JEE (Main) · 2024
JEE (Main) 2024 — 9 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
- Integral Calculus4
- Sequences and Series3
- Matrices and Determinants2
10 easy · 18 medium · 2 hard
Physics
30 Q
- Current Electricity3
- Kinematics2
- Laws of Motion2
23 easy · 7 medium · 0 hard
Chemistry
30 Q
- d- and f-Block Elements3
- Organic Chemistry: Some Basic Principles and Techniques3
- Aldehydes, Ketones and Carboxylic Acids3
18 easy · 12 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 | Sequences and SeriesArithmetic and geometric means | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic and geometric means, from Sequences and Series. NCERT: Class 11: Sequences and Series Since $\sin 3x+\cos 3x=\sqrt{2}\sin\left(3x+\frac{\pi}{4}\right)$, it takes every value in $[-\sqrt{2},\sqrt{2}]$, so the denominator runs over $[2-\sqrt{2},\,2+\sqrt{2}]$ (always positive).
Hence $a=\frac{1}{2+\sqrt{2}}$ and $b=\frac{1}{2-\sqrt{2}}$.
$a+b=\frac{(2-\sqrt{2})+(2+\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}=\frac{4}{2}=2$, so $\alpha=\frac{a+b}{2}=1$.
$ab=\frac{1}{4-2}=\frac{1}{2}$, so $\beta=\sqrt{ab}=\frac{1}{\sqrt{2}}$.
$\frac{\alpha}{\beta}=\frac{1}{1/\sqrt{2}}=\sqrt{2}$.
Answer: (B). | ||||
| 2 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $\frac{z-2i}{z+2i}=\frac{(z-2i)\overline{(z+2i)}}{|z+2i|^2}=\frac{(z-2i)(\bar z-2i)}{|z+2i|^2}$.
The numerator is $z\bar z-2i(z+\bar z)-4=(|z|^2-4)-2i(z+\bar z)$, whose real part is $|z|^2-4$.
So the real part is zero exactly when $|z|=2$ ($z\ne-2i$): $z$ lies on the circle of radius 2 about the origin.
The distance from $6+8i$ to the origin is $\sqrt{36+64}=10$, so the farthest point of the circle from $6+8i$ is at distance $10+2=12$ (the point $-\frac{6+8i}{5}$, which is not $-2i$).
Answer: (C). | ||||
| 3 | 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 Each root satisfies $x^2=\sqrt{2}x+\sqrt{3}$, so $x^{n+2}=\sqrt{2}x^{n+1}+\sqrt{3}x^n$. Subtracting the relations for $\alpha$ and $\beta$:
$P_{n+2}=\sqrt{2}P_{n+1}+\sqrt{3}P_n$.
Hence $11P_{12}=11\sqrt{2}P_{11}+11\sqrt{3}P_{10}$, and the expression becomes
$(11\sqrt{3}-10\sqrt{2}-11\sqrt{3})P_{10}+(11\sqrt{2}+10-11\sqrt{2})P_{11}=10\left(P_{11}-\sqrt{2}P_{10}\right)$.
By the recurrence, $P_{11}-\sqrt{2}P_{10}=\sqrt{3}P_9$, so the value is $10\sqrt{3}\,P_9$.
Answer: (C). | ||||
| 4 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants From $AB^{-1}=A^{-1}$, multiply on the left by $A$: $A^2B^{-1}=I$, so $A^2=B$.
From $BCB^{-1}=A$: $C=B^{-1}AB$, hence $C^2=B^{-1}A^2B=B^{-1}BB=B$ and $C^4=B^2$.
So the condition is $B^2+\alpha B+\beta I=O$.
For $B$, trace $=6$ and $\det B=5-3=2$; by the Cayley–Hamilton theorem $B^2-6B+2I=O$. Since $B$ is not a multiple of $I$, this is the only such relation, so $\alpha=-6$, $\beta=2$.
$2\beta-\alpha=4+6=10$.
Answer: (C). | ||||
| 5 | 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 $(1+2x)^{\frac{1}{2x}}=e^{\frac{\ln(1+2x)}{2x}}$ and $\frac{\ln(1+2x)}{2x}=\frac{2x-2x^2+\dots}{2x}=1-x+O(x^2)$.
So $(1+2x)^{\frac{1}{2x}}=e\cdot e^{-x+O(x^2)}=e(1-x)+O(x^2)$.
Then $\frac{e-(1+2x)^{1/(2x)}}{x}=\frac{ex+O(x^2)}{x}\to e$.
Answer: (A). | ||||
| 6 | 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 General term: $T_{r+1}={}^9C_r\left(x^{2/3}\right)^{9-r}\left(\frac{1}{2}x^{-2/5}\right)^r={}^9C_r\frac{1}{2^r}x^{\frac{2(9-r)}{3}-\frac{2r}{5}}$.
Exponent $=\frac{10(9-r)-6r}{15}=\frac{90-16r}{15}$.
For $x^{2/3}=x^{10/15}$: $90-16r=10$, $r=5$; coefficient ${}^9C_5/2^5=126/32=63/16$.
For $x^{-2/5}=x^{-6/15}$: $90-16r=-6$, $r=6$; coefficient ${}^9C_6/2^6=84/64=21/16$.
Sum $=\frac{63+21}{16}=\frac{84}{16}=\frac{21}{4}$.
Answer: (A). | ||||
| 7 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{a}{1-r}=57$ and $\frac{a^3}{1-r^3}=9747$ (with $|r|<1$).
Cubing the first: $\frac{a^3}{(1-r)^3}=57^3=185193$. Dividing: $\frac{(1-r)^3}{1-r^3}=\frac{9747}{185193}=\frac{1}{19}$.
Since $1-r^3=(1-r)(1+r+r^2)$: $\frac{(1-r)^2}{1+r+r^2}=\frac{1}{19}$, i.e. $19(1-2r+r^2)=1+r+r^2$, so $18r^2-39r+18=0$, i.e. $6r^2-13r+6=0$.
$r=\frac{13\pm5}{12}$, i.e. $r=\frac{3}{2}$ or $\frac{2}{3}$; convergence needs $r=\frac{2}{3}$.
Then $a=57\left(1-\frac{2}{3}\right)=19$ and $a+18r=19+12=31$.
Answer: (B). | ||||
| 8 | 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 $y=e^{3\sin^{-1}x}$, so $y'=\frac{3y}{\sqrt{1-x^2}}$, i.e. $(1-x^2)y'^2=9y^2$.
