JEE (Main) · 2025
JEE (Main) 2025 — 7 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
25 Q
- Matrices and Determinants3
- Permutations and Combinations2
- Conic Sections2
8 easy · 15 medium · 2 hard
Physics
25 Q
- Optics4
- Magnetic Effects of Current and Magnetism3
- Motion of System of Particles and Rigid Body2
19 easy · 5 medium · 1 hard
Chemistry
25 Q
- Some Basic Concepts of Chemistry2
- Electrochemistry2
- Chemical Kinetics2
11 easy · 14 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 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Put $A = \theta + \frac{\pi}{6}$, so $\theta + \frac{\pi}{3} = A + \frac{\pi}{6}$, and let $t = \tan A = \frac{y}{2}$.
$\frac{x}{3} = \tan\left(A + \frac{\pi}{6}\right) = \frac{t + \frac{1}{\sqrt3}}{1 - \frac{t}{\sqrt3}} = \frac{\sqrt3\,t + 1}{\sqrt3 - t}$.
So $x(\sqrt3 - t) = 3(\sqrt3\,t + 1)$. With $t = \frac{y}{2}$: $\sqrt3\,x - \frac{xy}{2} = \frac{3\sqrt3}{2}y + 3$.
Multiply by 2 and rearrange: $xy - 2\sqrt3\,x + 3\sqrt3\,y + 6 = 0$.
So $\alpha = -2\sqrt3$, $\beta = 3\sqrt3$, $\gamma = 6$ and $\alpha^2 + \beta^2 + \gamma^2 = 12 + 27 + 36 = 75$.
Answer: (A) $75$. | ||||
| 2 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | StatementsMedium | |
A medium statements question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations (S1): $\frac{z - i}{z + i}$ is real exactly when $z$, $i$ and $-i$ are collinear, that is when $z$ lies on the imaginary axis, $z = ki$. With $|z| = 1$, $z = i$ or $z = -i$; $z = -i$ is excluded. So the set is $\{i\}$ (where the ratio is $0$): one element, not two. (S1) is false.
(S2): write $z = \cos\phi + i\sin\phi$, $z \ne -1$. Then $\frac{z - 1}{z + 1} = \frac{e^{i\phi/2} - e^{-i\phi/2}}{e^{i\phi/2} + e^{-i\phi/2}} = i\tan\frac{\phi}{2}$, which is purely imaginary for every point of the circle other than $-1$. So the set is infinite. (S2) is true.
Answer: (D) only (S2) is correct. | ||||
| 3 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Both roots real and negative needs:
Real roots: $(p + 2)^2 - 4(2p + 9) \ge 0 \Rightarrow p^2 - 4p - 32 \ge 0 \Rightarrow (p - 8)(p + 4) \ge 0 \Rightarrow p \le -4$ or $p \ge 8$.
Sum negative: $p + 2 < 0 \Rightarrow p < -2$.
Product positive: $2p + 9 > 0 \Rightarrow p > -\frac{9}{2}$.
Together: $p \in \left(-\frac{9}{2}, -4\right]$, so $\alpha = -\frac{9}{2}$, $\beta = -4$.
$\beta - 2\alpha = -4 + 9 = 5$.
Answer: (C) $5$. | ||||
| 4 | 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 The first two equations are independent, so for infinitely many solutions the third must be a combination $a(\text{eq. 1}) + b(\text{eq. 2})$.
$y$: $3a + 3b = 3 \Rightarrow a + b = 1$. $x$: $2a + 7b = 12 \Rightarrow 2 + 5b = 12 \Rightarrow b = 2$, $a = -1$.
$z$: $-5 - 4 = -9 = -(4 + \lambda) \Rightarrow \lambda = 5$.
Right side: $-9 + 16 = 7 = 16 - \mu \Rightarrow \mu = 9$.
The radius is the distance of $(5, 9)$ from $4x - 3y = 0$: $\frac{|20 - 27|}{5} = \frac{7}{5}$.
Answer: (C) $\frac{7}{5}$. | ||||
| 5 | 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 For an $n \times n$ matrix $|adj\,A| = |A|^{n-1}$; with $n = 3$, $|adj\,A| = |A|^2$, $|adj(adj\,A)| = |A|^4$, $|adj(adj(adj\,A))| = |A|^8$.
$|A|^8 = 81 = 3^4 \Rightarrow |A|^2 = 3$. Then $|adj(adj\,A)| = |A|^4 = 9$.
The condition: $9^{\frac{(n-1)^2}{2}} = |A|^{3n^2 - 5n - 4}$, that is $3^{(n-1)^2} = 3^{\frac{3n^2 - 5n - 4}{2}}$.
$2(n - 1)^2 = 3n^2 - 5n - 4 \Rightarrow n^2 - n - 6 = 0 \Rightarrow n = 3$ or $n = -2$. (Both exponents of $|A|$ are even, so the sign of $|A|$ does not matter.)
$\left|A^{(n^2+n)}\right| = |A|^{n^2 + n}$: for $n = 3$, $|A|^{12} = 3^6 = 729$; for $n = -2$, $|A|^{2} = 3$.
Sum $= 729 + 3 = 732$.
Answer: (A) $732$. | ||||
| 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 Let the terms be $a, ar, ar^2, ar^3$. The numbers $a - 2,\ ar - 7,\ ar^2 - 9,\ ar^3 - 5$ are in AP.
First three: $2(ar - 7) = (a - 2) + (ar^2 - 9) \Rightarrow a(r - 1)^2 = -3$.
Last three: $2(ar^2 - 9) = (ar - 7) + (ar^3 - 5) \Rightarrow ar(r - 1)^2 = -6$.
Dividing: $r = 2$, and then $a = -3$.
The GP is $-3, -6, -12, -24$ (check: $-5, -13, -21, -29$ is an AP with difference $-8$).
$x_1x_2x_3x_4 = (-3)(-6)(-12)(-24) = 5184$, and $\frac{5184}{24} = 216$.
Answer: (A) $216$. | ||||
| 7 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $64 = 63 + 1 = 7 \times 9 + 1$, so $64 \equiv 1 \pmod 7$.
By the binomial theorem $(63 + 1)^N = 63k + 1$ for every positive integer $N$, so any positive integer power of 64 leaves remainder 1 on division by 7.
$\left(64^{64}\right)^{64} = 64^{64 \times 64}$ therefore leaves remainder $1$.
Answer: (A) $1$. | ||||
| 8 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The captain (a batsman) and vice-captain (a bowler) are already in, so 8 more players come from the remaining 6 batsmen and 5 bowlers.
