JEE (Main) · 2024
JEE (Main) 2024 — 31 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
- Quadratic Equations3
- Integral Calculus3
- Sets, Relations and Functions2
6 easy · 24 medium · 0 hard
Physics
30 Q
- Laws of Motion3
- Physical World and Measurement2
- Kinematics2
21 easy · 9 medium · 0 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Chemical Bonding and Molecular Structure2
- Electrochemistry2
24 easy · 6 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f(f(x)) = \dfrac{4\cdot\frac{4x+3}{6x-4}+3}{6\cdot\frac{4x+3}{6x-4}-4} = \dfrac{16x+12+18x-12}{24x+18-24x+16} = \dfrac{34x}{34} = x$.
So $g(x) = x$ on its domain, and $(g\circ g\circ g)(4) = 4$.
Answer: (B). | ||||
| 2 | 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 The denominator $x^2-8x+32 = (x-4)^2+16 > 0$ for every $x$.
So we need $ax^2+2(a+1)x+9a+4 < 0$ for every real $x$. A quadratic is negative everywhere only if its leading coefficient is negative, i.e. $a<0$.
No positive integer $a$ can work, so S is empty and has 0 elements.
Answer: (C). | ||||
| 3 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants At $x=0$: $f(0)=\begin{vmatrix}0&1&1\\2&0&6\\0&4&-2\end{vmatrix} = 0 - 1(-4-0) + 1(8-0) = 12$.
$f'(x)$ is the sum of three determinants, each with one row differentiated. The derivative rows at $x=0$ are $(0,0,3)$, $(0,2,0)$ and $(-1,0,0)$.
$D_1=\begin{vmatrix}0&0&3\\2&0&6\\0&4&-2\end{vmatrix} = 3(8-0) = 24$.
$D_2=\begin{vmatrix}0&1&1\\0&2&0\\0&4&-2\end{vmatrix} = 0$ (first column zero).
$D_3=\begin{vmatrix}0&1&1\\2&0&6\\-1&0&0\end{vmatrix} = -1\,(6-0) = -6$.
$f'(0) = 24+0-6 = 18$, so $2f(0)+f'(0) = 24+18 = 42$.
Answer: (C). | ||||
| 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 For infinitely many solutions $\Delta = 0$ and $\Delta_x=\Delta_y=\Delta_z=0$.
$\Delta_z=\begin{vmatrix}1&-2&-4\\2&\alpha&5\\3&-1&3\end{vmatrix} = (3\alpha+5)+2(6-15)-4(-2-3\alpha) = 15\alpha-5 = 0$, so $\alpha=\frac{1}{3}$.
$\Delta_y=\begin{vmatrix}1&-4&1\\2&5&3\\3&3&\beta\end{vmatrix} = (5\beta-9)+4(2\beta-9)+(6-15) = 13\beta-54 = 0$, so $\beta=\frac{54}{13}$.
Check: $\Delta = \alpha\beta+4\beta-3\alpha-17 = \frac{18}{13}+\frac{216}{13}-1-17 = 18-18 = 0$.
$12\alpha+13\beta = 4+54 = 58$.
Answer: (B). | ||||
| 5 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $n^4-3n^2+1 = (n^2-1)^2-n^2 = (n^2-n-1)(n^2+n-1)$.
So $T_n = \frac{n}{(n^2-n-1)(n^2+n-1)} = \frac{1}{2}\left[\frac{1}{n^2-n-1}-\frac{1}{n^2+n-1}\right]$.
Since $(n+1)^2-(n+1)-1 = n^2+n-1$, the sum telescopes:
$S_{10} = \frac{1}{2}\left[\frac{1}{1-1-1}-\frac{1}{100+10-1}\right] = \frac{1}{2}\left[-1-\frac{1}{109}\right] = -\frac{55}{109}$.
Answer: (C). | ||||
| 6 | 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 Let $u=|\sin x|\to0$. Then $e^{2u}-2u-1 = \frac{(2u)^2}{2}+O(u^3) = 2u^2+O(u^3)$.
So the limit is $\lim_{x\to0}\frac{2\sin^2x}{x^2} = 2$ (from both sides, since $u^2=\sin^2x$).
Answer: (A). | ||||
| 7 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Sum of coefficients: put $x=1$: $a=(1-2+2)^{2023}(3-4+2)^{2024} = 1$.
$b$: by L'Hospital's rule, $b=\lim_{x\to0}\frac{\log(1+x)}{(x^{2024}+1)\,2x} = \frac{1}{2}$.
The second equation is $x^2+x+4=0$, with discriminant $1-16<0$: its roots are a non-real conjugate pair.
A real quadratic sharing one non-real root also has its conjugate, so $cx^2+dx+e$ is a multiple of $x^2+x+4$: $c:d:e = 1:1:4$.
Hence $d:c:e = 1:1:4$.
Answer: (B). | ||||
| 8 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations The coefficients add to zero, so $x=1$ is a root; the other root is $\alpha=\frac{c+a-2b}{a+b-2c}$ (product of roots).
The denominator $a+b-2c>0$ since $a,b>c$. Also $\alpha<1 \iff c+a-2b<a+b-2c \iff c<b$, which holds.
If $b=\sqrt{ac}$, then $c+a-2b = (\sqrt a-\sqrt c)^2>0$, so $\alpha\in(0,1)$.
(I): if $\alpha\in(-1,0)$, then $\alpha<0$, which is impossible when $b$ is the G.M.; so $b$ cannot be the G.M. True.
(II): when $b$ is the G.M., $\alpha$ does lie in $(0,1)$, so with $\alpha\in(0,1)$ it is possible for $b$ to be the G.M. True.
