JEE (Main) · 2024
JEE (Main) 2024 — 31 January, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Limits, Continuity and Differentiability3
- Straight Lines3
- Integral Calculus3
13 easy · 15 medium · 2 hard
Physics
30 Q
- Current Electricity3
- Physical World and Measurement2
- Laws of Motion2
24 easy · 6 medium · 0 hard
Chemistry
30 Q
- Hydrocarbons4
- Amines3
- Chemical Bonding and Molecular Structure2
16 easy · 14 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Let $g(x) = x^3 - 3x + 1$. Then $g'(x) = 3(x^2 - 1) > 0$ for $x < -1$, so $g$ (and hence $f = e^{g}$) is increasing on $(-\infty, -1]$.
As $x \to -\infty$, $g \to -\infty$ and $f \to 0$; at $x = -1$, $g(-1) = -1 + 3 + 1 = 3$ and $f(-1) = e^3$.
So the range is $(0, e^3]$; for $f$ to be onto, $a = 0$, $b = e^3$.
Then $P = (2e^3 + 4, 2)$. Its distance from $x + e^{-3}y - 4 = 0$ is
$\frac{|2e^3 + 4 + 2e^{-3} - 4|}{\sqrt{1 + e^{-6}}} = \frac{2(e^3 + e^{-3})}{\sqrt{1 + e^{-6}}}$.
Multiplying numerator and denominator by $e^3$: $\frac{2(e^6 + 1)}{\sqrt{e^6 + 1}} = 2\sqrt{1 + e^6}$.
Answer: (C). | ||||
| 2 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Put $t = e^{\sin x} > 0$. The equation becomes $t - \frac{2}{t} = 2$, i.e. $t^2 - 2t - 2 = 0$, so $t = 1 \pm \sqrt{3}$.
Only $t = 1 + \sqrt{3} \approx 2.732$ is positive.
But $-1 \le \sin x \le 1$ gives $e^{-1} \le t \le e \approx 2.718$, and $1 + \sqrt{3} > e$.
So no real $x$ satisfies the equation: the number of solutions is 0.
Answer: (A). | ||||
| 3 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $s = z_1 + z_2 = 5$ and $p = z_1 z_2$.
$z_1^3 + z_2^3 = s^3 - 3ps = 125 - 15p = 20 + 15i$, so $15p = 105 - 15i$ and $p = 7 - i$.
$z_1^2 + z_2^2 = s^2 - 2p = 25 - 14 + 2i = 11 + 2i$.
$z_1^4 + z_2^4 = (z_1^2 + z_2^2)^2 - 2p^2 = (121 - 4 + 44i) - 2(49 - 1 - 14i) = (117 + 44i) - (96 - 28i) = 21 + 72i$.
$|21 + 72i| = \sqrt{441 + 5184} = \sqrt{5625} = 75$.
Answer: (D). | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Let $P = \begin{pmatrix}1&-1&0\\0&0&1\\1&1&0\end{pmatrix}$ (the three given column vectors) and $D = \begin{pmatrix}2&0&0\\0&4&0\\0&0&2\end{pmatrix}$. The three conditions together say $AP = PD$.
$\det P = 1(0 - 1) - (-1)(0 - 1) + 0 = -1 - 1 = -2 \ne 0$, so $P$ is invertible and $A = PDP^{-1}$.
Then $A - 3I = P(D - 3I)P^{-1}$, so $\det(A - 3I) = \det(D - 3I) = (2 - 3)(4 - 3)(2 - 3) = 1 \ne 0$.
A coefficient matrix with non-zero determinant gives exactly one solution: the system has a unique solution.
Answer: (C). | ||||
| 5 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Give each child 2 apples first; $21 - 6 = 15$ identical apples remain, to be shared freely among 3 children.
The number of non-negative integer solutions of $y_1 + y_2 + y_3 = 15$ is $\binom{15 + 2}{2} = \binom{17}{2} = 136$.
