JEE (Main) · 2023
JEE (Main) 2023 — 10 April, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Three Dimensional Geometry3
- Matrices and Determinants2
- Binomial Theorem2
12 easy · 16 medium · 2 hard
Physics
30 Q
- Oscillations and Waves2
- Gravitation2
- Properties of Bulk Matter2
23 easy · 7 medium · 0 hard
Chemistry
30 Q
- Some Basic Concepts of Chemistry2
- s-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 | 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 Write $a = \tan 1^\circ$, $b = \log_e 123$, $c = \log_e 1234$, so $f(x) = \frac{ax + b}{cx - a}$.
Then $f(f(x)) = \frac{a\frac{ax+b}{cx-a} + b}{c\frac{ax+b}{cx-a} - a} = \frac{a^2x + ab + bcx - ab}{acx + bc - acx + a^2} = \frac{(a^2 + bc)x}{a^2 + bc} = x$.
So $f(f(x)) + f\left(f\left(\frac{4}{x}\right)\right) = x + \frac{4}{x} \ge 2\sqrt{x \cdot \frac{4}{x}} = 4$ for $x > 0$ (AM ≥ GM), with equality at $x = 2$.
Least value = 4.
Answer: (C). | ||||
| 2 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $w = 2z = u + iv$ with $u = 2x$, $v = 2y$. Then $\frac{w - 3i}{w + i} = \frac{(w - 3i)(\bar{w} - i)}{|w + i|^2}$.
Numerator: $|w|^2 - iw - 3i\bar{w} - 3$. Its real part is $u^2 + v^2 - 3 + v - 3v = u^2 + v^2 - 2v - 3$.
Purely imaginary means this is 0: $4x^2 + 4y^2 - 4y - 3 = 0$.
With $x = -y^2$: $4y^4 + 4y^2 - 4y - 3 = 0$, so $y^4 + y^2 - y = \frac{3}{4}$.
Answer: (C). | ||||
| 3 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\Delta = \begin{vmatrix}2&-1&3\\3&2&-1\\4&5&\alpha\end{vmatrix} = 2(2\alpha + 5) + (3\alpha + 4) + 3(15 - 8) = 7\alpha + 35$, which is 0 only for $\alpha = -5$.
So for $\alpha \ne -5$ the solution is unique whatever $\beta$ is: (B) is correct, and (D) ($\alpha = -6$) cannot have infinitely many solutions.
For $\alpha = -5$: the third left side is $-1\cdot(2x - y + 3z) + 2(3x + 2y - z) = 4x + 5y - 5z$, so the system is consistent only if $\beta = -5 + 14 = 9$.
Hence $\alpha = -5, \beta = 9$ gives infinitely many solutions (C correct) and $\alpha = -5, \beta = 8$ is inconsistent (A correct).
The statement that is NOT correct is (D).
Answer: (D). | ||||
| 4 | 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 $|3A| = 3^3 \cdot 2 = 54$. Let $B = 54A^2$: $|B| = 54^3 |A|^2 = 54^3 \cdot 4$.
For a $3 \times 3$ matrix $|adj\,B| = |B|^2$, so $|3\,adj\,B| = 3^3 |B|^2 = 27 \cdot 54^6 \cdot 16$.
With $54 = 2 \cdot 3^3$: $3^3 \cdot 2^6 \cdot 3^{18} \cdot 2^4 = 2^{10} \cdot 3^{21} = 6^{10} \cdot 3^{11}$.
Answer: (D). | ||||
| 5 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem In $\left(ax - \frac{1}{bx^2}\right)^{13}$, $T_{r+1} = \binom{13}{r}(ax)^{13-r}\left(-\frac{1}{bx^2}\right)^r$ has power $13 - 3r$. For $x^7$, $r = 2$: coefficient $\binom{13}{2}\frac{a^{11}}{b^2} = \frac{78a^{11}}{b^2}$.
In $\left(ax + \frac{1}{bx^2}\right)^{13}$, $x^{-5}$ needs $13 - 3r = -5$, $r = 6$: coefficient $\binom{13}{6}\frac{a^7}{b^6} = \frac{1716a^7}{b^6}$.
Equating: $78a^{11}/b^2 = 1716a^7/b^6$, so $a^4b^4 = \frac{1716}{78} = 22$.
Answer: (B). | ||||
| 6 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $a^2(1 + r^2 + r^4) = 33033 = 3 \cdot 7 \cdot 11^2 \cdot 13$.
The only square factors are 1 and $11^2$. With $a = 1$, $r^4 + r^2 = 33032$ has no integer solution ($r^2 \approx 181.2$).
With $a = 11$: $1 + r^2 + r^4 = 273$, $r^4 + r^2 - 272 = 0$, $(r^2 - 16)(r^2 + 17) = 0$, so $r = 4$.
Terms 11, 44, 176; sum $= 11 \cdot 21 = 231$.
Answer: (A). | ||||
| 7 | 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 Cut squares of side $x$: $V = x(30 - 2x)^2$, $0 < x < 15$.
$V' = (30 - 2x)^2 - 4x(30 - 2x) = (30 - 2x)(30 - 6x) = 0$ gives $x = 5$ (a maximum; $x = 15$ gives zero volume).
