JEE (Main) · 2023
JEE (Main) 2023 — 15 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
- Integral Calculus3
- Three Dimensional Geometry3
- Sets, Relations and Functions2
6 easy · 19 medium · 5 hard
Physics
30 Q
- Oscillations and Waves2
- Properties of Bulk Matter2
- Gravitation2
25 easy · 5 medium · 0 hard
Chemistry
30 Q
- Coordination Compounds3
- Structure of Atom2
- s-Block Elements2
21 easy · 8 medium · 1 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 FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Each part must be defined.
Logarithm: $4x^2 + 11x + 6 = (4x + 3)(x + 2) > 0$, so $x < -2$ or $x > -\frac{3}{4}$.
$\sin^{-1}$: $-1 \le 4x + 3 \le 1$, so $-1 \le x \le -\frac{1}{2}$.
$\cos^{-1}$: $-3 \le 10x + 6 \le 3$, so $-\frac{9}{10} \le x \le -\frac{3}{10}$.
Intersection: $-\frac{3}{4} < x \le -\frac{1}{2}$, i.e. $(\alpha, \beta] = \left(-\frac{3}{4}, -\frac{1}{2}\right]$.
$36|\alpha + \beta| = 36 \cdot \frac{5}{4} = 45$.
Answer: (A). | ||||
| 2 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Case $x \ge 0$: $x^2 - 5(x + 2) + 6 = 0$, i.e. $x^2 - 5x - 4 = 0$, $x = \frac{5 \pm \sqrt{41}}{2}$. Only $\frac{5 + \sqrt{41}}{2} \approx 5.70$ is $\ge 0$: one root.
Case $-2 \le x < 0$: $-x^2 - 5(x + 2) + 6 = 0$, i.e. $x^2 + 5x + 4 = 0$, $x = -1$ or $-4$. Only $x = -1$ lies in $[-2, 0)$: one root.
Case $x < -2$: $-x^2 + 5(x + 2) + 6 = 0$, i.e. $x^2 - 5x - 16 = 0$, $x = \frac{5 \pm \sqrt{89}}{2}$. $\frac{5 - \sqrt{89}}{2} \approx -2.22 < -2$ is valid; the other is positive: one root.
Total: 3 real roots.
Answer: (B). | ||||
| 3 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z = 3 + iy$. Then $z - \bar{z} + z\bar{z} = 2iy + 9 + y^2$ and $2 - 3z + 5\bar{z} = 2 - 9 - 3iy + 15 - 5iy = 8(1 - iy)$.
$\frac{9 + y^2 + 2iy}{8(1 - iy)} = \frac{(9 + y^2 + 2iy)(1 + iy)}{8(1 + y^2)}$, whose real part is $\frac{9 + y^2 - 2y^2}{8(1 + y^2)} = \frac{9 - y^2}{8(1 + y^2)}$.
With $t = y^2 \ge 0$: $\frac{9 - t}{8(1 + t)} = \frac{1}{8}\left(\frac{10}{1 + t} - 1\right)$, which decreases from $\frac{9}{8}$ (at $t = 0$) towards $-\frac{1}{8}$ (never reached).
So the set is $\left(-\frac{1}{8}, \frac{9}{8}\right]$ and $24(\beta - \alpha) = 24 \cdot \frac{10}{8} = 30$.
Answer: (B). | ||||
| 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 From $n = 22 - 4m$: $17m + 88 - 16m = 93$, so $m = 5$, $n = 2$, and $|A| = 3$ with order 5.
$|mA| = 5^5 \cdot 3$. For a matrix $B$ of order 5, $|\mathrm{adj}(\mathrm{adj}\,B)| = |B|^{(5-1)^2} = |B|^{16}$.
$\det(2\,\mathrm{adj}(\mathrm{adj}(5A))) = 2^5 \cdot (5^5 \cdot 3)^{16} = 2^5 \cdot 3^{16} \cdot 5^{80}$.
Write $2^5 \cdot 3^{16} = 6^5 \cdot 3^{11}$: $c = 5$, $a = 11$, $b = 80$.
$a + b + c = 96$.
Answer: (B). | ||||
| 5 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Solve the first three equations. Second minus first: $y + 16z = 2$. Third minus twice the first: $-3y + 23z = -6$.
Substituting $y = 2 - 16z$: $-6 + 48z + 23z = -6$, so $z = 0$, $y = 2$, and $x = 2y - 9z - 7 = -3$.
The fourth equation must hold: $\lambda = -3(-3) + 2 + 0 = 11$.
Distance of $(-3, 2, 0)$ from $2x - 2y + z - 11 = 0$: $\frac{|-6 - 4 + 0 - 11|}{\sqrt{4 + 4 + 1}} = \frac{21}{3} = 7$.
Answer: (A). | ||||
| 6 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Remainders on division by 3: $1 \to 1$, $3 \to 0$, $5 \to 2$, $8 \to 2$. So one digit each of remainder 0 and 1, and two digits of remainder 2.
The digit sum is divisible by 3 when the remainders are (0, 0, 0), (1, 1, 1), (2, 2, 2) or (0, 1, 2) in some order.
(0,0,0): $1$ number. (1,1,1): $1$. (2,2,2): $2^3 = 8$. (0,1,2): $3! \times 1 \times 1 \times 2 = 12$.
Total $= 1 + 1 + 8 + 12 = 22$.
Answer: (D). | ||||
| 7 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $p_1 = 10a^9 b = 20$, so $a^9 b = 2$; with natural numbers, $a = 1$, $b = 2$.
$p_2 = 10a^9 c + \binom{10}{2} a^8 b^2 = 10c + 45 \cdot 4 = 10c + 180 = 210$, so $c = 3$.
$2(a + b + c) = 2 \cdot 6 = 12$.
Answer: (C). | ||||
| 8 | Sequences and SeriesArithmetic and geometric means | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic and geometric means, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the numbers be $a$ and $b$. Then $A_1 + A_2 = a + b$.
