JEE (Main) · 2022
JEE (Main) 2022 — 25 June, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Sequences and Series4
- Circles2
- Trigonometry2
5 easy · 23 medium · 2 hard
Physics
30 Q
- Properties of Bulk Matter3
- Electromagnetic Induction and Alternating Currents3
- Optics3
25 easy · 5 medium · 0 hard
Chemistry
30 Q
- Hydrocarbons3
- Surface Chemistry2
- Structure of Atom2
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 | 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 $L_1$ and $L_2$ are parallel, and the circle touches both, so its centre lies on the line midway between them and the radius is half the distance between them.
A line through the centre meets $L_1$ and $L_2$ at points symmetric about the centre, so the centre is the midpoint of $(-1, 2)$ and $(3, -6)$: $(1, -2)$.
$(-1, 2)$ on $L_1$: $-4 - 6 + K_1 = 0$, $K_1 = 10$. $(3, -6)$ on $L_2$: $12 + 18 + K_2 = 0$, $K_2 = -30$.
Distance between the lines $= \frac{|10 - (-30)|}{5} = 8$, so the radius is 4.
Circle: $(x-1)^2 + (y+2)^2 = 16$.
Answer: (C). | ||||
| 2 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t = \cos x$, $dt = -\sin x\,dx$; $x = 0 \to t = 1$, $x = \pi \to t = -1$:
$I = \int_{-1}^{1} \frac{e^{t}}{(1 + t^2)(e^{t} + e^{-t})}\,dt$.
Replacing $t$ by $-t$: $I = \int_{-1}^{1} \frac{e^{-t}}{(1 + t^2)(e^{t} + e^{-t})}\,dt$.
Adding: $2I = \int_{-1}^{1} \frac{dt}{1 + t^2} = \tan^{-1}1 - \tan^{-1}(-1) = \frac{\pi}{2}$.
So $I = \frac{\pi}{4}$.
Answer: (C). | ||||
| 3 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $a + b = 7k$, $b + c = 8k$, $c + a = 9k$. Adding, $a + b + c = 12k$, so $c = 5k$, $a = 4k$, $b = 3k$.
Sides $3k, 4k, 5k$: a right triangle (hypotenuse $5k$).
Circumradius $R = \frac{5k}{2}$ (half the hypotenuse). Inradius $r = \frac{\Delta}{s} = \frac{\frac{1}{2}\cdot 3k\cdot 4k}{6k} = k$.
$\frac{R}{r} = \frac{5}{2}$.
Answer: (A). | ||||
| 4 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let $g(x) = 2f(x)$. Then $g(x + y) = 4f(x)f(y) = g(x)g(y)$ and $g(1) = 4$, so $g(n) = 4^n$ and $f(n) = \frac{4^n}{2}$.
$\sum_{k=1}^{10} f(\alpha + k) = \frac{4^{\alpha}}{2}\sum_{k=1}^{10} 4^k = \frac{4^{\alpha}}{2}\cdot\frac{4(4^{10} - 1)}{3} = \frac{2\cdot 4^{\alpha}}{3}\left(2^{20} - 1\right)$.
Equating with $\frac{512}{3}(2^{20} - 1)$: $2\cdot 4^{\alpha} = 512$, $4^{\alpha} = 256 = 4^4$, so $\alpha = 4$.
Answer: (C). | ||||
| 5 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Write $e_1, e_2, e_3$ for the unit columns. $Ae_3 = (1, 1, 2)^T$.
$A(e_1 + e_3) = (-1, 0, 1)^T$ gives $Ae_1 = (-1, 0, 1)^T - (1, 1, 2)^T = (-2, -1, -1)^T$.
$A(e_1 + e_2) = (1, 1, 0)^T$ gives $Ae_2 = (1, 1, 0)^T - (-2, -1, -1)^T = (3, 2, 1)^T$.
So $A = \begin{pmatrix}-2&3&1\\-1&2&1\\-1&1&2\end{pmatrix}$ and $A - 2I = \begin{pmatrix}-4&3&1\\-1&0&1\\-1&1&0\end{pmatrix}$.
$\det(A - 2I) = -4(0 - 1) - 3(0 + 1) + 1(-1 - 0) = 4 - 3 - 1 = 0$, so there is no unique solution.
Row 1 $=$ Row 2 $+ 3\times$ Row 3: $(-1 - 3, 0 + 3, 1 + 0) = (-4, 3, 1)$. The right side obeys the same relation: $1 + 3\times 1 = 4$. So the first equation is a consequence of the other two, which are independent: the system is consistent with infinitely many solutions.
Answer: (B). | ||||
| 6 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Differentiating $f(g(x)) = x$: $f'(g(x))\,g'(x) = 1$.
$g(63) = t$ where $f(t) = 63$: $t^3 + t - 5 = 63$, $t^3 + t = 68$, $t = 4$.
$f'(x) = 3x^2 + 1$, $f'(4) = 49$.
$g'(63) = \frac{1}{49}$.
Answer: (A). | ||||
| 7 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter An implication is false only when its hypothesis is true and its conclusion false: p is T and $(\sim p) \vee q$ is F, which needs q F. So p = T, q = F.
P1: $p \to \sim q$ is $T \to T$ = T, so P1 $= \sim T$ = F.
P2: $p \wedge \sim q$ = T, $(\sim p) \vee q$ = F, so P2 = F.
Both are FALSE.
Answer: (C). | ||||
| 8 | 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 The r-th term is $\frac{1}{2^r 3^{11-r}}$, $r = 1, \ldots, 10$. Multiplying by $2^{10}3^{10}$:
$K = \sum_{r=1}^{10} 2^{10-r}3^{r-1} = 2^9 + 2^8\cdot 3 + \ldots + 3^9$, a G.P. with ratio $\frac{3}{2}$:
$K = 2^9\cdot\frac{(3/2)^{10} - 1}{3/2 - 1} = 3^{10} - 2^{10} = 59049 - 1024 = 58025$.
$58025 = 6\times 9670 + 5$.
Answer: (D). | ||||
| 9 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability f must have degree 5: $f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$. Then
$f' = 5x^4 + 4ax^3 + 3bx^2 + 2cx + d$, $f'' = 20x^3 + 12ax^2 + 6bx + 2c$.