Answer: (B). | ||||
| 6 | Permutations and CombinationsP(n, r) and C(n, r) | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on p(n, r) and c(n, r), from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Using Pascal's rule twice: ${}^6C_m + 2\,{}^6C_{m+1} + {}^6C_{m+2} = ({}^6C_m + {}^6C_{m+1}) + ({}^6C_{m+1} + {}^6C_{m+2}) = {}^7C_{m+1} + {}^7C_{m+2} = {}^8C_{m+2}$.
So ${}^8C_{m+2} > {}^8C_3 = 56$. The values of ${}^8C_r$ are $1, 8, 28, 56, 70, 56, 28, 8, 1$, so $m + 2 = 4$ and $m = 2$ (also $m + 2 \le 6$ is needed for ${}^6C_{m+2}$).
${}^nP_4 = n \cdot {}^{n-1}P_3$, so ${}^{n-1}P_3 : {}^nP_4 = 1 : n = 1 : 8$ and $n = 8$.
${}^nP_{m+1} + {}^{n+1}C_m = {}^8P_3 + {}^9C_2 = 336 + 36 = 372$.
Answer: (A). | ||||
| 7 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series With first term 1 and common difference $d \ne 0$: $T_2 = 1 + d$, $T_8 = 1 + 7d$, $T_{44} = 1 + 43d$.
In a G.P. the middle term squared equals the product of its neighbours: $(1 + 7d)^2 = (1 + d)(1 + 43d)$.
$1 + 14d + 49d^2 = 1 + 44d + 43d^2 \Rightarrow 6d^2 = 30d \Rightarrow d = 5$ (as $d \ne 0$).
$S_{20} = \frac{20}{2}[2 \cdot 1 + 19 \cdot 5] = 10 \times 97 = 970$.
Answer: (B). | ||||
| 8 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability For $x \ge 1$, $\log_e x \ge 0$ and $f(x) = e^{-\log_e x} = \frac{1}{x}$. For $0 < x < 1$, $|\log_e x| = -\log_e x$ and $f(x) = e^{\log_e x} = x$.
Both pieces are continuous and they agree at $x = 1$ (value 1), so $f$ is continuous everywhere: $m = 0$.
At $x = 1$: left derivative $= 1$, right derivative $= -\frac{1}{x^2}\big|_{x=1} = -1$, so $f$ is not differentiable there; elsewhere it is. $n = 1$.
$m + n = 1$.
Answer: (B). | ||||
| 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 Read $f : \mathbb{R} \to (0, \infty)$. For $x > 0$, $x < 5x < 7x$, and $f$ is strictly increasing, so $f(x) < f(5x) < f(7x)$.
Dividing by $f(x) > 0$: $1 < \frac{f(5x)}{f(x)} < \frac{f(7x)}{f(x)}$.
As $x \to \infty$ the right side tends to 1, so by the sandwich theorem $\frac{f(5x)}{f(x)} \to 1$.
Hence $\lim_{x \to \infty}\left[\frac{f(5x)}{f(x)} - 1\right] = 0$.
Answer: (D). | ||||
| 10 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines The centre of the circle is $(8, 2)$.
Let the line be $\frac{x}{a} + \frac{y}{b} = 1$ with $a, b > 0$, so $OA = a$, $OB = b$. It passes through $(8, 2)$: $\frac{8}{a} + \frac{2}{b} = 1$.
By the Cauchy–Schwarz inequality, $(a + b)\left(\frac{8}{a} + \frac{2}{b}\right) \ge (\sqrt{8} + \sqrt{2})^2 = (3\sqrt{2})^2 = 18$.
So $a + b \ge 18$, with equality when $\frac{a}{\sqrt 8} = \frac{b}{\sqrt 2}$, i.e. $a = 2b$: then $\frac{8}{2b} + \frac{2}{b} = 1$ gives $b = 6$, $a = 12$.
Minimum of $OA + OB = 18$.
Answer: (B). | ||||
| 11 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctEasy | |
A easy single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersections: $3(4x - x^2) = (x - 4)^2 \Rightarrow 12x - 3x^2 = x^2 - 8x + 16 \Rightarrow 4x^2 - 20x + 16 = 0 \Rightarrow x = 1, 4$.
