JEE (Main) · 2024
JEE (Main) 2024 — 27 January, 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
- Limits, Continuity and Differentiability4
- Integral Calculus3
- Sets, Relations and Functions2
11 easy · 17 medium · 2 hard
Physics
30 Q
- Electrostatics3
- Kinematics2
- Laws of Motion2
22 easy · 8 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Hydrocarbons3
- Chemical Bonding and Molecular Structure2
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 FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct 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: the empty set $\phi \in M$ and $\phi \cap \phi = \phi$, so $(\phi, \phi) \notin R$. R is not reflexive.
Symmetric: $A \cap B = B \cap A$, so $(A, B) \in R \Rightarrow (B, A) \in R$. R is symmetric.
Transitive: take $A = \{1\}$, $B = \{1, 2\}$, $C = \{2\}$. Then $A \cap B = \{1\} \neq \phi$ and $B \cap C = \{2\} \neq \phi$, but $A \cap C = \phi$. R is not transitive.
So R is symmetric only.
Answer: (C). | ||||
| 2 | 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 One-one? $f(2) = 2$ and $f(4) = 2$: two different inputs give the same value, so f is not one-one.
Onto? Every value of f is a prime. The natural numbers 1, 4, 6, … are not primes, so they are never taken; the range is the set of primes, not $\mathbf{N}$. f is not onto.
Answer: (D) neither one-one nor onto. | ||||
| 3 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|z - i| = |z + i|$: z is equidistant from $i$ and $-i$, so it lies on their perpendicular bisector, the real axis $y = 0$.
$|z + i| = |z - 1|$: z is equidistant from $-i$ $(0, -1)$ and $1$ $(1, 0)$; with $z = x + iy$: $x^2 + (y+1)^2 = (x-1)^2 + y^2 \Rightarrow 2y + 1 = -2x + 1 \Rightarrow y = -x$.
The two lines meet only at $z = 0$ (the circumcentre of the three points $i, -i, 1$). Check: $|0 - i| = |0 + i| = |0 - 1| = 1$.
So $n(S) = 1$.
Answer: (B). | ||||
| 4 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | StatementsEasy | |
A easy statements question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Multiply: the top-left $2\times2$ block of $f(x)f(y)$ is
$\begin{bmatrix} \cos x\cos y - \sin x\sin y & -\cos x\sin y - \sin x\cos y \\ \sin x\cos y + \cos x\sin y & -\sin x\sin y + \cos x\cos y \end{bmatrix} = \begin{bmatrix} \cos(x+y) & -\sin(x+y) \\ \sin(x+y) & \cos(x+y) \end{bmatrix}$,
and the third row and column stay $(0, 0, 1)$. So $f(x)f(y) = f(x+y)$: Statement II is true.
Then $f(x)f(-x) = f(0) = I$ and likewise $f(-x)f(x) = I$, so $f(-x) = f(x)^{-1}$: Statement I is true.
Answer: (A). | ||||
| 5 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Use ${}^{n}C_{r+1} = \frac{n}{r+1}\,{}^{n-1}C_r$. Then
$k^2 - 8 = \frac{{}^{n-1}C_r}{{}^{n}C_{r+1}} = \frac{r+1}{n}$.
For ${}^{n-1}C_r$ to be defined, $0 \le r \le n-1$, so $1 \le r + 1 \le n$ and $0 < \frac{r+1}{n} \le 1$.
Hence $0 < k^2 - 8 \le 1$, i.e. $8 < k^2 \le 9$.
Among the options (all with k positive) this is $2\sqrt{2} < k \le 3$.
Answer: (A). | ||||
| 6 | 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 The sum of all coefficients of a polynomial is its value at $x = 1$.
$A = (1 - 3 + 10)^n = 8^n = 2^{3n}$.
$B = (1 + 1)^n = 2^n$.
So $A = (2^n)^3 = B^3$.
Answer: (B). | ||||
| 7 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series First AP: $4 + 5(n-1)$; its $25^{\text{th}}$ term is $4 + 120 = 124$.
Second AP: $3n$; its $37^{\text{th}}$ term is $111$.
The common terms form an AP whose difference is $\mathrm{lcm}(5, 3) = 15$. The first common term is 9 (9 is in both). So the common terms are $9, 24, 39, \ldots$, i.e. $9 + 15k$.
They must not exceed $\min(124, 111) = 111$: $9 + 15k \le 111 \Rightarrow k \le 6.8$, so $k = 0, 1, \ldots, 6$.
Number of common terms = 7.
Answer: (B). | ||||
| 8 | 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 Left of 3: $x^2 - 7x + 12 = (x-3)(x-4)$. For $x$ slightly less than 3 both factors are negative, so $|x^2 - 7x + 12| = (x-3)(x-4)$, while $7x - 12 - x^2 = -(x-3)(x-4)$. Hence $f(x) = -\frac{a}{b}$ near $3^-$, and $f(3^-) = -\frac{a}{b}$.
Right of 3: for $3 < x < 4$, $[x] = 3$, so the exponent is $\frac{\sin(x-3)}{x-3} \to 1$, and $f(3^+) = 2^1 = 2$.
Continuity needs $f(3^-) = f(3^+) = f(3)$: $b = 2$ and $-\frac{a}{b} = 2 \Rightarrow a = -4$.
Exactly one pair, $(a, b) = (-4, 2)$.
Answer: (C) 1. | ||||
| 9 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For a: rationalise twice.
$\sqrt{1 + \sqrt{1+x^4}} - \sqrt{2} = \frac{\sqrt{1+x^4} - 1}{\sqrt{1 + \sqrt{1+x^4}} + \sqrt{2}} = \frac{x^4}{(\sqrt{1+x^4} + 1)(\sqrt{1 + \sqrt{1+x^4}} + \sqrt{2})}$.
Dividing by $x^4$ and letting $x \to 0$: $a = \frac{1}{2 \cdot 2\sqrt{2}} = \frac{1}{4\sqrt{2}}$.
For b: $\sqrt{2} - \sqrt{1 + \cos x} = \frac{1 - \cos x}{\sqrt{2} + \sqrt{1+\cos x}}$, so
$b = \lim \frac{\sin^2 x}{1 - \cos x}(\sqrt{2} + \sqrt{1 + \cos x}) = \lim (1 + \cos x)(\sqrt{2} + \sqrt{1+\cos x}) = 2 \cdot 2\sqrt{2} = 4\sqrt{2}$.
$ab^3 = \frac{(4\sqrt{2})^3}{4\sqrt{2}} = (4\sqrt{2})^2 = 32$.
Answer: (D). | ||||
| 10 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The centre of the circle is $C(2, 8)$.