With $b$ batsmen and $w$ bowlers in the team, $b + w = 10$, $b \ge 4$, $w \ge 4$, $b \le 7$, $w \le 6$: $(b, w) = (4, 6), (5, 5), (6, 4)$.
$(4, 6)$: $\binom{6}{3}\binom{5}{5} = 20 \times 1 = 20$.
$(5, 5)$: $\binom{6}{4}\binom{5}{4} = 15 \times 5 = 75$.
$(6, 4)$: $\binom{6}{5}\binom{5}{3} = 6 \times 10 = 60$.
Total $= 20 + 75 + 60 = 155$.
Answer: (C) $155$. | ||||
| 9 | 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 Wrong sum $= 100 \times 40 = 4000$; correct sum $= 4000 - 50 + 40 = 3990$, so $\mu = 39.9$.
Wrong $\sum x^2 = 100(5.1^2 + 40^2) = 100(26.01 + 1600) = 162601$.
Correct $\sum x^2 = 162601 - 2500 + 1600 = 161701$.
$\sigma^2 = \frac{161701}{100} - 39.9^2 = 1617.01 - 1592.01 = 25$, so $\sigma = 5$.
$10(\mu + \sigma) = 10(44.9) = 449$.
Answer: (B) $449$. | ||||
| 10 | Conic SectionsParabola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A point $(-2, y)$ is on P when its distance from the focus equals its distance from the directrix:
$|y - 1| = \frac{|2(-2) + y + 2|}{\sqrt5} = \frac{|y - 2|}{\sqrt5}$.
Squaring: $5(y - 1)^2 = (y - 2)^2 \Rightarrow 5y^2 - 10y + 5 = y^2 - 4y + 4 \Rightarrow 4y^2 - 6y + 1 = 0$.
Its discriminant $36 - 16 > 0$, so there are two such points, and the sum of their ordinates is $\frac{6}{4} = \frac{3}{2}$.
Answer: (C) $\frac{3}{2}$. | ||||
| 11 | 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 A: solve $3y - x = 2$, $x + y = 2$: $4y = 4$, $y = 1$, $x = 1$, so A$(1, 1)$.
B: AB meets $y = 0$ at $x = -2$, so B$(-2, 0)$. C: AC meets $y = 0$ at $x = 2$, so C$(2, 0)$.
Altitude from A is perpendicular to BC (the $x$-axis): $x = 1$.
Altitude from B is perpendicular to AC (slope $-1$), so has slope 1: $y = x + 2$.
They meet at P$(1, 3)$.
Area of PBC $= \frac{1}{2} \times BC \times 3 = \frac{1}{2} \times 4 \times 3 = 6$.
Answer: (D) $6$. | ||||
| 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 $C_1$ has centre $(-3, -3)$ and radius 3.
The distance between the centres is $\sqrt{(1 + 3)^2 + (3 + 3)^2} = \sqrt{52} = 2\sqrt{13}$.
For external contact the point of contact lies on the line of centres at distance 3 (the radius of $C_1$) from $(-3, -3)$:
$(\alpha, \beta) = (-3, -3) + \frac{3}{2\sqrt{13}}(4, 6) = \left(-3 + \frac{6}{\sqrt{13}},\ -3 + \frac{9}{\sqrt{13}}\right)$.
$\beta - \alpha = \frac{3}{\sqrt{13}}$, so $(\beta - \alpha)^2 = \frac{9}{13}$; $m = 9$, $n = 13$.
$m + n = 22$.
Answer: (C) $22$. | ||||
| 13 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium 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 first line: $(1 + 2s, -1 + 3s, 1 + 4s)$; of the second: $(3 + t, 4 + 2t, t)$.
From $(1, 1, 1)$ these are in the directions $(2s, 3s - 2, 4s)$ and $(2 + t, 3 + 2t, t - 1)$, which must be parallel.
First and third components: $\frac{2s}{2 + t} = \frac{4s}{t - 1} \Rightarrow t - 1 = 2(2 + t) \Rightarrow t = -5$ (for $s \ne 0$).
The direction is then $(-3, -7, -6)$, i.e. $(3, 7, 6)$. (Check: $(2s, 3s - 2, 4s) = k(-3, -7, -6)$ gives $s = -\frac{3k}{2}$ and $-\frac{9k}{2} - 2 = -7k$, $k = \frac{4}{5}$: consistent.)
L: $(1 + 3\lambda, 1 + 7\lambda, 1 + 6\lambda)$. At $\lambda = 2$: $(7, 15, 13)$.
The other points fail: $(4, 22, 7)$ needs $\lambda = 1$ from $x$ but $\lambda = 3$ from $y$; $(10, -29, -50)$ and $(5, 4, 3)$ likewise disagree.
Answer: (B) $(7, 15, 13)$. | ||||
| 14 | 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 $A(1, 2, 3)$, $B(0, 0, 5)$; directions $\vec{d_1} = (2, 3, 4)$, $\vec{d_2} = (1, \alpha, 1)$.
$\vec{d_1}\times\vec{d_2} = (3 - 4\alpha,\ 2,\ 2\alpha - 3)$, $\vec{AB} = (-1, -2, 2)$.
$\vec{AB}\cdot(\vec{d_1}\times\vec{d_2}) = -(3 - 4\alpha) - 4 + 2(2\alpha - 3) = 8\alpha - 13$.
$|\vec{d_1}\times\vec{d_2}|^2 = (3 - 4\alpha)^2 + 4 + (2\alpha - 3)^2 = 20\alpha^2 - 36\alpha + 22$.
$\frac{(8\alpha - 13)^2}{20\alpha^2 - 36\alpha + 22} = \frac{25}{6}$
$\Rightarrow 384\alpha^2 - 1248\alpha + 1014 = 500\alpha^2 - 900\alpha + 550 \Rightarrow 116\alpha^2 + 348\alpha - 464 = 0 \Rightarrow \alpha^2 + 3\alpha - 4 = 0$.
$\alpha = 1$ or $\alpha = -4$; the sum is $-3$.
Answer: (B) $-3$. | ||||
| 15 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\sin\theta = \frac{\sqrt{65}}{9}$ with $\theta$ acute, so $\cos\theta = \sqrt{1 - \frac{65}{81}} = \frac{4}{9}$ and $\hat{a}\cdot\hat{b} = \frac{4}{9}$.
$\hat{a}\times\hat{b}$ is perpendicular to both, so it adds nothing to the dot products.