On $[1, 4]$ the parabola $y = 4x - x^2$ lies above (at $x = 2$: $4$ against $\frac{4}{3}$).
Difference: $4x - x^2 - \frac{(x - 4)^2}{3} = \frac{-4x^2 + 20x - 16}{3} = \frac{4}{3}(x - 1)(4 - x)$.
Area $= \frac{4}{3}\int_1^4 (x - 1)(4 - x)\,dx = \frac{4}{3}\cdot\frac{(4 - 1)^3}{6} = \frac{4}{3}\cdot\frac{27}{6} = 6$.
Answer: (B). | ||||
| 12 | 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 The integrand of $f$ is even, so $f(x) = 2\int_0^x (t - t^2)e^{-t^2}dt$.
In $g$ put $t = u^2$ ($dt = 2u\,du$): $g(x) = \int_0^x u \cdot e^{-u^2}\cdot 2u\,du = 2\int_0^x u^2 e^{-u^2}du$.
Adding, the $t^2$ terms cancel: $f(x) + g(x) = 2\int_0^x t e^{-t^2}dt = \left[-e^{-t^2}\right]_0^x = 1 - e^{-x^2}$.
At $x = \sqrt{\log_e 9}$: $e^{-x^2} = \frac{1}{9}$, so $f + g = \frac{8}{9}$, and $9(f + g) = 8$.
Answer: (A). | ||||
| 13 | Differential EquationsSeparation of variables | Class 12: Differential Equations | Single correctEasy | |
A easy single correct question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separating: $\frac{dT}{T - 80} = -K\,dt \Rightarrow T - 80 = (160 - 80)e^{-Kt} = 80e^{-Kt}$.
$T(15) = 120$: $40 = 80e^{-15K}$, so $e^{-15K} = \frac{1}{2}$.
$T(45) = 80 + 80\left(e^{-15K}\right)^3 = 80 + 80 \cdot \frac{1}{8} = 90^\circ$ F.
Answer: (C). | ||||
| 14 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections The axis of the parabola is the perpendicular from the vertex $V(2, 3)$ to the directrix $2x + y - 6 = 0$. Its foot is
$N = (2, 3) - \frac{2 \cdot 2 + 3 - 6}{2^2 + 1^2}(2, 1) = (2, 3) - \frac{1}{5}(2, 1) = \left(\frac{8}{5}, \frac{14}{5}\right)$.
The vertex is the midpoint of $N$ and the focus $S$, so $S = 2V - N = \left(\frac{12}{5}, \frac{16}{5}\right)$.
For the ellipse, $e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2}$, so $b^2 = \frac{a^2}{2}$ and $E$ is $x^2 + 2y^2 = a^2$.
Through $S$: $a^2 = \frac{144}{25} + \frac{2 \cdot 256}{25} = \frac{656}{25}$.
Latus rectum $= \frac{2b^2}{a} = \frac{a^2}{a} = a$, so its square is $a^2 = \frac{656}{25}$.
Answer: (C). | ||||
| 15 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines The centroid divides the segment from the circumcentre to the orthocentre in the ratio $1 : 2$:
$G = \frac{2B + C}{3} = \left(\frac{6 - 6}{3}, \frac{8 - 8}{3}\right) = (0, 0)$, so $a = b = 0$ and $P = (3, 5)$.
The line $x - 2y - 1 = 0$ has direction $(2, 1)$, of length $\sqrt{5}$. Points along it from $P$: $(3 + 2t, 5 + t)$.
On $2x + 3y - 4 = 0$: $6 + 4t + 15 + 3t - 4 = 0 \Rightarrow 7t = -17 \Rightarrow t = -\frac{17}{7}$.
Distance $= |t|\sqrt{5} = \frac{17\sqrt{5}}{7}$.
Answer: (B). | ||||
| 16 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line is $F = (1 + 2t, 2 + 3t, 3 + 4t)$. For the foot of the perpendicular from $P(2, 3, 5)$:
$(P - F)\cdot(2, 3, 4) = 2(1 - 2t) + 3(1 - 3t) + 4(2 - 4t) = 13 - 29t = 0 \Rightarrow t = \frac{13}{29}$.