Differentiate: $(1-x^2)\cdot 2y'y''-2xy'^2=18yy'$. Dividing by $2y'$ ($y'\ne0$): $(1-x^2)y''-xy'=9y$.
At $x=\frac{1}{2}$: $\sin^{-1}\frac{1}{2}=\frac{\pi}{6}$, so $y=e^{\pi/2}$ and the value is $9e^{\pi/2}$.
Answer: (A). | ||||
| 9 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Both numerator and denominator tend to 0. Let $g(t)=\sin(2t^{1/3})+\cos(t^{1/3})$; then $g(x^3)=\sin 2x+\cos x$.
By L'Hôpital's rule and the fundamental theorem of calculus, the limit equals $\lim\frac{-g(x^3)\cdot 3x^2}{2\left(x-\frac{\pi}{2}\right)}=\lim\frac{-3x^2(\sin 2x+\cos x)}{2\left(x-\frac{\pi}{2}\right)}$.
At $x=\frac{\pi}{2}$, $\sin 2x+\cos x=0$, so this is again $\frac{0}{0}$. With $h(x)=\sin 2x+\cos x$, $\frac{h(x)}{x-\pi/2}\to h'\left(\frac{\pi}{2}\right)=2\cos\pi-\sin\frac{\pi}{2}=-3$.
Limit $=\frac{-3\cdot\frac{\pi^2}{4}\cdot(-3)}{2}=\frac{9\pi^2}{8}$.
Answer: (C). | ||||
| 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 Put $Y=\sqrt{3}\,y$. The ellipse becomes the circle $x^2+Y^2=18$ (radius $3\sqrt{2}$), and areas are multiplied by $\sqrt{3}$.
The line $y=x$ becomes $Y=\sqrt{3}x$, which makes $60^\circ$ with the $x$-axis. The region (first quadrant, below the line, inside the ellipse) becomes the circular sector $0\le\theta\le60^\circ$.
Its area in the $(x,Y)$ plane is $\frac{1}{2}\cdot18\cdot\frac{\pi}{3}=3\pi$.
Required area $=\frac{3\pi}{\sqrt{3}}=\sqrt{3}\pi$.
Answer: (A). | ||||
| 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 Let $x=\cos\theta$ with $\theta\in(0,\pi)$. Then $\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\tan\frac{\theta}{2}$, and $\cot^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\pi}{2}-\frac{\theta}{2}$.
So $\cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right)=\cos(\pi-\theta)=-\cos\theta=-x$.
$\int_{1/4}^{3/4}(-x)\,dx=-\frac{1}{2}\left(\frac{9}{16}-\frac{1}{16}\right)=-\frac{1}{4}$.
Answer: (B). | ||||
| 12 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $f(x)=\log_e(x+\sqrt{x^2+1})$ is odd, so $\int_{-1}^{1}f=0$ and the integral equals $\int_1^2 f(x)\,dx$.
By parts: $\int f\,dx=x f(x)-\int\frac{x}{\sqrt{x^2+1}}dx=x\log_e(x+\sqrt{x^2+1})-\sqrt{x^2+1}$.
$\left[\ \right]_1^2=2\log_e(2+\sqrt{5})-\sqrt{5}-\log_e(1+\sqrt{2})+\sqrt{2}$.
Since $(2+\sqrt{5})^2=9+4\sqrt{5}$, this is $\sqrt{2}-\sqrt{5}+\log_e\left(\frac{9+4\sqrt{5}}{1+\sqrt{2}}\right)$.
Answer: (A). | ||||
| 13 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Differentiating with respect to $x$: $\sqrt{1-y'^2}=y$, so $y'^2=1-y^2$.
Differentiating again: $2y'y''=-2yy'$, so wherever $y'\ne0$, $y''=-y$.
With $y(0)=0$ and $y\ge0$, the solution is $y=\sin x$ on $[0,3]$ (there $\sqrt{1-\cos^2x}=\sin x=y$ holds); at $x=2$, $y'=\cos 2\ne0$.
Hence $y''+y=0$ at $x=2$ and $y''+y+1=1$.
Answer: (A). | ||||
| 14 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines AP has slope $\frac{1-(-1)}{1-3}=-1$, and BC $\perp$ AP, so BC has slope 1 through B: $y=x+5$.
BP has slope $\frac{1-3}{1+2}=-\frac{2}{3}$, and AC $\perp$ BP, so AC has slope $\frac{3}{2}$ through A: $y=\frac{3}{2}x-\frac{11}{2}$.
Intersecting: $x+5=\frac{3}{2}x-\frac{11}{2}$ gives $x=21$, $y=26$. So C $=(21,26)$ and $\alpha+\beta=47$.
Circumcentre $(h,k)$ of PAB: equidistant from P and A gives $(h-1)^2+(k-1)^2=(h-3)^2+(k+1)^2$, i.e. $h-k=2$; from P and B gives $(h-1)^2+(k-1)^2=(h+2)^2+(k-3)^2$, i.e. $-6h+4k=11$.
Solving: $k=-\frac{23}{2}$, $h=-\frac{19}{2}$, so $h+k=-21$.
$(\alpha+\beta)+2(h+k)=47-42=5$.
Answer: (A). | ||||
| 15 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Ellipse: $a^2=100$, $b^2=75$, $e^2=1-\frac{75}{100}=\frac{1}{4}$, $e=\frac{1}{2}$; foci at distance $ae=5$ from the centre $(1,1)$ along the line $y=1$.
Hyperbola: eccentricity $2$ and the same foci, so its semi-transverse axis $A$ satisfies $2A=5$, $A=\frac{5}{2}$.
$B^2=A^2(e^2-1)=\frac{25}{4}\cdot3=\frac{75}{4}$.
$\alpha=2A=5$, $\beta^2=4B^2=75$. Then $3\alpha^2+2\beta^2=75+150=225$.
Answer: (D). | ||||
| 16 | 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 The given line passes through $Q\left(\frac{11}{3},\frac{11}{3},\frac{19}{3}\right)$ with direction $(2,1,2)$: points $Q+u(2,1,2)$.
L: $(1,2,3)+s(1,1,2)$.
Equating: $\frac{11}{3}+2u=1+s$, $\frac{11}{3}+u=2+s$, $\frac{19}{3}+2u=3+2s$. The first two give $u=-1$, then $s=\frac{2}{3}$; the third checks: $\frac{13}{3}=\frac{13}{3}$.