$\vec{c}\cdot\hat{a} = 3 + 6\cdot\frac{4}{9} = \frac{17}{3}$; $\vec{c}\cdot\hat{b} = 3\cdot\frac{4}{9} + 6 = \frac{22}{3}$.
$9\cdot\frac{17}{3} - 3\cdot\frac{22}{3} = 51 - 22 = 29$.
Answer: (C) $29$. | ||||
| 16 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability As $x \to 0^+$ each factor behaves like its first-order term:
$\tan\left(5x^{1/3}\right) \sim 5x^{1/3}$, $\log_e(1 + 3x^2) \sim 3x^2$, $\left(\tan^{-1}3\sqrt{x}\right)^2 \sim 9x$, $e^{5x^{4/3}} - 1 \sim 5x^{4/3}$.
Limit $= \frac{5x^{1/3}\cdot3x^2}{9x\cdot5x^{4/3}} = \frac{15x^{7/3}}{45x^{7/3}} = \frac{1}{3}$.
Answer: (B) $\frac{1}{3}$. | ||||
| 17 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 3x^2 + 2ax + \frac{b}{x}$.
$f'(-1) = 0$: $3 - 2a - b = 0$. $f'(2) = 0$: $12 + 4a + \frac{b}{2} = 0 \Rightarrow 24 + 8a + b = 0$.
Adding: $27 + 6a = 0$, $a = -\frac{9}{2}$, and $b = 3 - 2a = 12$.
$f(x) = x^3 - \frac{9}{2}x^2 + 12\log_e|x| + 1$ and $f'(x) = \frac{3x^3 - 9x^2 + 12}{x} = \frac{3(x + 1)(x - 2)^2}{x}$.
On $[-2, -1)$: $x + 1 < 0$, $x < 0$, so $f' > 0$; on $(-1, -\frac{1}{2}]$: $f' < 0$. The maximum is at $x = -1$; the minimum at an end.
$f(-1) = -1 - \frac{9}{2} + 0 + 1 = -4.5 = M$.
$f(-2) = -8 - 18 + 12(0.7) + 1 = -16.6$.
$f\left(-\frac{1}{2}\right) = -\frac{1}{8} - \frac{9}{8} - 12(0.7) + 1 = -8.65$.
So $m = -16.6$ and $|M + m| = |-21.1| = 21.1$.
Answer: (C) $21.1$. | ||||
| 18 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $J = \int_0^{\pi}\frac{\sin x}{1 + 3\cos^2 x}dx$. With $t = \cos x$: $J = \int_{-1}^{1}\frac{dt}{1 + 3t^2} = \frac{2}{\sqrt3}\tan^{-1}\sqrt3 = \frac{2}{\sqrt3}\cdot\frac{\pi}{3} = \frac{2\pi}{3\sqrt3}$.
For $K = \int_0^{\pi}\frac{x\sin x}{1 + 3\cos^2 x}dx$, replace $x$ by $\pi - x$: $K = \int_0^{\pi}\frac{(\pi - x)\sin x}{1 + 3\cos^2 x}dx = \pi J - K$, so $K = \frac{\pi}{2}J$.
Integral $= K + 3J = \left(\frac{\pi}{2} + 3\right)\frac{2\pi}{3\sqrt3} = \frac{\pi(\pi + 6)}{3\sqrt3}$.
Answer: (D) $\frac{\pi}{3\sqrt3}(\pi + 6)$. | ||||
| 19 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersections: $4 - \frac{x^2}{4} = \frac{x - 4}{2} \Rightarrow 16 - x^2 = 2x - 8 \Rightarrow x^2 + 2x - 24 = 0 \Rightarrow x = -6, 4$.
The parabola lies above the line between them:
$\alpha = \int_{-6}^{4}\left(6 - \frac{x}{2} - \frac{x^2}{4}\right)dx = \left[6x - \frac{x^2}{4} - \frac{x^3}{12}\right]_{-6}^{4}$
$= \left(24 - 4 - \frac{16}{3}\right) - \left(-36 - 9 + 18\right) = \frac{44}{3} + 27 = \frac{125}{3}$.
$6\alpha = 250$.
Answer: (D) $250$. | ||||
| 20 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Rewrite: $\frac{dy}{dx} + \frac{e^x(x - 2)}{x(x^2 + e^x)}y = \frac{x^2}{x^2 + e^x}$, a linear equation.
Let $u = 1 + \frac{e^x}{x^2} = \frac{x^2 + e^x}{x^2}$. Then $\frac{du}{dx} = \frac{e^x(x - 2)}{x^3}$ and $\frac{1}{u}\frac{du}{dx} = \frac{e^x(x - 2)}{x(x^2 + e^x)}$, exactly the coefficient of $y$.
So the integrating factor is $u = \frac{x^2 + e^x}{x^2}$, and
$y\cdot\frac{x^2 + e^x}{x^2} = \int\frac{x^2}{x^2 + e^x}\cdot\frac{x^2 + e^x}{x^2}dx = x + C$.
At $(1, 0)$: $0 = 1 + C$, $C = -1$.
$y(2) = (2 - 1)\cdot\frac{4}{4 + e^2} = \frac{4}{4 + e^2}$.
Answer: (C) $\frac{4}{4 + e^2}$. | ||||
| 21 | 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 Count the subsets of $\{1, 2, 3, 4, 5\}$ with no two consecutive elements by size.
Size 0: the empty set, 1.
Size 1: 5.
Size 2: choose 2 of 5 with a gap: $\binom{5 - 2 + 1}{2} = \binom{4}{2} = 6$ ($\{1,3\},\{1,4\},\{1,5\},\{2,4\},\{2,5\},\{3,5\}$).
Size 3: $\binom{3}{3} = 1$ ($\{1, 3, 5\}$).
Size 4 or more: impossible.
$n(S_5) = 1 + 5 + 6 + 1 = 13$. (In general $n(S_n)$ is a Fibonacci number: 3, 5, 8, 13 for $n = 2, 3, 4, 5$.)
Answer: 13. | ||||
| 22 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is singular when $ad = bc$. Entries may repeat.
List the ordered pairs from $\{2, 3, 6, 9\}$ by their product:
4: (2,2) — 1; 6: (2,3),(3,2) — 2; 9: (3,3) — 1; 12: (2,6),(6,2) — 2; 18: (2,9),(9,2),(3,6),(6,3) — 4; 27: (3,9),(9,3) — 2; 36: (6,6) — 1; 54: (6,9),(9,6) — 2; 81: (9,9) — 1. (Total 16.)