The shortest distance from C to the parabola is along a normal to the parabola through C. The normal at $(t^2, 2t)$ is $y = -tx + 2t + t^3$.
Through $(2, 8)$: $8 = -2t + 2t + t^3 \Rightarrow t^3 = 8 \Rightarrow t = 2$, the only real root.
Foot of the normal: $(4, 4)$.
$d^2 = (4 - 2)^2 + (4 - 8)^2 = 4 + 16 = 20$.
Answer: (B). | ||||
| 11 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Rationalise: $\frac{1}{\sqrt{3+x} + \sqrt{1+x}} = \frac{\sqrt{3+x} - \sqrt{1+x}}{(3+x) - (1+x)} = \frac{\sqrt{3+x} - \sqrt{1+x}}{2}$.
$\int_0^1 = \frac{1}{2}\cdot\frac{2}{3}\left[(3+x)^{3/2} - (1+x)^{3/2}\right]_0^1 = \frac{1}{3}\left[(8 - 2\sqrt{2}) - (3\sqrt{3} - 1)\right] = 3 - \frac{2}{3}\sqrt{2} - \sqrt{3}$.
So $a = 3$, $b = -\frac{2}{3}$, $c = -1$.
$2a + 3b - 4c = 6 - 2 + 4 = 8$.
Answer: (C). | ||||
| 12 | 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 Orthocentre of $P(1, 2)$, $Q(2, 3)$, $R(3, 1)$.
QR has slope $\frac{1-3}{3-2} = -2$; the altitude from P has slope $\frac{1}{2}$: $y - 2 = \frac{1}{2}(x - 1)$.
PQ has slope 1; the altitude from R has slope $-1$: $y = 4 - x$.
Solving: $4 - x - 2 = \frac{x-1}{2} \Rightarrow x = \frac{5}{3}$, $y = \frac{7}{3}$. So $a = \frac{5}{3}$, $b = \frac{7}{3}$, and $a + b = 4$.
Use $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ with $a + b = 4$: $4(4 - x) - (4 - x)^2 = 4x - x^2$, so
$I_1 = \int_a^b (4 - x)\sin(4x - x^2)\,dx = 4I_2 - I_1 \Rightarrow I_1 = 2I_2$.
$36\frac{I_1}{I_2} = 72$.
Answer: (A). | ||||
| 13 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separating variables: $x = 2e^{-at}$ and $y = e^{-bt}$.
$3y(1) = 2x(1)$: $3e^{-b} = 4e^{-a} \Rightarrow e^{a-b} = \frac{4}{3}$.
$x(t) = y(t)$: $2e^{-at} = e^{-bt} \Rightarrow 2 = e^{(a-b)t} = \left(\frac{4}{3}\right)^t$.
So $t = \log_{\frac{4}{3}} 2$.
Answer: (C). | ||||
| 14 | CirclesStandard and general form | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections The circle through $(0,0)$, $(1,0)$, $(0,1)$: $x^2 + y^2 + 2gx + 2fy = 0$ with $1 + 2g = 0$ and $1 + 2f = 0$, so $x^2 + y^2 - x - y = 0$.
Put $(2k, 3k)$: $4k^2 + 9k^2 - 2k - 3k = 0 \Rightarrow 13k^2 - 5k = 0$.
$k = 0$ gives $(0, 0)$ again, which the points being distinct excludes; so $k = \frac{5}{13}$.
Answer: (D). | ||||
| 15 | Straight LinesSection formula | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on section formula, from Straight Lines. NCERT: Class 11: Straight Lines The line meets the axes at $(5, 0)$ and $(0, 4)$. The points of trisection divide this segment in the ratios $1:2$ and $2:1$:
$\left(\frac{10}{3}, \frac{4}{3}\right)$ and $\left(\frac{5}{3}, \frac{8}{3}\right)$.
Slopes of $L_1$, $L_2$ through the origin: $m_1 = \frac{2}{5}$, $m_2 = \frac{8}{5}$.
$\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right| = \frac{6/5}{1 + 16/25} = \frac{6}{5}\cdot\frac{25}{41} = \frac{30}{41}$.
Answer: (D). | ||||
| 16 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Chord with a given mid-point: $T = S_1$.
$\frac{x}{25} + \frac{(2/5)y}{16} = \frac{1}{25} + \frac{4/25}{16}$. Multiplying by 400: $16x + 10y = 16 + 4 = 20$, i.e. $8x + 5y = 10$.
Intersect with $16x^2 + 25y^2 = 400$, using $5y = 10 - 8x$: $16x^2 + (10 - 8x)^2 = 400 \Rightarrow 80x^2 - 160x - 300 = 0 \Rightarrow 4x^2 - 8x - 15 = 0$.
$x_1 + x_2 = 2$ (mid-point $x = 1$, as it should be), $x_1x_2 = -\frac{15}{4}$, so $(x_1 - x_2)^2 = 4 + 15 = 19$.
On the chord $\Delta y = -\frac{8}{5}\Delta x$, so length$^2 = 19\left(1 + \frac{64}{25}\right) = \frac{19 \times 89}{25} = \frac{1691}{25}$.
Length $= \frac{\sqrt{1691}}{5}$.
Answer: (C). | ||||
| 17 | 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 Go from $P(7, -2, 11)$ in the direction $(2, -3, 6)$ (whose length is $\sqrt{4 + 9 + 36} = 7$): a general point is $(7 + 2t, -2 - 3t, 11 + 6t)$.
It lies on the first line when $y = 4$ and $x - 6 = \frac{z - 8}{3}$.
$-2 - 3t = 4 \Rightarrow t = -2$, giving $Q(3, 4, -1)$. Check: $x - 6 = -3$ and $\frac{z-8}{3} = \frac{-9}{3} = -3$. ✓
$PQ = |t| \times 7 = 14$.
Answer: (B). | ||||
| 18 | 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 $A(4, -1, 0)$, $B(\lambda, -1, 2)$; directions $\vec{d_1} = (1, 2, -3)$, $\vec{d_2} = (2, 4, -5)$.
$\vec{d_1}\times\vec{d_2} = (2\cdot(-5) - (-3)\cdot4,\ (-3)\cdot2 - 1\cdot(-5),\ 1\cdot4 - 2\cdot2) = (2, -1, 0)$, of length $\sqrt{5}$.
$\vec{AB} = (\lambda - 4, 0, 2)$, and $\vec{AB}\cdot(\vec{d_1}\times\vec{d_2}) = 2(\lambda - 4)$.
Shortest distance $= \frac{|2(\lambda - 4)|}{\sqrt{5}} = \frac{6}{\sqrt{5}} \Rightarrow |\lambda - 4| = 3 \Rightarrow \lambda = 7$ or $1$.
Sum $= 8$.