Comparing $f + f' + f''$ with $x^5 + 64$:
$x^4$: $a + 5 = 0$, $a = -5$. $x^3$: $b + 4a + 20 = 0$, $b = 0$. $x^2$: $c + 3b + 12a = 0$, $c = 60$. $x$: $d + 2c + 6b = 0$, $d = -120$. Constant: $e + d + 2c = 64$, $e = 64$.
$f(x) = x^5 - 5x^4 + 60x^2 - 120x + 64$, and $f(1) = 1 - 5 + 60 - 120 + 64 = 0$.
So the limit is $f'(1) = 5 - 20 + 120 - 120 = -15$.
Answer: (A). | ||||
| 10 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(E_1) = \frac{P(E_1 \cap E_2)}{P(E_2|E_1)} = \frac{1/8}{3/4} = \frac{1}{6}$; $P(E_2) = \frac{1/8}{1/2} = \frac{1}{4}$. So $P(E_1)P(E_2) = \frac{1}{24}$.
(A) $P(E_1 \cap E_2) = \frac{1}{8} \ne \frac{1}{24}$: false.
(B) $P(E_1' \cap E_2') = 1 - P(E_1 \cup E_2) = 1 - \left(\frac{1}{6} + \frac{1}{4} - \frac{1}{8}\right) = \frac{17}{24}$, while $P(E_1')P(E_2) = \frac{5}{6}\cdot\frac{1}{4} = \frac{5}{24}$: false.
(C) $P(E_1 \cap E_2') = P(E_1) - P(E_1 \cap E_2) = \frac{1}{6} - \frac{1}{8} = \frac{1}{24} = P(E_1)P(E_2)$: true.
(D) $P(E_1' \cap E_2) = \frac{1}{4} - \frac{1}{8} = \frac{1}{8} \ne \frac{1}{24}$: false.
Answer: (C). | ||||
| 11 | Matrices and DeterminantsTypes of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on types of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A^2 = \begin{bmatrix}-4&0\\0&-4\end{bmatrix} = -4I$, so $A^{2k} = (-4)^k I$ and $A^{2k-1} = (-4)^{k-1}A$.
$M = mI$ with $m = \sum_{k=1}^{10}(-4)^k = \frac{4(4^{10} - 1)}{5}$, and $N = sA$ with $s = \sum_{k=1}^{10}(-4)^{k-1} = \frac{1 - 4^{10}}{5}$.
$N^2 = s^2A^2 = -4s^2 I$, so $MN^2 = -4ms^2\,I$, a scalar matrix.
$-4ms^2 = -\frac{16(4^{10} - 1)^3}{125} \ne 1$, so $MN^2$ is a symmetric matrix that is not the identity.
Answer: (A). | ||||
| 12 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Differentiate the right side:
$\frac{d}{dx}\frac{x g}{e^x + 1} = \frac{(g + xg')(e^x + 1) - xge^x}{(e^x + 1)^2} = \frac{g(e^x + 1 - xe^x)}{(e^x + 1)^2} + \frac{xg'}{e^x + 1}$.
This must equal the integrand, so $xg'(x) = x(\cos x - \sin x)$, i.e. $g'(x) = \cos x - \sin x$ and $g(x) = \sin x + \cos x + C$.
(A) $g' > 0$ on $\left(0, \frac{\pi}{4}\right)$, so g is increasing there: false.
(B) $g'' = -\sin x - \cos x < 0$ there, so $g'$ is decreasing: false.
(C) $g + g' = 2\cos x + C$ is decreasing on $\left(0, \frac{\pi}{2}\right)$: false.
(D) $g - g' = 2\sin x + C$ is increasing on $\left(0, \frac{\pi}{2}\right)$: true.
Answer: (D). | ||||
| 13 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = \frac{2x}{x^2 + 1} + e^{-x}$. For $x \ge 0$ both terms are $\ge 0$ and $e^{-x} > 0$; for $x < 0$, $\frac{2x}{x^2 + 1} \ge -1$ while $e^{-x} > 1$. So $f' > 0$: f is increasing.
$g(x) = e^{-x} - 2e^{x}$, $g'(x) = -e^{-x} - 2e^x < 0$: g is decreasing.
So $f(g(u)) > f(g(v))$ exactly when $g(u) > g(v)$, i.e. when $u < v$:
$\frac{(\alpha - 1)^2}{3} < \alpha - \frac{5}{3}$, $(\alpha - 1)^2 < 3\alpha - 5$, $\alpha^2 - 5\alpha + 6 < 0$, $(\alpha - 2)(\alpha - 3) < 0$.
$\alpha \in (2, 3)$.
Answer: (A). | ||||
| 14 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctHard | |
A hard single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Equal angles with the axes and $a_i > 0$: $\vec{a} = t(\hat{i} + \hat{j} + \hat{k})$, $t > 0$.
Projection on $3\hat{i} + 4\hat{j}$: $\frac{3t + 4t}{5} = 7$, so $t = 5$ and $\vec{a} = 5\hat{i} + 5\hat{j} + 5\hat{k}$, $|\vec{a}| = 5\sqrt{3}$.
$\vec{b}$ lies in the plane of $\vec{a}$ and $\hat{i}$, so $\vec{b} = \lambda\vec{a} + \mu\hat{i}$; rotation by 90° makes $\vec{b}\perp\vec{a}$ and $|\vec{b}| = |\vec{a}|$.
$\vec{b}\cdot\vec{a} = 75\lambda + 5\mu = 0$ gives $\mu = -15\lambda$, so $\vec{b} = \lambda(-10\hat{i} + 5\hat{j} + 5\hat{k})$.
$|\vec{b}|^2 = 150\lambda^2 = 75$, $\lambda = \pm\frac{1}{\sqrt{2}}$.
Projection of $\vec{b}$ on $3\hat{i} + 4\hat{j}$: $\frac{\lambda(-30 + 20)}{5} = -2\lambda = \mp\sqrt{2}$.
The two directions of rotation give $\pm\sqrt{2}$; the value among the options is $\sqrt{2}$ (rotation taking $\lambda = -\frac{1}{\sqrt{2}}$).