Surface area of the open box = sheet minus the four corners $= 30^2 - 4 \cdot 5^2 = 900 - 100 = 800$ cm$^2$.
Answer: (D). | ||||
| 8 | Integral CalculusIntegral as an antiderivative | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{d}{dx}\left(\cos x\, e^{\sin^2 x}\right) = -\sin x\, e^{\sin^2 x} + \cos x\, e^{\sin^2 x}\cdot 2\sin x\cos x = e^{\sin^2 x}(\cos x\sin 2x - \sin x)$.
So $I(x) = \cos x\, e^{\sin^2 x} + C$; $I(0) = 1$ gives $C = 0$.
$I\left(\frac{\pi}{3}\right) = \frac{1}{2}e^{3/4}$.
Answer: (D). | ||||
| 9 | 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 Differentiate: $2xf + x^2f' - 1 = 4xf$, so $x^2f' - 2xf = 1$, i.e. $f' - \frac{2}{x}f = \frac{1}{x^2}$.
Integrating factor $\frac{1}{x^2}$: $\left(\frac{f}{x^2}\right)' = \frac{1}{x^4}$, so $\frac{f}{x^2} = -\frac{1}{3x^3} + C$.
$f(1) = \frac{2}{3}$: $C = 1$, so $f(x) = x^2 - \frac{1}{3x}$.
$f(3) = 9 - \frac{1}{9} = \frac{80}{9}$ and $18f(3) = 160$.
Answer: (B). | ||||
| 10 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $y = vx$: $v + xv' = \frac{1 + v^2}{2v}$, so $xv' = \frac{1 - v^2}{2v}$ and $\frac{2v\,dv}{1 - v^2} = \frac{dx}{x}$.
$-\ln|1 - v^2| = \ln x + c$, i.e. $x(1 - v^2) = k$, or $x^2 - y^2 = kx$.
$y(2) = 0$: $4 = 2k$, $k = 2$. So $y^2 = x^2 - 2x$.
At $x = 8$: $y^2 = 48$, $y = \pm 4\sqrt{3}$. Among the options, $4\sqrt{3}$.
Answer: (A). | ||||
| 11 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $\frac{x^2}{9} + y^2 = 1$: A = (3, 0), B = (0, 1). The major axis runs from $(-3, 0)$ to $(3, 0)$, so C: $x^2 + y^2 = 9$.
Line AB: $x = 3 - 3y$. Substituting: $(3 - 3y)^2 + y^2 = 9$, $10y^2 - 18y = 0$, so $y = 0$ (A) or $y = \frac{9}{5}$, $x = -\frac{12}{5}$. P = $\left(-\frac{12}{5}, \frac{9}{5}\right)$.
Area of $\triangle$AOP $= \frac{1}{2}\cdot OA \cdot (\text{height of P}) = \frac{1}{2}\cdot 3 \cdot \frac{9}{5} = \frac{27}{10}$.
$m - n = 27 - 10 = 17$.
Answer: (C). | ||||
| 12 | 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 With centre O, OA = OB = AB = $\lambda$: triangle OAB is equilateral, so the midpoint M of AB is at distance $\frac{\sqrt{3}}{2}\lambda$ from O, and OM ⟂ AB.
The point P with AP : PB = 2 : 3 has AP $= \frac{2}{5}\lambda$, so PM $= \frac{\lambda}{2} - \frac{2\lambda}{5} = \frac{\lambda}{10}$.
$OP^2 = \frac{3}{4}\lambda^2 + \frac{1}{100}\lambda^2 = \frac{76}{100}\lambda^2$, so OP $= \frac{2\sqrt{19}}{10}\lambda = \frac{\sqrt{19}}{5}\lambda$, a constant.
The locus is a circle about O of radius $\frac{\sqrt{19}}{5}\lambda$.
Answer: (B). | ||||
| 13 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct 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 Third vertex C = 3G − A − B = (6 − 2 − 0, 3 − 4 + 2, −3 − 6 + 5) = (4, 1, −4).
For the plane $x + 2y + 4z - 11 = 0$ at C: $4 + 2 - 16 - 11 = -21$; $1 + 4 + 16 = 21$.
Image $= C - 2\cdot\frac{-21}{21}(1, 2, 4) = (4, 1, -4) + (2, 4, 8) = (6, 5, 4)$.
$\alpha\beta + \beta\gamma + \gamma\alpha = 30 + 20 + 24 = 74$.
Answer: (C). | ||||
| 14 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line: $(3t - 3,\ t - 2,\ 1 - 2t)$. On the plane: $3t - 3 + t - 2 + 1 - 2t = 2$, so $2t = 6$, $t = 3$: P = (6, 1, −5).
Distance from $3x - 4y + 12z - 32 = 0$: $\frac{|18 - 4 - 60 - 32|}{\sqrt{9 + 16 + 144}} = \frac{78}{13} = 6$. So $q = 6$, $2q = 12$.
Sum 18, product 72: $x^2 - 18x + 72 = 0$.
Answer: (D). | ||||
| 15 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Points $(-2, 0, 5)$ and $(4, 1, -3)$; directions $\vec{b_1} = (1, -2, 2)$, $\vec{b_2} = (1, 2, 0)$.