Answer: (C). | ||||
| 19 | 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 Expand: $\vec{a}\cdot(\vec{c}\times\vec{b}) - \vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{c}$.
$\vec{a}\cdot(\vec{c}\times\vec{b}) = [\vec{a}\ \vec{c}\ \vec{b}] = (\vec{a}\times\vec{c})\cdot\vec{b} = \vec{b}\cdot\vec{b} = 9 \times 3 = 27$.
$\vec{a}\cdot\vec{b} = 3(1 - 2 + 1) = 0$, and $\vec{a}\cdot\vec{c} = 3$.
Value $= 27 - 0 - 3 = 24$.
(For completeness, $\vec{c} = \frac{(\vec{a}\cdot\vec{c})\vec{a} - \vec{a}\times\vec{b}}{|\vec{a}|^2} = \frac{(3, 6, 3) - (9, 0, -9)}{6} = (-1, 1, 2)$, which indeed gives $\vec{a}\times\vec{c} = (3, -3, 3)$.)
Answer: (B). | ||||
| 20 | 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 $\left(\sum a_k\right)^2 = \sum a_k^2 + 2\sum_{k<j} a_ka_j \Rightarrow 2500 = \sum a_k^2 + 2200 \Rightarrow \sum a_k^2 = 300$.
Variance $= \frac{300}{10} - \left(\frac{50}{10}\right)^2 = 30 - 25 = 5$.
Standard deviation $= \sqrt{5}$.
Answer: (C). | ||||
| 21 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations The roots of $x^2 + x + 1 = 0$ are the non-real cube roots of unity: $\alpha^3 = 1$ and $1 + \alpha + \alpha^2 = 0$.
$1 + \alpha = -\alpha^2$, so $(1 + \alpha)^7 = -\alpha^{14} = -\alpha^{12}\alpha^2 = -\alpha^2 = 1 + \alpha$.
So $A = 1$, $B = 1$, $C = 0$, and $5(3A - 2B - C) = 5(3 - 2 - 0) = 5$.
(Adding any multiple $t \ge 0$ of $1 + \alpha + \alpha^2 = 0$ gives $A = B = 1 + t$, $C = t$, and $3A - 2B - C = 1$ still, so the answer does not depend on the representation.)
Answer: 5. | ||||
| 22 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $AB = [AB_1, AB_2, AB_3] = C = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$, with $|C| = 3$.
$|A| = 2(1 - 0) - 0 + 1(0 - 1) = 1$, so $\alpha = |B| = \frac{|C|}{|A|} = 3$.
$A^{-1} = \mathrm{adj}A = \begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{bmatrix}$ (check: $AA^{-1} = I$), and $B = A^{-1}C$.
Diagonal of B:
$b_{11} = (1, 0, -1)\cdot(1, 0, 0) = 1$,
$b_{22} = (-1, 1, 1)\cdot(2, 3, 0) = 1$,
$b_{33} = (-1, 0, 2)\cdot(3, 2, 1) = -1$.
$\beta = 1 + 1 - 1 = 1$.
$\alpha^3 + \beta^3 = 27 + 1 = 28$.
Answer: 28. | ||||
| 23 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $g(x) = f(x) - \log_e x - x$. The condition says $g(x) - g(y) \ge 0$ for all x, y. Swapping x and y gives $g(y) - g(x) \ge 0$ too, so $g(x) = g(y)$: g is constant.
Hence $f(x) = \log_e x + x + c$ and $f'(x) = \frac{1}{x} + 1$.
$f'\left(\frac{1}{n^2}\right) = n^2 + 1$.
$\sum_{n=1}^{20}(n^2 + 1) = \frac{20 \cdot 21 \cdot 41}{6} + 20 = 2870 + 20 = 2890$.
Answer: 2890. | ||||
| 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 Split the series: $\sum_{k\ge1}\frac{3}{4^k} + p\sum_{k\ge1}\frac{k}{4^k}$.
$\sum_{k\ge1}\frac{3}{4^k} = 3\cdot\frac{1/4}{1 - 1/4} = 1$.
$\sum_{k\ge1}k r^k = \frac{r}{(1-r)^2}$ with $r = \frac{1}{4}$: $\frac{1/4}{9/16} = \frac{4}{9}$.
So $8 = 3 + 1 + \frac{4p}{9} \Rightarrow \frac{4p}{9} = 4 \Rightarrow p = 9$.
Answer: 9. | ||||
| 25 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalEasy | |
A easy numerical question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $f'(1)$, $f''(2)$, $f'''(3)$ are constants; call them a, b, c. Then $f(x) = x^3 + ax^2 + bx + c$.
$f'(x) = 3x^2 + 2ax + b$, $f''(x) = 6x + 2a$, $f'''(x) = 6$.
$c = f'''(3) = 6$.
$b = f''(2) = 12 + 2a$.
$a = f'(1) = 3 + 2a + b = 3 + 2a + 12 + 2a \Rightarrow -3a = 15 \Rightarrow a = -5$, so $b = 2$.
$f'(10) = 300 - 100 + 2 = 202$.
Answer: 202. | ||||
| 26 | 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 Integrate along y. For each $y \ge 0$: $x \ge 2y - 4$, $x \ge -2y^2$, $x \le 8 - 4y^2$.
Which lower bound is larger: $-2y^2 - (2y - 4) = -2(y + 2)(y - 1)$, positive for $0 \le y < 1$. So the left boundary is $x = -2y^2$ for $0 \le y \le 1$ and $x = 2y - 4$ for $y \ge 1$.
Where the strip closes: for $y \le 1$, $-2y^2 \le 8 - 4y^2$ holds (needs $y \le 2$). For $y \ge 1$, $2y - 4 \le 8 - 4y^2 \Rightarrow 2y^2 + y - 6 \le 0 \Rightarrow (2y - 3)(y + 2) \le 0 \Rightarrow y \le \frac{3}{2}$.
Area $= \int_0^1 (8 - 4y^2 + 2y^2)\,dy + \int_1^{3/2}(8 - 4y^2 - 2y + 4)\,dy$.
First: $\left[8y - \frac{2y^3}{3}\right]_0^1 = \frac{22}{3}$.
Second: $\left[12y - y^2 - \frac{4y^3}{3}\right]_1^{3/2} = 6 - \frac{5}{4} - \frac{4}{3}\cdot\frac{19}{8} = 6 - \frac{5}{4} - \frac{19}{6} = \frac{19}{12}$.
Area $= \frac{88}{12} + \frac{19}{12} = \frac{107}{12}$; 107 and 12 are coprime.
$m + n = 119$.
Answer: 119. | ||||
| 27 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Put $u = 2x + 3y$, so $\frac{du}{dx} = 2 + 3\frac{dy}{dx}$ and $\frac{dy}{dx} = -\frac{u - 2}{2u - 7}$.