Answer: (B). | ||||
| 15 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations $y' - \frac{y}{x + 1} = e^{3x}(x + 1)$. Integrating factor $\frac{1}{x + 1}$:
$\frac{d}{dx}\left(\frac{y}{x + 1}\right) = e^{3x}$, so $\frac{y}{x + 1} = \frac{e^{3x}}{3} + C$.
$y(0) = \frac{1}{3}$ gives $C = 0$: $y = \frac{(x + 1)e^{3x}}{3}$.
$y' = \frac{e^{3x}}{3} + (x + 1)e^{3x} = e^{3x}\left(x + \frac{4}{3}\right)$.
$y' = 0$ at $x = -\frac{4}{3}$, and $y'$ changes from negative to positive there.
So $x = -\frac{4}{3}$ is a point of local minimum.
Answer: (B). | ||||
| 16 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections A tangent to $y^2 = x$ (here $4a = 1$) is $y = mx + \frac{1}{4m}$.
It touches $x^2 + y^2 = 2$ when $c^2 = 2(1 + m^2)$: $\frac{1}{16m^2} = 2 + 2m^2$, i.e. $32m^4 + 32m^2 - 1 = 0$.
$m^2 = \frac{-32 + \sqrt{1024 + 128}}{64} = \frac{-32 + 24\sqrt{2}}{64} = \frac{-4 + 3\sqrt{2}}{8}$ (the other root is negative).
The two slopes are $\pm\sqrt{m^2}$, so $|m_1m_2| = m^2$ and $8|m_1m_2| = -4 + 3\sqrt{2}$.
Answer: (C). | ||||
| 17 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Image of P in $x + y + z = 5$: $Q = P - \frac{2(1 + 0 + 1 - 5)}{3}(1, 1, 1) = (1, 0, 1) + 2(1, 1, 1) = (3, 2, 3)$.
PQ is along $(1, 1, 1)$. L: $(1 + t, -1 + t, -1 + t)$. On S: $-1 + 3t = 5$, $t = 2$, so $R = (3, 1, 1)$.
$QR^2 = 0^2 + 1^2 + 2^2 = 5$.
Answer: (B). | ||||
| 18 | Differential EquationsHomogeneous equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on homogeneous equations, from Differential Equations. NCERT: Class 12: Differential Equations $\frac{dx}{dy} = -\frac{x^2 - xy + y^2}{y^2}$. Put $x = vy$: $v + y\frac{dv}{dy} = -(v^2 - v + 1)$, so $y\frac{dv}{dy} = -(v^2 + 1)$.
$\int\frac{dv}{1 + v^2} = -\int\frac{dy}{y}$: $\tan^{-1}\frac{x}{y} = -\log_e y + C$.
At (1, 1): $\frac{\pi}{4} = C$. So $\tan^{-1}\frac{x}{y} = \frac{\pi}{4} - \log_e y$.
At $(\alpha, \sqrt{3}\alpha)$: $\frac{x}{y} = \frac{1}{\sqrt{3}}$, $\tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}$, so $\log_e(\sqrt{3}\alpha) = \frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}$.
Answer: (C). | ||||
| 19 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections Eliminating t: $x^2 = 4t^2 = 12y$, a parabola with focus $S(0, 3)$ and axis the y-axis.
$A = \left(2t, \frac{t^2}{3}\right)$, $B = (0, b)$. $SA \perp BA$: $(2t)(2t) + \left(\frac{t^2}{3} - 3\right)\left(\frac{t^2}{3} - b\right) = 0$.
As $t \to 1$: $4 - \frac{8}{3}\left(\frac{1}{3} - b\right) = 0$, so $\frac{1}{3} - b = \frac{3}{2}$, $b = -\frac{7}{6}$.
Ordinate of the centroid: $k = \frac{3 + \frac{1}{3} - \frac{7}{6}}{3} = \frac{13/6}{3} = \frac{13}{18}$.
Answer: (D). | ||||
| 20 | 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 $|z_1| = |z_2| = |z_3| = 5$, so C is the circle $|z| = 5$, i.e. $x^2 + y^2 = 25$.
Line $z_2z_3$: from $(4, 3)$ to $(0, 5)$, slope $-\frac{1}{2}$. The perpendicular through $z_1 = (3, 4)$ has slope 2: $y = 2x - 2$.
$x^2 + (2x - 2)^2 = 25$: $5x^2 - 8x - 21 = 0$, $x = 3$ or $x = -\frac{7}{5}$.
$z \ne z_1$, so $z = -\frac{7}{5} - \frac{24}{5}i$, in the third quadrant.
$\arg z = -\pi + \tan^{-1}\frac{24/5}{7/5} = \tan^{-1}\left(\frac{24}{7}\right) - \pi$.
Answer: (B). | ||||
| 21 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalHard | |
A hard numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Left side: $\sum_{r=1}^{10}(2r - 1)\,r\,C_r = 2\sum r^2C_r - \sum rC_r$.
$\sum_{r=1}^{10} rC_r = 10\cdot 2^9 = 5120$ and $\sum_{r=1}^{10} r^2C_r = n(n + 1)2^{n-2} = 10\cdot 11\cdot 2^8 = 28160$.
Left side $= 56320 - 5120 = 51200$.
Bracket: $\sum_{r=0}^{10}\frac{C_r}{r + 1} = \int_0^1(1 + x)^{10}dx = \frac{2^{11} - 1}{11}$.
Taking the bracket as this complete sum (the intended reading, see note): $51200 = \frac{\alpha\cdot 2^{11}}{2^{\beta} - 1}\cdot\frac{2^{11} - 1}{11}$.
Matching the forms, $\beta = 11$ and $\alpha\cdot\frac{2^{11}}{11} = 51200$, so $\alpha = \frac{51200\times 11}{2048} = 275$.
$\alpha + \beta = 275 + 11 = 286$.
Answer: 286. | ||||
| 22 | Permutations and CombinationsFundamental principle of counting | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on fundamental principle of counting, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Write the number as abc with $a \in \{1, \ldots, 9\}$, $b \in \{0, \ldots, 9\}$, $c \in \{1, 3, 5, 7, 9\}$. The digit sum lies between 2 and 27, so it must be 7, 14 or 21.
For each c, count pairs (a, b) with $a + b = s - c$. The number of pairs with $a + b = m$ ($1 \le a \le 9$, $0 \le b \le 9$) is $m$ for $1 \le m \le 9$, and $19 - m$ for $10 \le m \le 18$.