$\vec{b_1}\times\vec{b_2} = (0 - 4,\ 2 - 0,\ 2 + 2) = (-4, 2, 4)$, of length 6.
Joining vector $(6, 1, -8)$; dot product $-24 + 2 - 32 = -54$.
Shortest distance $= \frac{54}{6} = 9$.
Answer: (D). | ||||
| 16 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\overrightarrow{AB} = (4, 3, 1)$, $\overrightarrow{AC} = (-2, 1, 2)$.
$\overrightarrow{AB}\times\overrightarrow{AC} = (6 - 1,\ -2 - 8,\ 4 + 6) = (5, -10, 10)$, of length 15.
Projection of $\overrightarrow{OP} = (-1, -2, 3)$: $\frac{-5 + 20 + 30}{15} = 3$.
Answer: (C). | ||||
| 17 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra Take O as origin, radius 1, P = (1, 0), Q = (0, 1). R, the midpoint of the arc, is at 45°: $\vec{v} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.
$(0, 1) = \alpha(1, 0) + \beta\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$: $\frac{\beta}{\sqrt{2}} = 1$, so $\beta = \sqrt{2}$; $\alpha + 1 = 0$, so $\alpha = -1$.
Roots $-1$ and $\beta^2 = 2$: sum 1, product $-2$: $x^2 - x - 2 = 0$.
Answer: (C). | ||||
| 18 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability N runs from 2 to 12. $N = 2$: $4 < 2$ false; $N = 3$: $8 < 6$ false; $N = 4$: $16 < 24$ true, and once true it stays true (each step multiplies $N!$ by more than 2).
So the event fails only for N = 2 (1 outcome) and N = 3 (2 outcomes): probability $= 1 - \frac{3}{36} = \frac{11}{12}$.
$4m - 3n = 44 - 36 = 8$.
Answer: (C). | ||||
| 19 | 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 With $\theta = \frac{\pi}{33}$: $\cos\theta\cos2\theta\cos4\theta\cos8\theta\cos16\theta = \frac{\sin 32\theta}{2^5\sin\theta}$ (multiply by $\sin\theta$ and use $2\sin x\cos x = \sin 2x$ five times).
$\sin\frac{32\pi}{33} = \sin\left(\pi - \frac{\pi}{33}\right) = \sin\frac{\pi}{33}$, so the product is $\frac{1}{32}$.
$96 \cdot \frac{1}{32} = 3$.
Answer: (C). | ||||
| 20 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $\sim[(p \vee q) \wedge (q \vee \sim r)] = (\sim p \wedge \sim q) \vee (\sim q \wedge r)$ (De Morgan).
Taking $\sim q$ out: $= \sim q \wedge (\sim p \vee r) = ((\sim p) \vee r) \wedge (\sim q)$.
Answer: (B). | ||||
| 21 | 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 $x = \ln a$, $y = \ln b$, $z = \ln c$, $k = \ln 2$. The conditions are $x(k + x) = y(y + z)$ and $ky = xz$.
From the second, $k = \frac{xz}{y}$ ($y \ne 0$, else $b = 1$ and then $c = 1$). Substituting: $\frac{x^2z}{y} + x^2 = y^2 + yz$, i.e. $\frac{x^2(y + z)}{y} = y(y + z)$, so $(y + z)(x^2 - y^2) = 0$.
Take $y + z = 0$, i.e. $bc = 1$. Then $ky = xz = -xy$ gives $x = -k$, $a = \frac{1}{2}$.
$6a + 5bc = 3 + 5 = 8$.
Answer: 8. | ||||
| 22 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalEasy | |
A easy numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $-6 < n^2 - 10n + 19 < 6$.
Right side: $n^2 - 10n + 13 < 0$, roots $5 \pm 2\sqrt{3} \approx 1.54, 8.46$: $n = 2, 3, \dots, 8$.
Left side: $n^2 - 10n + 25 > 0$, i.e. $(n - 5)^2 > 0$: $n \ne 5$.
Set $= \{2, 3, 4, 6, 7, 8\}$: 6 elements.
Answer: 6. | ||||
| 23 | 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 Total $7! = 5040$.
Containing 153: treat it as one block, 5 objects: $5! = 120$.
Containing 2467: one block, 4 objects: $4! = 24$.
Containing both: the two blocks use all seven digits: $2! = 2$.
Required $= 5040 - 120 - 24 + 2 = 4898$.
Answer: 4898. | ||||
| 24 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Let there be $n$ couples. A match: choose 2 men, $\binom{n}{2}$; choose 2 women who are not their wives, $\binom{n-2}{2}$; pair them into two teams in 2 ways (neither pairing is a couple).
$\binom{n}{2}\binom{n-2}{2}\cdot 2 = 840$, so $n(n - 1)(n - 2)(n - 3) = 1680 = 8\cdot7\cdot6\cdot5$, giving $n = 8$.
Persons $= 2n = 16$.
Answer: 16. | ||||
| 25 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem A term takes $1$ $p$ times, $(-x)$ $q$ times and $2x^3$ $s$ times with $p + q + s = 10$ and $q + 3s = 7$; its coefficient is $\frac{10!}{p!\,q!\,s!}(-1)^q 2^s$.