With common ratio $r$ ($r^4 = \frac{b}{a}$): $G_1 = ar$, $G_2 = ar^2$, $G_3 = ar^3$, and $G_1 G_3 = a^2 r^4 = ab$.
$G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2 = a^4(r^4 + r^8 + r^{12} + r^8) = a^4 r^4 (1 + r^4)^2 = ab \cdot (a + ar^4)^2 = ab(a + b)^2$.
So it equals $(A_1 + A_2)^2 G_1 G_3$.
Answer: (C). | ||||
| 9 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability On $[0, 1)$, $[x] = 0$: the three are $1 + x$, $2 + x$, $x$; the maximum is $2 + x$.
On $[1, 2)$, $[x] = 1$: the three are $2 + x$, $2 + x$, $x + 2$; the maximum is $2 + x$.
At $x = 2$: $5$, $4$, $6$; $f(2) = 6$.
So $f(x) = 2 + x$ on $[0, 2)$ and $f(2) = 6 \ne \lim_{x \to 2^-} f(x) = 4$.
$f$ is discontinuous only at $x = 2$: $m = 1$. On $(0, 2)$ $f(x) = 2 + x$ is differentiable everywhere: $n = 0$.
$(m + n)^2 + 2 = 3$.
Answer: (B). | ||||
| 10 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Replace $x$ by $1 - x$: $5 + 2(1 - x) - 2(1 - x)^2 = 5 + 2x - 2x^2$ (unchanged) and $e^{2 - 4x}$ becomes $e^{4x - 2}$. Since $\frac{1}{1 + e^{u}} + \frac{1}{1 + e^{-u}} = 1$, adding the two forms gives
$2I = \int_0^1 \frac{dx}{5 + 2x - 2x^2} = \frac{1}{2}\int_0^1 \frac{dx}{\frac{11}{4} - \left(x - \frac{1}{2}\right)^2}$.
With $a = \frac{\sqrt{11}}{2}$: $\frac{1}{2} \cdot \frac{1}{2a}\left[\log_e\frac{a + u}{a - u}\right]_{-1/2}^{1/2} = \frac{1}{2\sqrt{11}} \cdot 2\log_e\frac{\sqrt{11} + 1}{\sqrt{11} - 1}$.
So $I = \frac{1}{2\sqrt{11}}\log_e\frac{\sqrt{11} + 1}{\sqrt{11} - 1} = \frac{1}{\sqrt{11}}\log_e\sqrt{\frac{(\sqrt{11} + 1)^2}{10}} = \frac{1}{\sqrt{11}}\log_e\frac{\sqrt{11} + 1}{\sqrt{10}}$.
Hence $\alpha = \sqrt{11}$, $\beta = \sqrt{10}$, and $\alpha^4 - \beta^4 = 121 - 100 = 21$.
Answer: (D). | ||||
| 11 | 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 Let $L = \log_e(y + 2)$. The equation is linear in $x$:
$\frac{dx}{dy} + \frac{x}{2(y + 2)L} = \frac{2L - 4}{2(y + 2)L}$.
Integrating factor: $\exp\int\frac{dy}{2(y + 2)L} = e^{\frac{1}{2}\log_e L} = \sqrt{L}$.
$\frac{d}{dy}\left(x\sqrt{L}\right) = \frac{L - 2}{(y + 2)\sqrt{L}}$. With $t = L$, $dt = \frac{dy}{y + 2}$: $x\sqrt{L} = \int\frac{t - 2}{\sqrt{t}}dt = \frac{2}{3}L^{3/2} - 4\sqrt{L} + C$.
At $y = e^4 - 2$: $L = 4$, $x = 1$: $2 = \frac{16}{3} - 8 + C$, so $C = \frac{14}{3}$.
At $y = e^9 - 2$: $L = 9$: $3x = 18 - 12 + \frac{14}{3} = \frac{32}{3}$, so $x = \frac{32}{9}$.
Answer: (D). | ||||
| 12 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines Altitude from A is perpendicular to $BC$, direction $(5, 3)$: $5(x - 3) + 3(y + 7) = 0$, i.e. $5x + 3y + 6 = 0$.
Altitude from B is perpendicular to $AC$, direction $(1, 12)$: $(x + 1) + 12(y - 2) = 0$, i.e. $x + 12y - 23 = 0$.
Solving: $x = 23 - 12y$, $5(23 - 12y) + 3y + 6 = 0$, so $y = \frac{121}{57}$, $x = -\frac{141}{57}$.
$9\alpha - 6\beta + 60 = \frac{-1269 - 726}{57} + 60 = -35 + 60 = 25$.
Answer: (A). | ||||
| 13 | CirclesRadius and centre | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on radius and centre, from Circles. NCERT: Class 11: Conic Sections First circle: centre $\left(9, \frac{15}{2}\right)$, $r_1^2 = 81 + \frac{225}{4} - 131 = \frac{25}{4}$, $r_1 = \frac{5}{2}$.
Second circle: centre $(3, 3)$, $r_2^2 = 9 + 9 + 7 = 25$, $r_2 = 5$.
Distance between centres $= \sqrt{36 + \frac{81}{4}} = \frac{15}{2} = r_1 + r_2$: the circles touch externally, so there are 3 common tangents.
Answer: (C). | ||||
| 14 | 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 Two vectors in the plane: $(10, 5, 7)$ and $(10, 0, 4)$. Their cross product is $(20, 30, -50)$, so a normal is $(2, 3, -5)$.
Plane: $2(x + 1) + 3(y + 2) - 5(z + 3) = 0$, i.e. $2x + 3y - 5z - 7 = 0$.
For $P$: $6 - 6 + 45 - 7 = 38$. The foot is $Q = P - \frac{38}{38}(2, 3, -5) = (1, -5, -4)$.
$OQ = \sqrt{1 + 25 + 16} = \sqrt{42}$.
Answer: (D). | ||||
| 15 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Points and directions: $(\lambda, 3, -6)$, $\vec{b}_1 = (0, 4, 1)$; $(-\lambda, 0, 6)$, $\vec{b}_2 = (3, -4, 0)$.