Answer: (C). | ||||
| 9 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability As $x\to0^+$, the base $\frac{1+x}{2+x}\to\frac{1}{2}$ and the power $\frac{1}{x}\to\infty$, so the limit is 0. Continuity gives $g(0)=0$; $g$ linear, so $g(x)=kx$.
For $x>0$, $\ln f = \frac{1}{x}\ln\frac{1+x}{2+x}$, so $\frac{f'}{f} = -\frac{1}{x^2}\ln\frac{1+x}{2+x}+\frac{1}{x}\left(\frac{1}{1+x}-\frac{1}{2+x}\right)$.
At $x=1$: $f(1)=\frac{2}{3}$ and $\frac{f'(1)}{f(1)} = -\ln\frac{2}{3}+\frac{1}{6}$, so $f'(1) = \frac{1}{9}-\frac{2}{3}\ln\frac{2}{3}$.
$f(-1)=g(-1)=-k$, so $k = \frac{2}{3}\ln\frac{2}{3}-\frac{1}{9}$.
$g(3)=3k = 2\ln\frac{2}{3}-\frac{1}{3} = \ln\frac{4}{9}-\ln e^{1/3} = \log_e\left(\frac{4}{9e^{1/3}}\right)$.
Answer: (B). | ||||
| 10 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $y^2\le4x$ needs $x\ge0$, and $x<4$. For $x>0$ the sign of $\frac{(x-1)(x-2)}{(x-3)(x-4)}$ is: $+$ on $(0,1)$, $-$ on $(1,2)$, $+$ on $(2,3)$, $-$ on $(3,4)$.
So the region is the upper half of the parabola's interior ($y>0$) over $(0,1)\cup(2,3)$ and the lower half ($y<0$) over $(1,2)\cup(3,4)$.
Each half has height $2\sqrt{x}$ over its strip, so the area is $\int_0^4 2\sqrt{x}\,dx = \frac{4}{3}x^{3/2}\Big|_0^4 = \frac{32}{3}$.
Answer: (C). | ||||
| 11 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $\sec x-\sin x\tan x = \frac{1-\sin^2x}{\cos x} = \cos x$, so the denominator is $\sin x\cos x$.
$\frac{dy}{dx}-\frac{y}{\sin x\cos x} = \frac{\tan x}{\sin x\cos x} = \sec^2x$.
$\int\frac{dx}{\sin x\cos x} = \int\frac{\sec^2x}{\tan x}dx = \ln\tan x$, so the integrating factor is $\cot x$.
$\frac{d}{dx}(y\cot x) = \sec^2x\cot x = \frac{1}{\sin x\cos x}$, giving $y\cot x = \ln\tan x + C$.
At $x=\frac{\pi}{4}$: $2 = 0 + C$, so $C=2$.
At $x=\frac{\pi}{3}$: $\frac{y}{\sqrt3} = \ln\sqrt3+2$, so $y = \sqrt3(2+\log_e\sqrt3)$.
Answer: (C). | ||||
| 12 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $x=vy$: $\frac{dx}{dy} = v+y\frac{dv}{dy}$. The equation becomes $y\left(v+y\frac{dv}{dy}\right) = vy(\ln v+1)$, i.e. $y\frac{dv}{dy} = v\ln v$.
Separating: $\frac{dv}{v\ln v} = \frac{dy}{y}$, so $\ln|\ln v| = \ln y + C$, i.e. $|\ln v| = ky$.
At $(e,1)$: $v=e$, $|\ln e| = 1 = k$.
So $\left|\log_e\frac{x}{y}\right| = y$.
Answer: (D). | ||||
| 13 | 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 The given circle has centre $(5,-2)$ and radius $\sqrt{25+4-13} = 4$.
Solving $2x+3y=12$, $3x-2y=5$: $13x=39$, $x=3$, $y=2$, so C has centre $(3,2)$.
The diameter is a chord of C whose midpoint is $(5,-2)$, so the segment from $(3,2)$ to $(5,-2)$ is perpendicular to that chord.
Its length squared is $4+16=20$, and half the chord is 4. Hence $R^2 = 20+16 = 36$, $R=6$.
Answer: (B). | ||||
| 14 | 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 diagonals of a parallelogram bisect each other: the midpoint of AC is the midpoint of BD, $(1,1)$.
So $\alpha+\gamma = 2$ and $\beta+\delta = 2$, and $2(\alpha+\beta+\gamma+\delta) = 2\times4 = 8$.
Check: integer points on $3y=2x+1$ are $(1+3k,\,1+2k)$; $AB^2 = (3k)^2+(1+2k)^2 = 10$ gives $k=-1$, so $A(-2,-1)$ and $C(4,3)$, which indeed lie on the line.
Answer: (B). | ||||
| 15 | 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 Direction of the required line: $(2,3,5)\times(-1,3,2) = (6-15,\,-(4+5),\,6+3) = (-9,-9,9)\parallel(1,1,-1)$.
$\vec{PQ} = (-5,6,-5)$, $|\vec{PQ}|^2 = 86$.
Projection on $(1,1,-1)/\sqrt3$: $\frac{-5+6+5}{\sqrt3} = \frac{6}{\sqrt3}$, whose square is 12.
Distance $= \sqrt{86-12} = \sqrt{74}$.
Answer: (B). | ||||
| 16 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse: major axis along $y$, $e^2 = 1-\frac{9}{25} = \frac{16}{25}$, $e=\frac{4}{5}$, foci $(0,\pm4)$.