$s = 0, q = 7, p = 3$: $\frac{10!}{3!\,7!}(-1) = -120$.
$s = 1, q = 4, p = 5$: $\frac{10!}{5!\,4!\,1!}\cdot 2 = 1260 \cdot 2 = 2520$.
$s = 2, q = 1, p = 7$: $\frac{10!}{7!\,1!\,2!}(-1)\cdot 4 = 360 \cdot (-4) = -1440$.
Sum: $-120 + 2520 - 1440 = 960$.
Answer: 960. | ||||
| 26 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Terms $3 + 5k$, $k = 0, \dots, 74$: 75 terms, sum $= \frac{75}{2}(3 + 373) = 14100$.
$3 + 5k$ is divisible by 3 when $k$ is a multiple of 3: $k = 0, 3, \dots, 72$, the terms 3, 18, …, 363 (25 terms), sum $= \frac{25}{2}(3 + 363) = 4575$.
Required sum $= 14100 - 4575 = 9525$.
Answer: 9525. | ||||
| 27 | 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 Piece by piece: on $(-2, -1)$, $f = -2x$; on $[-1, 0)$, $f = -x$; on $[0, 1)$, $f = 0$; on $[1, 2)$, $f = x - 1$. All these values are $\ge 0$, so $|f| = f$.
At $x = -1$: left limit 2, value 1: not continuous (so not differentiable).
At $x = 0$: both sides give 0, continuous; slopes $-1$ and 0: not differentiable.
At $x = 1$: both sides give 0, continuous; slopes 0 and 1: not differentiable.
$m = 1$, $n = 3$ ($x = -1, 0, 1$), $m + n = 4$.
Answer: 4. | ||||
| 28 | 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 $p(x) = 1 - x^2$ (axis on the y-axis, through $(\pm1, 0)$ and $(0, 1)$). The circle has centre $(-1, 1)$, radius 1; it touches the x-axis at $(-1, 0)$ and its rightmost point is $(0, 1)$ — both on the parabola.
For $x < -1$ the parabola is below the x-axis, so no point of the disc is under it. For $-1 \le x \le 0$ the region runs from the lower arc $y = 1 - \sqrt{1 - (x + 1)^2}$ up to $y = 1 - x^2$ (which lies inside the disc).
$A = \int_{-1}^{0}\left[\sqrt{1 - (x + 1)^2} - x^2\right]dx = \frac{\pi}{4} - \frac{1}{3}$ (a quarter of the unit disc, minus $\int_{-1}^0 x^2dx$).
$\pi - 4A = \frac{4}{3}$, so $12(\pi - 4A) = 16$.
Answer: 16. | ||||
| 29 | Conic SectionsParabola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A tangent to $y^2 = 4x$: $y = mx + \frac{1}{m}$, touching at $\left(\frac{1}{m^2}, \frac{2}{m}\right)$.
Tangent to the circle (centre (4, 0), radius 4): $\frac{|4m + 1/m|}{\sqrt{1 + m^2}} = 4$, so $(4m^2 + 1)^2 = 16m^2(1 + m^2)$, $8m^2 = 1$, $m = \pm\frac{1}{2\sqrt{2}}$.
Take $m = \frac{1}{2\sqrt{2}}$: P = $(8, 4\sqrt{2})$; the line is $x - 2\sqrt{2}y + 8 = 0$.
Q is the foot of the perpendicular from (4, 0): $Q = (4, 0) - \frac{4 + 8}{9}(1, -2\sqrt{2}) = \left(\frac{8}{3}, \frac{8\sqrt{2}}{3}\right)$.
$PQ^2 = \left(\frac{16}{3}\right)^2 + \left(\frac{4\sqrt{2}}{3}\right)^2 = \frac{256 + 32}{9} = 32$ (the same for $m = -\frac{1}{2\sqrt{2}}$ by symmetry).
Answer: 32. | ||||
| 30 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mid-values 5, 15, 25, 35, 45. Mean: $\frac{10 + 45 + 25x + 175 + 180}{14 + x} = 28$, so $410 + 25x = 392 + 28x$, $x = 6$; N = 20.
$\sum f(m - 28)^2 = 2(529) + 3(169) + 6(9) + 5(49) + 4(289) = 1058 + 507 + 54 + 245 + 1156 = 3020$.
Variance $= \frac{3020}{20} = 151$.
Answer: 151. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 At $t = 0$: $A\sin\delta = \frac{A}{2}$, so $\sin\delta = \frac{1}{2}$: $\delta = \frac{\pi}{6}$ or $\frac{5\pi}{6}$.
Velocity $v = A\omega\cos(\omega t + \delta)$ must be positive at $t = 0$, so $\cos\delta > 0$: $\delta = \frac{\pi}{6}$.
Answer: (D). | ||||
| 32 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Matching listEasy | |
A easy matching list question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory A monoatomic molecule has only 3 translational degrees of freedom: (A)-(I).
A rigid diatomic molecule adds 2 rotational ones: (B)-(III).
A non-rigid diatomic molecule also vibrates along its one bond (1 vibrational mode): (C)-(IV).