Sum 7: $m = 6, 4, 2, 0, -2$ for $c = 1, 3, 5, 7, 9$: $6 + 4 + 2 = 12$.
Sum 14: $m = 13, 11, 9, 7, 5$: $6 + 8 + 9 + 7 + 5 = 35$.
Sum 21: $m = 20, 18, 16, 14, 12$: $0 + 1 + 3 + 5 + 7 = 16$.
Total $= 12 + 35 + 16 = 63$.
Answer: 63. | ||||
| 23 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $(\vec{a} - \vec{b})\times(\vec{a} + \vec{b}) = \vec{a}\times\vec{b} - \vec{b}\times\vec{a} = 2(\vec{a}\times\vec{b})$.
So the expression is $4|\vec{a}\times\vec{b}|^2 + 4(\vec{a}\cdot\vec{b})^2 = 4|\vec{a}|^2|\vec{b}|^2(\sin^2\theta + \cos^2\theta) = 4\cdot 16\cdot 9 = 576$.
Answer: 576. | ||||
| 24 | CirclesEquation from a diameter | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on equation from a diameter, from Circles. NCERT: Class 11: Conic Sections With $P(x_1, y_1)$, $Q(x_2, y_2)$, the circle on PQ as diameter is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$:
$x^2 - (x_1 + x_2)x + x_1x_2 + y^2 - (y_1 + y_2)y + y_1y_2 = 0$.
$x_1 + x_2 = \frac{r}{2}$, $x_1x_2 = \frac{p}{2}$, $y_1 + y_2 = s$, $y_1y_2 = -q$: $x^2 + y^2 - \frac{r}{2}x - sy + \frac{p}{2} - q = 0$.
Given circle: $x^2 + y^2 - \frac{11}{2}x - 7y - 11 = 0$. So $r = 11$, $s = 7$, $\frac{p}{2} - q = -11$, i.e. $p - 2q = -22$.
$2r + s - 2q + p = 22 + 7 - 22 = 7$.
Answer: 7. | ||||
| 25 | 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 $21 - 4\cos^2x = 17 + 4\sin^2x$. With $t = \sin^2x$ ($0 < t \le 1$): $\frac{14}{t} - 2t = 17 + 4t$, i.e. $6t^2 + 17t - 14 = 0$, $(3t - 2)(2t + 7) = 0$.
So $\sin^2x = \frac{2}{3}$, $\sin x = \pm\sqrt{\frac{2}{3}} \approx \pm 0.816$; let $\theta = \sin^{-1}\sqrt{\frac{2}{3}} \approx 54.7°$, which is more than 45°.
In $(45°, 315°)$: $x = \theta \approx 54.7°$, $180° - \theta \approx 125.3°$, $180° + \theta \approx 234.7°$, $360° - \theta \approx 305.3°$. All four lie in the interval.
Answer: 4. | ||||
| 26 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series 19 and 12 are the roots of $x^2 - 31x + 228 = 0$, so $19^{n+2} = 31\cdot 19^{n+1} - 228\cdot 19^n$, and likewise for 12. Subtracting:
$\alpha_{n+2} = 31\alpha_{n+1} - 228\alpha_n$. With $n = 8$: $31\alpha_9 - \alpha_{10} = 228\alpha_8$.
$\frac{228\alpha_8}{57\alpha_8} = 4$.
Answer: 4. | ||||
| 27 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f(x) = \left((2 - x^{25})(2 + x^{25})\right)^{1/50} = (4 - x^{50})^{1/50}$.
$f(1) = 3^{1/50}$; $f(f(1)) = (4 - 3)^{1/50} = 1$; $f(f(f(1))) = f(1) = 3^{1/50}$.
$g(1) = 3^{1/50} + 1$. Since $1 < 3^{1/50} < 2$ (about 1.022), $g(1) \approx 2.022$.
Greatest integer $\le g(1)$ is 2.
Answer: 2. | ||||
| 28 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Intersection: $\lambda = 1 + \mu$, $2\lambda = 3 + \mu$ give $\mu = 1$, $\lambda = 2$; check $3\lambda = 6 = 1 + 5\mu$. So $S = (2, 4, 6)$.
Normal to the plane: $(\hat{i} + 2\hat{j} + 3\hat{k})\times(\hat{i} + \hat{j} + 5\hat{k}) = (10 - 3)\hat{i} - (5 - 3)\hat{j} + (1 - 2)\hat{k} = 7\hat{i} - 2\hat{j} - \hat{k}$.
Matching $(a, b, -1)$: $a = 7$, $b = -2$. Through S: $14 - 8 - 6 + d = 0$, $d = 0$.
$a + b + d = 5$.
Answer: 5. | ||||
| 29 | 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 Let the 9 entries contain p ones and m minus-ones: $p - m = 5$, $p + m \le 9$.
$m = 0, p = 5$: $\frac{9!}{5!\,4!} = 126$.
$m = 1, p = 6$: $\frac{9!}{6!\,1!\,2!} = 252$.
$m = 2, p = 7$: $\frac{9!}{7!\,2!} = 36$.
($m = 3$ would need $p = 8$, 11 entries.) Total $= 126 + 252 + 36 = 414$.
Answer: 414. | ||||
| 30 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The n-th term is $\frac{3^n - 2^n}{3^n} = 1 - \left(\frac{2}{3}\right)^n$ (check: $\frac{3 - 2}{3}, \frac{9 - 4}{9}, \frac{27 - 8}{27}, \frac{81 - 16}{81}$).
Sum of 100 terms $= 100 - \sum_{n=1}^{100}\left(\frac{2}{3}\right)^n = 100 - 2\left(1 - \left(\frac{2}{3}\right)^{100}\right) = 98 + 2\left(\frac{2}{3}\right)^{100}$.
This is just above 98, so the greatest integer is 98.
Answer: 98. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement For a product of powers the maximum relative error is the sum of each relative error times the magnitude of its power; errors always add, even for a quantity in the denominator:
$\frac{\Delta Z}{Z} = 2\frac{\Delta A}{A} + 3\frac{\Delta B}{B} + 4\frac{\Delta C}{C}$.
Answer: (C). | ||||
| 32 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\vec{A}\cdot\vec{A} = |\vec{A}|^2 \ne 0$, so (A) is false.
