JEE (Main) · 2024
JEE (Main) 2024 — 5 April, Shift 2
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 Differentiability3
- Integral Calculus3
- Permutations and Combinations2
10 easy · 18 medium · 2 hard
Physics
30 Q
- Magnetic Effects of Current and Magnetism3
- Kinematics2
- Laws of Motion2
28 easy · 2 medium · 0 hard
Chemistry
30 Q
- Chemical Bonding and Molecular Structure3
- Hydrocarbons3
- Electrochemistry2
23 easy · 7 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 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations $S=\{2^1, 2^2, \ldots, 2^9\}$ has 9 elements, so each of A, B, C gets 3 elements.
The sets are named A, B, C, so a partition is an assignment of the 9 elements to the three named sets, 3 to each:
choose 3 for A, then 3 of the remaining 6 for B; the last 3 go to C.
Number $= \binom{9}{3}\binom{6}{3}\binom{3}{3} = 84\times 20\times 1 = 1680$ $\left(=\frac{9!}{3!\,3!\,3!}\right)$.
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 For $x\ge 0$: $g(x)=e^x\ge 1$, so $f(g(x))=e^x-1$, which takes every value in $[0,\infty)$.
For $x\le 0$: $g(x)=x+1\le 1$, so $f(g(x))=|x|=-x$, which also takes every value in $[0,\infty)$.
Not one-one: for example $f(g(-1))=1$ and $f(g(\ln 2))=2-1=1$.
Not onto: $f(g(x))\ge 0$ always, so no negative real number is attained, while the codomain is $\mathbf{R}$.
Answer: (D). | ||||
| 3 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $z=x+iy$. Multiply numerator and denominator by $1+\sqrt{3}i$:
$\frac{(x+1)+i(y-\sqrt3)}{1-\sqrt3 i}=\frac{[(x+1)+i(y-\sqrt3)](1+\sqrt3 i)}{4}$.
Its imaginary part is $\frac{\sqrt3(x+1)+(y-\sqrt3)}{4}=\frac{\sqrt3x+y}{4}$.
So $S_2$ is the half-plane $y\ge-\sqrt3x$, bounded by the line through the origin at angle $120^\circ$ (equivalently $-60^\circ$).
In polar form $z=re^{i\theta}$: $S_3$ is $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$, and $S_2$ is $-\frac{\pi}{3}\le\theta\le\frac{2\pi}{3}$.
Together: $-\frac{\pi}{3}\le\theta\le\frac{\pi}{2}$, a sector of angle $\frac{5\pi}{6}$ of the disc $|z|\le5$.
Area $=\frac12\cdot 5^2\cdot\frac{5\pi}{6}=\frac{125\pi}{12}$.
Answer: (D). | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For infinitely many solutions the coefficient determinant must vanish:
$\begin{vmatrix}1&1&1\\2&5&5\\1&2&m\end{vmatrix}=1(5m-10)-1(2m-5)+1(4-5)=3m-6=0$, so $m=2$.
With $m=2$ the third row must be a combination of the first two: $p(1,1,1)+q(2,5,5)=(1,2,2)$ gives $p+2q=1$, $p+5q=2$, so $q=\frac13$, $p=\frac13$.
Then $n=\frac13(4)+\frac13(17)=7$.
Check the options with $m=2$, $n=7$: $m^2+n^2-mn=4+49-14=39$.
Answer: (B). | ||||
| 5 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Compare cofactors of the first row of A with the first row of B.
$C_{11}=\alpha\cdot2\alpha-\beta\cdot\alpha=2\alpha^2-\alpha\beta=3\alpha\Rightarrow 2\alpha-\beta=3$ (as $\alpha\ne0$).
$C_{12}=-(\alpha\cdot2\alpha-\beta\cdot(-\beta))=-(2\alpha^2+\beta^2)=-9\Rightarrow 2\alpha^2+\beta^2=9$.
Substituting $\beta=2\alpha-3$: $6\alpha^2-12\alpha=0$, so $\alpha=2$, $\beta=1$.
Check: $C_{13}=\alpha^2+\alpha\beta=6=3\alpha$. ✓
$\det A=\beta C_{11}+\alpha C_{12}+3C_{13}=1(6)+2(-9)+3(6)=6$.
B is the cofactor matrix, so $B=(\operatorname{adj}A)^T$ and $\det B=\det(\operatorname{adj}A)=(\det A)^{2}=36$.
$\det(AB)=\det A\cdot\det B=6\times36=216$.
Answer: (C). | ||||
| 6 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Letters: B, B, H, J, O; in alphabetical order B, H, J, O.
Words beginning with B: the other letters B, H, J, O are distinct, $4!=24$ words (1st–24th).
Beginning with H: remaining B, B, J, O, $\frac{4!}{2!}=12$ words (25th–36th).
Beginning with J: 12 words (37th–48th).
Beginning with O: the 49th is OBBHJ and the 50th is OBBJH.
Answer: (B). | ||||
| 7 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are $6^3=216$ equally likely triples (a, b, c). "One real root bigger than the other" means two distinct real roots: $b^2>4ac$.
$b=1, 2$: $b^2\le4<4ac$, none.
$b=3$: $ac<\frac94$, $ac\in\{1,2\}$: (1,1), (1,2), (2,1) → 3.
$b=4$: $ac<4$, $ac\in\{1,2,3\}$: 1 + 2 + 2 = 5.
$b=5$: $ac<\frac{25}{4}$, $ac\le 6$: products 1 (1 way), 2 (2), 3 (2), 4 (3), 5 (2), 6 (4) → 14.
$b=6$: $ac<9$, $ac\le 8$: the 14 above plus $ac=8$ (2 ways: (2,4), (4,2)) → 16.
Favourable $=3+5+14+16=38$, so $p=\frac{38}{216}$ and $216p=38$.
Answer: (B). | ||||
| 8 | 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 For an A.P. the middle term is the mean: $K=4^{1+x}+4^{1-x}+16^x+16^{-x}$.
Let $u=4^x+4^{-x}\ge2$ (AM–GM, equality at $x=0$). Then $4^{1+x}+4^{1-x}=4u$ and $16^x+16^{-x}=u^2-2$.
$K=u^2+4u-2$, which increases with $u$ for $u\ge2$, so the least value is at $u=2$: $K=4+8-2=10$.
Answer: (A). | ||||
| 9 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1}=\binom{12}{r}\left(\frac{3^{1/5}}{x}\right)^{12-r}\left(\frac{2x}{5^{1/3}}\right)^{r}$, with power of $x$ equal to $2r-12$; the constant term has $r=6$.