Hyperbola: $e = \frac{15}{8}\cdot\frac{4}{5} = \frac{3}{2}$, foci $(0,\pm4)$, so it is $\frac{y^2}{A^2}-\frac{x^2}{B^2}=1$ with $Ae=4$, $A=\frac{8}{3}$, $B^2 = A^2(e^2-1) = \frac{64}{9}\cdot\frac{5}{4} = \frac{80}{9}$.
Check the point: $y^2 = \frac{196}{9}\cdot\frac{2}{5} = \frac{392}{45}$; $\frac{y^2}{A^2} = \frac{392}{320}$, $\frac{x^2}{B^2} = \frac{18}{80} = \frac{72}{320}$; difference $=1$.
Focal distances are $|e\,y \mp A|$; the smaller is $e\,y - A = \frac{3}{2}\cdot\frac{14}{3}\sqrt{\frac{2}{5}}-\frac{8}{3} = 7\sqrt{\frac{2}{5}}-\frac{8}{3}$.
Answer: (C). | ||||
| 17 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability 18 apples, 3 rotten; two drawn together (without replacement), $\binom{18}{2} = 153$ ways.
$P(0) = \frac{\binom{15}{2}}{153} = \frac{105}{153}$, $P(1) = \frac{3\cdot15}{153} = \frac{45}{153}$, $P(2) = \frac{3}{153}$.
$E(x) = \frac{45+6}{153} = \frac{1}{3}$, $E(x^2) = \frac{45+12}{153} = \frac{57}{153}$.
$\mathrm{Var}(x) = \frac{57}{153}-\frac{1}{9} = \frac{57-17}{153} = \frac{40}{153}$.
Answer: (B). | ||||
| 18 | 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 $(\vec p-\vec c)\times\vec b = \vec 0$, so $\vec p = \vec c+\lambda\vec b = (1+4\lambda,\,-3+\lambda,\,4+7\lambda)$.
$\vec p\cdot\vec a = 3(1+4\lambda)+(-3+\lambda)-2(4+7\lambda) = -8-\lambda = 0$, so $\lambda=-8$.
$\vec p = (-31,-11,-52)$ and $\vec p\cdot(1,-1,-1) = -31+11+52 = 32$.
Answer: (D). | ||||
| 19 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are 75 marbles; with replacement the draws are independent.
$P = \frac{10}{75}\cdot\frac{30}{75} = \frac{2}{15}\cdot\frac{2}{5} = \frac{4}{75}$.
Answer: (C). | ||||
| 20 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $A=\sin^{-1}\alpha$, $B=\sin^{-1}\beta$, $C=\sin^{-1}\gamma$, each in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, with $A+B+C=\pi$; this forces all three to be positive, so $A,B,C$ are the angles of a triangle and $\alpha,\beta,\gamma$ are $\sin A,\sin B,\sin C$.
The condition is $(\alpha+\beta)^2-\gamma^2 = 3\alpha\beta$, i.e. $\alpha^2+\beta^2-\gamma^2 = \alpha\beta$.
By the sine rule the sides are proportional to $\alpha,\beta,\gamma$, so $\cos C = \frac{\alpha^2+\beta^2-\gamma^2}{2\alpha\beta} = \frac{1}{2}$ and $C=\frac{\pi}{3}$.
$\gamma = \sin\frac{\pi}{3} = \frac{\sqrt3}{2}$.
Answer: (A). | ||||
| 21 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions In any equivalence relation containing R, 1 is related to 2 and to 4, and 2 to 3, so by symmetry and transitivity 1, 2, 3, 4 all fall in one class.
The smallest such S is therefore $A\times A$, with $4^2 = 16$ elements.
Answer: 16. | ||||
| 22 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $|1-i|^x = 2^{x/2} = 2^x$ only for $x=0$: $\alpha=1$.
$1-\sqrt\pi i = -i(\sqrt\pi+i)$, so the first fraction is $-i$; $\sqrt\pi-i = -i(1+\sqrt\pi i)$, so the second is $-i$. The bracket is $-2i$.
$(1+i)^4 = (2i)^2 = -4$. So $z = \frac{\pi}{4}(-4)(-2i) = 2\pi i$: $|z| = 2\pi$, $\arg z = \frac{\pi}{2}$, $\beta = 4$.
Distance of $(1,4)$ from $4x-3y-7=0$: $\frac{|4-12-7|}{5} = 3$.
Answer: 3. | ||||
| 23 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations DISTRIBUTION has I three times, T twice, and D, S, R, B, U, O, N once: 9 different letters.
All four different: $^9P_4 = 3024$.
One pair (I or T) and two other different letters: $2\cdot\binom{8}{2}\cdot\frac{4!}{2!} = 2\cdot28\cdot12 = 672$.
Two pairs (I, I, T, T): $\frac{4!}{2!\,2!} = 6$.
Three I's and one other letter: $8\cdot\frac{4!}{3!} = 32$.
Total $= 3024+672+6+32 = 3734$.
Answer: 3734. | ||||
| 24 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3} = \left(1+\frac{1}{x}\right)^3$, so the expression is $(1+x)(1-x^2)\frac{(1+x)^{15}}{x^{15}} = \frac{(1-x^2)(1+x)^{16}}{x^{15}}$.
Coefficient of $x^3$ = coefficient of $x^{18}$ in $(1-x^2)(1+x)^{16}$ $= 0-\binom{16}{16} = -1$.
Coefficient of $x^{-13}$ = coefficient of $x^2$ in $(1-x^2)(1+x)^{16}$ $= \binom{16}{2}-1 = 119$.
Sum $= 118$.
Answer: 118. | ||||
| 25 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = (e^x-1)^{11}(2x-1)^5(x-2)^7(x-3)^{12}(2x-10)^{61}$. It changes sign only at the zeros of odd order: $x = 0, \frac{1}{2}, 2, 5$ (at $x=3$ the power is even).