$\frac{du}{dx} = 2 - \frac{3(u-2)}{2u-7} = \frac{u - 8}{2u - 7}$.
Separate: $\frac{2u - 7}{u - 8}\,du = dx$, i.e. $\left(2 + \frac{9}{u - 8}\right)du = dx$.
$2u + 9\log_e|u - 8| = x + C \Rightarrow 4x + 6y + 9\log_e|2x + 3y - 8| = x + C$.
Divide by 3: $x + 2y + 3\log_e|2x + 3y - 8| = C'$.
$y(0) = 3$: $0 + 6 + 3\log_e 1 = C' \Rightarrow C' = 6$.
So $\alpha = 1$, $\beta = 2$, $\gamma = 8$, and $\alpha + 2\beta + 3\gamma = 1 + 4 + 24 = 29$.
Answer: 29. | ||||
| 28 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra The angle is acute when the dot product is positive (and the vectors are not parallel).
$(\alpha, -2, 2)\cdot(\alpha, 2\alpha, -2) = \alpha^2 - 4\alpha - 4 > 0$.
Roots of $\alpha^2 - 4\alpha - 4 = 0$: $\alpha = 2 \pm 2\sqrt{2} \approx 4.83, -0.83$. So we need $\alpha > 2 + 2\sqrt{2}$ or $\alpha < 2 - 2\sqrt{2}$.
The least positive integer is $\alpha = 5$ (dot product $25 - 20 - 4 = 1 > 0$; the vectors $(5, -2, 2)$ and $(5, 10, -2)$ are not parallel).
Answer: 5. | ||||
| 29 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(X = k) = \left(\frac{5}{6}\right)^{k-1}\frac{1}{6}$, and $P(X \ge k) = \left(\frac{5}{6}\right)^{k-1}$ (the first $k - 1$ tosses are not six).
$a = \left(\frac{5}{6}\right)^2\frac{1}{6} = \frac{25}{216}$.
$b = P(X \ge 3) = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$.
$c = \frac{P(X \ge 6)}{P(X \ge 4)} = \frac{(5/6)^5}{(5/6)^3} = \frac{25}{36}$.
$\frac{b + c}{a} = \frac{50/36}{25/216} = \frac{50}{36}\cdot\frac{216}{25} = 12$.
Answer: 12. | ||||
| 30 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalHard | |
A hard numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Equation: $1 - 2\sin^2 x + a\sin x = 2a - 7 \Rightarrow 2\sin^2 x - a\sin x + 2a - 8 = 0$.
Factor: $2(\sin^2 x - 4) - a(\sin x - 2) = (\sin x - 2)(2\sin x + 4 - a) = 0$.
$\sin x = 2$ is impossible, so $\sin x = \frac{a - 4}{2}$, which needs $-1 \le \frac{a-4}{2} \le 1$, i.e. $2 \le a \le 6$. So $p = 2$, $q = 6$.
$\frac{1}{\cot 63^\circ} = \tan 63^\circ$, so $r = (\tan 9^\circ + \tan 81^\circ) - (\tan 27^\circ + \tan 63^\circ)$.
$\tan\theta + \cot\theta = \frac{2}{\sin 2\theta}$: $r = \frac{2}{\sin 18^\circ} - \frac{2}{\sin 54^\circ}$.
$\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$, $\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5} + 1}{4}$, so $\frac{1}{\sin 18^\circ} = \sqrt{5} + 1$ and $\frac{1}{\sin 54^\circ} = \sqrt{5} - 1$.
$r = 2(\sqrt{5} + 1 - \sqrt{5} + 1) = 4$.
$pqr = 2 \cdot 6 \cdot 4 = 48$.
Answer: 48. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | StatementsEasy | |
A easy statements question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Planck's constant: $E = h\nu \Rightarrow [h] = \frac{[E]}{[\nu]} = \frac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}$.
Angular momentum: $L = mvr \Rightarrow ML^2T^{-1}$. Same: Statement I is true.
Linear momentum: $MLT^{-1}$. Moment of force (torque) $= rF$: $ML^2T^{-2}$. Different: Statement II is false.
Answer: (C). | ||||
| 32 | 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 $\vec{v} = \frac{d\vec{S}}{dt} = 4t\,\hat{j}$ (the $5\hat{k}$ term is constant).
At $t = 1$ s: $\vec{v} = 4\hat{j}$ m/s, i.e. 4 m/s in the y-direction.
Answer: (C). | ||||
| 33 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion No horizontal external force acts while the mass is added (the added mass has no horizontal velocity of its own), so horizontal momentum is conserved:
$1000 \times 6 = (1000 + 200)v \Rightarrow v = \frac{6000}{1200} = 5$ m/s.
Answer: (B). | ||||
| 34 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion For a banked track with no reliance on the flange force: $\tan\theta = \frac{v^2}{rg} = \frac{144}{400 \times 10} = 0.036$.
For a small angle, $\sin\theta \approx \tan\theta$, so the height of the outer rail is
$h = l\sin\theta \approx 1.5 \times 0.036 = 0.054$ m $= 5.4$ cm.
Answer: (C). | ||||
| 35 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $K = \frac{p^2}{2m} \Rightarrow p = \sqrt{2mK}$.
With equal K: $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{4}{25}} = \frac{2}{5}$.
Answer: (A) 2 : 5. | ||||
| 36 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g = \frac{GM}{R^2}$. Halving the diameter halves R, with M unchanged:
$g' = \frac{GM}{(R/2)^2} = 4\frac{GM}{R^2} = 4g$.
Answer: (A). | ||||
| 37 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | StatementsEasy | |
A easy statements question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Statement I: gases are far less viscous than liquids (air about $1.8\times10^{-5}$ Pa s, water about $10^{-3}$ Pa s). Statement I is incorrect.
Statement II: insoluble (or sparingly soluble) impurities such as oil, grease or detergent collect at the surface and lower the surface tension of the liquid; highly soluble impurities such as common salt raise it. Statement II is correct.
Answer: (D). | ||||
| 38 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics At constant volume no work is done, so all the heat goes into internal energy: $\Delta U = mc_v\Delta T$.
$\Delta U = 0.08 \times 0.17 \times 5 = 0.068$ kcal $= 68$ cal.
$68 \times 4.18 = 284.2$ J $\approx 284$ J.
Answer: (B). | ||||
| 39 | 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 Average kinetic energy of a monoatomic molecule: $\frac{3}{2}k_BT$.
$0.414$ eV $= 0.414 \times 1.6\times10^{-19} = 6.624\times10^{-20}$ J.