The pair $(a, d)$ and the pair $(b, c)$ must have the same product, so the count is the sum of the squares:
$1 + 4 + 1 + 4 + 16 + 4 + 1 + 4 + 1 = 36$.
Answer: 36. | ||||
| 23 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections The foci are $(\pm3, 0)$, so $a^2 + b^2 = 9$. The other focus is S$(3, 0)$, and its latus rectum has ends $\left(3, \pm\frac{b^2}{a}\right)$.
By symmetry each end makes $45°$ with PS at P, so the half-length equals PS: $\frac{b^2}{a} = 6$, $b^2 = 6a$.
$a^2 + 6a - 9 = 0 \Rightarrow a = 3(\sqrt2 - 1)$ (positive root), and $b^2 = 18(\sqrt2 - 1)$.
$a^2 = 9(3 - 2\sqrt2)$, so $a^2b^2 = 162(3 - 2\sqrt2)(\sqrt2 - 1) = 162(3\sqrt2 - 3 - 4 + 2\sqrt2) = 162(5\sqrt2 - 7) = 810\sqrt2 - 1134$.
$\alpha = 810$, $\beta = 1134$, $\alpha + \beta = 1944$.
Answer: 1944. | ||||
| 24 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalHard | |
A hard 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 Reflexive means $(1,1), (2,2), (3,3)$ are in; with $(1, 2)$ that is 4 elements, so at most 2 more from $(2,1), (1,3), (3,1), (2,3), (3,2)$.
4 elements: $\{(1,2)\}$ plus the diagonal is transitive and not symmetric: 1.
5 elements (one extra pair):
$(2,1)$: symmetric — no. $(1,3)$: transitive — yes. $(3,1)$: $(3,1),(1,2)$ needs $(3,2)$ — no. $(2,3)$: $(1,2),(2,3)$ needs $(1,3)$ — no. $(3,2)$: transitive — yes. That gives 2.
6 elements (two extra pairs, 10 choices):
$\{(1,3),(2,3)\}$: yes. $\{(1,3),(3,2)\}$: $(1,3),(3,2)$ gives $(1,2)$, present — yes. $\{(3,1),(3,2)\}$: $(3,1),(1,2)$ gives $(3,2)$, present — yes.
Every other pair fails: $\{(2,1),\cdot\}$ needs a missing pair in each case (e.g. $(2,1),(1,3) \to (2,3)$; $(3,2),(2,1) \to (3,1)$), $\{(1,3),(3,1)\}$ needs $(3,2)$, $\{(3,1),(2,3)\}$ needs $(2,1)$, $\{(2,3),(3,2)\}$ needs $(1,3)$. None of the three good ones is symmetric. That gives 3.
Total $= 1 + 2 + 3 = 6$.
Answer: 6. | ||||
| 25 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\left[\frac{x^2}{2}\right]$ jumps where $\frac{x^2}{2}$ is an integer: $x = \sqrt2, 2, \sqrt6, \sqrt8, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4$.
$[\sqrt{x}]$ jumps at $x = 1$ and $x = 4$.
At $x = 4$ both jump by 1: the left limit is $7 - 1 = 6$ and $f(4) = 8 - 2 = 6$, so $f$ is continuous there. ($x = 0$ is no jump: $f = 0$ just to its right.)
At each other point only one term jumps, so $f$ is discontinuous at $x = 1, \sqrt2, 2, \sqrt6, \sqrt8, \sqrt{10}, \sqrt{12}, \sqrt{14}$.
Number of points $= 8$.
Answer: 8. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | 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 $\epsilon_0\frac{d\Phi_E}{dt}$ is Maxwell's displacement current $I_d$, which enters the Ampere–Maxwell law alongside the conduction current.
Check with dimensions: by Gauss's law $\epsilon_0\Phi_E = q_{enc}$, so $\epsilon_0\Phi_E$ has the dimensions of charge [AT], and dividing by time gives [A], a current.
Answer: (C) electric current. | ||||
| 27 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For two thin lenses a distance $d$ apart: $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1f_2}$.
$\frac{1}{F} = \frac{1}{30} + \frac{1}{10} - \frac{10}{300} = \frac{1 + 3 - 1}{30} = \frac{1}{10}$ cm$^{-1}$, so $F = 10$ cm $= 0.1$ m.
$P = \frac{1}{0.1\text{ m}} = 10$ D.
Answer: (B) 10 D. | ||||
| 28 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $m = 1$ kg, so $\vec{a} = \vec{F} = 2t\hat{i} + 3t^2\hat{j}$.
Starting from rest at $t = 0$: $\vec{v} = \int_0^t\vec{a}\,dt = t^2\hat{i} + t^3\hat{j}$.
$P = \vec{F}\cdot\vec{v} = 2t\cdot t^2 + 3t^2\cdot t^3 = 2t^3 + 3t^5$ W.
Answer: (B) $(2t^3 + 3t^5)$ W. | ||||
| 29 | Motion of System of Particles and Rigid BodyCentre of mass | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on centre of mass, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Take the bend at the origin: the 3L arm along the $y$-axis and the 2L arm along the $x$-axis. A uniform rod's mass is proportional to length.
Vertical arm: mass $\frac{3}{5}M$, centre at $(0, 1.5L)$. Horizontal arm: mass $\frac{2}{5}M$, centre at $(L, 0)$.
$x_{cm} = \frac{2}{5}L = 0.4L = 4$ cm; $y_{cm} = \frac{3}{5}(1.5L) = 0.9L = 9$ cm.
Centre of mass: $4\hat{i} + 9\hat{j}$ (in cm).
Answer: (B). | ||||
| 30 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Along the incline: $ma = mg\sin60° - \mu_k mg\cos60°$.
$\frac{g}{2} = g\frac{\sqrt3}{2} - \mu_k g\frac{1}{2} \Rightarrow 1 = \sqrt3 - \mu_k \Rightarrow \mu_k = \sqrt3 - 1$.
Answer: (D) $\sqrt3 - 1$. | ||||
| 31 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $T = \frac{2u\sin\theta}{g}$, so $\frac{T_1}{T_2} = \frac{\sin(45° + \alpha)}{\sin(45° - \alpha)} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$.
(Their ranges are equal, but the times of flight are not.)
Answer: (A). | ||||
| 32 | 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 $\Delta L = \frac{FL}{AY}$ with $A = \frac{\pi d^2}{4}$; same $F$ and $Y$.