$F = \left(\frac{55}{29}, \frac{97}{29}, \frac{139}{29}\right)$, and the image is $2F - P = \left(\frac{52}{29}, \frac{107}{29}, \frac{133}{29}\right)$.
$2\alpha + 3\beta + 4\gamma = \frac{104 + 321 + 532}{29} = \frac{957}{29} = 33$.
Answer: (C). | ||||
| 17 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $L_2$ has direction $\vec{AB} = (3, 2, 0)$; it is indeed perpendicular to $(-2, 3, 1)$, since $-6 + 6 + 0 = 0$.
$L_1$ passes through $(1, -1, -4)$ with direction $(2, -3, 2)$.
$\vec{n} = (2, -3, 2) \times (3, 2, 0) = (-3 \cdot 0 - 2 \cdot 2,\; 2 \cdot 3 - 2 \cdot 0,\; 2 \cdot 2 - (-3) \cdot 3) = (-4, 6, 13)$, $|\vec{n}| = \sqrt{16 + 36 + 169} = \sqrt{221}$.
Joining vector: $A - (1, -1, -4) = (-5, 5, 7)$; its dot product with $\vec{n}$: $20 + 30 + 91 = 141$.
Shortest distance $= \frac{141}{\sqrt{221}}$.
Answer: (D). | ||||
| 18 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean: $a + b + 220 = 6 \times 55 = 330$, so $a + b = 110$.
Variance: $\frac{\sum x^2}{6} - 55^2 = 194$, so $\sum x^2 = 6 \times 3219 = 19314$.
$68^2 + 44^2 + 48^2 + 60^2 = 4624 + 1936 + 2304 + 3600 = 12464$, so $a^2 + b^2 = 6850$.
$2ab = 110^2 - 6850 = 5250$, and $(a - b)^2 = 6850 - 5250 = 1600$, so $a - b = 40$.
$a = 75$, $b = 35$, and $a + 3b = 75 + 105 = 180$.
Answer: (A). | ||||
| 19 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability $P(H) = 2P(T)$ and $P(H) + P(T) = 1$, so $P(H) = \frac{2}{3}$, $P(T) = \frac{1}{3}$.
The tosses are independent; the head can come in any of 3 positions:
$P = 3 \times \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{6}{27} = \frac{2}{9}$.
Answer: (D). | ||||
| 20 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $5$ radians lies between $\frac{3\pi}{2} \approx 4.71$ and $2\pi \approx 6.28$.
$5 - 2\pi \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\sin(5 - 2\pi) = \sin 5$, so $a = 5 - 2\pi$.
$2\pi - 5 \in [0, \pi]$ and $\cos(2\pi - 5) = \cos 5$, so $b = 2\pi - 5$.
$a^2 + b^2 = 2(2\pi - 5)^2 = 2(4\pi^2 - 20\pi + 25) = 8\pi^2 - 40\pi + 50$.
Answer: (D). | ||||
| 21 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalEasy | |
A easy numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $2x = 3y$ means $x = 3k$, $y = 2k$ for a natural number $k$; with $x \le 100$, $k = 1, 2, \ldots, 33$. So $R$ has 33 pairs $(3k, 2k)$.
A symmetric relation containing $R$ must also contain every $(2k, 3k)$: 33 more pairs, none of them already in $R$ (that would need $3k = 2k$).
The smallest such $R_1$ is $R$ together with these pairs, so the minimum $n = 66$.
Answer: 66. | ||||
| 22 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For a $3 \times 3$ matrix, $\det(\text{adj}\,M) = (\det M)^2$.
Applying adj 2024 times squares the determinant 2024 times: $n = 2^{2^{2024}}$.
Since $2^6 = 64 \equiv 1 \pmod 9$, we need $2^{2024} \bmod 6$. It is even, and $2^{2024} = 4^{1012} \equiv 1 \pmod 3$, so $2^{2024} \equiv 4 \pmod 6$.