$T_7=\binom{12}{6}\,3^{6/5}\,\frac{2^6}{5^2}=924\cdot\frac{64}{25}\cdot3\cdot3^{1/5}$.
Setting this equal to $\alpha\cdot2^8\cdot3^{1/5}$: $\alpha=\frac{924\times3\times64}{25\times256}=\frac{2772}{100}=\frac{693}{25}$.
$25\alpha=693$.
Answer: (D). | ||||
| 10 | 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 $2x^2+x$ is continuous, so only $[x^2]-[x]$ can jump: possibly where $x^2\in\{1,2,3,4\}$ or $x\in\{0,1,2\}$, i.e. at $x=-1, 0, 1, \sqrt2, \sqrt3, 2$.
$x=-1$: $f(-1)=2-1+1+1=3$; just to the right, $[x^2]=0$, $[x]=-1$, $f\to2-1+0+1=2$. Discontinuous.
$x=0$: $[x^2]=0$ on both sides, $[x]$ jumps from $-1$ to $0$. Discontinuous.
$x=1$: left: $3+0-0=3$; right and at 1: $3+1-1=3$. Both jumps cancel: continuous.
$x=\sqrt2$, $x=\sqrt3$: $[x^2]$ jumps by 1 while $[x]=1$ stays. Discontinuous.
$x=2$: $f(2)=10+4-2=12$; left: $10+3-1=12$. Continuous.
Points of discontinuity: $-1, 0, \sqrt2, \sqrt3$, i.e. 4.
Answer: (C). | ||||
| 11 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t=x^{10}$: $x=t^{1/10}$, $dx=\frac{1}{10}t^{\frac{1}{10}-1}dt$, limits stay 0 to 1.
$\int_0^1(1-x^{10})^{20}dx=\frac1{10}\int_0^1 t^{\frac1{10}-1}(1-t)^{21-1}dt=\frac1{10}\,\beta\!\left(\frac1{10}, 21\right)$.
So $a=\frac1{10}$, $b=\frac1{10}$, $c=21$ and $100(a+b+c)=100\times21.2=2120$.
Answer: (A). | ||||
| 12 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals For $x\ge0$: $y=x^2$ and $y=0$, which meet only at the origin, so no region is enclosed there.
For $x<0$: $y=-x^2$ and $y=2x$, which meet where $-x^2=2x$, i.e. $x=-2$ (and $x=0$).
On $[-2, 0]$, $-x^2\ge2x$ (at $x=-1$: $-1>-2$).
Area $=\int_{-2}^{0}(-x^2-2x)\,dx=\left[-\frac{x^3}{3}-x^2\right]_{-2}^{0}=0-\left(\frac83-4\right)=\frac43$.
Answer: (C). | ||||
| 13 | Differential EquationsOrder and degree | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on order and degree, from Differential Equations. NCERT: Class 12: Differential Equations Centre $(a, a)$, passing through the origin: $x^2+y^2-2ax-2ay=0$, so $2a=\frac{x^2+y^2}{x+y}$.
Differentiate: $2x+2yy'=2a(1+y')$, so $2a=\frac{2(x+yy')}{1+y'}$.
Equate: $(x^2+y^2)(1+y')=2(x+y)(x+yy')$.
$x^2+y^2+(x^2+y^2)y'=2x^2+2xy+(2xy+2y^2)y'$
$y'(x^2-y^2-2xy)=x^2-y^2+2xy$.
So $(x^2-y^2+2xy)dx=(x^2-y^2-2xy)dy$.
Answer: (A). | ||||
| 14 | CirclesStandard and general form | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on standard and general form, from Circles. NCERT: Class 11: Conic Sections Take A(0, 0), B(4, 0), C(4, 4), D(0, 4). Then E(2, 0) and F(2, 2), which lies on AC.
A circle touching BC ($x=4$) and CD ($y=4$) from inside the square has centre $(4-r, 4-r)$.
It passes through F: $(4-r-2)^2+(4-r-2)^2=r^2$, i.e. $2(2-r)^2=r^2$.
$2r^2-8r+8=r^2\Rightarrow r^2-8r+8=0$ (the usable root is $r=4-2\sqrt2$).
Answer: (B). | ||||
| 15 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections $C_1$: centre $(1, 1)$, radius 1. $C_2$: $(x+1)^2+y^2=4$, i.e. $x^2+y^2+2x-3=0$.
The circles do intersect: distance between centres $\sqrt5$ lies between $2-1$ and $2+1$.
Common chord: $C_1-C_2=0$: $-4x-2y+4=0$, i.e. $2x+y=2$.
On the y-axis, $x=0$: P(0, 2).
Square of the distance from (1, 1): $1^2+1^2=2$.
Answer: (B). | ||||
| 16 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines Line AB: slope $\frac23$, equation $2x-3y+5=0$; $AB=\sqrt{9+4}=\sqrt{13}$.
Area 10 means the distance of P(x, y) from AB is $\frac{20}{\sqrt{13}}$: $|2x-3y+5|=20$.
Points above the line (larger y) make $2x-3y+5$ negative, so $2x-3y+5=-20$, i.e. $-2x+3y=25$.
In the form $ax+by=15$: multiply by $\frac35$: $-\frac65x+\frac95y=15$, so $a=-\frac65$, $b=\frac95$.
$5a+2b=-6+\frac{18}{5}=-\frac{12}{5}$.
Answer: (D). | ||||
| 17 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line: $Q=(1+2t, -1+3t, 2+5t)$. For the foot of the perpendicular from P(8, 5, 7):
$(Q-P)\cdot(2,3,5)=0$: $2(2t-7)+3(3t-6)+5(5t-5)=38t-57=0$, $t=\frac32$.
Foot $M=(4, \frac72, \frac{19}{2})$. Image $=2M-P=(0, 2, 12)$.
$\alpha+\beta+\gamma=14$.
Answer: (D). | ||||
| 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 Let $\vec u=\vec c+\hat i$. The condition is $\vec u\times(\vec a+\vec b+\hat i)=\vec a\times\vec u=-\vec u\times\vec a$, so $\vec u\times(2\vec a+\vec b+\hat i)=\vec0$.
Hence $\vec u=\lambda(2\vec a+\vec b+\hat i)=\lambda(7\hat i+8\hat j)$ and $\vec c=(7\lambda-1)\hat i+8\lambda\hat j$.
