 0$ and $\Delta x_2 = -2$ cm.
The 30 kg block moves 2 cm in the negative direction, i.e. towards the 10 kg block.
Answer: (C). | ||||
| 36 | 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 The galvanometer and shunt are in parallel, so $I_g = I\frac{S}{G + S} = I\frac{8}{72 + 8} = \frac{I}{10}$, i.e. 10% of the total current.
Answer: (B). | ||||
| 37 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation Statement I: Newton's law of gravitation is universal; it applies to every pair of bodies. True.
Statement II: inside the earth $g' = g\left(1 - \frac{d}{R}\right)$, which is zero at the centre ($d = R$); so the weight $mg'$ is zero there. True.
Answer: (A). | ||||
| 38 | Work, Energy and PowerCollisions | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on collisions, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power In a head-on elastic collision with a body at rest, the fraction of kinetic energy transferred is $\frac{4m_1m_2}{(m_1 + m_2)^2} = \frac{4 \cdot m \cdot 5m}{(6m)^2} = \frac{20}{36} = 0.556$, i.e. 55.6%.
Answer: (C). | ||||
| 39 | Properties of Bulk MatterViscosity | 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 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 the net force is zero: $F_v = mg - F_B$. The ball's volume is $\frac{m}{d_1}$, so $F_B = \frac{m}{d_1}d_2 g$.
$F_v = mg\left(1 - \frac{d_2}{d_1}\right)$.
Answer: (B). | ||||
| 40 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $\mu = \mu_0(1 + \chi) = 4\pi \times 10^{-7} \times 100 = 4\pi \times 10^{-5}$ Wb/A-m.
Answer: (C). | ||||
| 41 | 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 The current first peaks when $120\pi t = \frac{\pi}{2}$, so $t = \frac{1}{240}$ s (a quarter of the period $T = \frac{1}{60}$ s).
Answer: (D). | ||||
| 42 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Ultraviolet rays kill microbes: sterilising surgical instruments (iii).
Microwaves are used in radar (iv).
Infrared radiation trapped by the atmosphere causes the greenhouse effect (ii).
X-rays are diffracted by crystals: study of crystal structure (i).
Answer: (A). | ||||
| 43 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{\sqrt{2mK}}$, so at equal K, $\frac{\lambda_\alpha}{\lambda_C} = \sqrt{\frac{m_C}{m_\alpha}} = \sqrt{\frac{12}{4}} = \sqrt{3}$.
Answer: (B). | ||||
| 44 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Between the plates the field is $\frac{\sigma}{\varepsilon_0}$, the sum of $\frac{\sigma}{2\varepsilon_0}$ from each plate. With one plate removed only $\frac{\sigma}{2\varepsilon_0}$ remains, so the force halves to 5 N.
Answer: (A). | ||||
| 45 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Starting from the mean position, $x = A\sin\omega t$. $\frac{A}{2} = A\sin 3\omega$ gives $3\omega = \frac{\pi}{6}$, so $\omega = \frac{\pi}{18}$ and $T = \frac{2\pi}{\omega} = 36$ s.
Answer: (D). | ||||
| 46 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $f' = f\frac{v + v_o}{v} = f\left(1 + \frac{1}{5}\right) = 1.2f$: an increase of 20%.
Answer: (A). | ||||
| 47 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Snell's law with $i = 2r$: $\sin 2r = \sqrt{2n}\sin r$, so $2\sin r\cos r = \sqrt{2n}\sin r$ and $\cos r = \sqrt{\frac{n}{2}}$.
$i = 2r = 2\cos^{-1}\left(\sqrt{\frac{n}{2}}\right)$.
Answer: (D). | ||||
| 48 | Atoms and NucleiBohr model | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on bohr model, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei 10.2 eV $= 13.6\left(1 - \frac{1}{4}\right)$ eV takes the electron from n = 1 to n = 2.
$L = \frac{nh}{2\pi}$, so $\Delta L = \frac{h}{2\pi} = \frac{6.6 \times 10^{-34}}{2 \times 3.14} = 1.05 \times 10^{-34}$ Js.
Answer: (B). | ||||
| 49 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Reading the waveforms interval by interval (between the dashed lines):
A = 0, B = 0: Y = 1
A = 0, B = 1: Y = 1
A = 1, B = 0: Y = 1
A = 1, B = 1: Y = 0
Y is 0 only when both inputs are 1: this is NAND, $Y = \overline{A \cdot B}$.