Hence $n \equiv 2^4 = 16 \equiv 7 \pmod 9$.
Answer: 7. | ||||
| 23 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem The sum is a G.P. of $n$ terms with first term $(x + 3)^{n-1}$ and ratio $\frac{x + 2}{x + 3}$:
$\sum_{k=0}^{n-1}(x + 3)^{n-1-k}(x + 2)^k = \frac{(x + 3)^n - (x + 2)^n}{(x + 3) - (x + 2)} = (x + 3)^n - (x + 2)^n$.
The sum of all its coefficients is its value at $x = 1$: $\sum_{r=0}^{n}\alpha_r = 4^n - 3^n$.
So $\beta = 4$, $\gamma = 3$ and $\beta^2 + \gamma^2 = 25$.
Answer: 25. | ||||
| 24 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Regroup: $(a^2x^2 - 2abx + b^2) + (b^2x^2 - 2bcx + c^2) = (ax - b)^2 + (bx - c)^2 = 0$.
For real $x$ both squares vanish: $x = \frac{b}{a} = \frac{c}{b}$, so the sides are $a, ax, ax^2$ ($x > 0$).
Triangle inequalities (divide by $a$): $1 + x > x^2$, $x + x^2 > 1$, $1 + x^2 > x$ (always true).
$x^2 - x - 1 < 0 \Rightarrow x < \frac{1 + \sqrt{5}}{2}$; $x^2 + x - 1 > 0 \Rightarrow x > \frac{\sqrt{5} - 1}{2}$.
So $\alpha = \frac{\sqrt{5} - 1}{2}$, $\beta = \frac{\sqrt{5} + 1}{2}$, and $\alpha^2 + \beta^2 = \frac{(6 - 2\sqrt{5}) + (6 + 2\sqrt{5})}{4} = 3$.
$12(\alpha^2 + \beta^2) = 36$.
Answer: 36. | ||||
| 25 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability The denominator behaves like $x^3$. Expand the numerator up to $x^3$:
$ax^2e^x = ax^2 + ax^3 + \ldots$; $-b\log_e(1 + x) = -bx + \frac{b}{2}x^2 - \frac{b}{3}x^3 + \ldots$; $cxe^{-x} = cx - cx^2 + \frac{c}{2}x^3 - \ldots$
For the limit to be 1: coefficient of $x$: $c - b = 0$; of $x^2$: $a + \frac{b}{2} - c = 0$; of $x^3$: $a - \frac{b}{3} + \frac{c}{2} = 1$.
So $c = b$, $a = \frac{b}{2}$, and $\frac{b}{2} - \frac{b}{3} + \frac{b}{2} = \frac{2b}{3} = 1$, giving $b = c = \frac{3}{2}$, $a = \frac{3}{4}$.
$16(a^2 + b^2 + c^2) = 16\left(\frac{9}{16} + \frac{9}{4} + \frac{9}{4}\right) = 9 + 36 + 36 = 81$.
Answer: 81. | ||||
| 26 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | NumericalHard | |
A hard numerical question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $f(x) = \frac{\sin x\cos x}{\sin^4 x + \cos^4 x}$. Then $f(\pi - x) = -f(x)$ and $f\left(\frac{\pi}{2} - x\right) = f(x)$.
First, $K = \int_0^{\pi/2} f\,dx$: divide by $\cos^4 x$ and put $t = \tan^2 x$, $dt = 2\tan x\sec^2 x\,dx$: $K = \frac{1}{2}\int_0^{\infty}\frac{dt}{1 + t^2} = \frac{\pi}{4}$.
$L = \int_0^{\pi/2} x f\,dx$: replacing $x$ by $\frac{\pi}{2} - x$ gives $2L = \frac{\pi}{2}K$, so $L = \frac{\pi^2}{16}$.
Now $I = \int_0^{\pi} x^2 f\,dx$. Replace $x$ by $\pi - x$: $I = -\int_0^{\pi}(\pi - x)^2 f\,dx = -\pi^2\int_0^{\pi} f + 2\pi\int_0^{\pi} x f - I$.