$\vec a\cdot\vec c=2(7\lambda-1)+40\lambda=54\lambda-2=-29\Rightarrow\lambda=-\frac12$.
$\vec c=-\frac92\hat i-4\hat j$.
$\vec c\cdot(-2\hat i+\hat j+\hat k)=9-4+0=5$.
Answer: (D). | ||||
| 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 $\vec a=\vec b\times\vec c$ is perpendicular to $\vec c$, so $|\vec c-\vec a|^2=|\vec c|^2+|\vec a|^2=|\vec c|^2+4$.
Also $|\vec a|=|\vec b||\vec c|\sin\alpha$: $2=3|\vec c|\sin\alpha$, so $|\vec c|^2=\frac{4}{9\sin^2\alpha}$ (and $\alpha\ne0$).
This is least when $\sin\alpha$ is greatest on $(0, \frac\pi3]$, i.e. $\sin^2\alpha=\frac34$: $|\vec c|^2=\frac{16}{27}$.
Minimum of $27|\vec c-\vec a|^2=27\left(\frac{16}{27}+4\right)=16+108=124$.
Answer: (B). | ||||
| 20 | Limits, Continuity and DifferentiabilityDerivatives up to order two | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on derivatives up to order two, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Write $c=\cos\theta$. Numerator: $2c+2c^2-1$. Denominator: $(4c^3-3c)+4(2c^2-1)+5c+2=4c^3+8c^2+2c-2=(2c^2+2c-1)(2c+2)$.
So $y=\frac{1}{2(1+\cos\theta)}=\frac14\sec^2\frac\theta2$.
$y'=\frac14\sec^2\frac\theta2\tan\frac\theta2$, $y''=\frac18\left(2\sec^2\frac\theta2\tan^2\frac\theta2+\sec^4\frac\theta2\right)$.
At $\theta=\frac\pi2$: $\sec^2\frac\pi4=2$, $\tan\frac\pi4=1$: $y=\frac12$, $y'=\frac12$, $y''=\frac18(4+4)=1$.
$y''+y'+y=2$.
Answer: (D). | ||||
| 21 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalMedium | |
A medium numerical question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Note $2+2x-x^2=3-(x-1)^2$. Let $s=\sin x$, $t=(x-1)^2$: $s^2+(3-t)s-3t=0$, i.e. $(s+3)(s-t)=0$.
$\sin x=-3$ is impossible, so $\sin x=(x-1)^2$.
Since $\sin x\le1$, we need $(x-1)^2\le1$, i.e. $0\le x\le2$.
Let $h(x)=\sin x-(x-1)^2$: $h(0)=-1<0$, $h(1)=\sin1>0$, $h(2)=\sin2-1<0$, so there is a root in $(0,1)$ and one in $(1,2)$.
$h''(x)=-\sin x-2<0$, so h is concave and has at most two roots.
Number of solutions $=2$.
Answer: 2. | ||||
| 22 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalHard | |
A hard numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Note $(\sqrt3-\sqrt2)^2=5-2\sqrt6$, $(\sqrt3-\sqrt2)^3=9\sqrt3-11\sqrt2$, $(\sqrt3-\sqrt2)^4=49-20\sqrt6$.
Let $x=\frac{\sqrt3-\sqrt2}{\sqrt3}=1-\sqrt{\frac23}$. Then the terms are
$\frac{x}{2},\ \frac{3x^2}{18}=\frac{x^2}{6},\ \frac{3\sqrt3\,x^3}{36\sqrt3}=\frac{x^3}{12},\ \frac{9x^4}{180}=\frac{x^4}{20},\ \ldots$, i.e. $\frac{x^n}{n(n+1)}$.
$\sum_{n\ge1}\frac{x^n}{n(n+1)}=\sum\frac{x^n}{n}-\sum\frac{x^n}{n+1}=-\ln(1-x)-\frac1x\left(-\ln(1-x)-x\right)=1+\frac{1-x}{x}\ln(1-x)$.
So the sum is $2+\frac{1-x}{x}\ln(1-x)$, with $1-x=\sqrt{\frac23}$ and $\frac{1-x}{x}=\frac{\sqrt2}{\sqrt3-\sqrt2}=\sqrt6+2$.
$\ln(1-x)=\frac12\ln\frac23$, so the sum is $2+\frac{\sqrt6+2}{2}\ln\frac23=2+\left(\sqrt{\frac32}+1\right)\ln\frac23$.
Hence $a=2$, $b=3$ (coprime) and $11a+18b=22+54=76$.
Answer: 76. | ||||
| 23 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalHard | |
A hard numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Roots of $2x^2+x-2=0$: $a=\frac{-1+\sqrt{17}}{4}$, $b=\frac{-1-\sqrt{17}}{4}$, with $ab=-1$.
If x is a root of $2x^2+x-2=0$ then $\frac1x$ is a root of $2+x-2x^2=0$, so $2+x-2x^2=-2\left(x-\frac1a\right)\left(x-\frac1b\right)$, which tends to 0.
Using $1-\cos u\approx\frac{u^2}{2}$ and $(1-ax)^2=a^2\left(x-\frac1a\right)^2$:
limit $=\frac{16\cdot\frac12\cdot4\left(x-\frac1a\right)^2\left(\frac1a-\frac1b\right)^2}{a^2\left(x-\frac1a\right)^2}=\frac{32}{a^2}\left(\frac1a-\frac1b\right)^2$.
$\frac1a-\frac1b=\frac{b-a}{ab}=\frac{-\sqrt{17}/2}{-1}=\frac{\sqrt{17}}{2}$, so the limit is $\frac{32\cdot17}{4a^2}=\frac{136}{a^2}$.
$a^2=\frac{18-2\sqrt{17}}{16}=\frac{9-\sqrt{17}}{8}$, so the limit $=\frac{1088}{9-\sqrt{17}}=\frac{1088(9+\sqrt{17})}{64}=17(9+\sqrt{17})=153+17\sqrt{17}$.
$\alpha+\beta=153+17=170$.
Answer: 170. | ||||
| 24 | 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 $8x-x^2-12=4-(x-4)^2\ge0$ requires $2\le x\le6$. Put $y=\sqrt{4-(x-4)^2}$: the point (x, y) runs over the upper half of the circle with centre (4, 0) and radius 2.
The expression is the square of the distance from (x, y) to the point (7, 4).