$T = \frac{2 \times 6.624\times10^{-20}}{3 \times 1.38\times10^{-23}} = \frac{13.248\times10^{-20}}{4.14\times10^{-23}} = 3200$ K.
Answer: (D). | ||||
| 40 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The potential of a point charge depends only on distance: $V = \frac{kq}{r}$.
$OP = \sqrt{3 + 3} = \sqrt{6}$ m and $OQ = \sqrt{6}$ m.
Equal distances give equal potentials, so $V_P - V_Q = 0$ V.
Answer: (C). | ||||
| 41 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity Resistance is proportional to length, so each piece has resistance $\frac{R}{5}$.
Five equal resistors in parallel: $R_{eq} = \frac{R/5}{5} = \frac{R}{25}$.
Answer: (B). | ||||
| 42 | 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 Velocity is unchanged only if the net Lorentz force $q(\vec{E} + \vec{v}\times\vec{B})$ is zero.
(A) $E = 0$, $B = 0$: no force. Possible.
(B) $E = 0$, $B \neq 0$: no force if $\vec{v}$ is parallel (or antiparallel) to $\vec{B}$. Possible.
(C) $E \neq 0$, $B = 0$: the force $q\vec{E}$ cannot vanish, so the velocity changes. Not possible.
(D) $E \neq 0$, $B \neq 0$: crossed fields with $\vec{E} = -\vec{v}\times\vec{B}$ (velocity selector) give zero force. Possible.
So (A), (B) and (D) only.
Answer: (C). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Area $A = 2.5 \times 2 = 5$ m$^2$. Taking 60° as the angle between the field and the normal to the loop:
$\Phi_i = BA\cos60^\circ = 4 \times 5 \times \frac{1}{2} = 10$ Wb, and $\Phi_f = 0$ once the loop is out of the field.
Average emf $= -\frac{\Delta\Phi}{\Delta t} = -\frac{0 - 10}{10} = +1$ V.
(Taking 60° between the field and the plane would give $\Phi_i = 20\sin60^\circ = 10\sqrt{3}$ Wb and an emf of $\sqrt{3}$ V, which is not offered.)
Answer: (B) +1 V. | ||||
| 44 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Wave speed $= \frac{\omega}{k} = \frac{1.5\times10^7}{0.05} = 3\times10^8$ m/s $= c$.
Intensity $I = \frac{1}{2}\epsilon_0 cE_0^2 = \frac{1}{2} \times 8.85\times10^{-12} \times 3\times10^8 \times (200)^2$.
$= \frac{1}{2} \times 8.85 \times 3 \times 4 \times 10^{-12+8+4} = 53.1$ W m$^{-2}$.
Answer: (A). | ||||
| 45 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics At minimum deviation $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\frac{A}{2}}$.
$\cot\frac{A}{2} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}$, so $\sin\left(\frac{A + \delta_m}{2}\right) = \cos\frac{A}{2} = \sin\left(\frac{\pi}{2} - \frac{A}{2}\right)$.
$\frac{A + \delta_m}{2} = \frac{\pi}{2} - \frac{A}{2} \Rightarrow \delta_m = \pi - 2A$.
Answer: (A). | ||||
| 46 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The photoelectric current is proportional to the number of photons per second reaching the cell, i.e. to the total light power gathered by the lens.
The light gathered from the source depends on the lens aperture (its diameter), which is unchanged. The shorter focal length makes the image smaller and brighter, but the same total light falls on the cell (the whole image still falls on the cell).
So the current stays I.
Answer: (B). | ||||
| 47 | 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 In Bohr's model $r_n \propto n^2$.
$\frac{r_4}{r_3} = \frac{16}{9} \Rightarrow r_4 = \frac{16}{9}R$.
Answer: (B). | ||||
| 48 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits A diode is reverse biased when its cathode (the bar) is at a higher potential than its anode (the triangle's base). In every circuit the triangle points to the right, so the left terminal is the anode.
(A) Anode at 0 V (ground), cathode towards −5 V: anode higher, forward biased.
(B) Anode towards +2 V, cathode at 0 V: forward biased.
(C) Anode at +2 V, cathode towards −10 V: forward biased.
(D) Anode at +2 V, cathode towards +4 V: cathode higher, reverse biased.
Answer: (D). | ||||
| 49 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter A spherometer is a screw-based instrument that measures small heights: the thickness of a thin plate, or the sagitta of a curved surface, from which the radius of curvature of a convex or concave surface follows ($R = \frac{l^2}{6h} + \frac{h}{2}$).
Specific rotation of a liquid is the rotation of the plane of polarised light, measured with a polarimeter, not a spherometer.
Answer: (C). | ||||
| 50 | Current ElectricityMetre bridge | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity Balance: $\frac{4.5}{X} = \frac{60}{40} \Rightarrow X = 3$ Ω.
$\rho = \frac{XA}{L} = \frac{3 \times \pi(\sqrt{7}\times10^{-4})^2}{0.1} = \frac{3 \times \pi \times 7\times10^{-8}}{0.1}$.
With $\pi = \frac{22}{7}$: $\rho = \frac{3 \times 22\times10^{-8}}{0.1} = 66\times10^{-7}$ Ω m.
So $R = 66$.
Answer: (C). | ||||
| 51 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane x-motion: $x = 5t + \frac{1}{2}(3)t^2 = 84 \Rightarrow 3t^2 + 10t - 168 = 0 \Rightarrow t = \frac{-10 + \sqrt{100 + 2016}}{6} = \frac{-10 + 46}{6} = 6$ s.
Velocity at $t = 6$ s: $v_x = 5 + 3(6) = 23$ m/s, $v_y = 0 + 2(6) = 12$ m/s.
Speed$^2 = 23^2 + 12^2 = 529 + 144 = 673$.
So $\alpha = 673$.
Answer: 673. | ||||
| 52 | 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 About an axis through one corner, perpendicular to the plane, the other three particles are at distances 2 m, 2 m (adjacent corners) and $2\sqrt{2}$ m (opposite corner); the particle on the axis contributes nothing.
$I = 1(2^2) + 1(2^2) + 1(2\sqrt{2})^2 = 4 + 4 + 8 = 16$ kg m$^2$.
Answer: 16. | ||||
| 53 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Pressure at the bottom (above atmospheric): $p = \rho gh = 1000 \times 10 \times 4000 = 4\times10^7$ N m$^{-2}$.
$\frac{\Delta V}{V} = \frac{p}{B} = \frac{4\times10^7}{2\times10^9} = 2\times10^{-2}$.
So $\alpha = 2$.
Answer: 2. | ||||
| 54 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v_{max} = A\omega \Rightarrow \omega = \frac{10}{4} = 2.5$ rad/s.