The meeting point is $Q-(2,1,2)$, so the distance is $|u|\cdot|(2,1,2)|=1\cdot3=3$.
Answer: (D). | ||||
| 17 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | StatementsMedium | |
A medium statements question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Statement I: $\vec{a}\times(\vec{r}-\vec{b})=\vec{0}$, so $\vec{r}=\vec{b}+\lambda\vec{a}$. Then $\vec{a}\cdot\vec{r}=\vec{a}\cdot\vec{b}+\lambda|\vec{a}|^2=7+14\lambda=0$, $\lambda=-\frac{1}{2}$.
$\vec{r}=(2,1,-1)-\frac{1}{2}(1,2,-3)=\left(\frac{3}{2},0,\frac{1}{2}\right)$, $|\vec{r}|=\sqrt{\frac{9}{4}+\frac{1}{4}}=\frac{\sqrt{10}}{2}\ne\sqrt{10}$. Statement I is incorrect.
Statement II: in a triangle, $\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C$. Also $\cos 2A+\cos 2B+\cos 2C=2\cos(A+B)\cos(A-B)+2\cos^2C-1=2\cos^2C-2\cos C\cos(A-B)-1\ge 2\cos^2C-2|\cos C|-1\ge-\frac{3}{2}$ (the quadratic $2t^2-2t-1$ has minimum $-\frac{3}{2}$ at $t=\frac{1}{2}$). Statement II is correct.
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 $\vec{d_1}=\vec{a}+\vec{b}=(1,\alpha,2)$, $\vec{d_2}=\vec{b}+\vec{c}=(-1,\beta,0)$.
$\vec{d_1}\times\vec{d_2}=(\alpha\cdot0-2\beta,\ 2\cdot(-1)-1\cdot0,\ \beta+\alpha)=(-2\beta,-2,\alpha+\beta)$.
Area $=\frac{1}{2}|\vec{d_1}\times\vec{d_2}|=\frac{\sqrt{21}}{2}$, so $4\beta^2+4+(\alpha+\beta)^2=21$, i.e. $4\beta^2+(\alpha+\beta)^2=17$.
Integer pairs with $\alpha\beta=-6$: $|\beta|=1$ needs $(\alpha+\beta)^2=13$ (no); $|\beta|=2$ needs $(\alpha+\beta)^2=1$: $(3,-2)$ and $(-3,2)$ work; $|\beta|\ge3$ makes $4\beta^2>17$.
So the pairs are $(3,-2)$ and $(-3,2)$; in either order $\alpha_1^2+\beta_1^2-\alpha_2\beta_2=13-(-6)=19$.
Answer: (B). | ||||
| 19 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics $N=7$. $\sum fx=c(2+2+3+4+5+6)=22c$, mean $=\frac{22c}{7}$.
$\sum fx^2=c^2(2+4+9+16+25+36)=92c^2$.
Variance $=\frac{92c^2}{7}-\frac{484c^2}{49}=\frac{644c^2-484c^2}{49}=\frac{160c^2}{49}$.
$\frac{160c^2}{49}=160$ gives $c=7$.
Answer: (B). | ||||
| 20 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability We need $x_1<x_2<x_3$. Each choice of 3 distinct faces gives exactly one such increasing sequence: ${}^6C_3=20$ outcomes out of $6^3=216$.
Probability $=\frac{20}{216}=\frac{5}{54}$.
Answer: (A). | ||||
| 21 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical 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 $2x=23-3y$ needs $y$ odd and $3y<23$: $y=1,3,5,7$ give $x=10,7,4,1$.
$A=\{(10,1),(7,3),(4,5),(1,7)\}$ has 4 elements and $B=\{10,7,4,1\}$ has 4 elements.
One-one functions from a 4-element set to a 4-element set: $4!=24$.
Answer: 24. | ||||
| 22 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants The system: $2x-5y=20$, $3x+my=m$. $|A|=2m+15$.
By Cramer's rule: $x=\frac{20m+5m}{2m+15}=\frac{25m}{2m+15}$, $y=\frac{2m-60}{2m+15}$.
$x<0$ requires $m$ and $2m+15$ of opposite signs: $-\frac{15}{2}<m<0$. There $2m+15>0$ and $2m-60<0$, so $y<0$ as well.
So $(a,b)=\left(-\frac{15}{2},0\right)$, on which $|A|=2m+15>0$.
$\int_{-15/2}^{0}(2m+15)\,dm=\left[m^2+15m\right]_{-15/2}^{0}=-\left(\frac{225}{4}-\frac{225}{2}\right)=\frac{225}{4}$.
$8\cdot\frac{225}{4}=450$.
Answer: 450. | ||||
| 23 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Three-digit numbers $\overline{abc}$ with $a\ge1$ and $a+b+c=14$. Put $a'=a-1$: $a'+b+c=13$ with $a'\le8$, $b,c\le9$.
Non-negative solutions: ${}^{15}C_2=105$.
Subtract $a'\ge9$: $a''+b+c=4$, ${}^6C_2=15$; $b\ge10$: ${}^5C_2=10$; $c\ge10$: $10$. No two of these can happen together (the sum would exceed 13).
Count $=105-15-10-10=70$ (100 and 1000 themselves have digit sum 1).
Answer: 70. | ||||
| 24 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\frac{1}{2k(2k-1)}=\frac{1}{2k-1}-\frac{1}{2k}$, so the second bracket is $1-\frac{1}{2}+\frac{1}{3}-\dots-\frac{1}{2024}=\sum_{j=1}^{2024}\frac{1}{j}-2\sum_{j=1}^{1012}\frac{1}{2j}=\sum_{j=1013}^{2024}\frac{1}{j}$.
The first bracket is $\sum_{j=\alpha+1}^{\alpha+1012}\frac{1}{j}$, also 1012 terms.
With $\alpha=1011$: $\sum_{1012}^{2023}\frac{1}{j}-\sum_{1013}^{2024}\frac{1}{j}=\frac{1}{1012}-\frac{1}{2024}=\frac{1}{2024}$. ✓
The left side decreases as $\alpha$ increases, so $\alpha=1011$ is the only solution.
Answer: 1011. | ||||
| 25 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $\sin^2 2x-\cos^2 2x=-\cos 4x$, so $f'(x)=4(p^2-6p+8)\sin 4x+2(2-p)$.