A non-linear polyatomic molecule has 3 rotational degrees and several vibrational modes: (D)-(II).
Answer: (B). | ||||
| 33 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Isothermal: $P_A = 8P$.
Adiabatic, $\gamma = \frac{5}{3}$: $P_B = P \cdot 8^{5/3} = 32P$.
$\frac{P_B}{P_A} = \frac{32}{8} = 4$.
Answer: (B). | ||||
| 34 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Orbital speed $v = \sqrt{\frac{GM}{r}}$, independent of the satellite's mass.
$\frac{v_1}{v_2} = \sqrt{\frac{3r}{r}} = \sqrt{3}$, i.e. $\sqrt{3} : 1$.
Answer: (B). | ||||
| 35 | 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 $\frac{\Delta P}{P} = 2\frac{\Delta a}{a} + 3\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{1}{2}\frac{\Delta d}{d} = 2(1) + 3(2) + 3 + \frac{1}{2}(4) = 13\%$.
Answer: (D). | ||||
| 36 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | StatementsEasy | |
A easy statements 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 Statement I: in a fluid at rest, pressure depends only on depth, so all points at the same level have the same pressure. True.
Statement II: Pascal's law — pressure applied to an enclosed liquid is transmitted undiminished in all directions. True.
Answer: (A). | ||||
| 37 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g_d = g\left(1 - \frac{d}{R}\right) = \frac{g}{2}$ at $d = \frac{R}{2}$.
Weight $= \frac{200}{2} = 100$ N.
Answer: (D). | ||||
| 38 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | StatementsEasy | |
A easy statements question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane x is the distance from the school. B's line ends at a larger x than A's, so B's home is farther: A lives closer (A true, B false).
B's line is steeper, so B travels faster (E true, D false).
A starts at t = 0 while B starts later and both arrive at about the same instant, so A takes more time, not less (C false).
Correct: (A) and (E).
Answer: (B). | ||||
| 39 | 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 $R = \frac{u^2\sin 2\theta}{g}$. At 15°: $50 = \frac{u^2}{g}\sin 30^\circ$, so $\frac{u^2}{g} = 100$ m.
At 45°: $R = \frac{u^2}{g}\sin 90^\circ = 100$ m.
Answer: (C). | ||||
| 40 | 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 Momentum is conserved: $mv = (m + 2m)V$, so $V = \frac{v}{3}$.
Answer: (C). | ||||
| 41 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The top and bottom wires to the right of the two horizontal capacitors are both joined to the right terminal, so they are at the same potential.
The two vertical capacitors therefore have both plates at the same potential: they are short-circuited and store no charge.
What remains is the top C and the bottom C, each between the left and right terminals: in parallel, $C_{eq} = 2C$.
Answer: (C). | ||||
| 42 | 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 $A_{max} = A_c + A_m = 18$ V, $A_{min} = A_c - A_m = 12$ V.
Ratio $= \frac{18}{12} = \frac{3}{2}$.
Answer: (D). | ||||
| 43 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Maximum zener current $I_{max} = \frac{P}{V_Z} = \frac{1.6}{8} = 0.2$ A.
The largest current flows at the highest input, 10 V (no load is shown): $R_s = \frac{10 - 8}{0.2} = 10$ Ω.
Answer: (A). | ||||
| 44 | 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 Bohr: $L_n = \frac{nh}{2\pi}$. Taking $L$ as the angular momentum in the first orbit, the second orbit has $2L$.
Change $= 2L - L = L$.
Answer: (B). | ||||
| 45 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | StatementsEasy | |
A easy statements question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Statement I: in a series LCR circuit $P = I_{rms}^2R$ is largest when the impedance is smallest, $Z = R$, at resonance. True.
Statement II: $P = V_{rms}I_{rms}\cos\phi$; in a pure resistor $\phi = 0$, so the power factor has its maximum value 1. True.
Answer: (A). | ||||
| 46 | 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 $\lambda = \frac{h}{\sqrt{3mkT}} \propto \frac{1}{\sqrt{T}}$.
Doubling T: $\lambda = \frac{\lambda_1}{\sqrt{2}}$.
Answer: (C). | ||||
| 47 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At first the image is 12 cm behind the mirror, 24 cm from the object.
After the mirror moves 4 cm towards the object, the object is 8 cm away and the image 8 cm behind the mirror, 16 cm from the object.
The image moves $24 - 16 = 8$ cm, towards the object and the mirror (twice the mirror's shift).
Answer: (C). | ||||
| 48 | 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 $E = E_0\sin(\omega t - kx)$, so the energy density $u \propto E^2 \propto \sin^2(\omega t - kx) = \frac{1}{2}[1 - \cos 2(\omega t - kx)]$.
It oscillates at $2\omega$: double the frequency of the wave.
Answer: (C). | ||||
| 49 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | StatementsMedium | |
A medium statements 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 $\frac{\theta}{I} = \frac{NAB}{k} \propto N$: doubling N doubles it. Statement I true.
Voltage sensitivity $\frac{\theta}{V} = \frac{NAB}{kR}$. Doubling the turns also doubles the wire length and so the coil resistance R; the voltage sensitivity stays the same. Statement II false.