$v = \omega\sqrt{A^2 - x^2}$: $5 = 2.5\sqrt{16 - x^2} \Rightarrow \sqrt{16 - x^2} = 2 \Rightarrow x^2 = 12$.
$x = \sqrt{12}$ cm, so $\alpha = 12$.
Answer: 12. | ||||
| 55 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Take a small element of the ring subtending angle $d\theta$ at the centre; its charge is $dQ = \frac{Q}{2\pi}d\theta$.
The central charge q pushes it outward with $dF = \frac{kq\,dQ}{R^2}$. The tension T at the two ends of the element pulls inward with net force $2T\sin\frac{d\theta}{2} \approx T\,d\theta$.
$T\,d\theta = \frac{kqQ}{2\pi R^2}d\theta \Rightarrow T = \frac{kqQ}{2\pi R^2}$.
$T = \frac{9\times10^9 \times 30\times10^{-12} \times 2\pi}{2\pi \times (0.3)^2} = \frac{0.27}{0.09} = 3$ N.
(The cross-sectional area is not needed for the tension; it would give the stress.)
Answer: 3. | ||||
| 56 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance In the steady state no current flows through the capacitor, so the circuit is two branches across the 10 V battery.
Branch through A: $R_1 + R_2 = 1 + 2 = 3$ Ω, current $\frac{10}{3}$ A; drop across $R_1$ = $\frac{10}{3}$ V, so $V_A = 10 - \frac{10}{3} = \frac{20}{3}$ V (taking the bottom as 0 V).
Branch through B: $R_3 + R_4 = 6 + 4 = 10$ Ω, current 1 A; drop across $R_3$ = 6 V, so $V_B = 4$ V.
$V_A - V_B = \frac{20}{3} - 4 = \frac{8}{3}$ V.
$Q = CV = 150 \times \frac{8}{3} = 400$ μC.
Answer: 400. | ||||
| 57 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy 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 Each wire is 2.5 cm from P. For currents in opposite directions, the fields of the two wires at a point between them point the same way, so they add.
$B = 2 \times \frac{\mu_0 I}{2\pi d} = \frac{\mu_0 I}{\pi d} = \frac{4\pi\times10^{-7} \times 10}{\pi \times 0.025} = 1.6\times10^{-4}$ T $= 160$ μT.
Answer: 160. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $e = -M\frac{di}{dt} = -Mi_0\omega\cos\omega t$, so $e_{max} = Mi_0\omega$.
$e_{max} = 0.002 \times 5 \times 50\pi = 0.5\pi = \frac{\pi}{2}$ V.
So $\alpha = 2$.
Answer: 2. | ||||
| 59 | 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 For near-normal viewing, each layer appears shortened by its refractive index: apparent depth $= \sum\frac{t_i}{\mu_i}$.
$= \frac{6}{3/2} + \frac{6}{8/5} = 4 + 3.75 = 7.75$ cm $= \frac{31}{4}$ cm.
So $\alpha = 31$.
Answer: 31. | ||||
| 60 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Energy released = (total binding energy of products) − (binding energy of the parent).
Products: two nuclides of $A \approx 118$: $2 \times 118 \times 8.6 = 236 \times 8.6 = 2029.6$ MeV.
Parent: $236 \times 7.6 = 1793.6$ MeV.
Released: $236 \times (8.6 - 7.6) = 236$ MeV.
Answer: 236. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{CCl_4}$: regular tetrahedron, four equal C–Cl dipoles cancel: non-polar.
$\mathrm{CO_2}$: linear O=C=O, the two dipoles cancel: non-polar.
$\mathrm{CH_2{=}CH_2}$: planar and symmetric: non-polar.
$\mathrm{CHCl_3}$: tetrahedral but with one C–H in place of a C–Cl, so the bond dipoles do not cancel ($\mu \approx 1.0$ D): polar.
Answer: (C). | ||||
| 62 | SolutionsRaoult's law | Class 12: Solutions | Single correctEasy | |
A easy single correct question on raoult's law, from Solutions. NCERT: Class 12: Solutions Negative deviation means the A–B attractions are stronger than A–A and B–B. Molecules escape less readily, so the vapour pressure is lower than Raoult's law predicts.
Lower vapour pressure means a higher temperature is needed for it to reach the external pressure: the boiling point rises (such mixtures can form maximum-boiling azeotropes).
Answer: (A). | ||||
| 63 | EquilibriumIonic equilibrium | Class 11: Equilibrium | StatementsMedium | |
A medium statements question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Statement II: for a salt of a weak acid and a weak base, $\mathrm{pH} = 7 + \frac{1}{2}(pK_a - pK_b)$: the solution is acidic if $K_a > K_b$ and basic if $K_b > K_a$. Statement II is correct.
Statement I: in ammonium carbonate the acid is $\mathrm{HCO_3^-}$ ($K_{a2}$ of carbonic acid $\approx 4.7\times10^{-11}$) and the base is $\mathrm{NH_3}$ ($K_b \approx 1.8\times10^{-5}$). Since $K_b \gg K_a$, carbonate hydrolysis ($K_h = K_w/K_{a2} \approx 2\times10^{-4}$) outweighs ammonium hydrolysis ($K_w/K_b \approx 5.6\times10^{-10}$): the solution is basic. Statement I is correct.
Answer: (A). | ||||
| 64 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | StatementsEasy | |
A easy statements question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Statement I: the lanthanoids and actinoids are placed in separate rows at the bottom so that elements of similar properties stay in single columns and the table keeps its structure; this is the reason given for the arrangement. True.
Statement II: s-block metals (alkali and alkaline earth metals) are highly reactive, with low ionisation enthalpies, and occur only in combined form (salts) in nature, never free. False.
Answer: (C). | ||||
| 65 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Fluorine is the most electronegative element and has no d orbitals in its valence shell, so in its compounds it shows only the −1 oxidation state.
Chlorine, bromine and iodine show −1 and also +1, +3, +5, +7 (using vacant d orbitals and bonding to more electronegative atoms such as O and F).
Answer: (A). | ||||
| 66 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Assertion and reasonEasy | |
A easy assertion and reason question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Boron melts at about 2453 K, far above Al (933 K), Ga (303 K), In and Tl: the assertion is correct.
Solid boron is a covalent network solid built of $\mathrm{B_{12}}$ icosahedra with very strong bonds throughout the crystal, unlike the metallic lattices of the other members. This very strong crystalline lattice is exactly why its melting point is so high.
Both are correct and (R) explains (A).
Answer: (A). | ||||
| 67 | 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 Spin-only moment $\mu = \sqrt{n(n+2)}$ B.M. grows with the number n of unpaired electrons (Hund's rule filling of five d orbitals):
$3d^8$: n = 2; $3d^6$: n = 4; $3d^3$: n = 3; $3d^7$: n = 3.