No critical point means $f'(x)\ne0$ for all $x$: $|2(2-p)|>4|p^2-6p+8|$ (for $p=2$, $f'\equiv0$ and every point is critical).
With $p^2-6p+8=(p-2)(p-4)$ and $p\ne2$: $2|p-2|>4|p-2||p-4|$, i.e. $|p-4|<\frac{1}{2}$, so $p\in\left(\frac{7}{2},\frac{9}{2}\right)$.
$16ab=16\cdot\frac{7}{2}\cdot\frac{9}{2}=252$.
Answer: 252. | ||||
| 26 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalEasy | |
A easy numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations The solution of $f'-3f=\alpha$ is $f(x)=Ce^{3x}-\frac{\alpha}{3}$.
As $x\to-\infty$, $f\to-\frac{\alpha}{3}=7$, so $\alpha=-21$ and $f(x)=Ce^{3x}+7$. $f(0)=1$ gives $C=-6$.
$f(-\log_e3)=-6e^{-3\log_e3}+7=-\frac{6}{27}+7=\frac{61}{9}$.
$9f(-\log_e3)=61$.
Answer: 61. | ||||
| 27 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections A chord of $C$ with mid-point $M(h,k)$ is $hx+ky=h^2+k^2$ (perpendicular to OM). The parabola bisects it when $M$ lies on $P$: $k^2=8h$, and $M$ must lie inside $C$: $h^2+k^2<4$.
The chord passes through $(\alpha,0)$: $h\alpha=h^2+k^2=h^2+8h$, so $h=0$ or $h=\alpha-8$.
$h=0$ gives $M=(0,0)$: the diameter along the $x$-axis, one chord through $(\alpha,0)$ for any $\alpha\ne0$.
$h=\alpha-8$ gives two more chords (with $k=\pm\sqrt{8h}$) when $h>0$ and $h^2+8h<4$, i.e. $0<h<-4+2\sqrt{5}$.
So three distinct chords exist exactly when $8<\alpha<4+2\sqrt{5}$: $p=8$, $q=4+2\sqrt{5}$.
$(2q-p)^2=(4\sqrt{5})^2=80$.
Answer: 80. | ||||
| 28 | Conic SectionsParabola | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $y^2=4ax$ with $a=\frac{3}{2}$. Take $A(at_1^2,2at_1)$, $B(at_2^2,2at_2)$, $C(at_3^2,2at_3)$; $L$ is $y=2at_3$.
$AM=2a|t_1-t_3|$, $BN=2a|t_2-t_3|$.
Chord $AB$: $2x-(t_1+t_2)y+2at_1t_2=0$. At $y=2at_3$: $x_D=a(t_1t_3+t_2t_3-t_1t_2)$.
$CD=|x_D-at_3^2|=a|t_3^2-t_1t_3-t_2t_3+t_1t_2|=a|t_3-t_1||t_3-t_2|$.
$\frac{AM\cdot BN}{CD}=\frac{4a^2|t_1-t_3||t_2-t_3|}{a|t_3-t_1||t_3-t_2|}=4a=6$, and its square is $36$.
Answer: 36. | ||||
| 29 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical 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: $(1+3t,\,2t,\,2+4t)$. For the foot of the perpendicular from $P(6,1,5)$: $(3t-5)\cdot3+(2t-1)\cdot2+(4t-3)\cdot4=0$, i.e. $29t-29=0$, $t=1$.
Foot $F=(4,2,6)$; image $=2F-P=(2,3,7)$.
Square of its distance from the origin: $4+9+49=62$.
Answer: 62. | ||||
| 30 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | NumericalEasy | |
A easy numerical question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions With $\sin^{-1}x=\frac{\pi}{2}-\cos^{-1}x$: $2\left(\frac{\pi}{2}-\cos^{-1}x\right)+3\cos^{-1}x=\pi+\cos^{-1}x=\frac{2\pi}{5}$.
Then $\cos^{-1}x=-\frac{3\pi}{5}$, impossible since $\cos^{-1}x\in[0,\pi]$. (Indeed the left side is at least $\pi$.)
Number of solutions: 0.
Answer: 0. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $h=\lambda\sqrt{2mE}$, so $[h]=[\mathrm{L}]\cdot\sqrt{[\mathrm{M}][\mathrm{ML^2T^{-2}}]}=[\mathrm{L}]\cdot[\mathrm{MLT^{-1}}]=[\mathrm{ML^2T^{-1}}]$.
Answer: (B). | ||||
| 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 Stopping distance of each car: $s=\frac{u^2}{2a}=\frac{20^2}{2\times2}=100$ m.
Together they close $200$ m, so the gap left is $300-200=100$ m.
Answer: (B). | ||||
| 33 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion The total momentum stays zero: $m_1v_1=m_2v_2$ with the velocities opposite.
$\frac{v_1}{v_2}=\frac{m_2}{m_1}=\frac{1}{2}$: the heavier (first) nucleus moves at half the speed of the lighter one.
Answer: (B). | ||||
| 34 | Laws of MotionEquilibrium of concurrent forces | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on equilibrium of concurrent forces, from Laws of Motion. NCERT: Class 11: Laws of Motion The lower half of the rope hangs vertically, so its tension is $T_2=mg=10$ N.
At the mid point, the upper half (tension $T_1$, at $45^\circ$ to the vertical) balances $F$ and $T_2$:
vertical: $T_1\cos45^\circ=T_2=10$ N; horizontal: $T_1\sin45^\circ=F$.
So $F=T_2\tan45^\circ=10$ N.
Answer: (A). | ||||
| 35 | 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 $E_2-E_1=-3.4-(-13.6)=10.2$ eV, so the electron is raised exactly to $n=2$.
From $n=2$ the only transition is $2\to1$: one spectral line.
Answer: (A). | ||||
| 36 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Take the temperature as $T=300$ K (27°C) along A→B, with $P=\frac{RT}{V^3}$.
$W_{AB}=\int_2^4\frac{RT}{V^3}dV=RT\left[-\frac{1}{2V^2}\right]_2^4=8\times300\left(\frac{1}{8}-\frac{1}{32}\right)=2400\times\frac{3}{32}=225$ J.
B→C at constant $P=10$ N/m$^2$ from $V=4$ to $2$ m$^3$: $W_{BC}=10\times(2-4)=-20$ J.