Write $x=4+2\cos\theta$, $y=2\sin\theta$, $0\le\theta\le\pi$: distance$^2=(3-2\cos\theta)^2+(4-2\sin\theta)^2=29-12\cos\theta-16\sin\theta$.
$12\cos\theta+16\sin\theta=20\sin(\theta+\phi)$ reaches its maximum 20 at $\tan\theta=\frac43$ (inside $[0,\pi]$), giving $m=9$.
On $[0,\pi]$ its least value is at the end $\theta=\pi$: $-12$ (at $\theta=0$ it is 12), giving $M=41$ (at x = 2).
$M^2-m^2=1681-81=1600$.
Answer: 1600. | ||||
| 25 | 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 Replace x by $\pi-x$ (the denominator is unchanged) and add: $f(t)=\pi\int_0^\pi\frac{dx}{1-k\sin^2x}$ with $k=\cos^2t$.
$\int_0^\pi\frac{dx}{1-k\sin^2x}=2\int_0^{\pi/2}\frac{\sec^2x\,dx}{1+(1-k)\tan^2x}=2\cdot\frac{1}{\sqrt{1-k}}\cdot\frac\pi2=\frac{\pi}{\sqrt{1-k}}=\frac{\pi}{\sin t}$ (as $\sin t>0$).
So $f(t)=\frac{\pi^2}{\sin t}$ and $\int_0^{\pi/2}\frac{\pi^2}{f(t)}dt=\int_0^{\pi/2}\sin t\,dt=1$.
Answer: 1. | ||||
| 26 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor: $\int\frac{2x}{(1+x^2)^2}dx=-\frac{1}{1+x^2}$, so I.F. $=e^{-\frac{1}{1+x^2}}$.
$\frac{d}{dx}\left(y\,e^{-\frac1{1+x^2}}\right)=xe^{\frac1{1+x^2}}e^{-\frac1{1+x^2}}=x$, so $y\,e^{-\frac1{1+x^2}}=\frac{x^2}{2}+C$; $y(0)=0$ gives $C=0$.
Thus $f(x)=\frac{x^2}{2}$. It meets $y=x+4$ where $x^2-2x-8=0$: $x=-2, 4$.
Area $=\int_{-2}^{4}\left(x+4-\frac{x^2}{2}\right)dx=\left[\frac{x^2}{2}+4x-\frac{x^3}{6}\right]_{-2}^{4}=\left(8+16-\frac{32}{3}\right)-\left(2-8+\frac43\right)=30-12=18$.
Answer: 18. | ||||
| 27 | Conic SectionsParabola | Class 11: Conic Sections | NumericalEasy | |
A easy numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections The tangent has slope $-\frac12$ (perpendicular to slope 2).
With $X=x-9$: $y^2=4X$ ($a=1$); the tangent of slope m touches at $\left(\frac{a}{m^2}, \frac{2a}{m}\right)=(4, -4)$ in (X, y).
So P = (13, −4). The circle's centre is (7, 4).
Distance $=\sqrt{6^2+8^2}=10$.
Answer: 10. | ||||
| 28 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical 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 of the lines: $P=(-2-3s, 2+4s, 5+2s)$, $Q=(-2-t, -6+2t, 1)$.
$\vec{QP}=(-3s+t, 8+4s-2t, 4+2s)$ must be perpendicular to both directions:
with $(-3,4,2)$: $29s-11t+40=0$; with $(-1,2,0)$: $11s-5t+16=0$.
Solving: $s=-1$, $t=1$. So P = (1, −2, 3) and Q = (−3, −4, 1).
The line PQ has direction (4, 2, 2) ∝ (2, 1, 1): points $(-3+2\lambda, -4+\lambda, 1+\lambda)$. For $x=-1$, $\lambda=1$: $(-1, -3, 2)$.
$\alpha=-3$, $\beta=2$, $(\alpha-\beta)^2=25$.
Answer: 25. | ||||
| 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 $K=1-\frac13-\frac16-\frac14=\frac14$.
$\mu=\frac\alpha3+\frac14-\frac34=\frac\alpha3-\frac12$; $E(X^2)=\frac{\alpha^2}{3}+\frac14+\frac94=\frac{\alpha^2}{3}+\frac52$.
$\sigma^2=E(X^2)-\mu^2=\frac{2\alpha^2}{9}+\frac\alpha3+\frac94$.
$\sigma=\mu+2=\frac\alpha3+\frac32$, so $\sigma^2=\frac{\alpha^2}{9}+\alpha+\frac94$.
Equating: $\frac{\alpha^2}{9}=\frac{2\alpha}{3}$, so $\alpha=0$ or $\alpha=6$.
$\alpha=0$ would repeat the value 0 already in the table, so $\alpha=6$: $\mu=\frac32$, $\sigma=\frac72$ (check: $\sigma^2=8+2+\frac94=\frac{49}{4}$ ✓).
$\sigma+\mu=5$.
Answer: 5. | ||||
| 30 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $x\ge-5$: $x(x+5)+2(x+7)-2=x^2+7x+12=0$, $x=-3$ or $-4$; both valid.
$-7\le x<-5$: $-x(x+5)+2(x+7)-2=0\Rightarrow x^2+3x-12=0$, $x=\frac{-3\pm\sqrt{57}}{2}$; only $\frac{-3-\sqrt{57}}{2}\approx-5.27$ lies in the interval.
$x<-7$: $-x(x+5)-2(x+7)-2=0\Rightarrow x^2+7x+16=0$, discriminant $49-64<0$: none.
Total: 3 real solutions.
Answer: 3. | ||||
| 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 Only like quantities can be added or subtracted.
$\frac{a}{V^2}$ has the dimensions of pressure, so $a=PV^2$; b has the dimensions of volume, so $b=V$.
$ab^{-1}=\frac{PV^2}{V}=PV$, which has the dimensions of energy: $[\mathrm{ML^{-1}T^{-2}}][\mathrm{L^3}]=[\mathrm{ML^2T^{-2}}]$.
Answer: (A). | ||||
| 32 | KinematicsUniform circular motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The monkey moves with the man on the same circle of radius 9 m.
Frequency: 120 revolutions in 180 s, $f=\frac23$ s$^{-1}$, so $\omega=2\pi f=\frac{4\pi}{3}$ rad/s.
$a_c=\omega^2r=\frac{16\pi^2}{9}\times9=16\pi^2$ m s$^{-2}$.
Answer: (C). | ||||
| 33 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion On a horizontal surface the normal reaction is $N=mg=50\times9.8=490$ N.