$\vec{b}_1 \times \vec{b}_2 = (4, 3, -12)$, of length 13. Joining vector: $(-2\lambda, -3, 12)$.
Shortest distance $= \frac{|-8\lambda - 9 - 144|}{13} = \frac{|8\lambda + 153|}{13} = 13$, so $8\lambda + 153 = \pm 169$.
$\lambda = 2$ or $\lambda = -\frac{322}{8}$. Sum $= 2 - \frac{161}{4} = -\frac{153}{4}$.
$8\left|\sum\lambda\right| = 8 \cdot \frac{153}{4} = 306$.
Answer: (C). | ||||
| 16 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Coplanar means the scalar triple product is zero:
$\begin{vmatrix}\lambda & -1 & 1\\ 1 & 2 & \mu\\ 3 & -4 & 5\end{vmatrix} = \lambda(10 + 4\mu) + (5 - 3\mu) + (-4 - 6) = 10\lambda + 4\lambda\mu - 3\mu - 5 = 0$.
Put $\lambda = \mu + 5$: $10\mu + 50 + 4\mu^2 + 20\mu - 3\mu - 5 = 4\mu^2 + 27\mu + 45 = 0$, so $\mu = -3$ or $\mu = -\frac{15}{4}$.
Pairs: $(2, -3)$ with $\lambda^2 + \mu^2 = 13$, and $\left(\frac{5}{4}, -\frac{15}{4}\right)$ with $\lambda^2 + \mu^2 = \frac{250}{16} = \frac{125}{8}$.
$80\left(13 + \frac{125}{8}\right) = 1040 + 1250 = 2290$.
Answer: (C). | ||||
| 17 | Vector AlgebraAddition of vectors | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on addition of vectors, from Vector Algebra. NCERT: Class 12: Vector Algebra With position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$:
$(\vec{b} - \vec{a}) - (\vec{c} - \vec{b}) + (\vec{d} - \vec{a}) - (\vec{c} - \vec{d}) = 2(\vec{b} + \vec{d}) - 2(\vec{a} + \vec{c}) = 4\left(\frac{\vec{b} + \vec{d}}{2} - \frac{\vec{a} + \vec{c}}{2}\right) = 4\,\overrightarrow{EF} = -4\,\overrightarrow{FE}$.
So $k = -4$.
Answer: (A). | ||||
| 18 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability By total probability, $P = \frac{1}{6}\sum_{k=1}^{6}\frac{\binom{6}{k}}{\binom{10}{k}}$.
Terms: $\frac{6}{10} = \frac{3}{5}$, $\frac{15}{45} = \frac{1}{3}$, $\frac{20}{120} = \frac{1}{6}$, $\frac{15}{210} = \frac{1}{14}$, $\frac{6}{252} = \frac{1}{42}$, $\frac{1}{210}$.
Over 210: $\frac{126 + 70 + 35 + 15 + 5 + 1}{210} = \frac{252}{210} = \frac{6}{5}$.
$P = \frac{1}{6} \cdot \frac{6}{5} = \frac{1}{5}$.
Answer: (A). | ||||
| 19 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Recorded: $\sum x = 200$, $\sum x^2 = 10(64 + 400) = 4640$.
Corrected: $\sum x = 200 - 50 + 40 = 190$, mean $19$; $\sum x^2 = 4640 - 2500 + 1600 = 3740$.
Variance $= \frac{3740}{10} - 19^2 = 374 - 361 = 13$.
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 $q \wedge \sim(p \wedge q) = q \wedge (\sim p \vee \sim q) = q \wedge \sim p$.
So the statement is $p \wedge q \wedge \sim p$, a contradiction; its negation is a tautology.
(A) $\sim(p \vee q)$ and (B) $p \vee q$ are not tautologies. (C) $\sim(p \wedge q) \wedge q = \sim p \wedge q$, not a tautology. (D) $\sim(p \wedge q) \vee p = \sim p \vee \sim q \vee p$, always true.
Answer: (D). | ||||
| 21 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $2a + 3b$ lies between 5 and 20. Values of $4c + 5d$ in that range: $9\,(1,1)$, $13\,(2,1)$, $14\,(1,2)$, $17\,(3,1)$, $18\,(2,2)$, $19\,(1,3)$.
Solutions of $2a + 3b = k$ with $a, b \in A$:
$k = 9$: $(3, 1)$. $k = 13$: $(2, 3)$. $k = 14$: $(4, 2), (1, 4)$. $k = 17$: $(4, 3)$. $k = 18$: $(3, 4)$. $k = 19$: none.
Total $= 1 + 1 + 2 + 1 + 1 + 0 = 6$.
Answer: 6. | ||||
| 22 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The digits come from $0, 1, \ldots, 7$, all different, and 7 is one of them. Let 7 be paired with $x$; the other pair is two different digits from $0$–$6$ (not $x$) with sum $7 + x$.
$x = 0$: sum 7: $\{1,6\}, \{2,5\}, \{3,4\}$ (3). $x = 1$: sum 8: $\{2,6\}, \{3,5\}$ (2). $x = 2$: sum 9: $\{3,6\}, \{4,5\}$ (2). $x = 3$: sum 10: $\{4,6\}$ (1). $x = 4$: sum 11: $\{5,6\}$ (1). $x = 5, 6$: none.
That is 9 choices of the two pairs. Each gives $2$ (which pair comes first) $\times 2 \times 2$ (order inside each pair) $= 8$ codes.
Total $= 9 \times 8 = 72$.
Answer: 72. | ||||
| 23 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $3^n \equiv 3 \pmod 7$ means $3^{n-1} \equiv 1 \pmod 7$. Powers of 3 mod 7: $3, 2, 6, 4, 5, 1$, so the order of 3 is 6 and $n - 1$ must be a multiple of 6: $n \equiv 1 \pmod 6$.
