JEE (Main) · 2023
JEE (Main) 2023 — 11 April, 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
- Three Dimensional Geometry3
- Conic Sections3
- Integral Calculus3
9 easy · 17 medium · 4 hard
Physics
30 Q
- Atoms and Nuclei2
- Optics2
- Electromagnetic Induction and Alternating Currents2
24 easy · 6 medium · 0 hard
Chemistry
30 Q
- Coordination Compounds3
- Equilibrium2
- p-Block Elements2
18 easy · 10 medium · 2 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsUnion, intersection and complement | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on union, intersection and complement, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Let a, b, c be the numbers of men with medals in exactly one, exactly two and all three events. Each man is counted once in $n(A \cup B \cup C)$, and the medal total counts him once per event he won (one medal per man per event):
$a + b + c = 60$ and $a + 2b + 3c = 48 + 25 + 18 = 91$, with $c = 5$.
Then $a + b = 55$ and $a + 2b = 76$, so $b = 21$.
(Equivalently, $91 - \sum n(\text{pairwise}) + 5 = 60$ gives $\sum n(\text{pairwise}) = 36$, and exactly two $= 36 - 3 \times 5 = 21$.)
Answer: (D). | ||||
| 2 | Three Dimensional GeometryCoordinates of a point in space | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on coordinates of a point in space, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry For the image of P in $2x + y - 3z - 6 = 0$:
$\frac{\alpha - 2}{2} = \frac{\beta - 3}{1} = \frac{\gamma - 5}{-3} = -2\cdot\frac{2(2) + 3 - 3(5) - 6}{2^2 + 1^2 + 3^2} = -2\cdot\frac{-14}{14} = 2$.
So $(\alpha, \beta, \gamma) = (6, 5, -1)$ and $\alpha + \beta + \gamma = 10$.
Answer: (B). | ||||
| 3 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections $E_k : \frac{x^2}{1/k} + \frac{y^2}{1/k^2} = 1$, so the semi-axes along x and y are $a = \frac{1}{\sqrt{k}}$ and $b = \frac{1}{k}$.
The four chords form a rhombus with vertices $(\pm a, 0)$, $(0, \pm b)$; the circle touching all four sides is centred at the origin, and its radius is the distance from O to the side $\frac{x}{a} + \frac{y}{b} = 1$:
$r_k = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$, so $\frac{1}{r_k^2} = \frac{1}{a^2} + \frac{1}{b^2} = k + k^2$.
$\sum_{k=1}^{20}(k^2 + k) = \frac{20 \cdot 21 \cdot 41}{6} + \frac{20 \cdot 21}{2} = 2870 + 210 = 3080$.
Answer: (B). | ||||
| 4 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations All non-negative solutions: $\binom{15 + 2}{2} = 136$.
Solutions with $x = y$: $2x + z = 15$, $x = 0, 1, \ldots, 7$: 8 solutions; likewise 8 with $y = z$ and 8 with $z = x$. Any two of these conditions together force $x = y = z = 5$, the single solution (5, 5, 5).
By inclusion–exclusion, solutions with at least two equal $= 3 \times 8 - 3 \times 1 + 1 = 22$.
Distinct triplets $= 136 - 22 = 114$.
Answer: (B). | ||||
| 5 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Conditions: base $x + \frac{7}{2} > 0$, $\ne 1$: $x > -\frac{7}{2}$, $x \ne -\frac{5}{2}$. Argument positive: $x \ne 7$, $x \ne \frac{3}{2}$.
Note $\left(\frac{x - 7}{2x - 3}\right)^2 \ge 1 \iff (x - 7)^2 \ge (2x - 3)^2 \iff (x - 7 - 2x + 3)(x - 7 + 2x - 3) \ge 0 \iff (x + 4)(3x - 10) \le 0 \iff -4 \le x \le \frac{10}{3}$.
Case 1, base > 1 ($x > -\frac{5}{2}$): need the argument $\ge 1$: $-\frac{5}{2} < x \le \frac{10}{3}$. Integers: $-2, -1, 0, 1, 2, 3$ (none excluded): 6.
Case 2, $0 <$ base $< 1$ ($-\frac{7}{2} < x < -\frac{5}{2}$): only $x = -3$, and we need the argument $\le 1$, i.e. $(x + 4)(3x - 10) \ge 0$; at $x = -3$ this is $1 \times (-19) < 0$. So $x = -3$ fails.
Number of integral solutions $= 6$.
Answer: (C). | ||||
| 6 | 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 Transpose $A' = \alpha A + I$: $A = \alpha A' + I = \alpha(\alpha A + I) + I$, so $(1 - \alpha^2)A = (1 + \alpha)I$ and, as $\alpha \ne \pm 1$, $A = \frac{1}{1 - \alpha}I = kI$ with $k = \frac{1}{1 - \alpha}$.
$A^2 - A = (k^2 - k)I$, so $\det(A^2 - A) = (k^2 - k)^2 = 4$, i.e. $k^2 - k = \pm 2$.
$k^2 - k + 2 = 0$ has no real root; $k^2 - k - 2 = 0$ gives $k = 2$ or $k = -1$.
$\alpha = 1 - \frac{1}{k}$: $k = 2 \Rightarrow \alpha = \frac{1}{2}$; $k = -1 \Rightarrow \alpha = 2$.
Sum $= \frac{5}{2}$.
Answer: (A). | ||||
| 7 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $|-x + [x]| = x - [x] = \{x\}$, the fractional part.
At $x = 0$: $f(0) = 0$. For $x \to 0^+$, $x^2 - x$ is slightly negative, so $[x^2 - x] = -1$, and $\{x\} \to 0$: limit $-1$. For $x \to 0^-$, $x^2 - x$ is slightly positive, so $[x^2 - x] = 0$, and $\{x\} \to 1$: limit $1$. Not continuous at 0.
At $x = 1$: $f(1) = 0 + 0 = 0$. For $x \to 1^+$, $x^2 - x = x(x - 1)$ is slightly positive: $0 + 0 = 0$. For $x \to 1^-$, it is slightly negative: $-1 + 1 = 0$. Continuous at 1.
Answer: (D). | ||||
| 8 | Three Dimensional GeometryDirection ratios and direction cosines | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on direction ratios and direction cosines, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The normal of the required plane is perpendicular to both normals (2, 4, 5) and (3, −2, 3):
$(2, 4, 5) \times (3, -2, 3) = (4 \cdot 3 - 5 \cdot (-2),\ 5 \cdot 3 - 2 \cdot 3,\ 2 \cdot (-2) - 4 \cdot 3) = (22, 9, -16)$.