Answer: (C). | ||||
| 50 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity The network is symmetric under exchanging a and b, so the top vertex T and the centre O are both at half the applied potential. The 4 Ω between T and O carries no current and can be removed.
Then three paths join a and b in parallel: a–T–b = 8 Ω, a–O–b = 8 Ω, and the 16 Ω.
$\frac{1}{R} = \frac{1}{8} + \frac{1}{8} + \frac{1}{16} = \frac{5}{16}$, so $R = 3.2$ Ω.
Answer: (D). | ||||
| 51 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The steel wire holds both blocks: $T = (2 + 1.14)\times 10 = 31.4$ N.
Area $A = \pi(2\times10^{-3})^2 = 1.256\times10^{-5}$ m$^2$.
$\Delta L = \frac{TL}{AY} = \frac{31.4 \times 1.6}{1.256\times10^{-5}\times 2\times10^{11}} = \frac{50.24}{2.512\times10^{6}} = 2.0\times10^{-5}$ m $= 20\times10^{-6}$ m.
Answer: 20. | ||||
| 52 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\omega = 6$ s$^{-1}$, $k = 0.003$ cm$^{-1}$.
$v = \frac{\omega}{k} = \frac{6}{0.003} = 2000$ cm/s $= 20$ m/s.
Answer: 20. | ||||
| 53 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Volume $\frac{1}{64}$ means radius $\frac{R}{4}$, so $I = \frac{2}{5}MR^2$ becomes $\frac{I}{16}$.
Angular momentum is conserved: $\omega$ becomes $16\omega$, period $\frac{24}{16}$ h.
$x = 16$.
Answer: 16. | ||||
| 54 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Work–energy theorem from the top (speed 22 m/s) to rest at the bottom: $W_{tube} + W_{gravity} = 0 - \frac{1}{2}(1)(22)^2 = -242$ J.
The block falls by $2R = 0.30$ m: $W_{gravity} = 1 \times 10 \times 0.3 = 3$ J.
$W_{tube} = -242 - 3 = -245$ J.
Answer: −245. | ||||
| 55 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Charges $q = 4\pi r^2\sigma$. Potential of X (inside all three): $V_X = \frac{\sigma}{\varepsilon_0}(a - b + c)$.
Potential of Z (outside the others): $V_Z = \frac{\sigma}{\varepsilon_0}\cdot\frac{a^2 - b^2 + c^2}{c}$.
Equal: $c(a - b) + c^2 = a^2 - b^2 + c^2$, so $c(a - b) = (a - b)(a + b)$ and $c = a + b = 5$ cm.
Answer: 5. | ||||
| 56 | 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 $N = \frac{10^{-6}}{60}\times 6\times10^{23} = 10^{16}$ nuclei.
Activity $= \lambda N = 1.5\times10^{-5}\times10^{16} = 1.5\times10^{11} = 15\times10^{10}$ Bq.
Answer: 15. | ||||
| 57 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics After the first polaroid: 16 W m$^{-2}$. With angle $\theta$ between the first two, the angle between the second and third is $90^\circ - \theta$.
$I = 16\cos^2\theta\sin^2\theta = 4\sin^2 2\theta = 3$, so $\sin^2 2\theta = \frac{3}{4}$, $2\theta = 60^\circ$, $\theta = 30^\circ$.
Answer: 30. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current emf $= Bvl = 0.15\times4\times1 = 0.6$ V; current $I = \frac{0.6}{5} = 0.12$ A.
Magnetic force on the rod $= BIl = 0.15\times0.12\times1 = 0.018$ N; at constant speed the applied force equals it.
$F = 18\times10^{-3}$ N.
Answer: 18. | ||||
| 59 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical 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 The solenoid must produce $H = nI$ equal to $2.4\times10^3$ A m$^{-1}$, with $n = \frac{60}{0.15} = 400$ m$^{-1}$.
$I = \frac{2400}{400} = 6$ A.
Answer: 6. | ||||
| 60 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Maximum: all in series, $10\times10 = 100$ Ω. Minimum: all in parallel, $\frac{10}{10} = 1$ Ω.
Ratio $= 100$.
Answer: 100. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry At STP (273.15 K, 1 bar) the molar volume is 22.7 L.
Moles $= \frac{2.8375}{22.7} = 0.125$ mol; molecules $= 0.125 \times 6.022\times10^{23} = 7.527\times10^{22}$.
Answer: (C). | ||||
| 62 | ThermodynamicsHess law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Target = (B) $- \frac{1}{2}$(A): $C + O_2 - CO - \frac{1}{2}O_2 \rightarrow CO_2 - CO_2$, i.e. $C + \frac{1}{2}O_2 \rightarrow CO$.
$\Delta H = -y - \frac{1}{2}(-x) = \frac{x - 2y}{2}$.
Answer: (D). | ||||
| 63 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter Adsorption lowers the surface energy and is always exothermic: $\Delta H_{ads} < 0$ (it has $\Delta S < 0$, so $\Delta H$ must be negative for it to be spontaneous).
Micelle formation is endothermic (slightly positive $\Delta H$) and is driven by the entropy gain as ordered water around the hydrocarbon tails is released: $\Delta H_{mic} > 0$.