Signs: $(-1,0)$: four negative factors, $+$; $(0,\frac12)$: $-$; $(\frac12,2)$: $+$; $(2,5)$: $-$; $(5,\infty)$: $+$.
Local maxima ($+$ to $-$) at $x=0$ and $x=2$: $p = 0+4 = 4$.
Local minima ($-$ to $+$) at $x=\frac12$ and $x=5$: $q = \frac{11}{2}$.
$p^2+2q = 16+11 = 27$.
Answer: 27. | ||||
| 26 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f(x)+f(1-x) = \frac{4^x}{4^x+2}+\frac{4}{4+2\cdot4^x} = \frac{4^x+2}{4^x+2} = 1$. So with $k=f(a)$ the limits are $k$ and $1-k$.
Replacing $x$ by $(k+1-k)-x = 1-x$ leaves $\sin^4(x(1-x))$ unchanged, so $M = \int (1-x)\sin^4(x(1-x))\,dx = N-M$, i.e. $N = 2M$ ($N\neq0$ since $a\neq\frac12$ makes the limits differ).
$\alpha M = 2\beta M$ gives $\alpha = 2\beta$; the least natural pair is $\beta=1$, $\alpha=2$, and $\alpha^2+\beta^2 = 5$.
Answer: 5. | ||||
| 27 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\sin2x = 2\sin x\cos x$. Put $t=\cos x$: the integral is $2\int_0^1 t^{13/2}(1+t^{5/2})^{1/2}dt$.
Put $u = t^{5/2}$, $du = \frac{5}{2}t^{3/2}dt$, and $t^{13/2}dt = u^2\cdot\frac{2}{5}du$: the integral is $\frac{4}{5}\int_0^1 u^2\sqrt{1+u}\,du$.
With $w=1+u$: $\int_1^2 (w-1)^2w^{1/2}dw = \left[\frac{2}{7}w^{7/2}-\frac{4}{5}w^{5/2}+\frac{2}{3}w^{3/2}\right]_1^2 = \frac{44\sqrt2}{105}-\frac{16}{105}$.
$525\cdot\frac{4}{5}\cdot\frac{44\sqrt2-16}{105} = 4(44\sqrt2-16) = 176\sqrt2-64$.
So $n = 176$.
Answer: 176. | ||||
| 28 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Foci: $a^2-b^2 = 25$. Latus rectum: $\frac{2b^2}{a} = 5\sqrt2$, so $b^2 = \frac{5\sqrt2}{2}a$.
$a^2-\frac{5\sqrt2}{2}a-25 = 0$ gives $a = 5\sqrt2$ (positive root), so $a^2=50$, $b^2=25$.
For $\frac{x^2}{b^2}-\frac{y^2}{a^2b^2}=1$: $e^2 = 1+\frac{a^2b^2}{b^2} = 1+a^2 = 51$.
Answer: 51. | ||||
| 29 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Line 1: $(t,t,1)$, direction $(1,1,0)$. Foot: $(a-t)+(a-t)=0$, $t=a$, so $Q=(a,a,1)$.
Line 2: $(s,-s,-1)$, direction $(1,-1,0)$. Foot: $(a-s)-(a+s)=0$, $s=0$, so $R=(0,0,-1)$.
$\vec{PQ} = (0,0,1-a)$, $\vec{PR} = (-a,-a,-1-a)$. Right angle: $(1-a)(-1-a) = a^2-1 = 0$, so $a^2=1$ (with $a=-1$; $a=1$ would make $Q=P$).
$12a^2 = 12$.
Answer: 12. | ||||
| 30 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $|\vec a\times\vec b|^2 = 1\cdot16-4 = 12$.
$\vec b\cdot\vec c = 0-3\cdot16 = -48$. $|\vec c|^2 = 4\cdot12+9\cdot16 = 192$ (the cross term vanishes since $\vec a\times\vec b\perp\vec b$).
$\cos\alpha = \frac{-48}{4\sqrt{192}}$, $\cos^2\alpha = \frac{144}{192} = \frac{3}{4}$, $\sin^2\alpha = \frac{1}{4}$.
$192\sin^2\alpha = 48$.
Answer: 48. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | 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 Each term has the dimensions of force, $[MLT^{-2}]$.
$a = \frac{F}{x^2} = [ML^{-1}T^{-2}]$; $b = \frac{F}{t^{1/2}} = [MLT^{-5/2}]$, so $b^2 = [M^2L^2T^{-5}]$.
$\frac{b^2}{a} = \frac{[M^2L^2T^{-5}]}{[ML^{-1}T^{-2}]} = [ML^3T^{-3}]$.
Answer: (D). | ||||
| 32 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\frac{dt}{dx} = 2\alpha x+\beta$, so $v = \frac{1}{2\alpha x+\beta}$.
$a = v\frac{dv}{dx} = v\cdot\left(-\frac{2\alpha}{(2\alpha x+\beta)^2}\right) = v\cdot(-2\alpha v^2) = -2\alpha v^3$.
Answer: (A). | ||||
| 33 | 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 Static friction supplies the centripetal force: $m\omega^2r\le\mu mg$.
$\omega_{max} = \sqrt{\frac{\mu g}{r}}$.
Answer: (B). | ||||
| 34 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion M is on the $53^\circ$ face and moves down it; m is on the $37^\circ$ face and moves up it. Kinetic friction opposes each motion.
$\sin53^\circ = 0.8$, $\cos53^\circ = 0.6$; $\sin37^\circ = 0.6$, $\cos37^\circ = 0.8$.