$\frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B}\left(\frac{d_B}{d_A}\right)^2 = \frac{1}{3}\cdot\frac{1}{4} = \frac{1}{12}$.
Answer: (B) 1:12. | ||||
| 33 | 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 $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$ for $f$ degrees of freedom.
Triatomic rigid (non-linear): 3 translational + 3 rotational, $f = 6$, $\gamma = \frac{4}{3}$ (III).
Diatomic non-rigid: 3 + 2 + 2 (one vibrational mode), $f = 7$, $\gamma = \frac{9}{7}$ (IV).
Monoatomic: $f = 3$, $\gamma = \frac{5}{3}$ (I).
Diatomic rigid: $f = 5$, $\gamma = \frac{7}{5}$ (II).
A-III, B-IV, C-I, D-II.
Answer: (B). | ||||
| 34 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $a\cos(1.5t)\cos(50.5t) = \frac{a}{2}\left[\cos(52t) + \cos(49t)\right]$: the two waves have $\omega_1 = 52$ rad/s and $\omega_2 = 49$ rad/s.
Beat angular frequency $= \omega_1 - \omega_2 = 3$ rad/s, so the beat period is $\frac{2\pi}{3} \approx 2.09$ s $\approx 2$ s.
Answer: (A) 2 s. | ||||
| 35 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $i_{rms}^2 = \langle i^2\rangle$. The cross term averages to zero over a cycle, so
$i_{rms}^2 = (5\sqrt2)^2 + \frac{10^2}{2} = 50 + 50 = 100$.
$i_{rms} = 10$ A.
Answer: (B) 10 Amp. | ||||
| 36 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity A triangular pyramid has 6 equal edges, so each segment has resistance $r = \frac{R}{6}$.
Between A and B: the direct edge AB ($r$), and two paths A–top–B and A–centre–B ($2r$ each). By symmetry the top and the centre vertex are at the same potential, so the edge joining them carries no current.
$\frac{1}{R_{AB}} = \frac{1}{r} + \frac{1}{2r} + \frac{1}{2r} = \frac{2}{r}$, so $R_{AB} = \frac{r}{2} = \frac{R}{12}$.
$n = 12$.
Answer: (B) 12. | ||||
| 37 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $r = \frac{mv}{qB} = \frac{p}{qB}$ and $p = \sqrt{2mE}$, so $r = \frac{\sqrt{2mE}}{qB} \propto \sqrt{E}$.
$r \propto \sqrt{E}$ starts at the origin and rises ever more slowly (concave down): the first graph.
Answer: (A). | ||||
| 38 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $q_1$ is at A, the centre of the arc, so its distance from $q_3$ stays 40 cm: its share of the energy does not change.
$q_2$ is at B, 30 cm from A on AD. At C: $BC = \sqrt{0.3^2 + 0.4^2} = 0.5$ m. At D: $BD = 0.4 - 0.3 = 0.1$ m.
$\Delta U = \frac{q_2q_3}{4\pi\epsilon_0}\left(\frac{1}{0.1} - \frac{1}{0.5}\right) = \frac{q_3}{4\pi\epsilon_0}(10 - 2)q_2 = \frac{q_3}{4\pi\epsilon_0}\cdot8q_2$.
$K = 8q_2$ (distances in metres).
Answer: (D) $8q_2$. | ||||
| 39 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter With the core, $B = \mu_0(1 + \chi)nI$, so $\frac{\Delta B}{B_0} = \chi = 1.2\times10^{-5}$.
Percentage increase $= 1.2\times10^{-5}\times100 = 1.2\times10^{-3}\% = \frac{6}{5}\times10^{-3}\%$.
Answer: (B). | ||||
| 40 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Take the particle entering region 2 perpendicular to the interface. In each region it moves on a semicircle of radius $r = \frac{mv}{qB}$: $r_2 = \frac{mv}{qB_2}$, $r_1 = \frac{mv}{qB_1}$.
Both fields point the same way (into the paper), so it turns in the same sense throughout. Going down into region 2 and turning, say, clockwise it comes back up a distance $2r_2$ to the left; going up into region 1 and turning clockwise again it comes back down a distance $2r_1$ to the right.
Net displacement along the interface $= 2r_1 - 2r_2 = \frac{2mv}{q}\left(\frac{1}{B_1} - \frac{1}{B_2}\right) = \frac{mv}{qB_1}\left(1 - \frac{B_1}{B_2}\right)\times2$.
Answer: (C). | ||||
| 41 | 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 Two oscillations of equal amplitude with phase difference $\phi = \frac{\pi}{3}$ (taken along the same direction):
$E_R = \sqrt{E_0^2 + E_0^2 + 2E_0^2\cos\frac{\pi}{3}} = \sqrt{3}E_0 \approx 1.73E_0$.
Answer: (A) $1.7E_0$. | ||||
| 42 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens maker's formula: $\frac{1}{f} = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
In air: $(1.6 - 1)K = \frac{1}{12}$. In water: $\left(\frac{1.6}{1.28} - 1\right)K = 0.25K$.
$\frac{f_w}{f_a} = \frac{0.6}{0.25} = 2.4$, so $f_w = 2.4\times12 = 28.8$ cm $= 288$ mm.
Answer: (B) 288 mm. | ||||
| 43 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei The largest wavelength of a series is its first line.
Lyman ($2 \to 1$): $\frac{1}{\lambda_L} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$.
Balmer ($3 \to 2$): $\frac{1}{\lambda_B} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$.
$\frac{\lambda_L}{\lambda_B} = \frac{5/36}{3/4} = \frac{5}{27}$.
Answer: (B) 5 : 27. | ||||
| 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 The second excited state is $n = 3$. $E_n = -\frac{13.6Z^2}{n^2}$ eV.
$\Delta E = 13.6Z^2\left(1 - \frac{1}{9}\right) = \frac{108.8}{9}Z^2$ eV.
$\frac{108.8}{9}Z^2 = 108.8 \Rightarrow Z^2 = 9$, $Z = 3$.
Answer: (B) 3. | ||||
| 45 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Without the Zener the 400 Ω branch would get $12\times\frac{400}{500} = 9.6$ V, above 4 V, so the Zener breaks down and holds the parallel branches at 4 V (an ideal voltmeter draws no current).
Ammeter current $= \frac{4}{400} = 0.01$ A $= 10$ mA.
(For interest: the 100 Ω resistor carries $\frac{12 - 4}{100} = 80$ mA, of which 70 mA goes through the Zener.)