(A) is NOR; (B) AND followed by NOT is NAND; (C) gives $\overline{\bar{A} + B} = A\bar{B}$; (D) gives $\overline{A\bar{B}}$, which is 0 for A = 1, B = 0.
Answer: (B). | ||||
| 50 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctMedium | |
A medium single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Total moles $n = \frac{PV}{RT} = \frac{10^5 \times 2 \times 10^{-3}}{8.3 \times 300} = \frac{200}{2490} \approx 0.0803$ mol.
With $n_1$ mol of $H_2$ and $n_2$ of $O_2$: $n_1 + n_2 = 0.0803$ and $2n_1 + 32n_2 = 0.76$.
Then $2(0.0803 - n_2) + 32n_2 = 0.76$, so $30n_2 = 0.599$, $n_2 \approx 0.0200$ and $n_1 \approx 0.0603$.
$n_1 : n_2 \approx 3 : 1$.
Answer: (B). | ||||
| 51 | ThermodynamicsSecond law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on second law, from Thermodynamics. NCERT: Class 11: Thermodynamics $T_1 = 527 + 273 = 800$ K, $T_2 = 200$ K. Efficiency $\eta = 1 - \frac{200}{800} = \frac{3}{4}$.
$Q_1 = \frac{W}{\eta} = \frac{12000\ kJ}{3/4} = 16000\ kJ = 16 \times 10^6$ J.
Answer: 16. | ||||
| 52 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Peak (full rated) brightness means the lamp runs at its rating: $I = \frac{P}{V} = \frac{5}{25} = 0.2$ A, and its resistance is $\frac{V^2}{P} = \frac{625}{5} = 125\ \Omega$.
Total resistance needed: $\frac{220}{0.2} = 1100\ \Omega$ (rms values), so $R = 1100 - 125 = 975\ \Omega$.
Answer: 975. | ||||
| 53 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The point opposite a slit is at $y = \frac{d}{2} = 0.3$ mm from the centre. The first dark fringe is at $y = \frac{\lambda D}{2d}$.
$\lambda = \frac{2dy}{D} = \frac{2 \times 0.6 \times 10^{-3} \times 0.3 \times 10^{-3}}{0.8} = 4.5 \times 10^{-7}$ m $= 450$ nm.
Answer: 450. | ||||
| 54 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei For $Li^{2+}$, Z = 3: $\Delta E = 13.6 Z^2\left(1 - \frac{1}{9}\right) = 13.6 \times 9 \times \frac{8}{9} = 108.8$ eV.
$\lambda = \frac{1242}{108.8}$ nm $= 11.415$ nm $= 114.15 \times 10^{-10}$ m, so x = 114 to the nearest integer.
Answer: 114. | ||||
| 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 With a shunt S, the terminal voltage balanced is $V = \frac{ES}{S + r}$, proportional to the balancing length.
$\frac{8/(8 + r)}{4/(4 + r)} = \frac{3}{2}$, so $16(4 + r) = 12(8 + r)$, $64 + 16r = 96 + 12r$ and $r = 8\ \Omega$.
Answer: 8. | ||||
| 56 | Current ElectricityDrift velocity and mobility | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity With uniform current density, $I = J \cdot \pi\left(R^2 - \frac{R^2}{4}\right) = 4 \times 10^6 \times \pi \times \frac{3}{4} \times 16 \times 10^{-6} = 48\pi$ A.
Answer: 48. | ||||
| 57 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Initial energy $\frac{1}{2}CV^2$. After sharing, each capacitor has $\frac{V}{2}$: total $2 \times \frac{1}{2}C\frac{V^2}{4} = \frac{1}{4}CV^2$.
Loss $= \frac{1}{4}CV^2 = \frac{1}{4} \times 50 \times 10^{-12} \times 10^4 = 1.25 \times 10^{-7}$ J $= 125$ nJ.
Answer: 125. | ||||
| 58 | Communication SystemsLine-of-sight communication | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on line-of-sight communication, from Communication Systems. NCERT: No matching NCERT chapter $d = \sqrt{2Rh_T} + \sqrt{2Rh_R} = \sqrt{2R}\left(\sqrt{25} + \sqrt{49}\right) = 12\sqrt{2R}$.
$\sqrt{2R} = \sqrt{128 \times 10^5} = \sqrt{1.28 \times 10^7} = 1600\sqrt{5}$ m (since $1600^2 \times 5 = 1.28 \times 10^7$).
$d = 12 \times 1600\sqrt{5} = 192\sqrt{5} \times 10^2$ m, so K = 192.