Plane: $22x + 9y - 16z + d = 0$ through $(-2, 3, 5)$: $-44 + 27 - 80 + d = 0$, $d = 97$, which matches the constant 97.
So $\alpha + \beta + \gamma = 22 + 9 - 16 = 15$.
Answer: (A). | ||||
| 9 | 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 The circle has centre (0, 2), radius 2; the region is inside it and on or below the parabola $y = \frac{x^2}{2}$.
Intersection: $x^2 = 2y$ in $x^2 + y^2 - 4y = 0$ gives $y^2 - 2y = 0$, so $y = 0$ or $y = 2$ ($x = 0$ or $x = \pm 2$).
For $-2 \le x \le 2$ the region runs from the lower semicircle $y = 2 - \sqrt{4 - x^2}$ up to the parabola (the parabola stays below the upper semicircle there).
Area $= \int_{-2}^{2}\left(\frac{x^2}{2} - 2 + \sqrt{4 - x^2}\right)dx = \frac{8}{3} - 8 + 2\pi = 2\pi - \frac{16}{3}$
(using $\int_{-2}^{2}\sqrt{4 - x^2}\,dx = 2\pi$, half the disc).
Answer: (B). | ||||
| 10 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | StatementsHard | |
A hard statements question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Let $g(x) = (x\log_e x)f(x)$. Then $g'(x) = (\log_e x + 1)f(x) + (x\log_e x)f'(x)$, which is exactly the left side, so $g'(x) \ge 1$ on [2, 4].
Statement (B): integrate from 2 to x: $g(x) - g(2) \ge x - 2$. With $g(2) = 2\log_e 2 \cdot \frac{1}{2} = \log_e 2$,
$f(x) \ge \frac{x - 2 + \log_e 2}{x\log_e x}$.
This bound decreases from $\frac{1}{2}$ at $x = 2$ to $\frac{2 + \log_e 2}{8\log_e 2} \approx 0.49$ at $x = 4$, so it is always above $\frac{1}{8}$: (B) holds.
Statement (A): integrate from x to 4: $g(4) - g(x) \ge 4 - x$. With $g(4) = 4\log_e 4 \cdot \frac{1}{4} = 2\log_e 2$,
$f(x) \le \frac{x - 4 + 2\log_e 2}{x\log_e x} < \frac{2\log_e 2}{x\log_e x} \le \frac{2\log_e 2}{2\log_e 2} = 1$ (since $x - 4 \le 0$ and $x\log_e x \ge 2\log_e 2$): (A) holds.
Both statements follow.
Answer: (C). | ||||
| 11 | 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 Normal of the first plane: $(\hat{i} + \hat{j}) \times (\hat{i} + \hat{k}) = \hat{i} - \hat{j} - \hat{k}$.
Normal of the second plane: $(\hat{i} - \hat{j}) \times (\hat{j} - \hat{k}) = \hat{i} + \hat{j} + \hat{k}$.
The line of intersection is along $(\hat{i} - \hat{j} - \hat{k}) \times (\hat{i} + \hat{j} + \hat{k}) = 0\hat{i} - 2\hat{j} + 2\hat{k}$, so $\vec{a} = \lambda(-\hat{j} + \hat{k})$.
$\vec{a}\cdot\vec{b} = \lambda(2 + 1) = 3\lambda = 6$, so $\lambda = 2$, $\vec{a} = -2\hat{j} + 2\hat{k}$, $|\vec{a}| = 2\sqrt{2}$, $|\vec{b}| = 3$.
$\cos\theta = \frac{6}{2\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}}$, $\theta = \frac{\pi}{4}$.
$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 6\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 6$.
Answer: (D). | ||||
| 12 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | StatementsEasy | |
A easy statements question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra Let $m = \max\{|a_1|, |a_2|, |a_3|\}$.
(A): $|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \ge m^2$, so $m \le |\vec{a}|$. True.
(B): $|\vec{a}|^2 \le 3m^2$, so $|\vec{a}| \le \sqrt{3}\,m \le 3m$. True.
(The condition $10|a_i| < 1$ plays no part.)
Answer: (C). | ||||
| 13 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Mean of an AP $= \frac{x_1 + x_{100}}{2} = 200$, so $x_{100} = 398$ and $d = \frac{398 - 2}{99} = 4$: $x_i = 4i - 2$.
$y_i = i(4i - 2 - i) = 3i^2 - 2i$.
$\sum_{i=1}^{100} y_i = 3 \cdot \frac{100 \cdot 101 \cdot 201}{6} - 2 \cdot \frac{100 \cdot 101}{2} = 1015050 - 10100 = 1004950$.
Mean $= 10049.50$.
Answer: (B). | ||||
| 14 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $w_1 = iz_1 = i(5 + 4i) = -4 + 5i$; $w_2 = -iz_2 = -i(3 + 5i) = 5 - 3i$.
$w_1 - w_2 = -9 + 8i$, which lies in the second quadrant.
Principal argument $= \pi - \tan^{-1}\frac{8}{9}$.
Answer: (C). | ||||
| 15 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t = e^x$, $dt = e^x dx$; the limits become $\frac{1}{2}$ and 2:
$I = \int_{1/2}^{2}\log_e\left(t + \sqrt{1 + t^2}\right)dt$.
By parts, since $\frac{d}{dt}\log_e(t + \sqrt{1 + t^2}) = \frac{1}{\sqrt{1 + t^2}}$:
$I = \left[t\log_e(t + \sqrt{1 + t^2}) - \sqrt{1 + t^2}\right]_{1/2}^{2}$.
At 2: $2\log_e(2 + \sqrt{5}) - \sqrt{5}$. At $\frac{1}{2}$: $\frac{1}{2}\log_e\frac{1 + \sqrt{5}}{2} - \frac{\sqrt{5}}{2}$.
$I = 2\log_e(2 + \sqrt{5}) - \frac{1}{2}\log_e\frac{1 + \sqrt{5}}{2} - \frac{\sqrt{5}}{2} = \log_e\frac{(2 + \sqrt{5})^2\sqrt{2}}{\sqrt{1 + \sqrt{5}}} - \frac{\sqrt{5}}{2}$.
Answer: (B). | ||||
| 16 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $u = \cos^2\theta$, so $\sin^2\theta = 1 - u$. The equation is $3u^2 - 5u + 2 = 2(1 - u)^3$.