In $[10, 100]$: $13, 19, \ldots, 97$, which is $\frac{97 - 13}{6} + 1 = 15$ numbers.
Answer: 15. | ||||
| 24 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The $n$-th bracket is $\sum_{k=0}^{n}\left(\frac{1}{2}\right)^{n-k}\left(-\frac{1}{3}\right)^{k} = \frac{\left(\frac{1}{2}\right)^{n+1} - \left(-\frac{1}{3}\right)^{n+1}}{\frac{1}{2} + \frac{1}{3}} = \frac{6}{5}\left[\left(\frac{1}{2}\right)^{n+1} - \left(-\frac{1}{3}\right)^{n+1}\right]$.
Summing over $n \ge 1$: $\sum\left(\frac{1}{2}\right)^{n+1} = \frac{1}{2}$ and $\sum\left(-\frac{1}{3}\right)^{n+1} = \frac{1/9}{1 + 1/3} = \frac{1}{12}$.
Sum $= \frac{6}{5}\left(\frac{1}{2} - \frac{1}{12}\right) = \frac{6}{5} \cdot \frac{5}{12} = \frac{1}{2}$.
$\alpha = 1$, $\beta = 2$, $\alpha + 3\beta = 7$.
Answer: 7. | ||||
| 25 | Straight LinesCartesian coordinates and distance formula | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines $AB = \sqrt{5}$ is fixed, so consider $g(t) = BC + CA = \sqrt{t^2 + 16} + \sqrt{(t - 2)^2 + 9}$.
$g(t)$ is the sum of distances from $(t, 0)$ to $(0, -4)$ and to $(2, 3)$, points on opposite sides of the x-axis. It is least where the segment joining them meets the x-axis: $y = -4 + 7s = 0$ gives $s = \frac{4}{7}$, $t = \frac{8}{7}$. So $\beta = \frac{8}{7}$.
$g$ is convex, so its maximum on $[0, 4]$ is at an end: $g(0) = 4 + \sqrt{13} \approx 7.61$, $g(4) = \sqrt{32} + \sqrt{13} \approx 9.26$. So $\alpha = 4$.
$6\alpha + 21\beta = 24 + 24 = 48$.
Answer: 48. | ||||
| 26 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Since $f(0) = 0$, $f(1) = \int_0^1\frac{dx}{(3 + 4x^2)\sqrt{4 - 3x^2}}$.
For $x > 0$ put $x = \frac{1}{t}$: the integrand becomes $-\frac{t\,dt}{(3t^2 + 4)\sqrt{4t^2 - 3}}$. Then $u^2 = 4t^2 - 3$, $t\,dt = \frac{u\,du}{4}$, $3t^2 + 4 = \frac{3u^2 + 25}{4}$, giving $-\int\frac{du}{3u^2 + 25} = -\frac{1}{5\sqrt{3}}\tan^{-1}\frac{\sqrt{3}u}{5}$, with $u = \frac{\sqrt{4 - 3x^2}}{x}$.
As $x \to 0^+$, $u \to \infty$ and the arctangent tends to $\frac{\pi}{2}$; at $x = 1$, $u = 1$. So
$f(1) = \frac{1}{5\sqrt{3}}\left[\frac{\pi}{2} - \tan^{-1}\frac{\sqrt{3}}{5}\right] = \frac{1}{5\sqrt{3}}\tan^{-1}\frac{5}{\sqrt{3}}$.
Thus $\alpha\beta = 5\sqrt{3}$, $\frac{\alpha}{\beta} = \frac{5}{\sqrt{3}}$: $\alpha = 5$, $\beta = \sqrt{3}$, and $\alpha^2 + \beta^2 = 25 + 3 = 28$.
Answer: 28. | ||||
| 27 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The parabola $x = \frac{2y^2}{3}$ meets $x + y = 3$ where $2y^2 + 3y - 9 = 0$, i.e. $y = \frac{3}{2}$ (above the axis), at $\left(\frac{3}{2}, \frac{3}{2}\right)$.
The region between the parabola, the line and $y = 0$ has area $\int_0^{3/2}\left(3 - y - \frac{2y^2}{3}\right)dy = \frac{9}{2} - \frac{9}{8} - \frac{3}{4} = \frac{21}{8}$.
The circle has centre $(3, 0)$, which is the corner where $x + y = 3$ meets $y = 0$. The angle of the region at that corner is $45^\circ$ (between the ray towards $-x$ and the line of slope $-1$). The circle does not reach the parabola (least distance from $(3,0)$ to the parabola is $\sqrt{3.9375} > \sqrt{2}$), so the part of the disc inside the region is a sector of angle $\frac{\pi}{4}$: area $\frac{1}{2} \cdot 2 \cdot \frac{\pi}{4} = \frac{\pi}{4}$.
$A = \frac{21}{8} - \frac{\pi}{4}$, so $4(\pi + 4A) = 4\left(\pi + \frac{21}{2} - \pi\right) = 42$.
Answer: 42. | ||||
| 28 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The minor axis subtends $60^\circ$ at a focus, so $\tan 30^\circ = \frac{b}{ae}$, i.e. $ae = \sqrt{3}b$.
$a^2 = b^2 + a^2e^2 = 4b^2$, so $a = 2b$.
Latus rectum $\frac{2b^2}{a} = \frac{2b^2}{2b} = b = \frac{1}{2}$, so $a = 1$.
Minor axis $2b = 1$, major axis $2a = 2$; $(1 + 2)^2 = 9$.
Answer: 9. | ||||
| 29 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalHard | |
A hard numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Plane: $(2x + y - z - 3) + \lambda(5x - 3y + 4z + 9) = 0$, normal $(2 + 5\lambda, 1 - 3\lambda, -1 + 4\lambda)$.
The line $\frac{x + 2}{2} = \frac{y - 3}{4} = \frac{z - 7}{5}$ has direction $(2, 4, 5)$. Parallel: $2(2 + 5\lambda) + 4(1 - 3\lambda) + 5(-1 + 4\lambda) = 3 + 18\lambda = 0$, $\lambda = -\frac{1}{6}$.