C→A at constant volume: $W_{CA}=0$.
Net work $=225-20+0=205$ J.
Answer: (C). | ||||
| 37 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Hydrogen: 2 kg is about 2000 mol of H atoms; each reaction uses 4 of them, so the number of reactions is $\frac{2000N_A}{4}=500N_A$ and $E_H=500N_A\times26.7=13350N_A$ MeV.
Uranium: 2 kg is $\frac{2000}{235}$ mol, so $E_U=\frac{2000}{235}N_A\times200\approx1702N_A$ MeV.
$\frac{E_H}{E_U}\approx\frac{13350}{1702}\approx7.8$, which is closest to 7.62 among the options.
Answer: (A). | ||||
| 38 | GravitationOrbital motion | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Total energy in a circular orbit: $E=-\frac{GMm}{2r}=-\frac{gR^2m}{2r}$.
$\Delta E=\frac{gR^2m}{2}\left(\frac{1}{2R}-\frac{1}{r}\right)=\frac{10\times10^3R^2}{2}\left(\frac{1}{2R}-\frac{1}{r}\right)=5000R^2\left(\frac{1}{2R}-\frac{1}{r}\right)$.
Setting this equal to $\frac{10^4R}{6}$: $\frac{1}{2R}-\frac{1}{r}=\frac{1}{3R}$, so $\frac{1}{r}=\frac{1}{6R}$, $r=6R$.
Answer: (D). | ||||
| 39 | Properties of Bulk MatterSurface tension | 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 surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter For a soap bubble $\Delta P=\frac{4T}{r}$, so $\Delta P_1=3\Delta P_2$ gives $r_1=\frac{r_2}{3}$.
$\frac{V_1}{V_2}=\left(\frac{r_1}{r_2}\right)^3=\frac{1}{27}$.
Answer: (C). | ||||
| 40 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctMedium | |
A medium 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 speed on entering the water, $\sqrt{2gh}$, must equal the terminal speed in water, $v_t=\frac{2r^2\rho g}{9\eta}$ (buoyancy of water neglected, as the ball is 100 times denser).
$v_t=\frac{2\times10^{-8}\times10^5\times9.8}{9\times9.8\times10^{-6}}=\frac{2\times10^{-3}}{9\times10^{-6}}=\frac{2000}{9}\approx222.2$ m/s.
$h=\frac{v_t^2}{2g}=\frac{(2000/9)^2}{2\times9.8}\approx\frac{49383}{19.6}\approx2520$ m, i.e. about 2518 m.
Answer: (D). | ||||
| 41 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Average translational kinetic energy $=\frac{3}{2}kT\propto T$ (in kelvin).
$T_1=-78+273=195$ K; doubling the energy needs $T_2=390$ K $=117^\circ$C.
Answer: (A). | ||||
| 42 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Only the charges enclosed by $S$ count (Gauss's law). From the figure, $+5q$, $q$ and $-2q$ are inside; $+3q$ and $-4q$ are outside.
$q_{\text{enc}}=5q+q-2q=4q$, so $\phi=\frac{4q}{\epsilon_0}$.
Answer: (A). | ||||
| 43 | Current ElectricityWheatstone bridge | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on wheatstone bridge, from Current Electricity. NCERT: Class 12: Current Electricity Between the left and right vertices of the diamond there are three branches of $2R$ each: through the top vertex, through the centre and through the bottom vertex; the vertical resistors join the top vertex, the centre and the bottom vertex.
By symmetry (the network looks the same with left and right exchanged and the current reversed), the top vertex, the centre and the bottom vertex are all at the potential midway between the left and right vertices, so no current flows in the two vertical resistors.
The diamond is then three $2R$ branches in parallel: $\frac{2R}{3}$.
Adding the two end resistors: $R_{AB}=R+\frac{2R}{3}+R=\frac{8R}{3}$.
Answer: (D). | ||||
| 44 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $r=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB}$. With equal $K$ and equal charge, $r\propto\sqrt{m}$.
$\frac{r_d}{r_p}=\sqrt{\frac{2}{1}}=\sqrt{2}$.
Answer: (A). | ||||
| 45 | 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 In 10 s the loop moves $2\times10=20$ cm. Its front edge is 20 cm into the field and its back edge 5 cm into it, so the whole loop is inside the 50 cm wide field (fully in from $t=7.5$ s until the front edge leaves at $t=25$ s).
The flux through the loop is then constant, so the induced emf is zero.
Answer: (A). | ||||
| 46 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics A ray parallel to the axis enters $L_1$, crosses the axis at its focus, and leaves $L_2$ parallel to the axis. So the focus of $L_1$ coincides with the focus of $L_2$.
Separation $=f_1+f_2=10+15=25$ cm.
Answer: (C). | ||||
| 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 Einstein's photoelectric equation: the photon energy $h\nu$ goes partly into freeing the electron (the work function $\phi$) and the rest is the largest kinetic energy an emitted electron can have.
$K_{\max}=h\nu-\phi=4.13-3.13=1$ eV.
Since $h\nu>\phi$, emission does occur; electrons from deeper levels come out with less than this.
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 The upper AND gate gives $AB$; the lower AND gate, fed with $\bar A$ and $\bar B$, gives $\bar A\bar B$; the NOR gate gives $E=\overline{AB+\bar A\bar B}$.
Check: $A=B=0$ gives $\bar A\bar B=1$, $E=0$; $A=B=1$ gives $AB=1$, $E=0$, as in the table.
$A=0,B=1$: $AB=0$, $\bar A\bar B=0$, $E=1$, so $X=1$. $A=1,B=0$: likewise $E=1$, so $Y=1$. (The circuit is an XOR gate.)
Answer: (D). | ||||
| 49 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The curve shows a small current in forward and reverse bias until, at a definite reverse voltage, the current rises sharply while the voltage stays nearly constant: reverse breakdown.
That is the characteristic of a zener diode, which is used as a voltage regulator.
Answer: (B). | ||||
| 50 | 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 The wave travels along the $x$-axis and $\vec B$ is along $y$, so $\vec E$ must be along $z$ (perpendicular to both). That rules out (A) and (B).
$E_0=cB_0=3\times10^8\times3.5\times10^{-7}=105$ V/m.
$E_z=105\sin(1.5\times10^3x+0.5\times10^{11}t)$ V m$^{-1}$.