$\vec{A}\times\vec{A} = |\vec{A}|^2\sin 0°\,\hat{n} = \vec{0}$: the cross product of a vector with itself is the zero vector, so (C) is true. A vector cannot be "less than" or "greater than" zero, so (B) and (D) are meaningless.
Answer: (C). | ||||
| 33 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $|\hat{A} + \hat{B}|^2 = 2 + 2\cos\theta = 4\cos^2\frac{\theta}{2}$, so $|\hat{A} + \hat{B}| = 2\cos\frac{\theta}{2}$.
$|\hat{A} - \hat{B}|^2 = 2 - 2\cos\theta = 4\sin^2\frac{\theta}{2}$, so $|\hat{A} - \hat{B}| = 2\sin\frac{\theta}{2}$.
Hence $|\hat{A} - \hat{B}| = |\hat{A} + \hat{B}|\tan\frac{\theta}{2}$.
Answer: (B). | ||||
| 34 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\vec{\tau} = \vec{r}\times\vec{F} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&2\\3&4&-2\end{vmatrix} = \hat{i}(1\cdot(-2) - 2\cdot 4) - \hat{j}(2\cdot(-2) - 2\cdot 3) + \hat{k}(2\cdot 4 - 1\cdot 3)$
$= -10\hat{i} + 10\hat{j} + 5\hat{k}$.
Answer: (B). | ||||
| 35 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $h = 2R$, so P is at distance $3R$ from the centre.
$g_P = \frac{GM}{(3R)^2} = \frac{g}{9}$.
Answer: (D). | ||||
| 36 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal velocity the weight (less buoyancy) balances Stokes' drag: $\frac{4}{3}\pi r^3(\rho - \sigma)g = 6\pi\eta r v_t$.
$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$, so $v_t \propto r^2$.
Answer: (C). | ||||
| 37 | 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 $v_{rms} = \sqrt{\frac{3RT}{M}}$ and $v_p = \sqrt{\frac{2RT}{M}}$ for any gas at any temperature.
$\frac{v_{rms}}{v_p} = \sqrt{\frac{3}{2}}$, so $v_{rms} = \sqrt{\frac{3}{2}}\,v_p$.
Answer: (B). | ||||
| 38 | 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 By Gauss's law, with a cylindrical surface crossing an infinite sheet, $E\cdot 2A = \frac{\sigma A}{\varepsilon_0}$, so $E = \frac{\sigma}{2\varepsilon_0}$, independent of the distance from the sheet.
So $E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$.
Answer: (C). | ||||
| 39 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Matching listEasy | |
A easy matching list question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current (A) An AC generator turns mechanical energy (rotation of the coil) into electrical energy: II.
(B) A galvanometer detects the presence of a current: I.
(C) A transformer steps an alternating voltage up or down: IV.
(D) A metal detector works on resonance in an AC (LC) circuit: the metal changes the inductance and detunes the circuit: III.
Answer: (A). | ||||
| 40 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Ampere's law on a circle of radius $r < R$: $B\cdot 2\pi r = \mu_0 I\frac{\pi r^2}{\pi R^2}$.
$B = \frac{\mu_0 I r}{2\pi R^2}$, so $B \propto r$.
Answer: (B). | ||||
| 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 Average power $P = V_{rms}I_{rms}\cos\phi$. The current is wattless when $\cos\phi = 0$, i.e. $\phi = 90°$.
In a purely inductive circuit the current lags the voltage by 90°, so no power is consumed. A resistive circuit has $\phi = 0$, and LCR or RC circuits with resistance have $\cos\phi \ne 0$.
Answer: (B). | ||||
| 42 | 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 $I = \frac{1}{2}\varepsilon_0 cE_0^2 = \frac{1}{2}\times 8.85\times 10^{-12}\times 3\times 10^8\times(56.5)^2$
$= 1.3275\times 10^{-3}\times 3192.25 \approx 4.24\ Wm^{-2}$.
Answer: (B). | ||||
| 43 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $I_R = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi = 10I + 6I\cos\phi$.
At P ($\phi = \pi/2$): $I_P = 10I$. At Q ($\phi = \pi$): $I_Q = 10I - 6I = 4I$.
$I_P - I_Q = 6I$.
Answer: (B). | ||||
| 44 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Refractive index $n = \sqrt{\varepsilon_r\mu_r} = \sqrt{4\times 1} = 2$.
Critical angle: $\sin C = \frac{1}{n} = \frac{1}{2}$, $C = 30°$.
All the light is reflected (total internal reflection) for an angle of incidence greater than 30°. Of the options only 60° exceeds it (at 30° the refracted ray just grazes the surface).
Answer: (D). | ||||
| 45 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | StatementsEasy | |
A easy statements question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Statement I: Davisson and Germer observed diffraction of an electron beam by a nickel crystal, with peaks at the angles predicted by the de Broglie wavelength. This established the wave nature of electrons: true.
Statement II: interference and diffraction are properties of waves, so electrons having wave nature can show them (as that very experiment shows): true.
Answer: (A). | ||||
| 46 | 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 $v_n = \frac{Z}{n}\times 2.19\times 10^6\ m/s$, so $v \propto \frac{Z}{n}$.
Same n = 3: $\frac{v_{He^+}}{v_H} = \frac{Z_{He}}{Z_H} = \frac{2}{1}$.
Answer: (D). | ||||
| 47 | 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 Light creates electron–hole pairs. This hardly changes the large number of majority carriers, but it changes the small number of minority carriers by a large fraction.
The reverse current is carried by minority carriers, so in reverse bias the fractional change in current with light intensity is large and easy to measure.
Answer: (D). | ||||
| 48 | Communication SystemsPropagation of signals and transmission media | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on propagation of signals and transmission media, from Communication Systems. NCERT: No matching NCERT chapter $\lambda = \frac{c}{f} = \frac{3\times 10^8}{10^{14}} = 3\ \mu m$, in the infrared: this is an optical signal.
Optical signals are carried efficiently by optical fibres (by total internal reflection). Copper cables suit frequencies far lower than this, and water absorbs infrared.
Answer: (B). | ||||
| 49 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $v_B = \frac{c}{n_B} = \frac{3\times 10^8}{1.47} = 2.041\times 10^8$ m/s.