Answer: (C) 10 mA. | ||||
| 46 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Mass M, radius R for each. The axis PQ is a diameter of disc A and passes through the point where B and C touch, so the centres of B and C are each R from the axis.
Disc about a diameter: $I = \frac{MR^2}{4}$.
Solid sphere B (parallel axes): $\frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
Shell C: $\frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$.
Total $= MR^2\left(\frac{1}{4} + \frac{7}{5} + \frac{5}{3}\right) = \frac{15 + 84 + 100}{60}MR^2 = \frac{199}{60}MR^2$.
With $MR^2 = 4I$: total $= \frac{199}{15}I$, so $x = 199$.
Answer: 199. | ||||
| 47 | Properties of Bulk MatterHeat transfer | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on heat transfer, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter $T = 1727 + 273 = 2000$ K, $T^4 = 1.6\times10^{13}$ K$^4$.
Radiating area (curved surface): $A = \pi dL = 3.14\times0.5\times10^{-3}\times0.1 = 1.57\times10^{-4}$ m$^2$.
$P = e\sigma AT^4$: $\sigma AT^4 = 6.0\times10^{-8}\times1.57\times10^{-4}\times1.6\times10^{13} = 150.72$ W.
$e = \frac{94.2}{150.72} = 0.625 = \frac{5}{8}$, so $x = 5$.
Answer: 5. | ||||
| 48 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics The work in a cycle is the area enclosed on the P–V diagram.
The cycle is a circle on the graph: P from 300 to 500 kPa (semi-axis $100\times10^3$ Pa), V from 150 to 350 cm$^3$ (semi-axis $100\times10^{-6}$ m$^3$).
Area $= \pi ab = 3.14\times(100\times10^3)\times(100\times10^{-6}) = 31.4$ J $= 314\times10^{-1}$ J.
Answer: 314. | ||||
| 49 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalHard | |
A hard numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Both branches are across $V_{in}$; take the bottom wire as zero. Left branch: R on top, C below, A between them. Right branch: C on top, R below, B between them. Let $Z = \frac{1}{j\omega C}$.
$V_A = V_{in}\frac{Z}{R + Z}$, $V_B = V_{in}\frac{R}{R + Z}$, so $V_B - V_A = V_{in}\frac{R - Z}{R + Z} = V_{in}\frac{j\omega CR - 1}{j\omega CR + 1}$.
This factor has magnitude 1 and phase $\pi - 2\tan^{-1}(\omega CR)$ (relative to $V_{in}$).
For a 90° difference, $2\tan^{-1}(\omega CR) = 90°$, so $\omega CR = 1$.
$\omega = \frac{1}{RC} = \frac{1}{10^5\times10^{-10}} = 10^5$ rad/s, so $x = 5$.
Answer: 5. | ||||
| 50 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics A layer of thickness $t$ and index $\mu$ shifts an object up by $t\left(1 - \frac{1}{\mu}\right)$ (near-normal viewing); shifts add.
$60\left(1 - \frac{1}{1.2}\right) + H\left(1 - \frac{1}{1.6}\right) = 40$
$10 + \frac{3H}{8} = 40 \Rightarrow H = 80$ cm.
Answer: 80. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | 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 Partial pressures are in the ratio of moles (Dalton's law), so the total pressure is not needed. Take 100 g of air:
$n(\mathrm{N_2}) = \frac{70.0}{28} = 2.5$ mol, $n(\mathrm{O_2}) = \frac{27.0}{32} = 0.84375$ mol, $n(\mathrm{Ar}) = \frac{3.0}{40} = 0.075$ mol.
(i) $\frac{p_{N_2}}{p_{O_2}} = \frac{2.5}{0.84375} = 2.96$.
(ii) $\frac{p_{O_2}}{p_{Ar}} = \frac{0.84375}{0.075} = 11.25 \approx 11.2$.
Answer: (B) 2.96, 11.2. | ||||
| 52 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | StatementsMedium | |
A medium statements question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Threshold wavelength: $\lambda_0 = \frac{c}{\nu_0} = \frac{3\times10^8}{5.16\times10^{14}} = 5.81\times10^{-7}$ m $= 581$ nm. Light of shorter wavelength than this ejects electrons.
A: yellow lies at about 570–580 nm on the spectrum shown, just below 581 nm, so electrons are ejected and current flows. True.
B: dimmer light means fewer photons, so fewer photoelectrons and a smaller current. True.
C: red light (about 650–750 nm) is beyond the threshold wavelength: no photoelectrons at all, so no current. False.
D: blue light (about 450–490 nm) is well inside the threshold. True.
E: white light contains the blue, green and yellow parts, so current flows. True.
Correct: A, B, D and E.
Answer: (D). | ||||
| 53 | ThermodynamicsHess law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Break the change into three steps (all in J):
1. Cool water from 10°C to 0°C: $\Delta H_1 = y(0 - 10) = -10y$.
2. Freeze at 0°C (the reverse of fusion): $\Delta H_2 = -x$ kJ $= -1000x$ J.
3. Cool ice from 0°C to –10°C: $\Delta H_3 = z(-10 - 0) = -10z$.
Total $= -1000x - 10y - 10z = -10(100x + y + z)$ J.
($-x - 10y - 10z$ mixes kJ with J.)
Answer: (B) $-10(100x + y + z)$. | ||||
| 54 | EquilibriumpH and buffers | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium pH 1.0 means $[\mathrm{H^+}] = 0.1$ M. Doubling the volume halves it: $[\mathrm{H^+}] = 0.05$ M.
pH $= -\log(5\times10^{-2}) = 2 - \log5 = 2 - (1 - \log2) = 1 + 0.30 = 1.3$.
Answer: (D) increase to 1.3. | ||||
| 55 | ElectrochemistryConductance | Class 12: Electrochemistry | StatementsMedium | |
A medium statements question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Mohr's salt is $\mathrm{FeSO_4\cdot(NH_4)_2SO_4\cdot6H_2O}$. In solution it gives $\mathrm{Fe^{2+}}$, $\mathrm{NH_4^+}$ and $\mathrm{SO_4^{2-}}$ only (the water is not ionic). Statement I is true.
By Kohlrausch's law, one formula unit gives 1 $\mathrm{Fe^{2+}}$, 2 $\mathrm{NH_4^+}$ and 2 $\mathrm{SO_4^{2-}}$:
$\Lambda_m^\infty = x_1 + 2x_2 + 2x_3$, not $x_1 + x_2 + 2x_3$. Statement II is false.