Multiplying by 6: $6(2x + y - z - 3) - (5x - 3y + 4z + 9) = 7x + 9y - 10z - 27 = 0$.
The line $\frac{x}{-3} = \frac{y - 5}{4} = \frac{z - 2}{12}$ has direction $(-3, 4, 12)$, of length 13. From A: $(8 - 3t, -1 + 4t, -19 + 12t)$ on the plane gives $56 - 21t - 9 + 36t + 190 - 120t - 27 = 210 - 105t = 0$, $t = 2$.
Distance $= 2 \times 13 = 26$.
Answer: 26. | ||||
| 30 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalHard | |
A hard numerical question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions At the common point $x = y = z = t$: $t(\sin A + \sin B + \sin C) = 18$ and $t(\sin 2A + \sin 2B + \sin 2C) = 9$.
In a triangle, $\sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$ and $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C = 32\prod\sin\frac{A}{2}\prod\cos\frac{A}{2}$.
Dividing: $\frac{32\prod\sin\frac{A}{2}\prod\cos\frac{A}{2}}{4\prod\cos\frac{A}{2}} = 8\prod\sin\frac{A}{2} = \frac{9}{18} = \frac{1}{2}$.
So $\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = \frac{1}{16}$ and $80 \cdot \frac{1}{16} = 5$.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | StatementsEasy | |
A easy statements question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves In SHM $F = -kx$ and $a = -\omega^2 x$.
(A) True: $|F| \propto x$.
(B) True: the minus sign means acceleration is opposite to displacement.
(C) True: $v = \omega\sqrt{A^2 - x^2}$ is greatest at $x = 0$.
(D) False: $|a| = \omega^2|x|$ is greatest at the extreme points, not least.
Answer: (A). | ||||
| 32 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory The average (translational) kinetic energy of a gas molecule is $\frac{3}{2}k_BT$, which depends only on the temperature, not on the mass of the molecule or the mass ratio of the mixture.
Both gases are at the same temperature, so $\frac{K_{argon}}{K_{hydrogen}} = 1$.
Answer: (B). | ||||
| 33 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics The cycle C(2 m³, 100 Pa) → D(2 m³, 400 Pa) → E(4 m³, 100 Pa) → C is a triangle on the P–V diagram, traversed clockwise, so the net work is positive and equal to the enclosed area.
$W = \frac{1}{2} \times (4 - 2) \times (400 - 100) = \frac{1}{2} \times 2 \times 300 = 300$ J.
Answer: (D). | ||||
| 34 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $\Delta L = \frac{FL}{\pi r^2 Y}$.
New wire: $\Delta L' = \frac{(2f)(2L)}{\pi(2r)^2 Y} = \frac{4fL}{4\pi r^2 Y} = l$.
Answer: (A). | ||||
| 35 | GravitationGravitational potential energy | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on gravitational potential energy, from Gravitation. NCERT: Class 11: Gravitation Energy conservation from $r = 2R$ to $r = R$:
$\frac{1}{2}mv^2 = GMm\left(\frac{1}{R} - \frac{1}{2R}\right) = \frac{GMm}{2R}$, so $v^2 = \frac{GM}{R} = gR$ (using $GM = gR^2$).
$v = \sqrt{gR}$.
Answer: (D). | ||||
| 36 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation The particles are at opposite ends of a diameter, $2a$ apart. Gravity provides the centripetal force:
$\frac{Gm^2}{(2a)^2} = m\omega^2 a$, so $\omega^2 = \frac{Gm}{4a^3}$.
$\omega = \sqrt{\frac{Gm}{4a^3}}$.
Answer: (B). | ||||
| 37 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $A\cos 30^\circ = 2\sqrt{3}$, so $A = 4$.
$x$-component $= A\sin 30^\circ = 4 \times \frac{1}{2} = 2$.
Answer: (A). | ||||
| 38 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $v = \frac{dx}{dt} = 10t - 4$. At $t = 2$ s: $v = 20 - 4 = 16$ m/s.
Answer: (B). | ||||
| 39 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $[LT^{-1}] = [L]^a[LT^{-2}]^b[ML^{-3}]^c$.
Mass: $c = 0$. Time: $-2b = -1$, so $b = \frac{1}{2}$. Length: $a + b - 3c = 1$, so $a = \frac{1}{2}$.
So $v = \sqrt{g\lambda}$: $a, b, c = \frac{1}{2}, \frac{1}{2}, 0$.
Answer: (C). | ||||
| 40 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $\vec{v} = 10\hat{i} + 30t\hat{j}$, $\vec{a} = 30\hat{j}$ m/s².
The net force $m\vec{a}$ is along the positive $y$-axis.
Answer: (B). | ||||
| 41 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance On the equatorial line of a short dipole, $E = \frac{1}{4\pi\epsilon_0}\frac{p}{r^3}$, so $E \propto \frac{1}{r^3}$.
Answer: (C). | ||||
| 42 | Communication SystemsRange of a TV tower | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on range of a tv tower, from Communication Systems. NCERT: No matching NCERT chapter $d_{max} = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
$\sqrt{2 \times 6.4 \times 10^6 \times 180} = \sqrt{2.304 \times 10^9} = 48$ km; $\sqrt{2 \times 6.4 \times 10^6 \times 245} = \sqrt{3.136 \times 10^9} = 56$ km.
$d_{max} = 48 + 56 = 104$ km.
Answer: (D). | ||||
| 43 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Take the battery's negative terminal (the left side of the circuit, joined to the left node) at 0 V; its positive terminal, the bottom-right corner, is at 10 V. Call the top-right corner T.
$D_3$ (10 Ω branch from the left node to the bottom-right corner) points towards the left node: it is forward biased and carries $\frac{10}{10} = 1$ A.
Path bottom-right → 10 Ω → T → $D_1$ → 10 Ω → left node: $D_1$ points from T to the left, so it conducts; with 20 Ω in series the current is $\frac{10}{20} = 0.5$ A and T is at 5 V.