$\int_0^{\pi} f = 0$ (odd about $\frac{\pi}{2}$), so $I = \pi\int_0^{\pi} x f\,dx$.
Split at $\frac{\pi}{2}$ and put $x = \pi - u$ on the second half: $\int_0^{\pi} x f = \int_0^{\pi/2}(2x - \pi) f\,dx = 2L - \pi K = \frac{\pi^2}{8} - \frac{\pi^2}{4} = -\frac{\pi^2}{8}$.
So $I = -\frac{\pi^3}{8}$ and $\left|\frac{120}{\pi^3} I\right| = \frac{120}{8} = 15$.
Answer: 15. | ||||
| 27 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalHard | |
A hard numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Put $u = \tan x$, $du = \sec^2 x\,dx$: $\frac{du}{dy} = -(e^{2y}u^2 + u)$.
Divide by $-u^2$ and put $v = \frac{1}{u}$ (so $\frac{dv}{dy} = -\frac{1}{u^2}\frac{du}{dy}$): $\frac{dv}{dy} = v + e^{2y}$, i.e. $\frac{dv}{dy} - v = e^{2y}$.
Integrating factor $e^{-y}$: $\frac{d}{dy}(ve^{-y}) = e^{y}$, so $ve^{-y} = e^{y} + C$ and $\cot x = e^{2y} + Ce^{y}$.
At $x = \frac{\pi}{4}$, $y = 0$: $1 = 1 + C$, so $C = 0$ and $\cot x = e^{2y}$.
At $x = \frac{\pi}{6}$: $e^{2\alpha} = \sqrt{3}$, so $e^{8\alpha} = (\sqrt{3})^4 = 9$.
Answer: 9. | ||||
| 28 | Straight LinesSection formula | Class 11: Straight Lines | NumericalEasy | |
A easy numerical question on section formula, from Straight Lines. NCERT: Class 11: Straight Lines The diagonals of a parallelogram bisect each other: $A + C = B + D$.
$(-2 + \alpha, -1 + \beta) = (1 + \gamma, \delta)$, so $\alpha = \gamma + 3$, $\beta = \delta + 1$.
$C$ on $2x - y = 5$: $2(\gamma + 3) - (\delta + 1) = 5 \Rightarrow 2\gamma - \delta = 0$.
$D$ on $3x - 2y = 6$: $3\gamma - 2\delta = 6$. With $\delta = 2\gamma$: $-\gamma = 6$, $\gamma = -6$, $\delta = -12$.
Then $\alpha = -3$, $\beta = -11$, and $|\alpha + \beta + \gamma + \delta| = |-32| = 32$.
Answer: 32. | ||||
| 29 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalEasy | |
A easy numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $\vec{AB} = (12, 4, 6)$, $|\vec{AB}| = 14$, unit vector $\frac{1}{7}(6, 2, 3)$.
Points at distance 21 from $A$: $A \pm 3(6, 2, 3)$, i.e. $(22, 0, 7)$ or $(-14, -12, -11)$. Only $(22, 0, 7)$ has non-negative coordinates.
$PQ = \sqrt{(22 - 4)^2 + (0 + 12)^2 + (7 - 3)^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$.
Answer: 22. | ||||
| 30 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalMedium | |
A medium numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a} + \vec{b} = (5, 1, 4)$; $\vec{a} \times \vec{b} = (2 \cdot 3 - 1 \cdot(-1),\; 1 \cdot 2 - 3 \cdot 3,\; 3 \cdot(-1) - 2 \cdot 2) = (7, -7, -7)$.
So $(\vec{a} + \vec{b}) \times \vec{c} = (14, -14, -14) + (0, 24, -6) = (14, 10, -20)$.
With $\vec{c} = (x, y, z)$: $(5, 1, 4) \times (x, y, z) = (z - 4y,\; 4x - 5z,\; 5y - x) = (14, 10, -20)$.
So $z = 14 + 4y$, $x = 5y + 20$ (the middle equation then holds automatically).
$\vec{a} - \vec{b} + \hat{i} = (2, 3, -2)$: $2(5y + 20) + 3y - 2(14 + 4y) = 5y + 12 = -3$, so $y = -3$, $x = 5$, $z = 2$.