Answer: (C) Statement I is true but Statement II is false. | ||||
| 56 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Let the initial pressure of A be $P_0$ and $p$ of it react: total $= (P_0 - p) + 2p + p = P_0 + 2p$.
At $t = \infty$ all of A has reacted: $3P_0 = 240$, so $P_0 = 80$ mm Hg. (First option correct.)
At 10 min: $80 + 2p = 160$, $p = 40$, so $p_A = 80 - 40 = 40$ mm Hg. (Second option correct.)
Half of A is gone in 10 min, so $t_{1/2} = 10$ min and $k = \frac{0.693}{10} = 0.0693$ min$^{-1}$, not 1.693 min$^{-1}$. (Third option incorrect.)
A first-order reaction only approaches completion: $p_A = P_0e^{-kt}$ never becomes exactly zero. (Fourth option is taken as correct.)
Answer: (C). | ||||
| 57 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Before: first-order growth, $\frac{dN}{dt} = kN$, so $\frac{N}{N_0} = e^{kt}$: it starts at 1 when $t = 0$ and rises exponentially (concave up).
After: $r = k'N^2$, a parabola through the origin, rising ever more steeply with N (concave up), not falling and not a straight line.
Only the fourth pair has both: $N/N_0$ starting at 1 and rising exponentially, and r against N an upward-curving parabola from the origin. (The third pair starts $N/N_0$ at 0; the first has r falling; the second has r linear.)
Answer: (D). | ||||
| 58 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listMedium | |
A medium matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{ICl_2^-}$: I has 7 + 1 = 8 valence electrons; 2 bond pairs and 3 lone pairs: 2 : 3 (III).
$\mathrm{H_2O}$: 2 bond pairs, 2 lone pairs: 2 : 2 (IV).
$\mathrm{SO_2}$: S forms two S=O double bonds, i.e. 4 bonding pairs, and keeps 1 lone pair: 4 : 1 (II).
$\mathrm{XeF_4}$: 4 bond pairs, 2 lone pairs: 4 : 2 (I).
A-III, B-IV, C-II, D-I.
Answer: (C). | ||||
| 59 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Enthalpies of atomisation (kJ mol$^{-1}$, approximately): Cr 397, Fe 416, Co 425, Ni 430. Cr is lowest.
Cr is $[\mathrm{Ar}]3d^54s^1$: valence electrons $= 5 + 1 = 6$.
Answer: (C) 6. | ||||
| 60 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter In group 14 the first ionisation enthalpies are C 1086, Si 786, Ge 761, Sn 708, Pb 715 kJ mol$^{-1}$, so A is Sn and B is Pb.
Going down the group the inert pair effect makes the +2 state more stable.
$\mathrm{Sn^{2+}}$ readily goes to the more stable $\mathrm{Sn^{4+}}$: a reducing agent.
$\mathrm{Pb^{4+}}$ readily goes to the more stable $\mathrm{Pb^{2+}}$: an oxidising agent.
Answer: (A) reducing and oxidising. | ||||
| 61 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In the 3d series vanadium has the highest enthalpy of atomisation (about 515 kJ mol$^{-1}$).
Its aquated ions: $\mathrm{V^{2+}}$ violet, $\mathrm{V^{3+}}$ green, $\mathrm{VO^{2+}}$ blue. So $\mathrm{M^{n+}}$ is $\mathrm{V^{3+}}$.
Its oxide is $\mathrm{V_2O_3}$; in the low oxidation state +3 the oxide is basic (while $\mathrm{V_2O_5}$ is amphoteric).
Answer: (A) basic. | ||||
| 62 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A gives AgCl, so $\mathrm{Cl^-}$ is outside the sphere: $\mathrm{[Co(NH_3)_5SO_4]Cl}$.
B gives $\mathrm{BaSO_4}$, so $\mathrm{SO_4^{2-}}$ is outside: $\mathrm{[Co(NH_3)_5Cl]SO_4}$.
Same composition, different ions in solution because a ligand and a counter-ion swap places: ionisation isomerism.
Answer: (C) Ionisation isomerism. | ||||
| 63 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter An ammonium salt warmed with NaOH gives ammonia: $\mathrm{NH_4^+ + OH^- \rightarrow NH_3 + H_2O}$. So X is $\mathrm{NH_3}$.
Ammonia with Nessler's reagent ($\mathrm{K_2HgI_4}$ in KOH) gives a brown precipitate (iodide of Millon's base).
HCl and $\mathrm{NH_4Cl}$ are not gases given off by NaOH, and HgO is not the reagent.
Answer: (D) X = $\mathrm{NH_3}$ and Y = $\mathrm{K_2HgI_4 + KOH}$. | ||||
| 64 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | StatementsMedium | |
A medium statements question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Statement I: the small dimethyl ether molecule forms hydrogen bonds with water through its oxygen and is freely soluble; in diethyl ether the larger hydrocarbon part makes it only slightly soluble (about 7 g per 100 mL). True.
Statement II: sodium does not react with an ether, so it removes traces of water from diethyl ether; it reacts with ethanol ($\mathrm{2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2}$), so it cannot be used to dry it. True.
Answer: (A) Both Statement I and Statement II are true. | ||||
| 65 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The structure is $\mathrm{CH_3-CH=C(CH_3)-CH_2Br}$.
Longest chain containing the double bond: 4 carbons. Number from the $\mathrm{CH_2Br}$ end so the double bond gets the lowest locant (2 either way) and then the substituents the lowest set: Br on C1, $\mathrm{CH_3}$ on C2, double bond C2=C3.
Name: 1-Bromo-2-methylbut-2-ene. (Numbering from the other end gives 4-bromo-3-methyl, a higher set; "prop-2-ene" ignores the four-carbon chain.)
Answer: (B). | ||||
| 66 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes A: 1-bromo-2-methylcyclohexane with NaOEt, a strong base: E2 elimination gives an alkene (1-methylcyclohexene). Can be applied.
B: 2-bromobutane with aqueous KOH: in water $\mathrm{OH^-}$ acts mainly as a nucleophile, giving butan-2-ol by substitution. Not a route to the alkene.
C: tert-butyl bromide with NaOMe: a 3° halide with a strong base eliminates to give 2-methylpropene. Can be applied.
D: phenol with $\mathrm{Na_2Cr_2O_7/H_2SO_4}$ is oxidised to benzoquinone: no elimination to an alkene.
E: a 3° alcohol over Cu at 573 K is dehydrated to the alkene (2-methylpropene). Can be applied.
Cannot be applied: B and D.