$v_A = v_B + 0.26\times 10^8 = 2.301\times 10^8$ m/s.
$\frac{n_B}{n_A} = \frac{v_A}{v_B} = \frac{2.301}{2.041} \approx 1.13$.
Answer: (C). | ||||
| 50 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium 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 As in the half-deflection method, a large resistance in series keeps the main current I almost unchanged when the shunt S is put across the galvanometer.
Without the shunt the deflection corresponds to I. With the shunt the galvanometer current is $I_g = I\frac{S}{G + S}$.
For one-third deflection $\frac{S}{G + S} = \frac{1}{3}$, so $G + S = 3S$ and $G = 2S$.
The method works, and G is twice the shunt resistance.
Answer: (B). | ||||
| 51 | Laws of MotionFriction | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Let x hang, with mass per length $\lambda$. At the limit the weight of the hanging part equals the maximum friction on the part on the table:
$\lambda xg = \mu\lambda(L - x)g$, so $x = \frac{\mu L}{1 + \mu} = \frac{0.5\times 6}{1.5} = 2$ m.
Answer: 2. | ||||
| 52 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | NumericalEasy | |
A easy numerical question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Kinetic energy lost $= \frac{1}{2}\times 0.5\times(12^2 - 6^2) = 0.25\times 108 = 27$ J, stored in the spring:
$\frac{1}{2}k(0.3)^2 = 27$, $k = \frac{54}{0.09} = 600\ Nm^{-1}$.
Answer: 600. | ||||
| 53 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical 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 Taking the velocity to fall uniformly from 10 m/s (36 km/h) at the top to zero at the bed, the shear stress is $\eta\frac{v}{d}$:
$10^{-3} = 10^{-2}\times\frac{10}{d}$, so $d = 100$ m.
Answer: 100. | ||||
| 54 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Condensing the steam at 100°C: $50\times 540 = 27000$ cal.
Cooling the water from 100°C to 20°C: $50\times 1\times 80 = 4000$ cal.
Total $= 31000$ cal $= 31\times 10^3$ cal per minute.
Answer: 31. | ||||
| 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 First overtone of an open pipe of length $L_o$: $\frac{2v}{2L_o} = \frac{v}{L_o}$. Fundamental of the closed pipe: $\frac{v}{4\times 20}$.
$\frac{v}{L_o} = \frac{v}{80}$, so $L_o = 80$ cm.
Answer: 80. | ||||
| 56 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Label the junctions along the line from A: A, 1 (after $C_1$), 2 (after $C_2$), 3 (after $C_3$), 4 (after $C_4$), 5 (after $C_5$), then $C_6$ to B. Every capacitor is 8 μF.
The upper wire on the left joins A to junction 2; the lower wire joins junction 1 to junction 3; the upper wire on the right joins junction 3 to junction 5.
So there are only these nodes: X = {A, 2}, Y = {1, 3, 5}, the point 4, and B.
$C_1$ (A–1), $C_2$ (1–2) and $C_3$ (2–3) each lie between X and Y: three in parallel, 24 μF.
$C_4$ (3–4) and $C_5$ (4–5) both lie between Y and point 4, which connects to nothing else: they take no charge.
$C_6$ lies between Y and B.
$C_{AB} = \frac{24\times 8}{24 + 8} = 6$ μF.
Answer: 6. | ||||
| 57 | 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 $H = I^2Rt$: $300 = 4\times R\times 15$, so $R = 5\ \Omega$.
With 3 A for 10 s: $H = 9\times 5\times 10 = 450$ J.
Answer: 450. | ||||
| 58 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity The outer ring is plain wire except for the battery (left) and one 5 Ω resistor (right). So the two upper points where the 5 Ω resistors meet the ring are both joined to the battery's upper terminal (call it P), and the bottom point where the 2.5 Ω resistor meets the ring is joined to the lower terminal (N).
Between P and N there are two paths:
(i) the two upper 5 Ω resistors in parallel (2.5 Ω) from P to the centre, then 2.5 Ω from the centre to N: 5 Ω;
(ii) the 5 Ω resistor on the right arc: 5 Ω.
These are in parallel: $R = 2.5\ \Omega$. Current $= \frac{5}{2.5} = 2$ A.
Answer: 2. | ||||
| 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 The energy spent goes into the magnetic field of the coil: $U = \frac{1}{2}LI^2 = \frac{1}{2}\times 2\times 2^2 = 4$ J (the coil has no stated resistance).
Answer: 4. | ||||
| 60 | Laws of MotionNewton's laws | Class 11: Laws of Motion | NumericalEasy | |
A easy numerical question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion $\vec{a} = \frac{\vec{F}}{m} = \frac{10\hat{i} + 5\hat{j}}{0.1} = 100\hat{i} + 50\hat{j}\ m/s^2$.
From rest (taking the start at the origin): $\vec{r} = \frac{1}{2}\vec{a}t^2 = 200\hat{i} + 100\hat{j}$ m.
$\frac{a}{b} = \frac{200}{100} = 2$.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | StatementsMedium | |
A medium statements question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Removing an electron strengthens the bond (raises the bond order) only when the electron comes from an antibonding orbital.
NO (15 electrons): the highest electron is in $\pi^*2p$; $NO^+$ has bond order 3 against 2.5. Stronger.
$N_2$: the highest electron is in bonding $\sigma 2p_z$; bond order falls from 3 to 2.5. Weaker.
$O_2$: the highest electrons are in $\pi^*2p$; $O_2^+$ has bond order 2.5 against 2. Stronger.
$C_2$: the highest electrons are in bonding $\pi 2p$; bond order falls from 2 to 1.5. Weaker.
$B_2$: the highest electrons are in bonding $\pi 2p$; bond order falls from 1 to 0.5. Weaker.
So (A) and (C) only.
Answer: (C). | ||||
| 62 | Surface ChemistryColloids and their preparation | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on colloids and their preparation, from Surface Chemistry. NCERT: No matching NCERT chapter The Tyndall effect (scattering of light by colloidal particles) is seen only when (i) the diameter of the dispersed particles is not much smaller than the wavelength of the light used, and (ii) the refractive indices of the dispersed phase and the medium differ greatly. So (A) is a correct condition and (B) is the incorrect statement.