Kinetic friction $f_k=\mu_kN=0.3\times490=147$ N.
Answer: (C). | ||||
| 34 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $\vec F=\frac{d\vec p}{dt}=-k\sin(kt)\,\hat i-k\cos(kt)\,\hat j$.
$\vec F\cdot\vec p=-k\sin(kt)\cos(kt)+k\cos(kt)\sin(kt)=0$.
So the force is perpendicular to the momentum (as expected, since $|\vec p|=1$ is constant): the angle is $\frac{\pi}{2}$.
Answer: (A). | ||||
| 35 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power With constant power P (starting from rest), $\frac12mv^2=Pt$, so $v=\sqrt{\frac{2Pt}{m}}\propto t^{1/2}$.
$s=\int_0^t v\,dt\propto\int_0^t t^{1/2}dt\propto t^{3/2}$.
Answer: (B). | ||||
| 36 | 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 20 VSD = 19 MSD, so 1 VSD $=\frac{19}{20}$ MSD.
Least count $=1\text{ MSD}-1\text{ VSD}=\frac{1}{20}$ MSD $=0.1$ mm.
So 1 MSD $=20\times0.1=2$ mm.
Answer: (B). | ||||
| 37 | GravitationKepler's laws | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on kepler's laws, from Gravitation. NCERT: Class 11: Gravitation For a circular orbit $T^2=\frac{4\pi^2r^3}{GM}$, so $r^3\propto MT^2$.
Earth's geostationary orbit: $T=24$ h, $r=4.2\times10^4$ km. For the planet: $M'=\frac{M}4$, $T'=6$ h.
$\frac{r'^3}{r^3}=\frac14\times\left(\frac{6}{24}\right)^2=\frac{1}{64}$, so $r'=\frac r4=1.05\times10^4$ km.
Answer: (C). | ||||
| 38 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Matching listEasy | |
A easy matching list 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 (A) The restoring force per unit area is stress → (III).
(B) Equal and opposite forces parallel to opposite faces produce shearing, measured by the shear modulus → (IV).
(C) A uniform normal force per unit area over the whole surface changes the volume: bulk modulus → (I).
(D) Equal and opposite forces normal to opposite faces stretch or compress the body lengthwise: Young's modulus → (II).
A→III, B→IV, C→I, D→II.
Answer: (D). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory A molecule of diameter d sweeps a collision cylinder of cross-section $\pi d^2$; allowing for the motion of the other molecules (relative speed $\sqrt2\,\bar v$), the number of collisions per unit length is $\sqrt2\,n\pi d^2$.
Mean free path $\lambda=\frac{1}{\sqrt2\,n\pi d^2}$.
Answer: (A). | ||||
| 40 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance A moving vehicle becomes charged by friction with the air (and the moving fluid). With rubber tyres the charge cannot leak away, and a spark could ignite the inflammable fluid.
A metal chain touching the ground earths the body and conducts the excess charge away, preventing sparking.
Answer: (C). | ||||
| 41 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $X_L=2\pi fL=2\times3.14\times50\times10\times10^{-3}=3.14\ \Omega$.
In a series circuit the same current flows through the inductor: $I=\frac{V_L}{X_L}=\frac{31.4}{3.14}=10$ A.
Answer: (A). | ||||
| 42 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity The 5 Ω and 10 Ω resistors are in parallel, so they have the same potential difference V across them.
Heat per second $P=\frac{V^2}{R}$, so $\frac{P_5}{P_{10}}=\frac{10}{5}=2:1$ (the 20 Ω resistor and the 10 V source do not affect the ratio).
Answer: (B). | ||||
| 43 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The electric force on a charge is $\vec F_1=q\vec E$, independent of its velocity.
The magnetic (Lorentz) force is $\vec F_2=q(\vec V\times\vec B)$; the order of the cross product matters, since $\vec B\times\vec V=-\vec V\times\vec B$.
Answer: (B). | ||||
| 44 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Matching listEasy | |
A easy matching list question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Infra-red lies between microwaves and visible light: 1 mm to 700 nm → (III).
Ultraviolet lies just below violet: 400 nm to 1 nm → (II).
X-rays: 1 nm to $10^{-3}$ nm → (IV).
Gamma rays: shorter than $10^{-3}$ nm → (I).
A→III, B→II, C→IV, D→I.
Answer: (B). | ||||
| 45 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | StatementsEasy | |
A easy statements question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Statement II: in a dispersive medium the refractive index depends on wavelength (it is larger for shorter wavelengths). True.
Statement I: red has the longest wavelength of the three, so the smallest refractive index and the smallest deviation; red bends less than yellow and violet. True.
Answer: (A). | ||||
| 46 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Full-scale current: $I_g=\frac{10}{100+400}=0.02$ A.
Shunt for a 10 A range: $S=\frac{I_gG}{I-I_g}=\frac{0.02\times100}{10-0.02}=\frac{2}{9.98}\approx0.20\ \Omega=20\times10^{-2}\ \Omega$.
So $x\approx20$.
Answer: (A). | ||||
| 47 | 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 $eV_0=K_{max}=h\nu-\phi$.
(A) True: $V_0$ rises with the frequency $\nu$.
(B) True: it depends on the work function $\phi$ of the emitter.
(C) Not true: intensity changes the number of photoelectrons (the photocurrent), not their maximum kinetic energy, so $V_0$ is unchanged.
(D) True: $V_0=\frac{K_{max}}{e}$.
Answer: (C). | ||||
| 48 | 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 the Bohr model $L=mvr$ and $\frac{mv^2}{r}=\frac{ke^2}{r^2}$, so $v=\sqrt{\frac{ke^2}{mr}}$.
$L=mr\sqrt{\frac{ke^2}{mr}}=\sqrt{ke^2m\,r}\propto\sqrt r$.
(Equivalently $L=\frac{nh}{2\pi}\propto n$ and $r\propto n^2$.)
Answer: (A). | ||||
| 49 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics For an adiabatic process of an ideal gas, $P^{1-\gamma}T^{\gamma}=$ constant, i.e. $P\propto T^{\frac{\gamma}{\gamma-1}}$.
Given $P\propto T^3$: $\frac{\gamma}{\gamma-1}=3\Rightarrow\gamma=3\gamma-3\Rightarrow\gamma=\frac32$.
Answer: (A). | ||||
| 50 | 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 The first OR gate gives $A+B$; the AND gate gives $B\cdot1=B$; the final OR gate gives $Y=(A+B)+B=A+B$.