Answer: 192. | ||||
| 59 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The load adds a pressure $\frac{25 \times 10}{0.5} = 500$ Pa on the surface.
Bernoulli: $\frac{1}{2}\rho v^2 = \rho g h + \Delta P$, so $v^2 = 2\left(10 \times 0.4 + \frac{500}{1000}\right) = 2(4.5) = 9$ and $v = 3$ m/s $= 300$ cm/s.
Answer: 300. | ||||
| 60 | Work, Energy and PowerMotion in a vertical circle | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on motion in a vertical circle, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power To just complete the vertical circle on a string, the bob needs $u = \sqrt{5gL} = \sqrt{5 \times 10 \times 2} = 10$ m/s at the bottom.
Momentum conservation: $0.075v = 0.075\frac{v}{3} + 0.05 \times 10$, so $0.05v = 0.5$ and $v = 10$ m/s.
Answer: 10. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | SolutionsConcentration terms | Class 12: Solutions | Single correctEasy | |
A easy single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Molality = moles of solute per kg of solvent. Both quantities are masses (or amounts), which do not change with temperature, so the molality of the solution is the same at 10°C and at $-21$°C. (The solution stays liquid at $-21$°C: its molality is about 6 mol/kg, and with i = 2 the freezing point is lowered by roughly $2 \times 1.86 \times 6 \approx 22$°C.)
So (A) is true, (R) is true, and (R) is exactly why (A) holds.
Answer: (A). | ||||
| 62 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $ClF_3$: 3 bond pairs + 2 lone pairs: T-shaped (A-III).
$BF_3$: 3 bond pairs: trigonal planar (B-IV).
$XeF_4$: 4 bond pairs + 2 lone pairs: square planar (C-I).
$SF_4$: 4 bond pairs + 1 lone pair: see-saw (D-II).
Answer: (B). | ||||
| 63 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics A spontaneous process at constant T and P has $\Delta G < 0$ (A-II).
$\Delta P = 0$ and $\Delta T = 0$ means isobaric and isothermal (B-III).
$\Delta H_{reaction} = \sum$ bond energies of reactants $- \sum$ bond energies of products (C-IV).
An exothermic process has $\Delta H < 0$ (D-I).
Answer: (B). | ||||
| 64 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Lyophilic sols protect lyophobic sols from coagulation: protective colloids (A-II).
An emulsion is a liquid dispersed in a liquid (B-I).
$FeCl_3$ added to hot water gives $Fe(OH)_3$ sol that adsorbs $Fe^{3+}$: positive (C-IV).
$FeCl_3$ added to NaOH gives a sol that adsorbs $OH^-$: negative (D-III).
Answer: (A). | ||||
| 65 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Both ions have 10 electrons, so (R) is true. But in an isoelectronic series the radius falls as the nuclear charge rises: $O^{2-}$ (Z = 8, about 140 pm) is much larger than $Mg^{2+}$ (Z = 12, about 72 pm). So (A) is false.
Answer: (D). | ||||
| 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 Gold is leached from its ore with dilute NaCN in air (A-IV).
Bauxite is leached with NaOH to dissolve alumina as aluminate (B-II).
Cresols and aniline act as froth stabilisers in froth flotation (C-I).
Blister copper gets its blistered look from escaping $SO_2$ (D-III).
Answer: (B). | ||||
| 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 Acidifying barium peroxide is a laboratory preparation of hydrogen peroxide: $BaO_2 + H_2SO_4 \to BaSO_4\downarrow + H_2O_2$.
Answer: (D). | ||||
| 68 | 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 Beryllium hydride is made by reducing $BeCl_2$ with lithium aluminium hydride: $2BeCl_2 + LiAlH_4 \to 2BeH_2 + LiCl + AlCl_3$.
Answer: (C). | ||||
| 69 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter $(CH_3)_4Si$ has no OH and cannot condense: it is tetramethylsilane, a silane (A-III).
$CH_3Si(OH)_3$ has three OH groups and condenses in more than one direction: a cross-linked (2D) silicone (B-IV).
$(CH_3)_2Si(OH)_2$ has two OH groups: a linear chain silicone (C-I).
$(CH_3)_3SiOH$ has one OH group: it condenses only to the dimer $(CH_3)_3Si-O-Si(CH_3)_3$ (D-II).
Answer: (D). | ||||
| 70 | 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 White phosphorus disproportionates in hot concentrated NaOH: $P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2$ (phosphine and sodium hypophosphite).