$D_2$ points from the left node towards T, but T (5 V) is above the left node (0 V): it is reverse biased and carries no current.
$I = 1 + 0.5 = 1.5$ A.
Answer: (A). | ||||
| 44 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei 15 years is 3 half-lives, so the fraction remaining is $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Fraction decayed $= 1 - \frac{1}{8} = \frac{7}{8}$.
Answer: (D). | ||||
| 45 | 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 $\lambda = \frac{h}{\sqrt{2mE}} \propto \frac{1}{\sqrt{E}}$. With $\frac{E}{4}$: $\lambda' = \lambda\sqrt{4} = 2\lambda$.
Answer: (B). | ||||
| 46 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics First minimum: $a\sin\theta = \lambda$, so $a = \frac{600 \text{ nm}}{\sin 30^\circ} = 1200$ nm $= 1.2\ \mu$m.
Answer: (B). | ||||
| 47 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Matching listEasy | |
A easy matching list question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Microwaves: 0.1 m to 1 mm (IV). Ultraviolet: 400 nm to 1 nm (I). X-rays: 1 nm to $10^{-3}$ nm (II). Infra-red: 1 mm to 700 nm (III).
So A→IV, B→I, C→II, D→III.
Answer: (C). | ||||
| 48 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Steady current $I = \frac{12}{6} = 2$ A, falling to zero in 1 ms.
$|e| = L\frac{\Delta I}{\Delta t}$: $20 = L \times \frac{2}{10^{-3}}$, so $L = 10^{-2}$ H $= 10$ mH.
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 Voltmeter: series resistance $R = \frac{V}{I_g} - G = \frac{50}{10^{-3}} - 54 = 49946\ \Omega \approx 50$ kΩ. (A) is true; (D) and (E) are false.
Ammeter: shunt $r = \frac{I_g G}{I - I_g} = \frac{1 \times 54}{10 - 1} = 6\ \Omega$. (C) is true; (B) is false.
Answer: (A). | ||||
| 50 | Current ElectricityResistivity | Class 12: Current Electricity | StatementsEasy | |
A easy statements question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Statement I is false: in series $R_{eq} = R_1 + R_2 + \ldots$, larger than the largest resistance (it is the parallel combination that is smaller than the least).
Statement II is false: resistivity depends on temperature, e.g. $\rho = \rho_0[1 + \alpha(T - T_0)]$ for metals.
Answer: (B). | ||||
| 51 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Length $L = \frac{18}{20} = 0.9$ m. In the fundamental mode $\lambda = 2L = 1.8$ m.
$v = f\lambda = 50 \times 1.8 = 90$ m/s.
Answer: 90. | ||||
| 52 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter An air bubble inside a liquid has one surface: $P_{in} = P_0 + \rho gh + \frac{2T}{r}$.
$\rho gh = 1000 \times 10 \times 0.1 = 1000$ Pa; $\frac{2T}{r} = \frac{2 \times 0.075}{10^{-3}} = 150$ Pa.
Excess $= 1150$ Pa.
Answer: 1150. | ||||
| 53 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Solid sphere: $k^2 = \frac{2}{5}R^2$. Solid cylinder about its axis: $k^2 = \frac{1}{2}R^2$.
$\frac{k_{sph}}{k_{cyl}} = \sqrt{\frac{2/5}{1/2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$, so $x = 5$.
Answer: 5. | ||||
| 54 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $W = \int_2^4 5x\,dx = \frac{5}{2}(16 - 4) = 30$ J.
Answer: 30. | ||||
| 55 | 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 The three capacitors are in parallel across 10 V: $C_{eq} = 2 + x + 3 = (5 + x)\ \mu$F.
$Q = C_{eq}V$: $100 = (5 + x) \times 10$, so $x = 5$.
Answer: 5. | ||||
| 56 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei First, second and third excited states are $n = 2, 3, 4$. $\lambda_1$: $3 \to 2$; $\lambda_2$: $4 \to 3$.
$\frac{1}{\lambda_1} \propto \frac{1}{4} - \frac{1}{9} = \frac{5}{36}$; $\frac{1}{\lambda_2} \propto \frac{1}{9} - \frac{1}{16} = \frac{7}{144}$.
$\frac{\lambda_1}{\lambda_2} = \frac{7/144}{5/36} = \frac{7}{20} = \frac{7}{4n}$, so $n = 5$.
Answer: 5. | ||||
| 57 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\mu = \frac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$ with $A = 60^\circ$: $\sin\frac{60^\circ + \delta_m}{2} = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$.
$\frac{60^\circ + \delta_m}{2} = 45^\circ$, so $\delta_m = 30^\circ$.
Answer: 30. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\omega = 210 \times \frac{2\pi}{60} = 7\pi = 22$ rad/s.
$e = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 0.2 \times 22 \times (0.2)^2 = 0.088$ V $= 88$ mV.
Answer: 88. | ||||
| 59 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalMedium | |
A medium numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The revolving electron is a current $I = \frac{ev}{2\pi r}$; the field at the centre is $B = \frac{\mu_0 I}{2r} = \frac{\mu_0 ev}{4\pi r^2}$.
$B = \frac{10^{-7} \times 1.6 \times 10^{-19} \times 6.76 \times 10^6}{(0.52 \times 10^{-10})^2} = \frac{1.0816 \times 10^{-19}}{2.704 \times 10^{-21}} = 40$ T.
Answer: 40. | ||||
| 60 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Between the left and right corners there are two branches: 2 Ω – A – 4 Ω and 4 Ω – B – 2 Ω, each 6 Ω, both across 9 V. Each carries $\frac{9}{6} = 1.5$ A.
From the left corner, the drop to A is $1.5 \times 2 = 3$ V and the drop to B is $1.5 \times 4 = 6$ V.
$|V_A - V_B| = 6 - 3 = 3$ V.