An OR output is 0 only when every input is 0: $A=0$ and $B=0$.
Answer: (A). | ||||
| 51 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $H=\frac{u^2\sin^2\theta}{2g}\propto u^2$ for the same angle of projection.
Halving u divides H by 4: $H'=\frac{64}{4}=16$ m.
Answer: 16. | ||||
| 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 Hollow sphere: $I=\frac23mR^2$. Rolling without slipping: $v=\omega R$.
$K_{rot}=\frac12\cdot\frac23mR^2\omega^2=\frac13mv^2$, $K_{trans}=\frac12mv^2$, total $=\frac56mv^2$.
$\frac{K_{rot}}{K_{total}}=\frac{1/3}{5/6}=\frac25$, so $x=2$.
Answer: 2. | ||||
| 53 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics First minimum: $a\sin\theta=\lambda$; on the screen $y=\frac{\lambda D}{a}$.
$y=\frac{550\times10^{-9}\times1}{0.20\times10^{-3}}=2.75\times10^{-3}$ m $=275\times10^{-5}$ m.
Answer: 275. | ||||
| 54 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Pascal's law: the pressure is the same on both surfaces (at the same level), $\frac{F_1}{A_1}=\frac{F_2}{A_2}$.
$F_2=F_1\left(\frac{d_2}{d_1}\right)^2=10\times\left(\frac{14}{1.4}\right)^2=10\times100=1000$ N.
Answer: 1000. | ||||
| 55 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Fundamental frequency $f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$, so at the same tension $fL$ is constant.
$L'=\frac{400\times90}{600}=60$ cm.
Answer: 60. | ||||
| 56 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Each part has resistance $\frac{20}{10}=2\ \Omega$.
Two parts in parallel: $\frac{2\times2}{2+2}=1\ \Omega$; the 10 parts make 5 such pairs.
The 5 pairs in series: $5\times1=5\ \Omega$.
Answer: 5. | ||||
| 57 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Inside a long solenoid $B=\mu_0\frac{m}{L}I$.
$m=\frac{BL}{\mu_0I}=\frac{6.28\times10^{-3}\times0.5}{4\pi\times10^{-7}\times5}=\frac{3.14\times10^{-3}}{6.28\times10^{-6}}=500$ (with $\pi=3.14$).
Answer: 500. | ||||
| 58 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance From the figure, p is on the axis at distance r from the centre O, and R is on the equatorial line at distance 2r from O.
Treating the dipole as short: axial field $E=\frac{2kp}{r^3}$; equatorial field at $2r$: $E_R=\frac{kp}{(2r)^3}=\frac{kp}{8r^3}$.
$\frac{E}{E_R}=\frac{2/r^3}{1/(8r^3)}=16$, so $E_R=\frac{E}{16}$ and $x=16$.
Answer: 16. | ||||
| 59 | 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 $|\varepsilon|=L\frac{dI}{dt}$, with $\frac{dI}{dt}=3$ A/s.
$L=\frac{12\times10^{-3}}{3}=4\times10^{-3}$ H $=4$ mH.
Answer: 4. | ||||
| 60 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Lyman series limit ($n=\infty\to1$): $\frac1{\lambda}=R$, so $\frac1R=915$ Å.
Longest Balmer line ($3\to2$): $\frac1{\lambda}=R\left(\frac14-\frac19\right)=\frac{5R}{36}$, so $\lambda=\frac{36}{5}\times915=6588$ Å.
Answer: 6588. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct 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 methane gives one mole of $\mathrm{CO_2}$.
Moles of $\mathrm{CO_2}=\frac{11}{44}=0.25$, so 0.25 mol of methane is needed.
Answer: (A). | ||||
| 62 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Assertion and reasonEasy | |
A easy assertion and reason question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) Both molecules are pyramidal (sp$^3$ N with one lone pair). $\mu(\mathrm{NH_3})\approx1.47$ D is greater than $\mu(\mathrm{NF_3})\approx0.23$ D. True.
(R) In $\mathrm{NH_3}$ the N–H bond dipoles point towards N (N is more electronegative), the same way as the lone-pair orbital dipole, so they add. In $\mathrm{NF_3}$ the N–F bond dipoles point towards the very electronegative F, opposite to the lone-pair dipole, so they largely cancel. True, and it is exactly why A holds.
Answer: (A). | ||||
| 63 | EquilibriumSolubility product | Class 11: Equilibrium | StatementsEasy | |
A easy statements question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium In a saturated solution the ionic product just equals $K_{sp}$. Passing HCl gas raises $[\mathrm{Cl^-}]$ (common ion), so the ionic product exceeds $K_{sp}$ and the salt crystallises out.
Statement I: $\mathrm{BaCl_2}$ separates as a white solid, seen as white turbidity. Correct.
Statement II: NaCl is precipitated by the common-ion effect (the basis of purifying common salt). Correct.
Answer: (A). | ||||
| 64 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry As written, M is the anode (left) and X the cathode (right), and the cell reaction is $\mathrm{M+X\to M^{2+}+X^{2-}}$.
$E^0_{cell}=E^0_{cathode}-E^0_{anode}=0.34-0.46=-0.12$ V.
A negative $E^0_{cell}$ means $\Delta G^0=-nFE^0>0$: the written reaction is not spontaneous, and its reverse $\mathrm{M^{2+}+X^{2-}\to M+X}$ (with $E^0=+0.12$ V) is.
Neither $\pm0.80$ V is the cell potential.
Answer: (B). | ||||
| 65 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Faraday's first law: $m=ZQ$. The electrochemical equivalent Z is the mass deposited by one coulomb.
So 1 C deposits 1 electrochemical equivalent of silver ($\frac{108}{96500}$ g ≈ 1.12 mg); one chemical equivalent needs 96500 C.
Answer: (C). | ||||
| 66 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsEasy | |
A easy statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Statement I: the metallic radius of Na is 186 pm (1.86 Å); $\mathrm{Na^+}$ (102 pm) is smaller because it loses its whole 3s shell. Correct.
Statement II: only cations are smaller than their atoms; anions are larger (e.g. $\mathrm{F^-}$ 133 pm against F 64 pm), since the added electron increases repulsion. False.
Answer: (C). | ||||
| 67 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure (A) ICl: a diatomic molecule, linear → (IV).
(B) $\mathrm{ICl_3}$: 3 bond pairs + 2 lone pairs (sp$^3$d), T-shaped → (I).