Answer: (C). | ||||
| 51 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $W=\int_2^4(3x^2+2x-5)\,dx=\left[x^3+x^2-5x\right]_2^4=(64+16-20)-(8+4-10)=60-2=58$ J.
Answer: 58. | ||||
| 52 | 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 The plate adds an optical path $(\mu-1)t$; a shift of 4 fringes means $(\mu-1)t=4\lambda$.
$t=\frac{4\times500\ \text{nm}}{0.5}=4000$ nm $=4$ μm.
Answer: 4. | ||||
| 53 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $Z=\sqrt{R^2+X_C^2}=\sqrt{16+48}=8\ \Omega$.
$V_{\text{rms}}=\frac{8\sqrt{2}}{\sqrt{2}}=8$ V, so $I_{\text{rms}}=1$ A.
$P=I_{\text{rms}}^2R=1\times4=4$ W.
Answer: 4. | ||||
| 54 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Straight: $M=m\ell=44$ Am$^2$, with pole strength $m$ and length $\ell$.
Bent into a semicircle: $\pi r=\ell$, and the poles are a diameter apart, $2r=\frac{2\ell}{\pi}$.
$M'=m\cdot\frac{2\ell}{\pi}=\frac{2M}{\pi}=\frac{2\times44\times7}{22}=28$ Am$^2$.
Answer: 28. | ||||
| 55 | Current ElectricityResistivity | Class 12: Current Electricity | NumericalEasy | |
A easy numerical question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity $R=R_{27}\left[1+\alpha(T-27)\right]$: $62=50\left[1+2.4\times10^{-4}(T-27)\right]$.
$2.4\times10^{-4}(T-27)=0.24$, so $T-27=1000$ and $T=1027$ °C.
Answer: 1027. | ||||
| 56 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalEasy | |
A easy numerical question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance From the figure the cube occupies $2\le x\le4$ m. Only the two faces perpendicular to $x$ carry flux, each of area 4 m$^2$.
Face at $x=4$: $E=8$ N/C outward, flux $+32$. Face at $x=2$: $E=4$ N/C inward, flux $-16$.
Net flux $=32-16=16$ N m$^2$/C.
Answer: 16. | ||||
| 57 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{0.5}{50}}=\frac{2\pi}{10}=\frac{44}{70}=\frac{22}{35}$ s.
So $x=22$.
Answer: 22. | ||||
| 58 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Slipping (no friction): $a_1=g\sin\theta$. Rolling disc: $a_2=\frac{g\sin\theta}{1+\frac{I}{mR^2}}=\frac{2}{3}g\sin\theta$.
For the same length $l=\frac{1}{2}at^2$, $t\propto\frac{1}{\sqrt a}$: $t_2=t\sqrt{\frac{a_1}{a_2}}=\left(\frac{3}{2}\right)^{1/2}t$.
So $\alpha=3$.
Answer: 3. | ||||
| 59 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $(\vec A+\vec B)\cdot\vec A=0$: $A^2+AB\cos\theta=0$, so $AB\cos\theta=-A^2$.
$|\vec A+\vec B|^2=\frac{B^2}{4}$: $A^2+B^2+2AB\cos\theta=\frac{B^2}{4}$, i.e. $B^2-A^2=\frac{B^2}{4}$, so $A=\frac{\sqrt3}{2}B$.
$\cos\theta=-\frac{A}{B}=-\frac{\sqrt3}{2}$, $\theta=150^\circ$.
Answer: 150. | ||||
| 60 | 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 voltmeter reads the voltage across $R$ in parallel with the 10 kΩ resistor, and the milliammeter reads the total current through that pair.
From the graph: 4 V at 2 mA, 6 V at 3 mA, 8 V at 4 mA, so $\frac{V}{I}=2$ kΩ is the parallel combination.
$\frac{1}{R}=\frac{1}{2000}-\frac{1}{10000}=\frac{4}{10000}$, so $R=2500\ \Omega$.
Answer: 2500. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry By the SI definition, the candela is the luminous intensity of a source emitting monochromatic radiation of frequency $540\times10^{12}$ Hz with a radiant intensity of $\frac{1}{683}$ W per steradian in that direction.
So A $=540$ and $\frac{1}{B}=\frac{1}{683}$, i.e. B $=683$.
Answer: (C). | ||||
| 62 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure BF$_3$: trigonal planar, no lone pair, 120°.
PF$_3$: trigonal pyramidal (one lone pair), about 98°.
ClF$_3$: T-shaped (two lone pairs in a trigonal bipyramid), F–Cl–F about 87.5°.
Increasing order: ClF$_3<$ PF$_3<$ BF$_3$.
Answer: (C). | ||||
| 63 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium $K_{sp}=[\mathrm{A^{2+}}][\mathrm{B^-}]^2=1.2\times10^{-4}\times(0.24\times10^{-3})^2=1.2\times10^{-4}\times5.76\times10^{-8}=6.91\times10^{-12}$.
Answer: (A). | ||||
| 64 | ElectrochemistryBatteries and fuel cells | Class 12: Electrochemistry | Matching listEasy | |
A easy matching list question on batteries and fuel cells, from Electrochemistry. NCERT: Class 12: Electrochemistry (A) Leclanche (dry) cell: zinc anode, Zn $\rightarrow$ Zn$^{2+}$ + 2e$^-$ → IV.
(B) Ni–Cd cell is a rechargeable (secondary) cell → III.
(C) A fuel cell converts the energy of combustion of a fuel (H$_2$, CH$_4$ …) directly into electrical energy → I.
(D) The mercury cell uses a paste electrolyte (no ions in solution in the cell reaction) and is used in hearing aids and watches → II.
A-IV, B-III, C-I, D-II.
Answer: (B). | ||||
| 65 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry For a weak electrolyte the degree of dissociation is $\alpha=\frac{\Lambda_m}{\Lambda_m^\circ}$, and Ostwald's dilution law gives $K_a=\frac{C\alpha^2}{1-\alpha}$.
$K_a=\frac{C\Lambda_m^2/\Lambda_m^{\circ2}}{1-\Lambda_m/\Lambda_m^\circ}=\frac{C\Lambda_m^2}{\Lambda_m^\circ(\Lambda_m^\circ-\Lambda_m)}$.
So $K_a\Lambda_m^{\circ2}-K_a\Lambda_m\Lambda_m^\circ=C\Lambda_m^2$, i.e. $\Lambda_m^2C-K_a\Lambda_m^{\circ2}+K_a\Lambda_m\Lambda_m^\circ=0$.