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Solid StateIonic radii and radius ratio | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ionic radii and radius ratio, from Solid State. NCERT: No matching NCERT chapter In CsCl, $Cl^-$ ions sit at the cube corners and $Cs^+$ at the body centre; the ions touch along the body diagonal.
Half the body diagonal $= \frac{\sqrt{3}a}{2} = r_{Cs^+} + r_{Cl^-}$.
Answer: (C). | ||||
| 62 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | StatementsEasy | |
A easy statements question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Statement I is correct: Bohr postulated $mvr = \frac{nh}{2\pi}$.
Statement II is correct: a definite orbit means the electron's position and momentum are known exactly at the same time, which the uncertainty principle forbids.
Answer: (A). | ||||
| 63 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | StatementsEasy | |
A easy statements question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) False: $NF_3$ has three bond pairs and one lone pair on N, so it is trigonal pyramidal.
(B) True: bond order of $N_2$ is 3 and of $O_2$ is 2, so $N_2$ (110 pm) is shorter than $O_2$ (121 pm).
(C) True: isoelectronic species fill the same molecular orbitals, e.g. $N_2$, $CO$, $NO^+$ all have bond order 3.
(D) False: O is more electronegative than S, so $H_2O$ (1.85 D) has the larger dipole moment than $H_2S$.
Answer: (B). | ||||
| 64 | EquilibriumpH and buffers | Class 11: Equilibrium | StatementsMedium | |
A medium statements question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium (A) False: an acid solution cannot have pH above 7. With water's own $10^{-7}$ M $H^+$, $[H^+] \approx 1.05 \times 10^{-7}$ M and pH $\approx 6.98$.
(B) True: removing $H^+$ from $H_2PO_4^-$ gives $HPO_4^{2-}$.
(C) True: ionisation of water is endothermic, so $K_w$ rises with temperature.
(D) False: at half neutralisation $[HA] = [A^-]$, so $pH = pK_a$, not $\frac{1}{2}pK_a$.
Answer: (C). | ||||
| 65 | Isolation of MetalsRoasting and calcination | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on roasting and calcination, from Isolation of Metals. NCERT: No matching NCERT chapter Calcination is heating in a limited supply of air to drive off water or $CO_2$ (options A, B, C).
Heating a sulphide ore in excess air to convert it to the oxide, $2PbS + 3O_2 \rightarrow 2PbO + 2SO_2$, is roasting.
Answer: (D). | ||||
| 66 | 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 shift reaction: $CO(g) + H_2O(g) \xrightarrow[\text{catalyst}]{673\ K} CO_2(g) + H_2(g)$.
Carbon monoxide (C at +2) is oxidised to carbon dioxide (C at +4).
Answer: (B). | ||||
| 67 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter Assertion is false: Be and Mg (and their salts) do not colour the flame.
Reason is true: the electrons of Be and Mg are held strongly (small atoms, high ionisation enthalpy), so the energy needed to excite them is too high for a flame.
(The Reason explains why the Assertion is false.)
Answer: (D). | ||||
| 68 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter For good quality cement the ratio of silica ($SiO_2$) to alumina ($Al_2O_3$) should be between 2.5 and 4.
Of the options, only 3 lies in this range.
Answer: (B). | ||||
| 69 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter $H_4P_2O_7$ (pyrophosphoric acid): two $PO_4$ units joined by one bridging O: 1 P–O–P bond.
$(HPO_3)_3$ (cyclotrimetaphosphoric acid): a six-membered ring of alternating P and O: 3 P–O–P bonds.
$P_4O_{10}$: each of the six edges of the $P_4$ tetrahedron is bridged by an O: 6 P–O–P bonds.
Answer: (D). | ||||
| 70 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctHard | |
A hard single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All four are $M^{3+}$ ions with the same ligand, water, so $\Delta_0$ depends on the metal ion. The measured values are approximately:
$[Ti(H_2O)_6]^{3+}$ ($d^1$): 20 300 cm⁻¹; $[Cr(H_2O)_6]^{3+}$ ($d^3$): 17 400 cm⁻¹; $[Mn(H_2O)_6]^{3+}$ ($d^4$): 21 000 cm⁻¹; $[Fe(H_2O)_6]^{3+}$ ($d^5$): 13 700 cm⁻¹.
The largest is for $[Mn(H_2O)_6]^{3+}$.
Answer: (C). | ||||
| 71 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Photochemical smog needs strong sunlight, warm weather and nitrogen oxides and hydrocarbons from traffic. It forms in warm, sunny, dry conditions.
Srinagar in January is cold with weak sunlight, so smog formation is least likely there.
Answer: (B). | ||||
| 72 | 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 The starting compound is indane-1,2,3-trione. Its central carbonyl (C2) lies between two other carbonyl groups, whose electron withdrawal makes it very electrophilic and destabilises it.
Water adds to that central C=O to give a stable gem-diol, 2,2-dihydroxyindane-1,3-dione (ninhydrin). The gem-diol is stabilised by intramolecular hydrogen bonding with the two flanking C=O groups.
Answer: (D). | ||||
| 73 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Electron-donating groups activate the ring; electron-withdrawing groups deactivate it.
$-NMe_2$ (strong +R) > $-OCH_3$ (+R, weaker than N) > $-CH_3$ (+I, hyperconjugation) > H > $-CF_3$ (strong −I, deactivating).
So e > d > a > b > c.
Answer: (B). | ||||
| 74 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Cl deactivates the ring (−I) but directs ortho/para (+R). With $CH_3COCl$ and anhydrous $AlCl_3$, the bulky acyl group goes mainly para; the acyl group then deactivates the ring, so there is no second acylation.
Major product: 4-chloroacetophenone.
Answer: (B). | ||||
| 75 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Target: 2-bromo-4-chlorobenzoic acid from 4-nitrotoluene. The $CH_3$ becomes COOH, the $NO_2$ becomes Cl, and Br goes ortho to the old $CH_3$.