$|\vec{c}|^2 = 25 + 9 + 4 = 38$.
Answer: 38. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Only like quantities can be subtracted, so $[B] = [x^2] = L^2$.
Then $[A] = \frac{[B]}{[E][t]} = \frac{L^2}{ML^2T^{-2}\cdot T} = M^{-1}T$.
$[AB] = M^{-1}T \cdot L^2 = L^2M^{-1}T^1$.
Answer: (B). | ||||
| 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 $\left|\vec{A} + \vec{B}\right|^2 = R^2 + R^2 + 2R^2\cos\theta = 2R^2(1 + \cos\theta) = 4R^2\cos^2\frac{\theta}{2}$, so $\left|\vec{A} + \vec{B}\right| = 2R\cos\frac{\theta}{2}$ (C true, A false).
$\left|\vec{A} - \vec{B}\right|^2 = 2R^2(1 - \cos\theta) = 4R^2\sin^2\frac{\theta}{2}$, so $\left|\vec{A} - \vec{B}\right| = 2R\sin\frac{\theta}{2}$ (B and D false).
Answer: (C). | ||||
| 33 | 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 For two masses over a light smooth fixed pulley, $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$ (taking $m_1$ as the heavier).
$\frac{m_1 - m_2}{m_1 + m_2} = \frac{1}{8} \Rightarrow 8m_1 - 8m_2 = m_1 + m_2 \Rightarrow 7m_1 = 9m_2$.
$\frac{m_1}{m_2} = \frac{9}{7}$.
Answer: (D). | ||||
| 34 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Take both forces along the incline ($\theta = 30^\circ$, $\mu = 0.1$, $m = 5$ kg).
Just moving up: friction acts down the plane, $F_1 = mg\sin\theta + \mu mg\cos\theta$.
Just preventing sliding down: friction acts up the plane, $F_2 = mg\sin\theta - \mu mg\cos\theta$.
$F_1 - F_2 = 2\mu mg\cos\theta = 2 \times 0.1 \times 50 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ N $\approx 8.66$ N.
This value is not among the options: (A), $\frac{5\sqrt{3}}{2}$ N, is only $\mu mg\cos\theta$, half the difference, and nothing in the question as printed gives it.
No option is correct as printed. | ||||
| 35 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $\vec{a} = \frac{\vec{F}}{m} = 3t\hat{i} + 3t^2\hat{j}$. Starting from rest, $\vec{v} = \int_0^t \vec{a}\,dt = \frac{3t^2}{2}\hat{i} + t^3\hat{j}$.
$P = \vec{F}\cdot\vec{v} = 6t \cdot \frac{3t^2}{2} + 6t^2 \cdot t^3 = 9t^3 + 6t^5$ W.
Answer: (D). | ||||
| 36 | GravitationEscape velocity | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation $v_e = \sqrt{\frac{2GM}{R}}$, so $\frac{v_{moon}}{v} = \sqrt{\frac{M_m}{M_p}\cdot\frac{R_p}{R_m}} = \sqrt{\frac{1}{144}\times 16} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
$v_{moon} = \frac{v}{3}$.
Answer: (D). | ||||
| 37 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter With the medium's density negligible there is no buoyancy, so at terminal velocity the weight balances Stokes' drag: $mg = 6\pi\eta r v$.
Hence $v = \frac{mg}{6\pi\eta r} \propto \frac{1}{r}$ for the same mass.
Doubling the radius halves the terminal velocity: $\frac{v}{2}$.
Answer: (A). | ||||
| 38 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $v = \sqrt{\frac{\gamma RT}{M}}$ with $T = 273$ K and $M = 32$ g/mol $= 0.032$ kg/mol.
$v = \sqrt{\frac{1.4 \times 8.3 \times 273}{0.032}} = \sqrt{\frac{3172.26}{0.032}} = \sqrt{99133} \approx 315$ m/s.
Of the options, 310 m/s is the closest.