(The forms with $AC^{1/2}$ are the Kohlrausch law for strong electrolytes, $\Lambda_m=\Lambda_m^\circ-AC^{1/2}$.)
Answer: (C). | ||||
| 66 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Rn has 86 electrons; Es (99) has 13 more: 7s$^2$ and 11 in 5f.
Es: [Rn] 5f$^{11}$ 6d$^0$ 7s$^2$.
Answer: (B). | ||||
| 67 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | Matching listEasy | |
A easy matching list question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom N (7): [He] 2s$^2$ 2p$^3$ → III. S (16): [Ne] 3s$^2$ 3p$^4$ → II. Br (35): [Ar] 3d$^{10}$ 4s$^2$ 4p$^5$ → I. Kr (36): [Ar] 3d$^{10}$ 4s$^2$ 4p$^6$ → IV.
A-III, B-II, C-I, D-IV.
Answer: (C). | ||||
| 68 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Group 13 data (approximate): melting points B 2453, Al 933, Ga 303, In 430, Tl 576 K; M$^{3+}$ radii B 27, Al 53.5, Ga 62, In 80, Tl 88.5 pm; $\Delta_iH_1$ B 801, Al 577, Ga 579, In 558, Tl 589 kJ mol$^{-1}$; atomic radii B 85, Al 143, Ga 135, In 167, Tl 170 pm.
(A) Melting point: B $>$ Al $>$ Tl $>$ In $>$ Ga → IV (gallium melts near room temperature).
(B) M$^{3+}$ radius increases down the group: Tl $>$ In $>$ Ga $>$ Al $>$ B → I.
(C) First ionisation enthalpy: B $>$ Tl $>$ Al $\approx$ Ga $>$ In → II.
(D) Atomic radius: Tl $>$ In $>$ Al $>$ Ga $>$ B → III (Ga is smaller than Al because of poor d-electron shielding).
A-IV, B-I, C-II, D-III.
Answer: (A). | ||||
| 69 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | StatementsMedium | |
A medium statements question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Statement I: in the d-block, higher oxidation states become more stable down a group (Mo(VI), W(VI) are more stable than Cr(VI)); in the p-block the lower state becomes more stable down the group (inert pair effect). True.
Statement II: $E^\circ(\mathrm{Cu^{2+}/Cu})=+0.34$ V is positive, so copper cannot reduce H$^+$ to H$_2$ from (non-oxidising) acids, weak ones included. True.
Answer: (A). | ||||
| 70 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds EDTA$^{4-}$ is hexadentate: two N and four carboxylate O atoms bind the metal, giving six donor atoms around Ca$^{2+}$ in an octahedral arrangement (this is the basis of EDTA titration of Ca$^{2+}$ in hardness of water).
Answer: (C). | ||||
| 71 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Matching listEasy | |
A easy matching list question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds (A) [Ni(CN)$_4$]$^{2-}$: Ni$^{2+}$ d$^8$, strong-field CN$^-$ pairs the electrons, square planar, dsp$^2$ → III.
(B) [Ni(CO)$_4$]: Ni(0), 3d$^{10}$, tetrahedral, sp$^3$ → I.
(C) [Co(NH$_3$)$_6$]$^{3+}$: Co$^{3+}$ d$^6$ low spin, inner orbital, d$^2$sp$^3$ → IV.
(D) [CoF$_6$]$^{3-}$: weak-field F$^-$, high spin, outer orbital, sp$^3$d$^2$ → II.
A-III, B-I, C-IV, D-II.
Answer: (C). | ||||
| 72 | Principles Related to Practical ChemistryDetection of elements and functional groups | No matching NCERT chapter | Matching listMedium | |
A medium matching list question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter (A) Bromine water (test for unsaturation): its reddish orange colour disappears → II.
(B) Ceric ammonium nitrate (test for alcohols): a red colour appears → III.
(C) Neutral FeCl$_3$ (test for phenols/enols): blue, green, violet or red colour → IV.
(D) 2,4-DNP (test for carbonyl compounds): yellow, orange or orange-red precipitate of the hydrazone → I.
A-II, B-III, C-IV, D-I.
Answer: (A). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques III is the neutral structure (every atom with an octet, no charge separation): most stable.
II separates charge with the negative charge on the more electronegative oxygen and the positive charge on carbon: a reasonable contributor.
I puts the positive charge on oxygen (with an incomplete octet) and the negative charge on carbon: charges on the wrong atoms, least stable.
Order: III $>$ II $>$ I.
Answer: (A). | ||||
| 74 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Ethyne, H–C≡C–H, is linear with both carbons sp hybridised: (A) and (C) are correct.
The triple bond (one σ, two π) is shorter (120 pm against 134 pm) and stronger (about 823 against 681 kJ mol$^{-1}$) than the C=C bond of ethene: (D) is correct and (B) is incorrect.
Answer: (B). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes The substrate is Ph–CH(CH$_3$)–CH(OCH$_3$)Br: the bromine sits on the carbon that also carries the methoxy group (an α-bromo ether).
Loss of Br$^-$ gives a carbocation stabilised by the lone pair of the adjacent oxygen (an oxocarbenium ion, CH=O$^+$CH$_3$), so no rearrangement is needed; cyanide attacks that same carbon.
P = Ph–CH(CH$_3$)–CH(OCH$_3$)–CN, i.e. option (B).
Answer: (B). | ||||
| 76 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The structure is CH$_3$C(Br)=CH–CH$_2$–CH(Br)–C(CH$_3$)=C(CH$_3$)Br.
C1=C2: C1 carries CH$_3$ and Br, C2 carries H and CH$_2$ — geometrical isomerism possible.
C4 carries H, Br, CH$_2$– and –C(CH$_3$)=C(CH$_3$)Br — a stereocentre.
C5=C6: C5 carries CH$_3$ and the CHBr chain, C6 carries CH$_3$ and Br — geometrical isomerism possible.
The three stereogenic units are independent and the molecule has no symmetry, so the number of stereoisomers is $2^3=8$.
Answer: (A). | ||||
| 77 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Tollens' reagent is reduced to a silver mirror by aldehydes (aliphatic and aromatic): formaldehyde and benzaldehyde give it.
Formic acid, HCOOH, has an aldehyde-like H–C=O group and also reduces Tollens' reagent.