Block M: $Mg\sin53^\circ-\mu Mg\cos53^\circ-T = Ma$: $80-15-T = 20$, so $T = 45$ N.
Block m: $T-mg\sin37^\circ-\mu mg\cos37^\circ = ma$: $45-6m-2m = 2m$, so $m = 4.5$ kg.
Answer: (A). | ||||
| 35 | 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 Momentum is conserved, so the gun and the shell have equal and opposite momenta $p$.
$K = \frac{p^2}{2M}$, so $\frac{K_{gun}}{K_{shell}} = \frac{p^2/2M_1}{p^2/2M_2} = \frac{M_2}{M_1}$.
Answer: (B). | ||||
| 36 | GravitationNewton's law of gravitation | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on newton's law of gravitation, from Gravitation. NCERT: Class 11: Gravitation Let the side be $s$. The two neighbours each pull with $\frac{Gm^2}{s^2}$ at right angles: resultant $\sqrt2\frac{Gm^2}{s^2}$ along the diagonal.
The opposite corner pulls with $\frac{Gm^2}{2s^2}$ along the same diagonal.
Total $= \left(\sqrt2+\frac12\right)\frac{Gm^2}{s^2} = \frac{2\sqrt2+1}{2}\cdot\frac{Gm^2}{s^2}$.
Setting this equal to $\frac{2\sqrt2+1}{32}\cdot\frac{Gm^2}{L^2}$ gives $s^2 = 16L^2$, $s = 4L$.
Answer: (B). | ||||
| 37 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter The ball starts from rest. As it speeds up, the viscous drag (proportional to $v$ by Stokes' law) grows, so the acceleration $\frac{mg-F_B-6\pi\eta rv}{m}$ keeps falling.
The velocity therefore rises with decreasing slope and approaches the terminal velocity: a curve rising from the origin and levelling off.
Answer: (C). | ||||
| 38 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory For an isobaric process $V = \frac{nR}{P}T$, a straight line through the origin with slope $\frac{nR}{P}$.
The $P_2$ line is steeper, so $\frac{1}{P_2}>\frac{1}{P_1}$, i.e. $P_1>P_2$.
Answer: (A). | ||||
| 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 The average translational kinetic energy of a molecule is $\frac{3}{2}kT$, the same for every gas at a given temperature; mass, speed and momentum depend on the gas.
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 For like charges the field vanishes between them, at distance $x$ from $q$:
$\frac{kq}{x^2} = \frac{k\,3q}{(r-x)^2}$, so $r-x = \sqrt3\,x$ and $x = \frac{r}{1+\sqrt3}$.
Answer: (C). | ||||
| 41 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $R_1 = R_0(1+\alpha_1t)$, $R_2 = R_0(1+\alpha_2t)$.
Series: $R_s = 2R_0\left(1+\frac{\alpha_1+\alpha_2}{2}t\right)$, coefficient $\frac{\alpha_1+\alpha_2}{2}$.
Parallel: $R_p = \frac{R_0(1+\alpha_1t)(1+\alpha_2t)}{2+(\alpha_1+\alpha_2)t}\approx\frac{R_0}{2}\left(1+\frac{\alpha_1+\alpha_2}{2}t\right)$ to first order in $t$, coefficient $\frac{\alpha_1+\alpha_2}{2}$.
Answer: (A). | ||||
| 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 The figure shows the field out of the page ($+\hat k$, the dots), perpendicular to the wire; the current goes up the left straight section, round the semicircle at the top and down the right section.
The two straight sections in the field have equal lengths and opposite currents, so their forces ($\pm iBl\hat i$) cancel.
The semicircle acts like a straight wire joining its ends: $\vec l = 2R\hat i$ (left to right). $\vec F = i\vec l\times\vec B = i(2R\hat i)\times(B\hat k) = -2iBR\hat j$.
Answer: (B). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $\varepsilon = NA\frac{\Delta B}{\Delta t}$ with $A = \pi(0.01)^2 = \pi\times10^{-4}\ \mathrm{m^2}$ and $\frac{\Delta B}{\Delta t} = \frac{2000}{2} = 1000$ T/s.
$22 = N\cdot\pi\times10^{-4}\cdot1000 = 0.1\pi N$, so $N = \frac{220}{\pi} = 70$ (with $\pi = \frac{22}{7}$).
Answer: (D). | ||||
| 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 Electric and magnetic parts each average $\frac14\varepsilon_0E_0^2$, so the total is $u = \frac12\varepsilon_0E_0^2$.
$u = \frac12\times8.85\times10^{-12}\times2500 = 1.106\times10^{-8}\ \mathrm{Jm^{-3}}$.
Answer: (A). | ||||
| 45 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\mu = \frac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}} = \cot\frac{A}{2} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}$.
So $\sin\frac{A+\delta_m}{2} = \cos\frac{A}{2} = \sin\left(90^\circ-\frac{A}{2}\right)$, giving $\frac{A+\delta_m}{2} = 90^\circ-\frac{A}{2}$ and $\delta_m = 180^\circ-2A$.
Answer: (A). | ||||
| 46 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\frac{hc}{\lambda}-\phi = 8$ eV and $\frac{hc}{3\lambda}-\phi = 2$ eV.
Subtracting: $\frac{2}{3}\frac{hc}{\lambda} = 6$ eV, so $\frac{hc}{\lambda} = 9$ eV and $\phi = 1$ eV.
$\lambda_0 = \frac{hc}{\phi} = 9\lambda$.
Answer: (D). | ||||
| 47 | 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 First member ($2\to1$): $\frac{1}{\lambda} = R\left(1-\frac14\right) = \frac{3R}{4}$. Second ($3\to1$): $\frac{1}{\lambda'} = R\left(1-\frac19\right) = \frac{8R}{9}$.