Answer: (D). | ||||
| 64 | Isolation of MetalsFroth flotation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on froth flotation, from Isolation of Metals. NCERT: No matching NCERT chapter In froth flotation, pine oils, fatty acids and xanthates act as collectors (they make the ore particles non-wettable), while cresols and aniline are froth stabilisers.
Answer: (C). | ||||
| 65 | 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 Water gas is made by passing steam over coke at about 1270 K ($T_1$).
The water-gas shift reaction is run at about 673 K over an iron chromate catalyst ($T_2$).
So $T_1 > T_2$.
Answer: (A). | ||||
| 66 | 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 $CaO + H_2O \rightarrow Ca(OH)_2$ (exothermic): A is calcium hydroxide, slaked lime, sparingly soluble.
Its aqueous solution (lime water) turns milky with $CO_2$: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3\downarrow$ (B) $+ H_2O$; more $CO_2$ gives soluble $Ca(HCO_3)_2$.
'A' itself is slaked lime; lime water is its solution.
Answer: (C). | ||||
| 67 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter $PbEt_4$ (tetraethyl lead, the old anti-knock agent) exists; $(NH_4)_2BeF_4$ is used to make $BeF_2$; $BeH_2$ is a known polymeric hydride.
Sodium burnt in air gives the peroxide $Na_2O_2$, not the superoxide; superoxides $MO_2$ are formed by K, Rb and Cs. $NaO_2$ is the one taken as not existing.
Answer: (C). | ||||
| 68 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | StatementsMedium | |
A medium statements question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements (A) $E^\circ(Fe^{3+}/Fe^{2+}) = +0.77$ V, $E^\circ(Mn^{3+}/Mn^{2+}) = +1.57$ V: iron's is smaller. False.
(B) Oxide ion stabilises high oxidation states (e.g. $Mn_2O_7$, $MnO_4^-$, $CrO_4^{2-}$) through its small size and multiple bonding. True.
(C) $Cr^{2+}$ is a strong reducing agent ($E^\circ(Cr^{3+}/Cr^{2+}) = -0.41$ V) and liberates $H_2$ from dilute acid. True.
(D) $V^{2+}$ is $d^3$: $\mu = \sqrt{15} \approx 3.87$ BM, not 4.4–5.2. False.
Answer: (B). | ||||
| 69 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[Co(NH_3)_6]^{3+}$: $Co^{3+}$, $d^6$; $NH_3$ pairs the electrons, $t_{2g}^6e_g^0$, octahedral, low spin and diamagnetic.
$[CoF_6]^{3-}$ and $[CoCl_6]^{3-}$ are high spin (weak-field ligands, 4 unpaired electrons); $[NiCl_4]^{2-}$ is tetrahedral and paramagnetic.
Answer: (C). | ||||
| 70 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Steel plants produce slag (III); thermal power plants, fly ash (II); fertilizer industries, gypsum (I); paper mills, biodegradable wastes (IV).
Answer: (A). | ||||
| 71 | 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 The less strongly adsorbed component moves faster and elutes first. B, eluted later, is more strongly adsorbed on the stationary phase and travels less: low $R_f$.
Answer: (D). | ||||
| 72 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure trans-But-2-ene, 1,4-dichlorobenzene and benzene are symmetric with zero dipole moment, so (A), (C) and (D) each contain a non-polar compound.
$CH_2Cl_2$ and $CHCl_3$ are both tetrahedral with unequal substituents: both polar.
Answer: (B). | ||||
| 73 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | StatementsMedium | |
A medium statements question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes (A) $S_N2$: the primary bromide (isobutyl) is much less hindered than the tertiary (tert-butyl) one. Correct.
(B) $S_N1$: benzyl bromide gives a resonance-stabilised benzyl cation; 2-phenylethyl bromide would give a primary cation. Correct.
(C) Electrophilic substitution: the $-NO_2$ group strongly deactivates the ring, so chlorobenzene reacts faster. Correct.
(D) Nucleophilic aromatic substitution: a para $-NO_2$ group stabilises the Meisenheimer intermediate, so p-nitrobromobenzene reacts faster than bromobenzene. Correct.
All four are correct.
Answer: (D). | ||||
| 74 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Hot alkaline $KMnO_4$ oxidises every side chain that carries a benzylic hydrogen down to $-COOH$ on the ring, whatever its length.
$-CH_2CH_3 \rightarrow -COOH$; $-CH=CH_2 \rightarrow -COOH$; and $-CH_2COOCH_3$ (benzylic $CH_2$; the ester is hydrolysed in the alkaline medium) $\rightarrow -COOH$.
Acidification ($H_3O^+$) frees the three carboxylic acids: the naphthalene tricarboxylic acid with all three COOH groups on the ring.
Answer: (D). | ||||
| 75 | Alcohols, Phenols and EthersWilliamson synthesis | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on williamson synthesis, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Williamson synthesis needs a good nucleophile and an $S_N2$-reactive halide: sodium phenoxide with methyl bromide gives $PhOCH_3$.
MeOH has no leaving group; aryl bromide does not undergo $S_N2$ with methoxide; benzene with MeBr would need a Lewis acid and gives toluene, not an ether.