$3u^2 - 5u + 2 = (u - 1)(3u - 2)$ and $2(1 - u)^3 = -2(u - 1)^3$, so
$(u - 1)\left[(3u - 2) + 2(u - 1)^2\right] = 0$, i.e. $(u - 1)(2u^2 - u) = 0$: $u = 1$, $u = 0$ or $u = \frac{1}{2}$.
$\cos^2\theta = 1$: $\theta = 0, \pi, 2\pi$ (3 values).
$\cos^2\theta = 0$: $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ (2 values).
$\cos^2\theta = \frac{1}{2}$: $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ (4 values).
Total $= 9$.
Answer: (B). | ||||
| 17 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability $|S| = 3^4 = 81$. M is singular when $a_{11}a_{22} = a_{12}a_{21}$.
Number of ordered pairs (p, q) from {0, 1, 2} with product: 0 — 5 pairs; 1 — 1 pair; 2 — 2 pairs; 4 — 1 pair.
Singular matrices $= 5^2 + 1^2 + 2^2 + 1^2 = 31$; invertible $= 81 - 31 = 50$.
$P(A) = \frac{50}{81}$.
Answer: (D). | ||||
| 18 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctMedium | |
A medium single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Shifting does not change variance. The five shifted elements of A have mean 2, variance 12, so mean of squares $12 + 4 = 16$. The five shifted elements of B have mean 10, variance 20, so mean of squares $20 + 100 = 120$.
For C: mean $= \frac{5 \cdot 2 + 5 \cdot 10}{10} = 6$; mean of squares $= \frac{16 + 120}{2} = 68$; variance $= 68 - 36 = 32$.
Sum $= 6 + 32 = 38$.
Answer: (A). | ||||
| 19 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections AB cuts off the right triangle OAB of area $\frac{1}{2}\alpha\beta$ from the rectangle of area 10. Since $\alpha \le 2$, $\beta \le 5$, the triangle is at most half the rectangle, so it is the smaller part: $\frac{1}{2}\alpha\beta = \frac{1}{5} \times 10 = 2$, i.e. $\alpha\beta = 4$.
Mid-point $(h, k) = \left(\frac{\alpha}{2}, \frac{\beta}{2}\right)$: $4hk = 4$, so $hk = 1$, i.e. the locus $xy = 1$ — a (rectangular) hyperbola.
Answer: (C). | ||||
| 20 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctHard | |
A hard single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations $y\,dx + x\,dy = d(xy)$. With $u = xy$: $\frac{du}{1 - u^2} = dx$, so $\frac{1}{2}\log_e\left|\frac{1 + u}{1 - u}\right| = x + C$.
At $x = 1$, $y = 2$: $u = 2$, $\frac{1}{2}\log_e 3 = 1 + C$.
At $x = 2$: $\frac{1}{2}\log_e\left|\frac{1 + u}{1 - u}\right| = 2 + \frac{1}{2}\log_e 3 - 1$, so $\left|\frac{1 + u}{1 - u}\right| = 3e^2$.
At $x = 1$, $\frac{1 + u}{1 - u} = -3 < 0$, and along the solution curve this ratio cannot change sign (it would have to pass through 0 or infinity, where the left side is infinite), so $\frac{1 + u}{1 - u} = -3e^2$.
$1 + u = -3e^2 + 3e^2u$, $u = \frac{3e^2 + 1}{3e^2 - 1}$, and $\alpha = \frac{u}{2} = \frac{1 + 3e^2}{2(3e^2 - 1)}$.
Answer: (C). | ||||
| 21 | Binomial TheoremBinomial theorem for a positive integral index | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on binomial theorem for a positive integral index, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Coefficient of $x^r$: $\binom{9}{r}2^{9 - r}$. All coefficients add to $(2 + 1)^9 = 19683$.
Remove $r = 0$: $2^9 = 512$; $r = 8$: $\binom{9}{8} \cdot 2 = 18$; $r = 9$: 1.
Sum for $r = 1$ to 7 $= 19683 - 531 = 19152$; mean $= \frac{19152}{7} = 2736$.
Answer: 2736. | ||||
| 22 | 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 Direction of l: $(1, 2, 3) \times (2, 2, 1) = (2 - 6, 6 - 1, 2 - 4) = (-4, 5, -2)$, so l: $t(-4, 5, -2)$.
Meeting $l_1$: $1 + \lambda = -4t$, $-11 + 2\lambda = 5t$, $-7 + 3\lambda = -2t$. The first two give $t = -1$, $\lambda = 3$, and the third checks ($2 = 2$). So $P = (4, -5, 2)$.
A point of $l_2$: $(-1 + 2\mu, 2\mu, 1 + \mu)$. $\vec{PQ} = (-5 + 2\mu, 5 + 2\mu, -1 + \mu)$ must be perpendicular to (2, 2, 1):
$-10 + 4\mu + 10 + 4\mu - 1 + \mu = 9\mu - 1 = 0$, $\mu = \frac{1}{9}$.
$Q = \left(-\frac{7}{9}, \frac{2}{9}, \frac{10}{9}\right)$, $\alpha + \beta + \gamma = \frac{5}{9}$, and $9(\alpha + \beta + \gamma) = 5$.
Answer: 5. | ||||
| 23 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter An implication is false only when its premise is true and its conclusion false.
Conclusion $q \vee r$ false means $q = r = F$. Then the premise is $(p \vee F) \wedge (p \vee F) = p$, which is true only for $p = T$.
So exactly one of the $2^3 = 8$ triplets, (T, F, F), makes the statement false; it is true for $8 - 1 = 7$.
Answer: 7. | ||||
| 24 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $a + b = 7$, $ab = -1$, so $\frac{1}{a} = -b$ and $a^{-2} = b^2$; likewise $b^{-2} = a^2$.
$a^{21} + a^{17} = a^{19}(a^2 + a^{-2}) = a^{19}(a^2 + b^2)$, and $b^{21} + b^{17} = b^{19}(b^2 + a^2)$.
So the ratio is $a^2 + b^2 = (a + b)^2 - 2ab = 49 + 2 = 51$.
Answer: 51. | ||||
| 25 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | NumericalMedium | |
A medium numerical question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $S = \sum_{k=0}^{108}\frac{109 - k}{5^k}$. Then $\frac{S}{5} = \sum_{k=1}^{109}\frac{110 - k}{5^k}$, and subtracting term by term:
$S - \frac{S}{5} = 109 - \left(\frac{1}{5} + \frac{1}{5^2} + \ldots + \frac{1}{5^{109}}\right) = 109 - \frac{1}{4}\left(1 - \frac{1}{5^{109}}\right)$.