$\lambda' = \frac{3/4}{8/9}\lambda = \frac{27}{32}\lambda$.
Answer: (A). | ||||
| 48 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Each input passes through a NOT gate, and the two results feed a NAND gate.
$Y = \overline{\bar A\cdot\bar B} = A+B$ (De Morgan), which is OR.
Answer: (B). | ||||
| 49 | 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 Closed pipe fundamental: $\frac{v}{4L_c}$. Open pipe first overtone: $\frac{2v}{2L_o} = \frac{v}{L_o}$.
$\frac{v}{4L_c} = \frac{v}{L_o}$ gives $L_c = \frac{L_o}{4} = 15$ cm.
Answer: (D). | ||||
| 50 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $R = \frac{\rho L}{\pi d^2/4}$, so $\frac{\Delta R}{R} = \frac{\Delta L}{L}+2\frac{\Delta d}{d} = 0.1\%+0.2\% = 0.3\%$.
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 Falling from $H$ to $h$ takes $t_1 = \sqrt{\frac{2(H-h)}{g}}$. After the impact the velocity is horizontal, so the fall through $h$ takes $t_2 = \sqrt{\frac{2h}{g}}$.
$T(h) = \sqrt{\frac{2}{g}}\left(\sqrt{H-h}+\sqrt h\right)$. $\frac{dT}{dh} = 0$ gives $\frac{1}{\sqrt h} = \frac{1}{\sqrt{H-h}}$, so $h = \frac{H}{2}$ (a maximum).
$\frac{H}{h} = 2$.
Answer: 2. | ||||
| 52 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Rolling disc: $K = \frac12mv^2+\frac12\cdot\frac12mR^2\cdot\frac{v^2}{R^2} = \frac34mv^2 = \frac34\times50\times0.16 = 6$ J.
By the work-energy theorem the work needed to stop it is 6 J in magnitude.
Answer: 6. | ||||
| 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 $\Delta P = B\frac{\Delta V}{V} = 9\times10^8\times2\times10^{-4} = 1.8\times10^5$ Pa.
$h = \frac{\Delta P}{\rho g} = \frac{1.8\times10^5}{10^4} = 18$ m.
Answer: 18. | ||||
| 54 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At $x = \frac{2A}{3}$: $v = \omega\sqrt{A^2-\frac{4A^2}{9}} = \frac{\sqrt5}{3}\omega A$. The new speed is $3v = \sqrt5\,\omega A$, with $\omega$ unchanged.
New amplitude: $A'^2 = x^2+\frac{(3v)^2}{\omega^2} = \frac{4A^2}{9}+5A^2 = \frac{49A^2}{9}$, so $A' = \frac{7A}{3}$.
$n = 7$.
Answer: 7. | ||||
| 55 | 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 With the battery connected, $Q\propto C$, so $C' = 1.25\,C$.
$C = \frac{\varepsilon_0A}{d}$, $C' = \frac{\varepsilon_0A}{d-t+t/K}$, so $d-t+\frac{t}{K} = \frac{d}{1.25} = 4$ mm.
$5-2+\frac{2}{K} = 4$ gives $K = 2$.
Answer: 2. | ||||
| 56 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity The top line runs from the left end through $R_2$ and $R_4$ to A; the bottom line is a plain wire to B. At the far left a plain wire joins the top-left corner to the bottom line, so the left end of $R_2$ is at B's potential, and $R_1$ (6 Ω), joined across that same pair of points, is shorted and carries no current.
$R_2$ (2 Ω) and $R_3$ (2 Ω) both connect the junction after $R_2$ to B: in parallel, 1 Ω.
In series with $R_4$ (2 Ω): 3 Ω.
That 3 Ω is in parallel with $R_5$ (3 Ω): 1.5 Ω; and with $R_6$ (3 Ω): $\frac{1.5\times3}{4.5} = 1$ Ω.
Answer: 1. | ||||
| 57 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical 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 $\vec u\times\vec B = (3\hat i+5\hat j)\times(B_0\hat i+2B_0\hat j) = (6B_0-5B_0)\hat k = B_0\hat k$.
With $e$ the electron's charge, $\vec F = e(\vec u\times\vec B) = eB_0\hat k = 5e\hat k$, so $B_0 = 5$ T.
Answer: 5. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Field at the centre of a square loop of side $L$ carrying $I$: four sides, each $\frac{\mu_0I}{4\pi(L/2)}(\sin45^\circ+\sin45^\circ)$, total $B = \frac{2\sqrt2\mu_0I}{\pi L}$.
The small loop sees this nearly uniform field: $\Phi = Bl^2$, so $M = \frac{2\sqrt2\mu_0l^2}{\pi L}$.
With $L = l^2$: $M = \frac{2\sqrt2\mu_0}{\pi} = 2\sqrt2\times4\times10^{-7} = 8\sqrt2\times10^{-7} = \sqrt{128}\times10^{-7}$ H.
$x = 128$.
Answer: 128. | ||||
| 59 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Let the intensities be $I$ and $9I$.
Incoherent: $I_1 = I+9I = 10I$.
Coherent, $60^\circ$: $I_2 = I+9I+2\sqrt{9I^2}\cos60^\circ = 10I+3I = 13I$.
$\frac{I_1}{I_2} = \frac{10}{13}$, $x = 13$.
Answer: 13. | ||||
| 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 $E = \Delta m\,c^2 = 0.4\times10^{-3}\times9\times10^{16} = 3.6\times10^{13}$ J.
1 kWh $= 3.6\times10^6$ J, so $E = 10^7$ kWh and $n = 1$.