(i) $Br_2/Fe$: $CH_3$ (o/p director) and $NO_2$ (meta director) both send Br to the position ortho to $CH_3$: 2-bromo-4-nitrotoluene.
(ii) $Fe/H^+$ reduces $NO_2$ to $NH_2$; (iii) HONO gives the diazonium salt; (iv) CuCl (Sandmeyer) replaces it by Cl; (v) $KMnO_4$ oxidises $CH_3$ to COOH.
(D) brominates last, when COOH (meta) and Cl (o/p) both send Br ortho to Cl, not ortho to COOH. (A) and (B) never replace N by Cl.
Answer: (C). | ||||
| 76 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (i), (ii) Hydroboration–oxidation adds water anti-Markovnikov: propene → propan-1-ol, $CH_3CH_2CH_2OH$.
(iii) PCC oxidises the primary alcohol to the aldehyde: propanal, $CH_3CH_2CHO$.
(iv) $CH_3MgBr$ adds to the aldehyde and, after work-up, gives the secondary alcohol butan-2-ol, $CH_3CH_2CH(OH)CH_3$.
Answer: (B). | ||||
| 77 | AminesDiazonium salts | Class 12: Amines | Single correctEasy | |
A easy single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Aniline with $NaNO_2$ (and acid) at 0–5 °C gives benzenediazonium chloride, A.
A couples with N,N-dimethylaniline at its para position (electrophilic substitution) to give the azo dye p-(dimethylamino)azobenzene, $C_6H_5-N=N-C_6H_4-N(CH_3)_2$.
Answer: (C). | ||||
| 78 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Tetrafluoroethene → Teflon (III). Acrylonitrile → polyacrylonitrile, Orlon (I). Caprolactam → Nylon-6 (IV). Isoprene → natural rubber, cis-polyisoprene (II).
A→III, B→I, C→IV, D→II.
Answer: (A). | ||||
| 79 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Arginine, like other amino acids, exists as a dipolar ion (zwitterion). So it is a crystalline solid with a fairly high melting point, soluble in water but not in non-polar solvents such as benzene.
It has several ionisable groups ($-COOH$, $\alpha-NH_3^+$, guanidinium), so it has more than one $pK_a$.
The untrue statement is "high solubility in benzene".
Answer: (B). | ||||
| 80 | 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 Paper chromatography is partition chromatography. The paper (cellulose) only holds water; the water trapped in its pores is the stationary phase, and the solvent moving up the paper is the mobile phase.
Answer: (B). | ||||
| 81 | Coordination CompoundsWerner theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on werner theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[Cr(H_2O)_5Cl]Cl_2$ ionises as $[Cr(H_2O)_5Cl]^{2+} + 2Cl^-$; the Cl inside the coordination sphere is not precipitated.
Ionisable $Cl^-$ $= 20 \times 0.01 \times 2 = 0.4$ mmol, needing 0.4 mmol $Ag^+$.
Volume $= \frac{0.4}{0.1} = 4$ mL.
Answer: 4. | ||||
| 82 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics At the melting point $\Delta G = 0$, so $T = \frac{\Delta H}{\Delta S} = \frac{30400}{28.4} = 1070.4$ K.
Nearest integer: 1070.
Answer: 1070. | ||||
| 83 | SolutionsRaoult's law | Class 12: Solutions | NumericalMedium | |
A medium numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions Take 100 mL of solution: mass $= 120$ g, glucose $= 30$ g ($\frac{30}{180} = \frac{1}{6}$ mol), water $= 90$ g ($5$ mol).
Raoult's law: $p = p^\circ x_{water} = 24 \times \frac{5}{5 + \frac{1}{6}} = 24 \times \frac{30}{31} = 23.2$ mm Hg.
Nearest integer: 23.
Answer: 23. | ||||
| 84 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry (A) Correct: conductivity (conductance of unit volume) falls on dilution for both kinds, because there are fewer ions per unit volume.
(B) Incorrect: the number of ions per unit volume decreases on dilution.
(C) Correct: molar conductivity rises as concentration falls.
(D) Correct: for strong electrolytes $\Lambda_m$ rises slowly and linearly with $\sqrt{c}$; for weak electrolytes it rises steeply at low concentration.
(E) Incorrect: the rise is due to an increase in the degree of dissociation.
Correct statements: A, C, D = 3.
Answer: 3. | ||||
| 85 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics With the same pre-exponential factor, equal rates mean $\frac{E_{a,cat}}{T_1} = \frac{E_a}{T_2}$:
$E_{a,cat} = 300 \times \frac{300}{600} = 150$ kJ/mol (forward, catalysed).
$E_{a,backward} = E_{a,forward} - \Delta H = 150 - 20 = 130$ kJ/mol.
Answer: 130. | ||||
| 86 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter Coagulating (flocculation) value is the millimoles of electrolyte needed to coagulate one litre of the sol in about 2 hours.
NaCl used $= 20 \times 0.5 = 10$ mmol for 200 mL of sol, i.e. $\frac{10}{0.2} = 50$ mmol per litre.
Answer: 50. | ||||
| 87 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom Electrons: $O^{2-}$ 10, $F^-$ 10, Al 13, $Mg^{2+}$ 10, $Na^+$ 10, $O^+$ 7, Mg 12, $Al^{3+}$ 10, F 9.
The 10-electron species $O^{2-}, F^-, Mg^{2+}, Na^+, Al^{3+}$ are isoelectronic: 5.
Answer: 5. | ||||
| 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 $2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$.
Mn goes from +7 to +2: a change of 5.
Answer: 5. | ||||
| 89 | 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 $[Co(H_2O)_6]^{2+}$: $Co^{2+}$ is $3d^7$ and water is a weak-field ligand, so the complex is high spin: $t_{2g}^5 e_g^2$.
$t_{2g}^5$ has one unpaired electron.
Answer: 1. | ||||
| 90 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions Chromyl chloride is $CrO_2Cl_2$: $x + 2(-2) + 2(-1) = 0$, so $x = +6$.
Answer: 6. | ||||
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