$\frac{4S}{5} = 109 - \frac{1}{4} + \frac{1}{4 \cdot 5^{109}}$, so $16S = 20 \cdot 109 - 5 + \frac{1}{5^{108}} = 2175 + 5^{-108}$.
Since $(25)^{-54} = 5^{-108}$, $16S - (25)^{-54} = 2175$.
Answer: 2175. | ||||
| 26 | Integral CalculusIntegration by parts | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by parts, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals $\frac{d}{dt}\left[t^{11}(1 + 3t)^6\right] = 11t^{10}(1 + 3t)^6 + 18t^{11}(1 + 3t)^5$.
So $11\alpha(10, 6) + 18\alpha(11, 5) = \left[t^{11}(1 + 3t)^6\right]_0^2 = 2^{11} \cdot 7^6 = 2^5 \cdot (2 \cdot 7)^6 = 32 \cdot 14^6$.
$p = 32$.
Answer: 32. | ||||
| 27 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | NumericalEasy | |
A easy numerical question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $T_{r+1} = \binom{680}{r}3^{\frac{680 - r}{2}}5^{\frac{r}{4}}$. It is an integer when $\frac{r}{4}$ and $\frac{680 - r}{2}$ are whole numbers, i.e. when r is a multiple of 4 (then $680 - r$ is even automatically).
$r = 0, 4, 8, \ldots, 680$: $\frac{680}{4} + 1 = 171$ terms.
Answer: 171. | ||||
| 28 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $e^2 = 1 + \frac{3 + n}{1 + n} = \frac{2n + 4}{n + 1} = \frac{2(n + 1) + 2}{n + 1}$.
For even n, $n + 1$ is odd and $\gcd(2n + 4, n + 1) = \gcd(2, n + 1) = 1$, so e is rational only if $n + 1 = m^2$ and $2n + 4 = 2(m^2 + 1)$ are both perfect squares, with m odd.
$m = 3$: 20; $m = 5$: 52 — not squares. $m = 7$: $2 \times 50 = 100 = 10^2$. ($m = 1$ gives $n = 0 \notin \mathbb{N}$.)
So $k = 48$: $a^2 = 49$, $b^2 = 51$, $e = \frac{10}{7}$.
$l = \frac{2b^2}{a} = \frac{102}{7}$, and $21l = 306$.
Answer: 306. | ||||
| 29 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalHard | |
A hard numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A^3 = A$ means every eigenvalue $\lambda$ satisfies $\lambda^3 = \lambda$, so $\lambda \in \{0, 1, -1\}$, and A is diagonalisable. The trace is 0, and the eigenvalues cannot all be 0 (A would then be the zero matrix), so they are $1, -1, 0$ and the characteristic polynomial is $\lambda^3 - \lambda$.
Characteristic polynomial: $\lambda^3 - (\text{tr}A)\lambda^2 + (M_{11} + M_{22} + M_{33})\lambda - \det A$, with
$M_{11} = -3c$, $M_{22} = -2$, $M_{33} = -a$, and $\det A = -1(0 - 3) + 2(ac - 0) = 3 + 2ac$.
Matching with $\lambda^3 - \lambda$: $-3c - 2 - a = -1$, i.e. $a + 3c = -1$; and $3 + 2ac = 0$, i.e. $ac = -\frac{3}{2}$.
Then $c(-1 - 3c) = -\frac{3}{2}$: $6c^2 + 2c - 3 = 0$, $c = \frac{-1 \pm \sqrt{19}}{6}$, and $a = -1 - 3c = \frac{-1 \mp \sqrt{19}}{2}$.
Positive value: $a = \frac{\sqrt{19} - 1}{2} \approx 1.68 \in (1, 2]$. (Conversely, with these values the characteristic polynomial is $\lambda^3 - \lambda$, so $A^3 = A$ by the Cayley–Hamilton theorem.)
$n = 2$.
Answer: 2. | ||||
| 30 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalEasy | |
A easy numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations This is the number of derangements of 5 objects:
$D_5 = 5!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right) = 120 - 120 + 60 - 20 + 5 - 1 = 44$.
Answer: 44. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Communication SystemsRange of a TV tower | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on range of a tv tower, from Communication Systems. NCERT: No matching NCERT chapter With the transmitting antenna at ground level, the line-of-sight distance is set by the receiving antenna alone: $d = \sqrt{2Rh}$.
$h = \frac{d^2}{2R} = \frac{(4 \times 10^3)^2}{2 \times 6.4 \times 10^6} = \frac{16 \times 10^6}{12.8 \times 10^6} = 1.25$ m $= 125 \times 10^{-2}$ m.
$x = 125$.
Answer: (C). | ||||
| 32 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Each input goes to both terminals of a NOR gate; a NOR gate with its inputs joined is a NOT gate. So the first stage gives $\bar{a}$ and $\bar{b}$.
The output gate is a NOR: $Y = \overline{\bar{a} + \bar{b}} = a \cdot b$ (De Morgan).
This is the AND gate.
Answer: (C). | ||||
| 33 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Half-life of A: $\frac{\ln 2}{\lambda_A}$. Mean life of B: $\frac{1}{\lambda_B}$.
$\frac{\ln 2}{\lambda_A} = \frac{1}{\lambda_B}$ gives $\lambda_A = \lambda_B \ln 2$.
Answer: (B). | ||||
| 34 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\frac{hc}{\lambda} = \phi + eV_0$ and $\frac{hc}{2\lambda} = \phi + \frac{eV_0}{4}$.
Subtracting: $\frac{hc}{2\lambda} = \frac{3}{4}eV_0$, so $eV_0 = \frac{2hc}{3\lambda}$.
$\phi = \frac{hc}{\lambda} - \frac{2hc}{3\lambda} = \frac{hc}{3\lambda}$, so the threshold wavelength $\lambda_0 = \frac{hc}{\phi} = 3\lambda$.
Answer: (C). | ||||
| 35 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics $\sin C = \frac{n_{rarer}}{n_{denser}} = \frac{v_{denser}}{v_{rarer}}$.
$v_{denser} = 3 \times 10^8 \times \sin 45^\circ = \frac{3 \times 10^8}{\sqrt{2}} \approx 2.12 \times 10^8$ m/s.