Answer: 1. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | StatementsEasy | |
A easy statements question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Conditions for LCAO: the combining orbitals must have the same or nearly the same energy (A), the same symmetry about the molecular axis (C), and must overlap to the maximum extent (so B is wrong). D contradicts C.
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 Positive deviation occurs when A–B attractions are weaker than A–A and B–B.
Acetone + $\mathrm{CS_2}$: the dipolar attractions between acetone molecules are weakened by non-polar $\mathrm{CS_2}$, so the mixture shows positive deviation.
Chloroform + acetone (H-bonding between them), acetone + aniline, and chloroform + benzene show negative deviation.
Answer: (D). | ||||
| 63 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Each species has coefficient 1, so $K_C = \frac{[\mathrm{FeSCN^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}$.
Answer: (C). | ||||
| 64 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry Electrolytic conductance depends on the nature of the electrolyte, the solvent (and its viscosity), the concentration and the temperature. It does not depend on the electrodes used.
Answer: (B). | ||||
| 65 | 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 For the standard first-order gas reaction $\mathrm{A(g)\to B(g)+C(g)}$: if the pressure of A falls by $x$, the total is $P_t = P_i+x$, so the pressure of A left is $P_i-x = 2P_i-P_t$.
$k = \frac{2.303}{t}\log\frac{P_i}{P_A} = \frac{2.303}{t}\log\frac{P_i}{2P_i-P_t}$.
Answer: (B). | ||||
| 66 | Classification of Elements and PeriodicityElectron gain enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctMedium | |
A medium single correct question on electron gain enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Electron gain enthalpies (kJ/mol): F $-328$, Br $-325$, S $-200$, Ar $+96$ (a noble gas does not take an electron readily).
Ordered by how negative (how much energy is released): F > Br > S > Ar, i.e. C > B > D > A.
Answer: (B). | ||||
| 67 | 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 $\mathrm{SiO_2}$ and $\mathrm{GeO_2}$ are acidic, $\mathrm{CO}$ is neutral, and $\mathrm{GeO}$ is distinctly acidic; $\mathrm{SnO_2}$ and $\mathrm{PbO_2}$ are amphoteric.
Answer: (D). | ||||
| 68 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | StatementsMedium | |
A medium statements question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements A: false — $\mathrm{CrO_4^{2-}}$ is tetrahedral.
B: true — dichromates are made by acidifying chromates (from chromite ore).
C: false — $\mathrm{MnO_4^{2-}}$ has Mn(VI), $d^1$, one unpaired electron: paramagnetic.
D: true — $3\mathrm{MnO_4^{2-}}+4\mathrm{H^+}\to2\mathrm{MnO_4^-}+\mathrm{MnO_2}+2\mathrm{H_2O}$.
E: true — higher oxidation states give more covalent, more acidic oxides.
Answer: (A). | ||||
| 69 | 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 Mn (as $\mathrm{MnO_2}$ in dry cells), Ni and Cd (nickel–cadmium cells) are the d-block metals used in the battery industry. Cr is not; Fe is not listed among battery metals in the syllabus texts.
Answer: (D). | ||||
| 70 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | StatementsMedium | |
A medium statements question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A: false — crystal field theory cannot explain why anionic ligands such as halides lie at the weak end of the spectrochemical series (a known limitation).
B: true — a stated limitation of VBT: no quantitative interpretation of thermodynamic or kinetic stability.
C: true — $\mathrm{Ni^{2+}}$ ($d^8$) with strong-field $\mathrm{CN^-}$ is square planar, $\mathrm{dsp^2}$.
D: false — cis-$\mathrm{[PtCl_2(en)_2]^{2+}}$ is chiral and exists as two optical isomers.
Answer: (D). | ||||
| 71 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Chromatography (for example column or thin-layer adsorption chromatography) separates components by their different adsorption on a stationary phase.
Answer: (A). | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Matching listEasy | |
A easy matching list question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A. Simple distillation separates liquids with very different boiling points: chloroform (334 K) and aniline (457 K) → IV.
B. Fractional distillation separates close-boiling fractions: crude oil → III.
C. Steam distillation suits steam-volatile, water-immiscible liquids: aniline–water → II.
D. Reduced pressure lets glycerol be distilled from spent lye without decomposing → I.
Answer: (D). | ||||
| 73 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter Noble gases are monoatomic and held only by weak dispersion forces, so they liquefy at very low temperatures and have very low boiling points.
Statement I is false. Statement II wrongly calls the forces strong and concludes high boiling points: false.
Answer: (B). | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques A carbocation has a positively charged carbon with only six valence electrons, so it is electron-deficient and acts as an electrophile. A free radical has seven electrons and a carbanion eight.
Answer: (D). | ||||
| 75 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsEasy | |
A easy statements question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The chain has 7 carbons: $\mathrm{HOCH_2}$, three $\mathrm{CH_2}$, $\mathrm{CH_2}$, $\mathrm{C{=}O}$, $\mathrm{CH_3}$.
The ketone outranks the alcohol, so it is the suffix and gets the lower locant: heptan-2-one, with OH at C-7 as the prefix hydroxy: 7-hydroxyheptan-2-one. Statement I is correct.
'2-oxoheptan-7-ol' treats the alcohol as principal group, which is wrong. Statement II is incorrect.
Answer: (C). | ||||
| 76 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | Single correctEasy | |
A easy single correct question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Alcoholic KOH: dehydrohalogenation, A = propene $\mathrm{CH_3CH{=}CH_2}$.
HBr (Markovnikov): B = 2-bromopropane.
Aqueous KOH: substitution, C = propan-2-ol.