Acetone, a ketone, does not.
So A, B and C.
Answer: (D). | ||||
| 78 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The iodoform test needs a CH$_3$CO– group (or CH$_3$CH(OH)–) when the hypoiodite acts.
(A) Pentan-3-one has no CH$_3$CO– group: negative.
(B) Aqueous KOH opens 1,2-epoxybutane to butane-1,2-diol, which has no CH$_3$CH(OH)– unit: negative.
(C) Aqueous KOH hydrolyses the gem-dichloride 2,2-dichlorobutane to butan-2-one, CH$_3$COCH$_2$CH$_3$, which gives iodoform: positive.
(D) Butanal, CH$_3$CH$_2$CH$_2$CHO, has no CH$_3$CO– group: negative.
Answer: (C). | ||||
| 79 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids With excess CH$_3$MgBr both groups react.
The ester –CO$_2$CH$_3$ takes two equivalents (via the ketone) and gives the tertiary alkoxide –C(CH$_3$)$_2$O$^-$, i.e. –C(CH$_3$)$_2$OH after work-up.
The nitrile –CN adds one equivalent to give the imine salt –C(CH$_3$)=NMgBr, which is hydrolysed by H$_3$O$^+$ to the ketone –COCH$_3$ (the imine salt does not react further).
Product: a benzene ring with –COCH$_3$ and –C(CH$_3$)$_2$OH para to each other, option (D).
Answer: (D). | ||||
| 80 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules (A) Glucose is a six-carbon aldose: correct.
(B) Its solubility in water comes from its five –OH groups hydrogen bonding with water, not from the aldehyde group (which is mostly tied up in the ring form): incorrect.
(C) In solution it exists as the α- and β-pyranose anomers in equilibrium with a little open-chain form: correct.
(D) Sucrose is made of glucose and fructose: correct.
Answer: (B). | ||||
| 81 | 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 Take $\Delta x=10^{-15}$ m (the diameter). $\Delta v=\frac{h}{4\pi m\Delta x}=\frac{6.626\times10^{-34}}{4\times3.14\times9.1\times10^{-31}\times10^{-15}}=\frac{6.626\times10^{-34}}{1.143\times10^{-44}}\approx5.80\times10^{10}$ m s$^{-1}$.
$=57.97\times10^9$ m s$^{-1}$, i.e. 58 to the nearest integer.
Answer: 58. | ||||
| 82 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure O$_2$ (16 electrons): $\pi^*2p_x^1\,\pi^*2p_y^1$, 2 electrons in $\pi^*$.
O$_2^+$ (15): 1 electron in $\pi^*$. O$_2^-$ (17): 3 electrons in $\pi^*$.
Total $=2+1+3=6$.
Answer: 6. | ||||
| 83 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics At the boiling point (1 atm) liquid and vapour are in equilibrium: $\Delta G=0$, so $T=\frac{\Delta H_{vap}}{\Delta S_{vap}}=\frac{30000}{75}=400$ K.
Answer: 400. | ||||
| 84 | SolutionsRaoult's law | Class 12: Solutions | NumericalEasy | |
A easy numerical question on raoult's law, from Solutions. NCERT: Class 12: Solutions Equimolar liquid: $x=0.5$ each. Partial pressures: benzene $0.5\times80=40$ Torr, methyl benzene $0.5\times24=12$ Torr; total 52 Torr.
Mole fraction of methyl benzene in the vapour $=\frac{12}{52}=0.2308=23.08\times10^{-2}$, i.e. 23.
Answer: 23. | ||||
| 85 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Initially only A at $P_0$. When it has fallen by $x$: total $=P_0-x+2x+x=P_0+2x$. At completion the total is $3P_0=300$, so $P_0=100$ torr.
At 23 s: $100+2x=200$, $x=50$, so half of A has reacted: $t_{1/2}=23$ s.
$k=\frac{2.303\log_{10}2}{t_{1/2}}=\frac{2.303\times0.301}{23}=0.0301$ s$^{-1}=3.01\times10^{-2}$ s$^{-1}$.
Answer: 3. | ||||
| 86 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | NumericalMedium | |
A medium numerical question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Cr is [Ar] 3d$^5$ 4s$^1$; Cr$^+$ is 3d$^5$ (half-filled, stable), so removing the second electron is hardest: Cr has the highest second ionisation enthalpy of these.
Cr$^+$ (3d$^5$) has 5 unpaired electrons: $\mu=\sqrt{5\times7}=\sqrt{35}\approx5.92$ BM, i.e. 6.
Answer: 6. | ||||
| 87 | Principles of Qualitative AnalysisGroup analysis of cations | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on group analysis of cations, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Group IV cations giving black sulphides are Co$^{2+}$ and Ni$^{2+}$; the black sulphide dissolves in aqua regia to give the chloride B.
The test with KNO$_2$ and acetic acid is the confirmatory test for cobalt: a yellow precipitate of potassium hexanitritocobaltate(III), K$_3$[Co(NO$_2$)$_6$] = C.
In C, Co$^{3+}$ is d$^6$ with the strong-field NO$_2^-$ ligand: low spin $t_{2g}^6$, no unpaired electrons, $\mu=0$ BM.
Answer: 0. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques From the figure the plate is 10 cm long, the spots start on a line 1 cm above the bottom and the solvent front is 1 cm below the top: the solvent has moved $10-1-1=8$ cm.
A has moved $5-1=4$ cm: $R_f(A)=\frac{4}{8}=0.5$. B has moved $7-1=6$ cm: $R_f(B)=\frac{6}{8}=0.75$.
$\frac{R_f(B)}{R_f(A)}=\frac{0.75}{0.5}=1.5=15\times10^{-1}$, so $x=15$.
Answer: 15. | ||||
| 89 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter Fuming sulphuric acid (oleum) is H$_2$S$_2$O$_7$ (H$_2$SO$_4$ + SO$_3$), which has 7 oxygen atoms.
Answer: 7. | ||||
| 90 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Friedel–Crafts reactions fail on strongly deactivated rings (–NO$_2$) and on amines, whose –NH$_2$ group binds the Lewis acid AlCl$_3$ and deactivates the ring.
Cannot: nitrobenzene, aniline, $m$-nitroaniline, $m$-dinitrobenzene — 4.
Can: toluene, xylene, cumene (activated) and chlorobenzene (weakly deactivated, still gives o/p products).
Answer: 4. | ||||
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