Answer: (A). | ||||
| 36 | 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 Curve A falls steadily as f increases, like $\frac{1}{f}$: this is $X_C = \frac{1}{2\pi fC}$.
Curve B is a straight line through the origin: this is $X_L = 2\pi fL$.
(R would be a horizontal line; Z of the series LCR circuit has a minimum at resonance.)
Answer: (A). | ||||
| 37 | 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 average energy density of an EM wave (electric and magnetic parts together) is $u = \frac{1}{2}\varepsilon_0 E_0^2$.
$U = \frac{1}{2}\varepsilon_0 E_0^2 V = \frac{1}{2} \times 8.85 \times 10^{-12} \times 400 \times 5 \times 10^{-4} = 8.85 \times 10^{-13}$ J.
Answer: (A). | ||||
| 38 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $B = \mu_0(1 + \chi)nI$, against $B_0 = \mu_0 nI$ in free space.
Fractional increase $= \chi = 2 \times 10^{-2}$, i.e. 2%.
Answer: (A). | ||||
| 39 | 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 Current sensitivity $S_I = \frac{NAB}{k}$; voltage sensitivity $S_V = \frac{S_I}{G}$, where G is the coil resistance.
G is kept constant, so $S_V$ changes in the same proportion as $S_I$: +25%.
Answer: (A). | ||||
| 40 | Current ElectricityOhm's law | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity At fixed V, heat in time t is $\frac{V^2t}{R_{eq}}$.
Parallel: $R_{eq} = \frac{R}{2}$; series: $R_{eq} = 2R$.
Ratio parallel : series $= \frac{2R}{R/2} = 4 : 1$.
Answer: (B). | ||||
| 41 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $E_1 = \frac{1}{2}(2)V^2 = V^2$, charge $Q = 2V$.
Shared between two identical capacitors in parallel: total capacitance 4 F, common potential $\frac{2V}{4} = \frac{V}{2}$.
$E_2 = \frac{1}{2}(4)\left(\frac{V}{2}\right)^2 = \frac{V^2}{2}$.
$\frac{E_2}{E_1} = \frac{1}{2}$, i.e. 1 : 2.
Answer: (A). | ||||
| 42 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $KE = \frac{1}{2}k(A^2 - x^2)$: largest ($\frac{1}{2}kA^2$) at the mean position, zero at the extreme position, and falling as a downward parabola (slope 0 at x = 0, steepest at x = A).
That is the fourth graph.
Answer: (D). | ||||
| 43 | 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 $v_{rms} = \sqrt{\frac{3RT}{M}}$, so at the same temperature it falls as the molar mass rises.
Ne: 20, $\text{Cl}_2$: 71, $\text{UF}_6$: 352 g/mol.
So $v_{rms}$(mono) > $v_{rms}$(dia) > $v_{rms}$(poly).
Answer: (A). | ||||
| 44 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Heat absorbed: $Q = mL = 2257$ kJ.
Work done by the system: $W = P\Delta V = 10^5 \times (1.671 - 0.001) = 1.67 \times 10^5$ J $= 167$ kJ.
$\Delta U = Q - W = 2257 - 167 = +2090$ kJ.
Answer: (B). | ||||
| 45 | ThermodynamicsZeroth law and temperature | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on zeroth law and temperature, from Thermodynamics. NCERT: Class 11: Thermodynamics Linear scales: $\frac{X - (-15)}{65 - (-15)} = \frac{F - 32}{212 - 32}$.
For $X = -95$: $\frac{-80}{80} = -1 = \frac{F - 32}{180}$, so $F = 32 - 180 = -148$°F.
Answer: (D). | ||||
| 46 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation $g = \frac{GM}{R^2} = \frac{4}{3}\pi G\rho R$, so $g \propto \rho R$.
$\frac{g_A}{g_B} = \frac{\rho \cdot R}{\frac{\rho}{3} \cdot 4R} = \frac{3}{4}$.
Answer: (C). | ||||
| 47 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Areas under the graph:
0–5 s: $\frac{1}{2} \times 5 \times 10 = 25$ m; 5–10 s: $5 \times 10 = 50$ m; 10–15 s: $\frac{1}{2}(10 + 20) \times 5 = 75$ m; 15–20 s: $\frac{1}{2} \times 5 \times 20 = 50$ m; 20–25 s: $\frac{1}{2} \times 5 \times 20 = 50$ m below the axis (moving backwards).
Distance $= 25 + 50 + 75 + 50 + 50 = 250$ m; displacement $= 200 - 50 = 150$ m.
Ratio $= \frac{250}{150} = \frac{5}{3}$.
Answer: (C). | ||||
| 48 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion At the point of slipping, friction supplies the centripetal force: $\mu mg = m\omega^2 r$, so $r = \frac{\mu g}{\omega^2} \propto \frac{1}{\omega^2}$.
Halving $\omega$ makes r four times: $4 \times 1 = 4$ cm.
Answer: (C). | ||||
| 49 | Laws of MotionNewton's laws | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on newton's laws, from Laws of Motion. NCERT: Class 11: Laws of Motion Force = rate of change of momentum = $nmv$, where n is the number of bullets per second.
$n = \frac{F}{mv} = \frac{125}{0.01 \times 250} = 50$.
Answer: (C). | ||||
| 50 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | StatementsEasy | |
A easy statements question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Statement I: all three are units of astronomical distance. Correct.
Statement II: 1 AU $\approx 1.5 \times 10^{11}$ m, 1 ly $\approx 9.46 \times 10^{15}$ m, 1 pc $\approx 3.08 \times 10^{16}$ m (about 3.26 ly). So AU < ly < pc, not AU < pc < ly. Incorrect.
Answer: (C). | ||||
| 51 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Six lines: $\frac{n(n - 1)}{2} = 6$, so the atoms are excited to $n = 4$.
Energy absorbed: $13.6\left(1 - \frac{1}{16}\right) = 12.75$ eV.
$f = \frac{E}{h} = \frac{12.75}{4.25 \times 10^{-15}} = 3 \times 10^{15}$ Hz.
$x = 3$.
Answer: 3. | ||||
| 52 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics By the lens maker's formula the power is proportional to $(\mu_{rel} - 1)$ for the same surfaces.
In air: $1.8 - 1 = 0.8$. In the liquid: $\frac{1.8}{1.5} - 1 = 0.2$.
Ratio $= \frac{0.8}{0.2} = 4$, i.e. 4 : 1.
Answer: 4. | ||||
| 53 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current From the graph, B rises linearly from 0, reaching 4 mT at 2 s and 8 mT at 4 s: $\frac{dB}{dt} = \frac{8 - 4}{4 - 2} = 2$ mT/s.