Answer: (D). | ||||
| 71 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[Ti(H_2O)_6]^{3+}$: $d^1$, CFSE $= -0.4\Delta_o$.
$[Co(H_2O)_6]^{2+}$: $d^7$ high spin ($t_{2g}^5e_g^2$), CFSE $= -0.8\Delta_o$, with the small $\Delta_o$ of water.
$[Co(CN)_6]^{3-}$: $d^6$ low spin with the strong-field $CN^-$ ($t_{2g}^6$), CFSE $= -2.4\Delta_o$, and $\Delta_o$ is large for $Co^{3+}$ with $CN^-$.
$[Cu(NH_3)_4]^{2+}$: $d^9$, one hole in the upper level; its stabilisation is far less.
The cyanide complex has by far the largest stabilisation.
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 Statement I: classical (London) smog forms in cool, humid weather from smoke, fog and sulphur dioxide, and is chemically reducing. Correct.
Statement II: photochemical smog contains ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate (PAN). Correct.
Answer: (A). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A separating funnel is a pear-shaped (or globular) glass vessel with a stopper at the top and a stopcock on its stem, so the lower of two immiscible liquid layers can be run off. That is figure (A). (B) is an ordinary funnel; (C) is a funnel with a perforated plate (a filtering funnel); (D) is a bent adapter.
Answer: (A). | ||||
| 74 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Cyclohexa-1,3-diene (C1=C2, C3=C4): cutting both double bonds gives C2–C3 as OHC–CHO (glyoxal) and C4–C5–C6–C1 as OHC–$CH_2CH_2$–CHO (ethane-1,2-dicarbaldehyde). So A is cyclohex-1,3-diene.
1-Methylcyclopent-1-ene: cutting C1=C2 opens the ring to $CH_3CO-CH_2CH_2CH_2-CHO$, which is 5-oxohexanal. So B is 1-methylcyclopent-1-ene (cyclopentene would give pentanedial).
Answer: (D). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $PhS^-$ is a strong, soft nucleophile and DMF is a polar aprotic solvent: conditions for $S_N2$.
The secondary benzylic C–I is attacked from the back, so I is replaced by SPh with inversion of configuration (wedge becomes hash).
The ring F is not replaced: it is meta to the $NO_2$ group (F at C-2, $NO_2$ at C-4 relative to the side chain), so the ring is not activated for nucleophilic aromatic substitution at that carbon.
That is product (A).
Answer: (A). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (A) $SOCl_2$ gives benzoyl chloride; Rosenmund reduction ($H_2$, Pd/$BaSO_4$, poisoned with quinoline) stops at benzaldehyde. Yes.
(B) Chromic acid ($CrO_3/H_2SO_4$) oxidises a primary alcohol through the aldehyde to benzoic acid. No.
(C) $NaBH_4$ does not reduce an ester, so methyl benzoate is unchanged and PCC has nothing to oxidise. No.
(D) Etard-type oxidation: $CrO_3$ in acetic anhydride gives benzylidene diacetate, $C_6H_5CH(OCOCH_3)_2$, which acid hydrolysis turns into benzaldehyde. Yes.
So (A) and (D).
Answer: (C). | ||||
| 77 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Statement I: in the Hofmann bromamide degradation an aryl group migrates just as an alkyl group does (benzamide gives aniline), so 'only an alkyl group' is wrong. Incorrect.
Statement II: after loss of bromide the nitrogen is electron-deficient (nitrene-like), and the group migrates from the carbonyl carbon to this nitrogen, giving the isocyanate. Correct.
Answer: (D). | ||||
| 78 | 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 Bakelite, a hard thermoset, is used for electrical switches (A-II).
Glyptal is used in paints and lacquers (B-III).
PVC is used for water pipes (C-IV).
Polystyrene is used for radio and television cabinets (D-I).
Answer: (A). | ||||
| 79 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules $C_4H_8O_4$ reducing Tollens' reagent, with three OH groups (triacetate), is an aldotetrose: CHO–CHOH–CHOH–$CH_2OH$.
$HNO_3$ oxidises both ends to COOH. The product is optically inactive only if it is meso-tartaric acid, i.e. both OH on the same side: the sugar is erythrose (threose would give optically active tartaric acid).
Bromine water oxidises only the CHO, giving erythronic acid, which is optically active (its ends differ).
The L-isomer has the OH on the lowest chiral carbon on the left; for erythrose both OH are then on the left: structure (A).