$3d^6$ has the most unpaired electrons, $\mu = \sqrt{24} \approx 4.90$ B.M.
Answer: (B). | ||||
| 68 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Xe has 54 electrons, so Nd has 6 more. Among the lanthanoids, only La, Ce, Gd and Lu keep a 5d electron; Nd fills 4f: [Xe] $4f^4\,6s^2$ (4 + 2 = 6).
(A) is the configuration of Ce (Z = 58), (C) of Sm (Z = 62), and (D) uses 5f/7s, which belong to the actinoids.
Answer: (B). | ||||
| 69 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds All three ligands ($\mathrm{F^-}$, $\mathrm{H_2O}$) are weak field, so the complexes are high spin.
P: $\mathrm{Fe^{3+}}$, $3d^5$, 5 unpaired, $\mu = \sqrt{35} \approx 5.92$ B.M.
Q: $\mathrm{V^{2+}}$, $3d^3$, 3 unpaired, $\mu = \sqrt{15} \approx 3.87$ B.M.
R: $\mathrm{Fe^{2+}}$, $3d^6$ high spin, 4 unpaired, $\mu = \sqrt{24} \approx 4.90$ B.M.
Order: Q < R < P.
Answer: (B). | ||||
| 70 | 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 ring carries two methyl groups on one carbon and an ethyl group two carbons away.
Lowest locants: numbering from the gem-dimethyl carbon gives {1, 1, 3}; numbering from the ethyl carbon gives {1, 3, 3}. At the first difference, 1 < 3, so {1, 1, 3} wins: methyls on C1, ethyl on C3.
Prefixes go in alphabetical order (ethyl before methyl; "di" is ignored): 3-Ethyl-1,1-dimethylcyclohexane.
(A) has the right locants but the wrong order of prefixes; (B) and (C) have higher locant sets.
Answer: (D). | ||||
| 71 | 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 Cyclohexene is a ring of six carbon atoms with one C=C double bond. It is cyclic, so not acyclic; it has no benzene ring, so it is not benzenoid; and with only one double bond it is not aromatic.
Carbocyclic compounds that behave like open-chain aliphatic compounds are called alicyclic. Cyclohexene is alicyclic.
Answer: (B). | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques α-Hydrogens next to a C=O group are acidic because the carbanion left behind is stabilised by resonance with the carbonyl (enolate).
In (B), the $\mathrm{CH_2}$ sits between two C=O groups (a 1,3-dicarbonyl, active methylene). Its carbanion delocalises the negative charge over both oxygens, so these hydrogens are by far the most acidic ($pK_a \approx 9$, against about 19–20 for a simple ketone).
(A) has only ordinary α-H of a monoketone. In (C) each $\mathrm{CH_2}$ is α to only one C=O. In (D) the α-$\mathrm{CH_2}$ is next to one C=O only (the second C=O is not on the same carbon's other side).
Answer: (B). | ||||
| 73 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes $\mathrm{S_N1}$: the leaving group departs first to give a planar carbocation, which the nucleophile can attack from either face, giving (largely) a racemic mixture: racemisation.
$\mathrm{S_N2}$: the nucleophile attacks from the side opposite the leaving group in one step, so the configuration is inverted (Walden inversion).
Answer: (D). | ||||
| 74 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | StatementsEasy | |
A easy statements question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Statement I: at the para position, $\mathrm{-NO_2}$ stabilises the phenoxide by both −I and −R effects, so p-nitrophenol ($pK_a \approx 7.1$) is the most acidic. In o-nitrophenol intramolecular hydrogen bonding holds the O–H proton ($pK_a \approx 7.2$), and at the meta position only the −I effect acts ($pK_a \approx 8.3$). True.
Statement II: Lucas reagent (conc. HCl + anhydrous $\mathrm{ZnCl_2}$) gives immediate turbidity only with tertiary alcohols; secondary alcohols take about five minutes, and primary alcohols such as ethanol give no turbidity at room temperature. False.
Answer: (C). | ||||
| 75 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers The compounds: (A) butanol, (B) 4-nitrophenol, (C) 4-methoxyphenol, (D) phenol, (E) 2,4-dinitrophenol.
An alcohol is much less acidic than any phenol, since its alkoxide has no resonance stabilisation: (A) is least acidic.
Among phenols, electron-donating groups destabilise the phenoxide and electron-withdrawing groups stabilise it. $\mathrm{-OCH_3}$ at the para position donates by +R, so (C) is less acidic than phenol (D). One $\mathrm{-NO_2}$ at para (B) raises acidity, and two (E, at ortho and para) raise it further.
Order: (A) < (C) < (D) < (B) < (E).
Answer: (A). | ||||
| 76 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Enol content is high when the enol form gains extra stability.
Cyclohexane-1,3,5-trione (C) has a $\mathrm{CH_2}$ between every pair of C=O groups. Enolising all three gives 1,3,5-trihydroxybenzene (phloroglucinol), an aromatic ring; this aromatic stabilisation makes the enol form overwhelmingly favoured.
Cyclohexane-1,3-dione (D) is substantially enolised (1,3-dicarbonyl, conjugation and H-bonding) but its enol is not aromatic. The 1,4-dione (B) and the 1,2,3-trione (A) have no such stabilisation to the same extent.
Answer: (C). | ||||
| 77 | AminesBasicity of amines | Class 12: Amines | Single correctEasy | |
A easy single correct question on basicity of amines, from Amines. NCERT: Class 12: Amines A Brønsted base accepts a proton with the lone pair on nitrogen; the more available the lone pair, the stronger the base.
Pyrrolidine (C) is a secondary aliphatic amine: its $sp^3$ nitrogen lone pair is localised and pushed up by two alkyl groups ($pK_b \approx 2.7$). Strongest.
Aniline (A) and diphenylamine (B): the lone pair is delocalised into one or two benzene rings, so they are weak bases (diphenylamine weaker still).
Pyrrole (D): the lone pair is part of the aromatic sextet; protonating N would destroy aromaticity, so pyrrole is hardly basic.
Answer: (C). | ||||
| 78 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules In a nucleic acid chain, the phosphate group links the 5′-carbon of one nucleotide's sugar to the 3′-carbon of the next, forming two ester bonds on one phosphate: a phosphodiester linkage.
(Glycosidic bonds join the base to the sugar within a nucleotide, peptide bonds join amino acids, and disulphide bridges join cysteine residues.)
Answer: (C). | ||||
| 79 | Principles Related to Practical ChemistryQualitative salt analysis | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on qualitative salt analysis, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter Lead is amphoteric: $\mathrm{PbCrO_4}$ dissolves in hot NaOH as the plumbite (tetrahydroxidoplumbate(II)) ion:
$\mathrm{PbCrO_4 + 4OH^- \rightarrow [Pb(OH)_4]^{2-} + CrO_4^{2-}}$.