$|\varepsilon| = A\frac{dB}{dt} = 4 \times 2 \times 10^{-3} = 8 \times 10^{-3}$ V $= 8$ mV.
Answer: 8. | ||||
| 54 | Current ElectricityCells, emf and internal resistance | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on cells, emf and internal resistance, from Current Electricity. NCERT: Class 12: Current Electricity In the figure the two cells are joined with their positive (longer) plates facing each other, so their emfs oppose: net emf $= 8 - 4 = 4$ V.
Total resistance: $r_1 + r_2 + 4.5 + (3 \parallel 6) = 0.5 + 1 + 4.5 + 2 = 8$ Ω.
Current from the cells: $I = \frac{4}{8} = 0.5$ A.
It divides between 3 Ω and 6 Ω in the ratio 2 : 1: current in 3 Ω $= 0.5 \times \frac{6}{9} = \frac{1}{3}$ A.
$\frac{x}{3} = \frac{1}{3}$, so $x = 1$.
Answer: 1. | ||||
| 55 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The lower charge sits at the foot of the incline (inclination 30°); the upper charge is on the incline at height h, so their separation along the incline is $r = \frac{h}{\sin 30^\circ} = 2h$.
For the upper charge, along the incline: $\frac{kq_0^2}{r^2} = mg\sin 30^\circ$, i.e. $\frac{kq_0^2}{4h^2} = \frac{mg}{2}$.
$h^2 = \frac{kq_0^2}{2mg} = \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{2 \times 0.02 \times 10} = \frac{0.036}{0.4} = 0.09$, so $h = 0.3$ m $= 300 \times 10^{-3}$ m.
$x = 300$.
Answer: 300. | ||||
| 56 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\omega = 2\pi \times 160$ rad/s and $k = 2\pi \times 0.5$ rad/m, so $v = \frac{\omega}{k} = \frac{160}{0.5} = 320$ m/s.
In km/h: $320 \times \frac{18}{5} = 1152$ km h⁻¹.
Answer: 1152. | ||||
| 57 | Properties of Bulk MatterElasticity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalMedium | |
A medium numerical question on elasticity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter By Hooke's law the extension is proportional to the tension: $l_1 = L + 100c$, $l_2 = L + 120c$, where L is the natural length.
$10l_2 = 11l_1$: $10L + 1200c = 11L + 1100c$, so $L = 100c$.
Then $l_1 = 200c = 2L$, so $L = \frac{1}{2}l_1$ and $x = 2$.
Answer: 2. | ||||
| 58 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion About a diameter: $I_d = \frac{2}{5}MR^2 = \frac{2}{5} \times 0.5 \times (0.05)^2 = 5 \times 10^{-4}$ kg m². Angular momentum $L = I_d\omega = 5 \times 10^{-3}$ kg m² s⁻¹.
About a tangent (parallel axes): $I_t = \frac{7}{5}MR^2 = 1.75 \times 10^{-3}$ kg m².
$\frac{I_t}{L} = \frac{1.75 \times 10^{-3}}{5 \times 10^{-3}} = 0.35 = 35 \times 10^{-2}$ (in SI units).
$x = 35$.
Answer: 35. | ||||
| 59 | 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 $W = \int_0^4 (2 + 3x)\,dx = \left[2x + \frac{3x^2}{2}\right]_0^4 = 8 + 24 = 32$ J.
Answer: 32. | ||||
| 60 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane The two times at which a projectile is at the same height are the roots of $u\sin\theta\,t - \frac{1}{2}gt^2 = h$; their sum is the time of flight: $t_1 + t_2 = \frac{2u\sin\theta}{g}$.
$3 + 5 = \frac{2u \times \frac{1}{2}}{10} = \frac{u}{10}$, so $u = 80$ m s⁻¹.
Answer: 80. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | Assertion and reasonMedium | |
A medium assertion and reason question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom A: Photoemission is instantaneous — there is no time lag once light above the threshold frequency falls on the surface. Correct.
R: A photon gives its energy to an electron in a single, all-or-nothing event, and an electron bound in an atom takes up a photon only when the photon's energy suits a permitted transition (or exceeds the binding energy, as in photoemission). A photon of "any" energy does not transfer its energy to the electron. Incorrect.
Answer: (C). | ||||
| 62 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Matching listEasy | |
A easy matching list question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure A. $\text{H}_3\text{O}^+$: 3 bond pairs + 1 lone pair on O — pyramidal (III).
B. Acetylide $\text{C}_2^{2-}$ ($[\text{C}\equiv\text{C}]^{2-}$): sp carbons — linear (II).
C. $\text{NH}_4^+$: 4 bond pairs — tetrahedral (I).
D. $\text{ClO}_2^-$: 2 bond pairs + 2 lone pairs on Cl — bent (IV).
Answer: (A). | ||||
| 63 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Moles: $\text{Ag}^+$ = 25 mmol, $\text{I}^-$ = 26.25 mmol, $\text{K}^+$ = 26.25 mmol, $\text{NO}_3^-$ = 25 mmol, in 50 mL.
AgI precipitates almost completely: $\text{I}^-$ left over = 1.25 mmol (0.025 M), and $\text{Ag}^+$ remains only to the extent allowed by $K_{sp}$ of AgI (about $10^{-17}/0.025$ M — negligible).
$\text{K}^+$ (0.525 M) and $\text{NO}_3^-$ (0.5 M) stay in solution in large amounts.
So the ions present in very small quantity are $\text{Ag}^+$ and $\text{I}^-$.
Answer: (B). | ||||
| 64 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across period 2 the ionisation enthalpy generally rises, with two exceptions: B < Be (the 2p electron of B is easier to remove than a 2s electron of Be) and O < N (N has a stable half-filled 2p subshell).
Values (kJ/mol): Li 520, B 801, Be 899, C 1086, O 1314, N 1402, F 1681.
Order: Li < B < Be < C < O < N < F.
Answer: (B). | ||||
| 65 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter These are the self-reduction steps in the Bessemer converter. The copper formed solidifies with a blistered surface as the dissolved $\text{SO}_2$ escapes, and is called blister copper (about 98% pure). (Copper matte is the $\text{Cu}_2\text{S}$–FeS mixture fed to the converter.)
Answer: (C). | ||||
| 66 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: $\text{CH}_4 + \text{H}_2\text{O} \xrightarrow[\text{Ni}]{1270\text{ K}} \text{CO} + 3\text{H}_2$ (steam reforming, water-gas). Correct.