Answer: (A). | ||||
| 76 | 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 P: Gabriel synthesis gives only primary aliphatic amines (not arylamines): 2-phenylethylamine $C_6H_5CH_2CH_2NH_2$ fits. This excludes (A) (secondary amine) and (C) (an arylamine).
Q: an insoluble Hinsberg product means a secondary amine. In (D), Q is N,N-dimethylaniline, a tertiary amine that does not react; so (D) fails.
R: diazotisation and coupling with $\beta$-naphthol to a red dye needs a primary aromatic amine: 4-ethylaniline in (B).
Only (B) fits all three tests.
Answer: (B). | ||||
| 77 | PolymersBiodegradable polymers | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on biodegradable polymers, from Polymers. NCERT: No matching NCERT chapter Nylon-2-nylon-6 is a biodegradable polyamide (II); Buna-N is a synthetic rubber (III); urea-formaldehyde resin is thermosetting (I); Dacron is a polyester (IV).
Answer: (B). | ||||
| 78 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Secondary and tertiary structures are held by hydrogen bonds, disulphide (–S–S–) bridges, ionic and van der Waals forces. There is no peroxide (–O–O–) linkage in proteins.
Answer: (B). | ||||
| 79 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | StatementsEasy | |
A easy statements question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Statement I: $Na_2Cr_2O_7$ is hygroscopic, so it cannot be weighed accurately; $K_2Cr_2O_7$ is the primary standard. True.
Statement II: $K_2Cr_2O_7$ is less soluble than $Na_2Cr_2O_7$ (that is how it is crystallised out from the sodium salt with KCl). False.
Answer: (C). | ||||
| 80 | Principles Related to Practical ChemistryPreparation of inorganic and organic compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on preparation of inorganic and organic compounds, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter On prolonged heating in air, $Fe^{2+}$ is oxidised to $Fe^{3+}$ (the solution turns yellow), spoiling Mohr's salt. Heating is kept short to prevent this oxidation.
Answer: (D). | ||||
| 81 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Boyle's law: $P_1V_1 = P_2V_2$ with $V_2 = 0.6V_1$.
$P_2 = \frac{940.3}{0.6} = 1567.2$ mm Hg ≈ 1567.
Answer: 1567. | ||||
| 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 (A) A black body absorbs and emits electromagnetic radiation. Correct.
(B) Its spectrum depends only on its temperature. Correct.
(C) The intensity–frequency curve rises to a maximum and falls. Correct.
(D) The maximum shifts to higher frequency as the temperature rises (Wien). Correct.
No statement is incorrect.
Answer: 0. | ||||
| 83 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $N_3^-$: linear (16 valence electrons, like $CO_2$). $I_3^-$: linear (3 lone pairs on the central I, $sp^3d$).
$NO_2^-$, $O_3$ and $SO_2$: central atom with 2 bond pairs and 1 lone pair: bent.
Answer: 3. | ||||
| 84 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $HA \rightleftharpoons H^+ + A^-$: $i = 1 + \alpha = 1.3$.
$\Delta T_f(observed) = i\,\Delta T_f(theoretical) = 1.3\,\Delta T_f(theoretical)$: 30% more than expected.
Answer: 30. | ||||
| 85 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium (A) Correct: equilibrium in physical processes is attained only in a closed system at a given temperature.
(B) Correct: it is dynamic, with forward and reverse processes at equal rates.
(C) Incorrect: the measurable properties become constant, not equal to each other.
(D) Correct: at a given temperature a solid has a fixed solubility.
Answer: 3. | ||||
| 86 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Add free energies, not potentials: $FeO_4^{2-} \rightarrow Fe^{3+}$ is a 3-electron change (Fe +6 to +3), $Fe^{3+} \rightarrow Fe^{2+}$ a 1-electron change.
$4E^\circ = 3(2.2) + 1(0.70) = 7.3$, so $E^\circ = 1.825$ V $= 1825 \times 10^{-3}$ V.
Answer: 1825. | ||||
| 87 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Parallel first-order reactions: $k = k_1 + k_2$, so $\frac{1}{t_{1/2}} = \frac{1}{12} + \frac{1}{3} = \frac{5}{12}$.
$t_{1/2} = 2.4$ min ≈ 2 min.
Answer: 2. | ||||
| 88 | 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 $Na_2O + H_2O \rightarrow 2NaOH$: X = NaOH, 1 O.
$Cl_2O_7 + H_2O \rightarrow 2HClO_4$: Y = $HClO_4$, 4 O.
Total = 5.
Answer: 5. | ||||
| 89 | 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 I has 7 valence electrons. In $IF_5$: 5 bond pairs + 1 lone pair (square pyramidal). In $IF_7$: 7 bond pairs, no lone pair (pentagonal bipyramidal).
Sum = 1 + 0 = 1.
Answer: 1. | ||||
| 90 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $K_4[Fe(CN)_6]$: $Fe^{2+}$, $d^6$; $CN^-$ is a strong-field ligand, so the configuration is $t_{2g}^6e_g^0$.
Six electrons in $t_{2g}$ = 3 pairs.
Answer: 3. | ||||
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