(C) The beam of the projector in a cinema hall is visible because of the Tyndall effect: correct.
(D) True solutions do not show it and colloids do, so it distinguishes them: correct.
Answer: (B). | ||||
| 63 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom $Al^{3+}$ has $13 - 3 = 10$ electrons.
$Br^-$ 36, $Be^{2+}$ 2; $Cl^-$ 18, $Li^+$ 2; $S^{2-}$ 18, $K^+$ 18; $O^{2-}$ 10, $Mg^{2+}$ 10.
Only $O^{2-}$ and $Mg^{2+}$ both have 10 electrons.
Answer: (D). | ||||
| 64 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter $4Au + 8CN^- + 2H_2O + O_2 \rightarrow 4[Au(CN)_2]^- + 4OH^-$: [A] is $[Au(CN)_2]^-$.
Zinc displaces gold: $2[Au(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Au$: [B] is $[Zn(CN)_4]^{2-}$.
Answer: (C). | ||||
| 65 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter An electron-deficient molecule has a central atom with fewer than eight electrons in its valence shell (or too few electrons for normal two-centre bonds).
$B_2H_6$: has only 12 valence electrons for 8 B–H links, so it uses 3-centre–2-electron bonds: electron deficient.
$BCl_3$: boron has only 6 valence electrons: electron deficient.
$PH_3$ and $NH_3$ are electron rich (lone pair); $CCl_4$ is electron precise; LiH is an ionic hydride, not an electron-deficient molecule.
Number $= 2$.
Answer: (C). | ||||
| 66 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry In water, smaller ions are more heavily hydrated, so their hydrated radius is larger and they move more slowly.
Hydration falls down the group: $Be^{2+} > Mg^{2+} > Ca^{2+} > Sr^{2+}$, so the hydrated size is smallest for $Sr^{2+}$, which has the highest ionic mobility.
Answer: (D). | ||||
| 67 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter $AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]Cl$, the soluble diamminesilver(I) chloride (Ag(I) forms linear two-coordinate complexes).
Answer: (C). | ||||
| 68 | d- and f-Block ElementsLanthanoids and actinoids | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on lanthanoids and actinoids, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The common, most stable oxidation state of the lanthanoids is +3. Ce(IV) ($E°_{Ce^{4+}/Ce^{3+}} = +1.74$ V), though it has the noble-gas configuration, readily gains an electron to become $Ce^{3+}$.
So Ce(IV) is a good oxidising agent.
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 $E°(M^{3+}/M^{2+})$: Mn +1.57 V, Fe +0.77 V, Ti −0.37 V, Cr −0.41 V.
$Mn^{3+}$ ($d^4$) gains an electron to reach the stable half-filled $d^5$ of $Mn^{2+}$, so it has the highest reduction potential: the strongest oxidising agent.
Answer: (A). | ||||
| 70 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Excess nutrients (phosphates, nitrates) cause dense algal growth. When the algae die, their decomposition uses up the dissolved oxygen (the BOD rises), and fish and other aquatic life die.
So eutrophication leads to loss of biodiversity.
Answer: (A). | ||||
| 71 | Organic Chemistry: Some Basic Principles and TechniquesPurification methods | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on purification methods, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Dilute $HNO_3$ gives o-nitrophenol and p-nitrophenol.
o-Nitrophenol has intramolecular hydrogen bonding and is steam volatile; p-nitrophenol is held by intermolecular hydrogen bonds and is not. Steam distillation separates them, and works on a large scale.
Answer: (C). | ||||
| 72 | HydrocarbonsConformations of ethane | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on conformations of ethane, from Hydrocarbons. NCERT: Class 11: Hydrocarbons These are Newman projections of butane about C2–C3.
(B) and (D) are eclipsed (the back groups sit just behind the front groups), so they are not staggered.
(A) is staggered with the two $CH_3$ groups 60° apart (gauche).
(C) is staggered with the two $CH_3$ groups 180° apart (anti): the maximum dihedral angle.
Answer: (C). | ||||
| 73 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons In acid, 2-methylpropene is protonated to the tert-butyl cation $(CH_3)_3C^+$.
This cation adds to a second molecule of 2-methylpropene at its $CH_2$ end, giving the tertiary cation $(CH_3)_3C-CH_2-C^+(CH_3)_2$.
Isobutane $(CH_3)_3C-H$ transfers a hydride to this cation, giving $(CH_3)_3C-CH_2-CH(CH_3)_2$ (2,2,4-trimethylpentane, "isooctane") and a new tert-butyl cation that continues the chain.
Answer: (B). | ||||
| 74 | 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 Ethylidene is $CH_3CH=$; ethylidene chloride is $CH_3CHCl_2$, both chlorines on the same carbon: 1,1-dichloroethane. (Ethylene dichloride is the 1,2-isomer.)
Answer: (D). | ||||
| 75 | 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 tert-butyl chloride, a tertiary halide, and potassium tert-butoxide is a strong, bulky base.
Substitution (Williamson) is impossible at a tertiary carbon; the base removes a β-hydrogen (E2), giving $(CH_3)_2C=CH_2$, 2-methylprop-1-ene.
Answer: (D). | ||||
| 76 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers This is the Reimer–Tiemann reaction. In aqueous NaOH, $CHCl_3$ gives dichlorocarbene ($:CCl_2$), and phenol is present as sodium phenoxide.
The phenoxide attacks $:CCl_2$ at the ortho position; after proton transfer the product is the sodium phenoxide carrying an ortho $-CHCl_2$ group (X).
NaOH then hydrolyses $-CHCl_2$ to $-CHO$ (through the gem-diol), and acid frees the phenol, giving salicylaldehyde.
So X is the o-(dichloromethyl)phenoxide.
Answer: (C). | ||||
| 77 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide, $C_6H_5C(CH_3)_2-O-O-H$.
With dilute acid it rearranges and is hydrolysed to phenol and acetone: $C_6H_5C(CH_3)_2OOH \xrightarrow{H^+/H_2O} C_6H_5OH + CH_3COCH_3$.
A = phenol, B = acetone.