Statement II: $\text{NH}_4\text{Cl} + \text{NaNO}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O} + \text{NaCl}$, the laboratory preparation of nitrogen. Correct.
Answer: (A). | ||||
| 67 | s-Block ElementsBiological importance of sodium, potassium, magnesium and calcium | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on biological importance of sodium, potassium, magnesium and calcium, from s-Block Elements. NCERT: No matching NCERT chapter A. K: $\text{K}^+$ (with $\text{Na}^+$) is moved across cell membranes by the sodium–potassium pump (III).
B. KCl: used as a potash fertilizer (II).
C. KOH: absorbs $\text{CO}_2$ ($2\text{KOH} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O}$) (IV).
D. Li: lithium (as $^6$Li, a source of tritium) is used in thermonuclear reactions (I).
Answer: (D). | ||||
| 68 | 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 $\text{GaAlCl}_4$ is the salt $\text{Ga}^+[\text{AlCl}_4]^-$: Ga is in the +1 state (inert pair effect), and all four Cl are bonded to Al in the tetrahedral $[\text{AlCl}_4]^-$ anion.
(A) Ga is +1, not +3 — false. (B) Ga is not coordinated by Cl; it is the free cation — false. (D) Cl is bonded only to Al — false.
(C) Ga (1.81) is more electronegative than Al (1.61) — the d-block contraction — and Ga is the cation. True.
Answer: (C). | ||||
| 69 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Facial and meridional isomers arise for octahedral complexes of the type $[\text{MA}_3\text{B}_3]$. $[\text{Co(NH}_3)_3(\text{NO}_2)_3]$ is of this type, so it has a mer isomer.
$[\text{Co(en)}_3]$ shows only optical isomerism; $[\text{Co(en)}_2\text{Cl}_2]$ shows cis/trans (and optical); square planar $[\text{Pt(NH}_3)_2\text{Cl}_2]$ shows cis/trans.
Answer: (A). | ||||
| 70 | Coordination CompoundsNomenclature | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on nomenclature, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Ambidentate ligands can bind through either of two different atoms: $\text{NO}_2^-$ (N or O) and $\text{NCS}^-$ (N or S).
Sets (A), (B) and (D) each contain $\text{NCS}^-$ or $\text{NO}_2^-$. Set (C) — oxalate (bidentate through two O), ethylenediamine (bidentate through two N) and water — has none.
Answer: (C). | ||||
| 71 | Environmental ChemistryAir and water pollution | No matching NCERT chapter | StatementsEasy | |
A easy statements question on air and water pollution, from Environmental Chemistry. NCERT: No matching NCERT chapter Statement I: clean water has BOD below 5 ppm, and dissolved oxygen around 8–10 ppm (fish need more than about 6 ppm). BOD 4 ppm with DO 8 ppm is good water. Correct.
Statement II: the permissible limits in drinking water are about 5 ppm for zinc and 50 ppm for nitrate, so 5 ppm of each is acceptable. Correct.
Answer: (A). | ||||
| 72 | 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 On the TLC plate A has travelled farthest from the base line, then C, then B: $R_f$(A) > $R_f$(C) > $R_f$(B).
A higher $R_f$ means weaker adsorption on silica (less polar), and the least strongly adsorbed component is eluted first from a silica gel column.
Order of elution: A, C, B.
Answer: (B). | ||||
| 73 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons c (1-tetralone): the C=O is conjugated with the ring and withdraws electrons — deactivated, slowest.
a (1-tetralol): the OH is on the saturated ring; the ring carries only alkyl substituents — weakly activated.
b (5,6,7,8-tetrahydro-1-naphthol): OH directly on the aromatic ring (+R) — strongly activated.
d (chroman-8-ol): OH and the ring ether O are both on the aromatic ring — two +R donors, fastest.
Increasing rate: c < a < b < d.
Answer: (A). | ||||
| 74 | 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 (I) 4-Methoxybenzyl chloride: the p-OMe group (+R) strongly stabilises the benzyl carbocation, so substitution goes by $S_N1$ — first order.
(II) 4-Nitrobenzyl chloride: the p-$\text{NO}_2$ group (−R, −I) destabilises the carbocation, so $S_N1$ is suppressed and the reaction goes by $S_N2$ — second order.
Answer: (A). | ||||
| 75 | 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 Hot $\text{KMnO}_4$ cleaves the cyclic ketone next to the carbonyl, on the side of the more substituted α-carbon (the one carrying R). That carbon becomes a ketone carbonyl and the original C=O becomes $\text{CO}_2\text{H}$:
A = $\text{R–CO–(CH}_2)_4\text{–CO}_2\text{H}$ (a keto acid).
Wolff–Kishner reduction ($\text{NH}_2\text{NH}_2$, KOH, then $\text{H}_3\text{O}^+$) turns the ketone C=O into $\text{CH}_2$ and leaves the acid:
B = $\text{R–CH}_2\text{–(CH}_2)_4\text{–CO}_2\text{H}$.
Answer: (C). | ||||
| 76 | Principles of Qualitative AnalysisCharacteristic precipitates and colours | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on characteristic precipitates and colours, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter (A) Basic iron(III) acetate complex: the deep red colour seen in the acetate test with neutral $\text{FeCl}_3$ is this complex in solution — it is soluble (on boiling it hydrolyses to a brown-red precipitate).
(B) $\text{K}_3[\text{Co(NO}_2)_6]$: the yellow precipitate of the test for $\text{K}^+$ — insoluble.
(C) $\text{Fe}_4[\text{Fe(CN)}_6]_3$: Prussian blue precipitate — insoluble.
(D) Ammonium arsenomolybdate: the yellow precipitate of the test for arsenic — insoluble.
Answer: (A). | ||||
| 77 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines One $\text{NH}_2$ of o-phenylenediamine is diazotised by $\text{HNO}_2$. The neighbouring $\text{NH}_2$ then attacks the diazonium nitrogen intramolecularly, closing a five-membered ring of three N atoms with loss of $\text{H}^+$.
X = 1H-benzotriazole.
Answer: (B). | ||||
| 78 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Polymerising ethene with the Ziegler–Natta catalyst ($\text{Et}_3\text{Al} + \text{TiCl}_4$) at low pressure gives linear, closely packed chains: high density polythene.
(Low density polythene is made at high pressure with a free-radical initiator and is branched.)