(C) $\mathrm{ClF_5}$: 5 bond pairs + 1 lone pair (sp$^3$d$^2$), square pyramidal → (II).
(D) $\mathrm{IF_7}$: 7 bond pairs (sp$^3$d$^3$), pentagonal bipyramidal → (III).
A→IV, B→I, C→II, D→III.
Answer: (D). | ||||
| 68 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) False: Ga (135 pm) is smaller than Al (143 pm) because of poor shielding by its 3d electrons; the order is Tl > In > Al > Ga > B.
(B) False: electronegativity falls from B to Al and then rises slightly (Ga, In, Tl), so it does not decrease steadily down the group.
(C) True: $\mathrm{2Al+6HCl\to2AlCl_3+3H_2}$, while concentrated $\mathrm{HNO_3}$ forms a thin protective oxide film that makes Al passive.
(D) False: a stable +1 state is shown only by the heavier members (Tl especially, from the inert pair effect), not by B or Al.
(E) True: in the octahedral $\mathrm{[Al(H_2O)_6]^{3+}}$ aluminium uses 3s, 3p and 3d orbitals: sp$^3$d$^2$.
Correct: (C) and (E) only.
Answer: (B). | ||||
| 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 The $\mathrm{M^{3+}/M^{2+}}$ standard potentials are negative for all three: Ti ≈ −0.37 V, V −0.26 V, Cr −0.41 V.
So each $\mathrm{M^{2+}}$ is a strong reducing agent that can reduce $\mathrm{H^+}$: $\mathrm{2M^{2+}+2H^+\to2M^{3+}+H_2}$.
All three, i.e. 3, liberate hydrogen from a dilute acid.
Answer: (D). | ||||
| 70 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds sp$^3$ hybridisation gives a tetrahedral complex. In a tetrahedron every pair of positions is adjacent (all bond angles 109.5°), so there is no cis/trans relation and no geometrical isomer, whatever the ligands.
(Mabcd in a tetrahedron is chiral and shows optical isomerism, which is not asked.)
Answer: (A). | ||||
| 71 | 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 five are tetrahedral; there the e set lies lower and the $t_2$ set higher, and the small splitting keeps them high spin.
$\mathrm{TiCl_4}$: Ti(IV), $d^0$ → no $t_2$ electrons.
$\mathrm{[MnO_4]^-}$: Mn(VII), $d^0$ → none.
$\mathrm{[FeO_4]^{2-}}$: Fe(VI), $d^2$ → $e^2t_2^0$, none.
$\mathrm{[FeCl_4]^-}$: Fe(III), $d^5$ → $e^2t_2^3$, has $t_2$ electrons.
$\mathrm{[CoCl_4]^{2-}}$: Co(II), $d^7$ → $e^4t_2^3$, has $t_2$ electrons.
Count: 3.
Answer: (B). | ||||
| 72 | Principles Related to Practical ChemistryPreparation of inorganic and organic compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on preparation of inorganic and organic compounds, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter $\mathrm{Fe^{2+}}$ hydrolyses in water ($\mathrm{FeSO_4+2H_2O\rightleftharpoons Fe(OH)_2+H_2SO_4}$), giving a turbid solution and aiding oxidation to Fe(III).
A little dilute $\mathrm{H_2SO_4}$ shifts this equilibrium back and so prevents the hydrolysis of ferrous sulphate.
Answer: (C). | ||||
| 73 | 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 Principal group: the carboxylic acid outranks the aldehyde, so the name ends in '-oic acid' and the –CHO becomes the prefix 'formyl'.
The chain containing the COOH carbon and the C=C: C1 (COOH) – C2 (bears CHO) – C3 – C4 (bears OH) – C5 – C6=C7.
Seven carbons, double bond from C6: hept-6-enoic acid, with 2-formyl and 4-hydroxy.
Name: 2-formyl-4-hydroxyhept-6-enoic acid. (Numbering from C1 = COOH, a double bond can never be '7-ene' in a seven-carbon chain.)
Answer: (C). | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Matching listEasy | |
A easy matching list question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques (A) Propan-1-ol and propan-2-ol differ only in the position of –OH: position isomerism → (III).
(B) $\mathrm{CH_3OC_3H_7}$ and $\mathrm{C_2H_5OC_2H_5}$ differ in the alkyl groups on either side of the same functional group: metamerism → (I).
(C) A ketone and an aldehyde ($\mathrm{C_3H_6O}$): functional isomerism → (IV).
(D) Neopentane and isopentane ($\mathrm{C_5H_{12}}$) differ in the carbon skeleton: chain isomerism → (II).
A→III, B→I, C→IV, D→II.
Answer: (D). | ||||
| 75 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Step 1 (Friedel–Crafts acylation with succinic anhydride / $\mathrm{AlCl_3}$): benzene gives A = 4-oxo-4-phenylbutanoic acid, $\mathrm{C_6H_5CO(CH_2)_2COOH}$.
Step 2 (Clemmensen reduction, Zn–Hg/HCl): the ketone C=O becomes $\mathrm{CH_2}$, the acid is untouched: B = 4-phenylbutanoic acid, $\mathrm{C_6H_5(CH_2)_3COOH}$.
Step 3 ($\mathrm{H^+}$): intramolecular Friedel–Crafts acylation closes the six-membered ring to α-tetralone, the product shown.
Option (B) wrongly makes A without the ketone; (C) and (D) start from a diketone, which neither step 1 nor the final ketone allows.
Answer: (A). | ||||
| 76 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (A) Not possible: the C–O bond of phenol has partial double-bond character (resonance with the ring) and the $sp^2$ carbon resists substitution, so HCl cannot replace –OH by –Cl.
(B) Possible: chlorobenzene with NaOH at high temperature and pressure gives sodium phenoxide, and acidification gives phenol (Dow process).
(C) Possible: –OCH$_3$ is an o/p-directing activator; $\mathrm{Cl_2/AlCl_3}$ gives p-chloroanisole as the major product.
(D) Possible: anisole with HBr cleaves at the methyl–O bond to give phenol and $\mathrm{CH_3Br}$.
Answer: (A). | ||||
| 77 | 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 The substrate is a tertiary bromide (1-bromo-1-methylcyclopentane); with $\mathrm{OH^-}$ in ethanol it undergoes dehydrohalogenation.
By Saytzeff's rule the more substituted alkene is the major product: the ring double bond, 1-methylcyclopentene (trisubstituted), rather than the exocyclic methylenecyclopentane (disubstituted).