Answer: (C) B & D Only. | ||||
| 67 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | StatementsMedium | |
A medium statements question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Statement I: $\mathrm{CH_3CH=CHCH_3}$ (cis or trans) is cut at the double bond into two $\mathrm{CH_3CHO}$, ethanal. True.
Statement II: 3,6-dimethyloct-4-ene is $\mathrm{CH_3CH_2CH(CH_3)-CH=CH-CH(CH_3)CH_2CH_3}$. Ozonolysis gives two molecules of $\mathrm{CH_3CH_2CH(CH_3)CHO}$ (2-methylbutanal), whose C2 carries H, $\mathrm{CH_3}$, $\mathrm{C_2H_5}$ and CHO: a chiral carbon. False.
Answer: (C) Statement I is true but Statement II is false. | ||||
| 68 | 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 $\mathrm{NaHCO_3}$ releases $\mathrm{CO_2}$ only with an acid stronger than carbonic acid ($\mathrm{p}K_a \approx 6.4$).
4-Nitrobenzoic acid ($\mathrm{p}K_a \approx 3.4$): reacts.
Anilinium chloride, $\mathrm{PhNH_3^+Cl^-}$ ($\mathrm{p}K_a \approx 4.6$): reacts.
Picric acid, 2,4,6-trinitrophenol ($\mathrm{p}K_a \approx 0.4$): reacts.
3-Nitrophenol ($\mathrm{p}K_a \approx 8.4$): weaker than carbonic acid, so no effervescence. (A meta nitro group cannot withdraw by resonance.)
Answer: (C) 3-nitrophenol. | ||||
| 69 | AminesClassification and identification of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines The carbylamine test (warming with $\mathrm{CHCl_3}$ and alcoholic KOH, giving a foul-smelling isocyanide) is given only by primary amines.
A: aniline, $\mathrm{C_6H_5NH_2}$, primary: positive.
B: dimethylamine, secondary: negative.
C: methylamine, primary: positive.
D: trimethylamine, tertiary: negative.
E: N-methylaniline, secondary: negative.
Answer: (C) A and C Only. | ||||
| 70 | BiomoleculesCarbohydrates | Class 12: Biomolecules | StatementsMedium | |
A medium statements question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Statement I: the fructose in sucrose is D-(–)-fructose, which is laevorotatory. Sucrose is made from D-(+)-glucose and D-(–)-fructose, and its hydrolysis gives D-(+)-glucose and D-(–)-fructose. Writing D-(+) fructose makes the statement false.
Statement II: hydrolysis turns dextrorotatory sucrose ($+66.5°$) into a laevorotatory mixture (glucose $+52.5°$, fructose $-92.4°$); the sign of rotation is inverted, so the product is called invert sugar. True.
Answer: (D) Statement I is false but Statement II is true. | ||||
| 71 | ElectrochemistryNernst equation | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Electrolysis deposits metal and removes the ions from solution:
1 F deposits $\frac{1}{2}$ mol Cu: $[\mathrm{Cu^{2+}}] = 1.5 - 0.5 = 1.0$ M.
0.1 F deposits 0.1 mol Ag: $[\mathrm{Ag^+}] = 0.2 - 0.1 = 0.1$ M.
Cell: $\mathrm{Cu|Cu^{2+}(1.0\,M)||Ag^+(0.1\,M)|Ag}$, reaction $\mathrm{Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag}$, $n = 2$.
$E^\circ = 0.80 - 0.34 = 0.46$ V.
$E = 0.46 - \frac{0.06}{2}\log\frac{[\mathrm{Cu^{2+}}]}{[\mathrm{Ag^+}]^2} = 0.46 - 0.03\log\frac{1.0}{0.01} = 0.46 - 0.06 = 0.40$ V $= 400$ mV.
Answer: 400. | ||||
| 72 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\mathrm{MX_3 \rightarrow M^{3+} + 3X^-}$: each formula unit gives $n = 4$ ions.
$i = 1 + (n - 1)\alpha = 1 + 3\alpha = 2 \Rightarrow \alpha = \frac{1}{3}$.
Percentage dissociation $= 33.3\% \approx 33$.
Answer: 33. | ||||
| 73 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $d^2sp^3$ (inner-orbital) needs a strong-field ligand that pairs the d electrons.
$\mathrm{[FeF_6]^{3-}}$: $\mathrm{Fe^{3+}}$ $d^5$, weak $\mathrm{F^-}$: $sp^3d^2$, 5 unpaired. Not counted.
$\mathrm{[Fe(CN)_6]^{3-}}$: $d^5$ low spin, $d^2sp^3$, 1 unpaired: paramagnetic. Counted.
$\mathrm{[Mn(CN)_6]^{3-}}$: $\mathrm{Mn^{3+}}$ $d^4$ low spin, $d^2sp^3$, 2 unpaired: paramagnetic. Counted.
$\mathrm{[Co(C_2O_4)_3]^{3-}}$: $\mathrm{Co^{3+}}$ $d^6$ low spin, $d^2sp^3$, 0 unpaired: diamagnetic. Not counted.
$\mathrm{[MnCl_6]^{3-}}$: weak $\mathrm{Cl^-}$, $sp^3d^2$, 4 unpaired. Not counted.
$\mathrm{[CoF_6]^{3-}}$: weak $\mathrm{F^-}$, $sp^3d^2$, 4 unpaired. Not counted.
Number $= 2$.
Answer: 2. | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Carbon in the $\mathrm{CO_2}$: $\frac{12}{44}\times220 = 60$ mg.
Percentage of carbon $= \frac{60}{500}\times100 = 12\%$.
Answer: 12. | ||||
| 75 | Some Basic Concepts of ChemistryEmpirical and molecular formula | Class 11: Some Basic Concepts of Chemistry | NumericalMedium | |
A medium numerical question on empirical and molecular formula, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Count the atoms in the structure: two benzene rings (12 C) plus the $\mathrm{CH_2-CH(NH_2)-COOH}$ chain (3 C): 15 C; H: 2 + 2 ring H, 1 OH, 2 $\mathrm{CH_2}$, 1 CH, 2 $\mathrm{NH_2}$, 1 COOH = 11 H; 4 I; 1 N; O: 1 (OH) + 1 (ether) + 2 (COOH) = 4.
Formula $\mathrm{C_{15}H_{11}I_4NO_4}$; molar mass $= 180 + 11 + 508 + 14 + 64 = 777$ g mol$^{-1}$.
Percentage of iodine $= \frac{508}{777}\times100 = 65.4\% \approx 65$.
Answer: 65. | ||||
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