Answer: (C). | ||||
| 79 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctHard | |
A hard single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules A positive Schiff's test needs a free aldehyde group, so X is an open-chain aldotetrose (this rules out the two cyclic forms). The aldotetroses have two chiral carbons; L means the OH on the lowest chiral carbon (C3) is on the left.
Its three OH groups give a triacetate, as stated.
$\text{NaBH}_4$ reduces CHO to $\text{CH}_2\text{OH}$. L-Erythrose (both OH on the left) would give erythritol, which has a mirror plane (meso) and is achiral. L-Threose (OH right at C2, left at C3) gives L-threitol, which is chiral.
Since B is chiral, X is L-threose. ($\text{HNO}_3$ oxidises it to the chiral tartaric acid A.)
Answer: (C). | ||||
| 80 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter Brown ring with freshly prepared $\text{FeSO}_4$ and conc. $\text{H}_2\text{SO}_4$: $[\text{Fe(H}_2\text{O)}_5\text{NO}]^{2+}$ — the test for nitrate.
Deep red colour with neutral $\text{FeCl}_3$ that gives a brown-red precipitate of basic iron(III) acetate on boiling: the test for acetate.
The mixture contains $\text{CH}_3\text{COO}^-$ and $\text{NO}_3^-$.
Answer: (B). | ||||
| 81 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Sugar: $0.25 \times 200 + 0.40 \times 500 = 50 + 200 = 250$ g in $700$ g of solution.
Mass % $= \frac{250}{700} \times 100 = 35.71\% \approx 36$.
Answer: 36. | ||||
| 82 | Solid StatePacking in fcc, bcc and hcp | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on packing in fcc, bcc and hcp, from Solid State. NCERT: No matching NCERT chapter $\rho = \frac{ZM}{a^3N_A}$, so $Z = \frac{\rho a^3N_A}{M}$, with $a = 300$ pm $= 3 \times 10^{-8}$ cm, $a^3 = 2.7 \times 10^{-23}$ cm³.
$Z = \frac{3.0 \times 2.7 \times 10^{-23} \times 6.02 \times 10^{23}}{12} = \frac{48.76}{12} = 4.06 \approx 4$.
Answer: 4. | ||||
| 83 | ThermodynamicsEnthalpy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on enthalpy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$ (the 1 : 2 mole ratio).
$\Delta H = -1700 - (-840) = -860$ kJ per mole of $\text{Fe}_2\text{O}_3$.
Mass of mixture: $160 + 2 \times 27 = 214$ g.
Heat per gram $= \frac{860}{214} = 4.02 \approx 4$ kJ.
Answer: 4. | ||||
| 84 | Solutionsvan't Hoff factor | Class 12: Solutions | NumericalEasy | |
A easy numerical question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions Isotonic: $i \times 0.004 = 0.01$, so $i = 2.5$.
$\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-}$: $i = 1 + (3 - 1)\alpha = 1 + 2\alpha = 2.5$, $\alpha = 0.75$.
Percentage dissociation $= 75$.
Answer: 75. | ||||
| 85 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium 40% of the water reacts: 0.4 mol. At equilibrium: CO 0.6, $\text{H}_2\text{O}$ 0.6, $\text{CO}_2$ 0.4, $\text{H}_2$ 0.4 mol (in 10 L; the volume cancels since $\Delta n = 0$).
$K_c = \frac{0.4 \times 0.4}{0.6 \times 0.6} = \frac{4}{9} = 0.444$.
$K_c \times 10^2 = 44.4 \approx 44$.
Answer: 44. | ||||
| 86 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Use free energies, which add: $\Delta G^\circ = -nFE^\circ$.
$\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$: $\Delta G_1 = -2Fm$.
$\text{Pb}^{4+} + 4e^- \rightarrow \text{Pb}$: $\Delta G_2 = -4Fn$.
$\text{Pb}^{4+} + 2e^- \rightarrow \text{Pb}^{2+}$ = (2nd) − (1st): $\Delta G_3 = -4Fn + 2Fm$, so $E^\circ_{(\text{Pb}^{4+}/\text{Pb}^{2+})} = \frac{-\Delta G_3}{2F} = 2n - m$.
The couple written $\text{Pb}^{2+}/\text{Pb}^{4+}$ (the oxidation $\text{Pb}^{2+} \rightarrow \text{Pb}^{4+}$) has the opposite sign: $E^\circ = m - 2n$.
$x = 2$.
Answer: 2. | ||||
| 87 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Rate of disappearance of $\text{FeSO}_4$: $\frac{10 - 8.8}{30 \times 60} = \frac{1.2}{1800} = 6.67 \times 10^{-4}$ mol L⁻¹ s⁻¹.
By stoichiometry, $\frac{d[\text{Fe}_2(\text{SO}_4)_3]}{dt} = \frac{3}{6} \times 6.67 \times 10^{-4} = 3.33 \times 10^{-4} = 333 \times 10^{-6}$ mol L⁻¹ s⁻¹.
Answer: 333. | ||||
| 88 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Both contain $\text{Cr}^{3+}$, $3d^3$: in an octahedral field the three electrons occupy $t_{2g}^3$ whatever the ligand strength, so both have 3 unpaired electrons.
$\mu = \sqrt{3(3 + 2)} = \sqrt{15}$ BM for each; the ratio is 1.
Answer: 1. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalHard | |
A hard numerical question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Protonation of the OH and loss of water gives a secondary carbocation on the five-membered ring carbon that carried the OH. The neighbouring ring-fusion carbon carries the methyl group and no H, so a 1,2-methyl shift moves the positive charge to the ring-fusion carbon, giving a more stable tertiary carbocation; $\text{Br}^-$ then attacks it to give A.
Hyperconjugation structures = number of C–H bonds on the carbons adjacent to the cationic carbon:
the former carbinol carbon (now CH bearing the methyl): 1 H; the other ring-fusion carbon (CH): 1 H; the $\text{CH}_2$ of the six-membered ring: 2 H.
Total $= 1 + 1 + 2 = 4$.
Answer: 4. | ||||
| 90 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalEasy | |
A easy numerical question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The substrate, 4-hydroxybutanal, has an acidic O–H. The first mole of MeMgBr is used up by it (giving $\text{CH}_4$ and the alkoxide); the second adds to the aldehyde C=O. Hydrolysis then gives pentane-1,4-diol, the product shown.
So 2 mol of MeMgBr are needed per mole of substrate: $\frac{x}{y} = 2$.
Answer: 2. | ||||
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