(B) has the lower OH on the left but the upper on the right: L-threose. (C) is D-erythrose. (D) is D-threose.
Answer: (A). | ||||
| 80 | Chemistry in Everyday LifeCleansing agents | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on cleansing agents, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter (A) Cetyltrimethylammonium bromide is a cationic detergent, used in hair conditioners (A-IV).
(B) Sodium dodecylbenzenesulphonate is an anionic detergent, used in toothpastes and household cleaning (B-II).
(C) Sodium stearate with sodium carbonate and sodium rosinate is laundry soap (C-III).
(D) The polyethylene glycol ester of stearic acid is a non-ionic detergent, used in dishwashing (D-I).
Answer: (B). | ||||
| 81 | Solid StatePoint defects | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on point defects, from Solid State. NCERT: No matching NCERT chapter Let x be the number of $Fe^{2+}$ per formula unit, so $0.93 - x$ are $Fe^{3+}$. Charge balance with one $O^{2-}$: $2x + 3(0.93 - x) = 2$, so $x = 0.79$.
Percentage of the iron ions that are $Fe^{2+}$: $\frac{0.79}{0.93} \times 100 = 84.9\% \approx 85$.
Answer: 85. | ||||
| 82 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $\Delta x \cdot m\Delta v = \frac{h}{4\pi}$, so $m = \frac{h}{4\pi\,\Delta x\,\Delta v} = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 10^{-7} \times 2.4 \times 10^{-26}} = \frac{6.626 \times 10^{-34}}{3.016 \times 10^{-32}} = 0.0220$ kg.
That is 21.97 g, i.e. 22 g.
Answer: 22. | ||||
| 83 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\Delta T_b = K_b m$, and the molality is the same in both solvents (same solute mass, same solvent mass). So $\Delta T_b(A) : \Delta T_b(B) = 1 : 8$, giving $y = 8$ (with $x = 1$).
Answer: 8. | ||||
| 84 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium At equilibrium: [NO] = 0.4, so 0.4 mol NOCl reacted: [NOCl] = 1.6 and $[Cl_2]$ = 0.2 mol/L.
$K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2} = \frac{0.16 \times 0.2}{2.56} = 0.0125 = 125 \times 10^{-4}$.
Answer: 125. | ||||
| 85 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Kohlrausch's law: $\Lambda^\circ_{AgI} = \Lambda^\circ_{AgNO_3} + \Lambda^\circ_{NaI} - \Lambda^\circ_{NaNO_3} = 13.3 + 12.7 - 12.0 = 14.0$ mS $m^2$ $mol^{-1}$.
Answer: 14. | ||||
| 86 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Compare with $\ln k = \ln A - \frac{E_a}{RT}$: $\frac{E_a}{R} = 2.0 \times 10^4$ K.
$E_a = 2.0 \times 10^4 \times 8.3 = 1.66 \times 10^5$ J $mol^{-1}$ = 166 kJ $mol^{-1}$.
Answer: 166. | ||||
| 87 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements (A) Cu(II) is $3d^9$: one unpaired electron in every complex, so always paramagnetic. Correct.
(B) Cu(I) is $3d^{10}$: no d–d transitions, so its complexes are generally colourless. Correct.
(C) Cu(I) is unstable in aqueous solution and is readily oxidised to Cu(II) (it also disproportionates, $2Cu^+ \to Cu^{2+} + Cu$). Correct.
(D) Fehling's solution contains Cu(II) complexed by tartrate; Cu(I) appears only in the red $Cu_2O$ precipitate formed. Incorrect.
Three statements are correct.
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 $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$.
$Mn^{2+}$ is $3d^5$ with 5 unpaired electrons: $\mu = \sqrt{5(5 + 2)} = \sqrt{35} = 5.92 \approx 6$ B.M.
Answer: 6. | ||||
| 89 | 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 Equal moles (0.15) of $A_2B$ and $AB_3$ have equal mass, so their molar masses are equal: $2A + B = A + 3B$, giving $A = 2B$.
Answer: 2. | ||||
| 90 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques 1,2-Dimethylcyclopentane: cis (a meso form, achiral) and trans (a pair of enantiomers): 3 stereoisomers.
1,3-Dimethylcyclopentane: likewise cis (meso) and a trans pair of enantiomers: 3 stereoisomers.
1,1-Dimethylcyclopentane has no stereocentre and shows no stereoisomerism.
Total stereoisomers $= 3 + 3 = 6$.
Answer: 6. | ||||
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