Answer: (C). | ||||
| 78 | AminesClassification and identification of amines | Class 12: Amines | Single correctMedium | |
A medium single correct question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hofmann bromamide degradation: $RCONH_2 \xrightarrow{Br_2/KOH} RCONHBr \xrightarrow{OH^-} RCON^-Br \rightarrow R-N=C=O$ (the alkyl group migrates from carbon to nitrogen as $Br^-$ leaves) $\xrightarrow{2OH^-} RNH_2 + CO_3^{2-}$.
The isocyanate $R-N=C=O$ is the key intermediate whose hydrolysis gives the amine.
Answer: (C). | ||||
| 79 | Surface ChemistryEmulsions, surfactants and micelles | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on emulsions, surfactants and micelles, from Surface Chemistry. NCERT: No matching NCERT chapter Soap cleans by forming micelles that trap grease in their hydrophobic interior. Micelles form only above the critical micelle concentration (CMC).
With very little soap the concentration stays below the CMC, no micelles form, and the grease is not removed.
Answer: (C). | ||||
| 80 | Chemistry in Everyday LifeChemicals in food | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on chemicals in food, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Alitame is an artificial sweetener (about 2000 times as sweet as sucrose).
Bithional is an antiseptic, Salvarsan an arsenic-based antimicrobial drug, and lactose a natural sugar.
Answer: (B). | ||||
| 81 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Molar mass of $C_7H_5N_3O_6 = 7(12) + 5(1) + 3(14) + 6(16) = 227$ g/mol.
Moles $= \frac{681}{227} = 3$; each molecule has 3 N atoms, so 9 mol of N atoms.
$9\times 6.02\times 10^{23} = 5.418\times 10^{24} = 5418\times 10^{21}$.
Answer: 5418. | ||||
| 82 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | NumericalMedium | |
A medium numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter NaCl has the rock-salt (fcc) structure with 4 formula units per unit cell; molar mass 58.5 g/mol.
$\rho = \frac{ZM}{N_Aa^3}$: $a^3 = \frac{4\times 58.5}{6.02\times 10^{23}\times 43.1} = \frac{234}{2.595\times 10^{25}} = 9.02\times 10^{-24}\ cm^3$.
$a = 2.08\times 10^{-8}$ cm $= 2.08\times 10^{-10}$ m.
The $Na^+$–$Cl^-$ distance is $\frac{a}{2} = 1.04\times 10^{-10}$ m, i.e. 1 to the nearest integer.
Answer: 1. | ||||
| 83 | Structure of AtomBohr model | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom Li ($1s^2 2s^1$) loses its outermost electron, which is in n = 2. Treating it with the hydrogen-like formula with Z = 3:
$E = 2.2\times 10^{-18}\times\frac{Z^2}{n^2} = 2.2\times 10^{-18}\times\frac{9}{4} = 4.95\times 10^{-18}$ J.
$\lambda = \frac{hc}{E} = \frac{6.63\times 10^{-34}\times 3\times 10^8}{4.95\times 10^{-18}} = 4.02\times 10^{-8}$ m.
Answer: 4. | ||||
| 84 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics At equilibrium $\Delta G = \Delta H - T\Delta S = 0$ (taking $\Delta H$ and $\Delta S$ independent of temperature):
$T = \frac{\Delta H}{\Delta S} = \frac{-165000}{-550} = 300$ K.
Answer: 300. | ||||
| 85 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Half of the solution carries $0.01$ mmol of $H_2SO_4$; dilution does not change the amount, so solution (A) has 0.01 mmol.
Adding 0.01 mmol gives $0.02$ mmol in all $= 20\times 10^{-3}$ mmol.
Answer: 20. | ||||
| 86 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalMedium | |
A medium numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $N_2O_4 \rightleftharpoons 2NO_2$. From 1 mol with $\alpha = 0.5$: $N_2O_4$ 0.5 mol, $NO_2$ 1.0 mol, total 1.5 mol, at P = 1 atm.
$K_p = \frac{(1.0/1.5)^2}{0.5/1.5}\times 1 = \frac{4/9}{1/3} = \frac{4}{3} = 1.33$.
$\Delta G° = -2.3RT\log K_p = -2.3\times 8.31\times 300\times 0.1239 = -710.4\ J\ mol^{-1}$.
x = 710.
Answer: 710. | ||||
| 87 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry $Fe^{3+}/Fe^{2+}$ has the higher reduction potential, so it is reduced and iodide is oxidised: $2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$.
$E°_{cell} = 0.77 - 0.54 = 0.23$ V $= 23\times 10^{-2}$ V.
Answer: 23. | ||||
| 88 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate of appearance of C $= \frac{20 - 10}{10} = 1$ mmol $dm^{-3}s^{-1}$.
Rate of D = 9; rate of B $= \frac{9}{1.5} = 6$; rate of A $= \frac{6}{2} = 3$.
The rates are in the ratio of the coefficients: $\gamma_1 : \gamma_2 : \gamma_3 : \gamma_4 = 3 : 6 : 1 : 9$, so the equation is $3A + 6B \rightarrow C + 9D$.
Rate of reaction $= \frac{1}{\gamma_3}\frac{d[C]}{dt} = \frac{1}{1}\times 1 = 1$ mmol $dm^{-3}s^{-1}$ (equally $\frac{9}{9}$, $\frac{6}{6}$, $\frac{3}{3}$).
Answer: 1. | ||||
| 89 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Energy absorbed by the four-coordinate complex: $E = \frac{hc}{\lambda} = \frac{6.63\times 10^{-34}\times 3.08\times 10^8}{600\times 10^{-9}} = 3.403\times 10^{-19}$ J. Taking it as the tetrahedral splitting $\Delta_t$:
$\Delta_t = \frac{4}{9}\Delta_o$, so $\Delta_o = \frac{9}{4}\times 3.403\times 10^{-19} = 7.66\times 10^{-19}$ J $= 766\times 10^{-21}$ J.
Answer: 766. | ||||
| 90 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Pent-1-ene $C_5H_{10}$: $M = 70$ g/mol; one C=C adds one $Br_2$ (160 g/mol).
Moles of pentene $= \frac{5.0}{70} = 0.0714$; mass of $Br_2 = 0.0714\times 160 = 11.43$ g $= 1143\times 10^{-2}$ g.
Answer: 1143. | ||||
Sit this paper free.
Timed like the exam or chapter by chapter, with a solution for each question. Run by a non-profit trust.