Answer: (D). | ||||
| 77 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Assertion and reasonEasy | |
A easy assertion and reason question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers A: true — phenol $\mathrm{pK_a}\approx10.0$, ethanol $\approx15.9$.
R: false — a lower $\mathrm{pK_a}$ means a stronger acid, so phenol (its phenoxide ion stabilised by resonance) is the stronger acid.
Answer: (C). | ||||
| 78 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Assertion and reasonEasy | |
A easy assertion and reason question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers A: true — the O–H oxygen attacks electrophiles (alcohol as nucleophile), while protonated alcohols are attacked at carbon by nucleophiles (alcohol as electrophile).
R: true — $2\mathrm{ROH}+2\mathrm{Na}\to2\mathrm{RONa}+\mathrm{H_2}$. But this shows the acidic character of alcohols (O–H cleavage), not their dual nucleophile/electrophile behaviour, so it does not explain A.
Answer: (B). | ||||
| 79 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Matching listEasy | |
A easy matching list question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules A. Glucose has no carboxyl group, so $\mathrm{NaHCO_3}$ does not react → II.
B. Nitric acid oxidises both ends: saccharic (glucaric) acid → IV.
C. Prolonged heating with HI reduces it to n-hexane → III.
D. Bromine water oxidises the aldehyde group only: gluconic acid → I.
Answer: (D). | ||||
| 80 | 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 Ammonium arsenomolybdate is a yellow precipitate, lead iodide is yellow, and ammonium sulphide is a yellow solution; lead sulphate is a white precipitate.
Answer: (C). | ||||
| 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 $\mathrm{CH_4+2O_2\to CO_2+2H_2O}$: 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 $= 50\times10^{-2}$ mol of methane.
Answer: 50. | ||||
| 82 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom Energy per photon: $E = \frac{hc}{\lambda} = \frac{6.626\times10^{-34}\times3\times10^8}{242\times10^{-9}} = 8.214\times10^{-19}$ J.
Per mole: $8.214\times10^{-19}\times6.022\times10^{23} = 4.946\times10^5$ J mol$^{-1}$ $= 494.6$ kJ mol$^{-1}$.
Nearest integer: 495.
Answer: 495. | ||||
| 83 | Chemical Bonding and Molecular StructureHybridisation | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on hybridisation, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mathrm{NH_3}$ $sp^3$; $\mathrm{SO_2}$ $sp^2$; $\mathrm{SiO_2}$ $sp^3$ (each Si bonded tetrahedrally to four O in the network); $\mathrm{BeCl_2}$ $sp$; $\mathrm{CO_2}$ $sp$; $\mathrm{H_2O}$ $sp^3$; $\mathrm{CH_4}$ $sp^3$; $\mathrm{BF_3}$ $sp^2$.
$sp^3$: $\mathrm{NH_3, SiO_2, H_2O, CH_4}$ — 4.
Answer: 4. | ||||
| 84 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta_rG^\ominus = -RT\ln K_P = -8.314\times298\times\ln(2.47\times10^{-29})$.
$\ln(2.47\times10^{-29}) = 0.904-29\times2.3026 = -65.87$.
$\Delta_rG^\ominus = 2477.6\times65.87\ \mathrm{J} = 1.632\times10^5$ J $= 163.2$ kJ, i.e. 163.
Answer: 163. | ||||
| 85 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $\mathrm{Cu^{2+}+2e^-\to Cu}$: 2 F deposit one gram atom, so 1 F deposits 0.5 gram atom $= 5\times10^{-1}$.
Answer: 5. | ||||
| 86 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{Ni^{2+}}$ is $3d^8$. In an octahedral field $t_{2g}^6e_g^2$ leaves 2 unpaired electrons whatever the ligand.
$\mu = \sqrt{n(n+2)} = \sqrt8 = 2.83$ BM $= 28.3\times10^{-1}$ BM, i.e. 28.
Answer: 28. | ||||
| 87 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Kolbe electrolysis gives the radicals $\mathrm{CH_3^\bullet}$ and $\mathrm{C_2H_5^\bullet}$, which pair in three ways: ethane ($\mathrm{CH_3-CH_3}$), propane ($\mathrm{CH_3-C_2H_5}$) and butane ($\mathrm{C_2H_5-C_2H_5}$).
Answer: 3. | ||||
| 88 | Haloalkanes and HaloarenesElimination reactions | Class 12: Haloalkanes and Haloarenes | NumericalEasy | |
A easy numerical question on elimination reactions, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes With NaOH in ethanol (alcoholic alkali) bromoethane undergoes elimination: A = ethene $\mathrm{CH_2{=}CH_2}$, 4 H.
With NaOH in water, substitution: B = ethanol $\mathrm{C_2H_5OH}$, 6 H.
Total $4+6 = 10$.
Answer: 10. | ||||
| 89 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The phenolic O–H of 4-hydroxybenzaldehyde is acidic, and one equivalent of PhMgBr is used up deprotonating it (giving benzene and the magnesium phenoxide) before it can add to the C=O group.
Aqueous $\mathrm{NH_4Cl}$ then re-protonates the phenoxide, so P is 4-hydroxybenzaldehyde again, with one hydroxyl group.
Answer: 1. | ||||
| 90 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter With hot concentrated $\mathrm{H_2SO_4}$: NaBr gives reddish-brown $\mathrm{Br_2}$ vapour, $\mathrm{NaNO_3}$ gives brown $\mathrm{NO_2}$, KI gives violet $\mathrm{I_2}$ vapour, while $\mathrm{CaF_2}$ gives colourless HF.
Molar mass of $\mathrm{CaF_2} = 40+2\times19 = 78$ g mol$^{-1}$.
Answer: 78. | ||||
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