Answer: (B). | ||||
| 78 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (i) Jones reagent ($\mathrm{CrO_3/H_2SO_4}$) oxidises the primary alcohol ethanol to ethanoic acid.
(ii) $\mathrm{KMnO_4}$ leaves ethanoic acid as it is (it is already fully oxidised).
(iii) NaOH, CaO, heat (soda-lime decarboxylation of the salt): $\mathrm{CH_3COONa+NaOH\xrightarrow{CaO,\,\Delta}CH_4+Na_2CO_3}$.
P = methane.
Answer: (B). | ||||
| 79 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Hot acidic $\mathrm{KMnO_4}$ cleaves the C=C bond of cyclohexene; each =CH– carbon becomes –COOH.
The ring opens to a straight six-carbon diacid: $\mathrm{HOOC(CH_2)_4COOH}$, adipic acid.
Answer: (C). | ||||
| 80 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Heating disrupts the hydrogen bonds and other forces that hold the secondary and tertiary structure; the globular albumin unfolds and coagulates. This is denaturation.
The primary structure (peptide bonds) is not broken, while the secondary structure and the biological activity are lost.
Answer: (D). | ||||
| 81 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom For $n=4$, $l=0,1,2,3$. $|m_l|=1$ means $m_l=+1$ or $-1$, possible for $l=1,2,3$: $3\times2=6$ orbitals.
Each orbital holds exactly one electron with $m_s=-\frac12$.
Total: 6.
Answer: 6. | ||||
| 82 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Zero dipole moment: $\mathrm{H_2}$ (non-polar bond), $\mathrm{CO_2}$ and $\mathrm{BeF_2}$ (linear), $\mathrm{BF_3}$ (trigonal planar), $\mathrm{CH_4}$ and $\mathrm{SiF_4}$ (tetrahedral, identical bonds): 6.
Polar: HF, $\mathrm{H_2S}$ and $\mathrm{H_2O}$ (bent), $\mathrm{NH_3}$ (pyramidal), $\mathrm{CHCl_3}$ (unsymmetrical tetrahedron).
Answer: 6. | ||||
| 83 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta_cH^0=\sum\Delta_fH^0(\text{products})-\sum\Delta_fH^0(\text{reactants})$.
For 1 mol: $6(-393.5)+3(-286)-48.5-0=-2361-858-48.5=-3267.5$ kJ.
For 2 mol: $-6535$ kJ, so $x=6535$.
Answer: 6535. | ||||
| 84 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalMedium | |
A medium numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Mass of acetic acid $=5\times1.2=6$ g $=0.1$ mol; mass of water $=1000$ g $=1$ kg, so molality $m=0.1$ mol kg$^{-1}$ (taking the concentration as 0.1 M too).
Degree of dissociation: $\alpha=\sqrt{\frac{K_a}{C}}=\sqrt{\frac{6.25\times10^{-5}}{0.1}}=\sqrt{6.25\times10^{-4}}=0.025$.
van't Hoff factor $i=1+\alpha=1.025$.
$\Delta T_f=iK_fm=1.025\times1.86\times0.1=0.1907$ K.
The solution freezes at $-0.19$ °C $=-19.07\times10^{-2}$ °C, so $x=19$.
Answer: 19. | ||||
| 85 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics A single-step (elementary) reaction follows its stoichiometry: rate $=kP_A^2P_B$.
When $P_C=0.5$ atm: $P_A=1.5-2(0.5)=0.5$ atm, $P_B=0.7-0.5=0.2$ atm.
$\frac{r_1}{r_2}=\frac{(1.5)^2(0.7)}{(0.5)^2(0.2)}=\frac{1.575}{0.05}=31.5=315\times10^{-1}$.
Answer: 315. | ||||
| 86 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements $\mathrm{4FeCr_2O_4+8Na_2CO_3+7O_2\to8Na_2CrO_4+2Fe_2O_3+8CO_2}$.
A = $\mathrm{Na_2CrO_4}$: Cr(VI), $3d^0$, no unpaired electron, $\mu=0$.
B = $\mathrm{Fe_2O_3}$: Fe(III), $3d^5$, 5 unpaired electrons, $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92$ B.M.
Sum $=5.92\approx6$ B.M.
Answer: 6. | ||||
| 87 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalEasy | |
A easy numerical question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids $\mathrm{2C_6H_5CHO+CH_3COCH_3\to C_6H_5CH{=}CHCOCH{=}CHC_6H_5+2H_2O}$.
Dibenzalacetone $\mathrm{C_{17}H_{14}O}$: $M=234$ g mol$^{-1}$, so $\frac{351}{234}=1.5$ mol; acetone $\frac{87}{58}=1.5$ mol, consistent.
Benzaldehyde needed $=2\times1.5=3$ mol $=3\times106=318$ g.
Answer: 318. | ||||
| 88 | Principles Related to Practical ChemistryDetection of elements and functional groups | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter $R_f=\frac{\text{distance moved by the sample}}{\text{distance moved by the solvent front}}$, both from the base line.
$R_f(A)=\frac{5.0}{12.5}=0.40$; $R_f(C)=\frac{10.0}{12.5}=0.80$.
Ratio $=\frac{0.40}{0.80}=0.50=50\times10^{-2}$.
Answer: 50. | ||||
| 89 | 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 Hot alkaline $\mathrm{KMnO_4}$ oxidises the whole side chain of propylbenzene to –COO$^-$: A = potassium benzoate.
$\mathrm{H_3O^+}$: B = benzoic acid.
$\mathrm{Br_2/FeBr_3}$: –COOH is meta-directing, C = 3-bromobenzoic acid.
π bonds in C: 3 in the benzene ring + 1 in C=O = 4.
Answer: 4. | ||||
| 90 | AminesDiazonium salts | Class 12: Amines | NumericalEasy | |
A easy numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines $\mathrm{C_2H_5NH_2+NaNO_2+2HCl\to C_2H_5N_2^+Cl^-+NaCl+2H_2O}$, then the unstable aliphatic diazonium salt hydrolyses: $\mathrm{C_2H_5N_2^+Cl^-+H_2O\to C_2H_5OH+N_2+HCl}$.
One mole of HCl is liberated per mole of ethanamine, so 0.2 mol NaOH ↔ 0.2 mol HCl ↔ 0.2 mol ethanamine.
$X=0.2\times45=9$ g.
Answer: 9. | ||||
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