$[\mathrm{Pb(OH)_4}]^{2-}$ carries a 2− charge and has four $\mathrm{OH^-}$ ligands: a dianionic complex with coordination number four.
Answer: (B). | ||||
| 80 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter This is the chromyl chloride test for chloride.
$\mathrm{4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2KHSO_4 + 4NaHSO_4 + 2CrO_2Cl_2 + 3H_2O}$: the reddish-brown fumes (B) are chromyl chloride, $\mathrm{CrO_2Cl_2}$.
$\mathrm{CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O}$: the yellow solution (C) is sodium chromate.
Answer: (B). | ||||
| 81 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$: one mole of $\mathrm{CH_4}$ gives one mole of $\mathrm{CO_2}$.
22 g $\mathrm{CO_2}$ = $\frac{22}{44} = 0.5$ mol, so 0.5 mol $\mathrm{CH_4}$ = $0.5 \times 16 = 8$ g.
Answer: 8. | ||||
| 82 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom For n = 4, l = 0, 1, 2, 3: the 4s, 4p, 4d and 4f subshells, with 1 + 3 + 5 + 7 = 16 orbitals.
When completely filled, each orbital holds two electrons, one with $s = +\frac{1}{2}$ and one with $s = -\frac{1}{2}$.
Electrons with $s = +\frac{1}{2}$: 16.
Answer: 16. | ||||
| 83 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure CO and $\mathrm{NO^+}$ are isoelectronic with $\mathrm{N_2}$ (14 electrons):
$\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2\,\pi2p_x^2 = \pi2p_y^2\,\sigma2p_z^2$.
Bond order $= \frac{10 - 4}{2} = 3$ for each.
Sum = 3 + 3 = 6.
Answer: 6. | ||||
| 84 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics Irreversible expansion against a constant external pressure: $w = -p_{ext}\Delta V = -80\times10^3\ \mathrm{Pa} \times (45 - 30)\times10^{-3}\ \mathrm{m^3} = -1200$ J.
Isothermal process of an ideal gas: $\Delta U = 0$, so $q = \Delta U - w = +1200$ J (heat absorbed by the gas).
Answer: 1200. | ||||
| 85 | Redox ReactionsOxidation number | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions $\mathrm{SO_3}$: $x - 6 = 0$, +6.
$\mathrm{H_2SO_3}$: $2 + x - 6 = 0$, +4.
$\mathrm{SOCl_2}$: $x - 2 - 2 = 0$, +4.
$\mathrm{SF_4}$: $x - 4 = 0$, +4.
$\mathrm{BaSO_4}$: $2 + x - 8 = 0$, +6.
$\mathrm{H_2S_2O_7}$: $2 + 2x - 14 = 0$, +6.
Three compounds have S in +4: $\mathrm{H_2SO_3}$, $\mathrm{SOCl_2}$, $\mathrm{SF_4}$.
Answer: 3. | ||||
| 86 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry At S.T.P. one mole of gas occupies 22.4 L, so 5600 mL $\mathrm{O_2}$ = $\frac{5.6}{22.4} = 0.25$ mol.
$\mathrm{2H_2O \rightarrow O_2 + 4H^+ + 4e^-}$: each mole of $\mathrm{O_2}$ needs 4 F, so the charge is $4 \times 0.25 = 1$ F.
$\mathrm{Ag^+ + e^- \rightarrow Ag}$: 1 F deposits 1 mol Ag = 108 g.
Answer: 108. | ||||
| 87 | Chemical KineticsOrder of reaction | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on order of reaction, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate $= k[\mathrm{HI}]^n$.
From 1 to 2 the concentration doubles (0.005 → 0.01) and the rate goes from $7.5\times10^{-4}$ to $3.0\times10^{-3}$, a factor of 4.
$2^n = 4 \Rightarrow n = 2$. (From 2 to 3 the same: doubling gives $\times4$.)
Answer: 2. | ||||
| 88 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Meta directors withdraw electrons from the ring by resonance (−R) and have no lone pair to donate on the atom attached to the ring: $\mathrm{-NO_2}$, $\mathrm{-CN}$, $\mathrm{-COR}$, $\mathrm{-COOH}$. That is 4.
Ortho/para directors: $\mathrm{-OCH_3}$, $\mathrm{-OH}$, $\mathrm{-NHCOCH_3}$ (+R), $\mathrm{-CH_3}$ (+I, hyperconjugation), and $\mathrm{-Cl}$ (deactivating but ortho/para directing through its lone pairs).
Answer: 4. | ||||
| 89 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | NumericalHard | |
A hard numerical question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons A compound is aromatic if it contains a cyclic, planar, fully conjugated ring with $(4n + 2)$ π electrons.
(A) As drawn: four double bonds and a single bond between the two ring-fusion carbons, i.e. 4a,8a-dihydronaphthalene. The fusion carbons are sp³, so the rings are not fully conjugated: not aromatic. (Read as naphthalene, with one double bond missing from the drawing, it would be aromatic, 10 π electrons.)
(B) Benzyl-substituted cyclohexene: it contains a benzene ring (6 π electrons), so it is an aromatic compound (the cyclohexene ring is not aromatic).
(C) The carbene lone pair completes a planar, fully conjugated bicyclic system of 10 π electrons ($n = 2$), as in the indenyl anion: aromatic.
(D) Fluorenyl anion: the carbanion lone pair completes a cyclopentadienyl anion fused to two benzene rings; the whole planar system is conjugated with 14 π electrons ($n = 3$). Aromatic.
Total = 3 as drawn (4 if (A) is read as naphthalene).
Answer: 3 or 4. | ||||
| 90 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 3-Methylhex-2-ene: $\mathrm{CH_3-CH{=}C(CH_3)-CH_2CH_2CH_3}$.
With peroxide, HBr adds by the free-radical (anti-Markovnikov) route: $\mathrm{Br^\bullet}$ adds to C2 so that the more stable tertiary radical forms at C3, which then takes H from HBr.
Product (A): $\mathrm{CH_3-CHBr-CH(CH_3)-CH_2CH_2CH_3}$, 2-bromo-3-methylhexane.
C2 (H, Br, $\mathrm{CH_3}$, $\mathrm{CH(CH_3)C_3H_7}$) and C3 (H, $\mathrm{CH_3}$, $\mathrm{C_3H_7}$, $\mathrm{CHBrCH_3}$) are both chiral, and the molecule is unsymmetrical, so there is no meso form.
Stereoisomers $= 2^2 = 4$.
Answer: 4. | ||||
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