Answer: (D). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Argon is monatomic (3 degrees of freedom): $U_1 = 8 \times \frac{3}{2}RT = 12RT$.
Oxygen is diatomic; without vibration it has 5 degrees of freedom: $U_2 = 6 \times \frac{5}{2}RT = 15RT$.
Total $U = 27RT$.
Answer: (B). | ||||
| 40 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $F = \frac{q_1q_2}{4\pi\varepsilon_0 r^2}$. In the medium at distance $\frac{r}{5}$: $F' = \frac{q_1q_2}{4\pi\varepsilon_0 K (r/5)^2} = \frac{25}{K}F = \frac{25}{5}F = 5F$.
Answer: (B). | ||||
| 41 | 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 Illumination is proportional to the power $P = I^2R$, with the lamp's resistance taken as constant.
New current $= 0.8I$, so new power $= 0.64P$: a decrease of $36\%$.
Answer: (A). | ||||
| 42 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Area $A = 0.20 \times 0.10 = 0.02$ m$^2$, so $|\vec{m}| = IA = 5 \times 0.02 = 0.1$ A m$^2$.
Seen from the negative $X$ side the current is anticlockwise, so by the right-hand rule the area vector points towards that viewer: $\vec{m} = -0.1\hat{i}$ A m$^2$.
$\vec{\tau} = \vec{m} \times \vec{B} = (-0.1\hat{i}) \times (2 \times 10^{-3}\hat{j}) = -2 \times 10^{-4}\hat{k}$ N m.
Magnitude $2 \times 10^{-4}$ N m, along the negative $Z$-direction.
Answer: (B). | ||||
| 43 | 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 $P_{avg} = V_{rms}I_{rms}\cos\phi = \frac{V_0I_0}{2}\cos\phi$ with phase difference $\phi = \frac{\pi}{3}$.
$P_{avg} = \frac{20 \times 10}{2} \times \frac{1}{2} = 50$ W.
Answer: (C). | ||||
| 44 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | StatementsEasy | |
A easy statements question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Statement I: the average energy densities of the two fields are equal, $\frac{1}{2}\varepsilon_0E^2 = \frac{B^2}{2\mu_0}$ (since $E = cB$), so the energy is shared equally. Correct.
Statement II: an electromagnetic wave carries momentum ($p = U/c$), so it exerts radiation pressure on a surface it strikes. Correct.
Both statements are correct.
Answer: (A). | ||||
| 45 | OpticsPolarisation | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on polarisation, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Complete polarization of the reflected ray means $60^\circ$ is the Brewster angle, at which the reflected and refracted rays are perpendicular: $i_p + r = 90^\circ$.
$r = 90^\circ - 60^\circ = 30^\circ$.
Answer: (A). | ||||
| 46 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter The new frequency is $\frac{1.5\nu_0}{2} = 0.75\nu_0$, below the threshold frequency $\nu_0$.
Below threshold no photoelectrons are emitted, whatever the intensity. The number is zero.
Answer: (A). | ||||
| 47 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Nuclear radius $R = R_0A^{1/3}$, so $A \propto R^3$.
Half the radius gives $\frac{1}{8}$ the mass number: $A = \frac{192}{8} = 24$.
Answer: (C). | ||||
| 48 | Electronic DevicesLogic gates | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on logic gates, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits Trace the wiring: the upper OR gate gets $A$ and $\bar{B}$ (B through a NOT gate); the lower OR gate gets $B$ and $\bar{A}$ (the line from A crosses the B line and passes through a NOT gate).
Upper output: $A + \bar{B}$. Lower output: $B + \bar{A}$.
The NOR gate gives $Y = \overline{(A + \bar{B}) + (B + \bar{A})} = \overline{(A + \bar{A}) + (B + \bar{B})} = \overline{1} = 0$.
So $Y = 0$ for every input: the truth table with all outputs 0, option (B).
Answer: (B). | ||||
| 49 | Current ElectricityMetre bridge | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on metre bridge, from Current Electricity. NCERT: Class 12: Current Electricity Balance condition: $\frac{X}{25} = \frac{r \times 40}{r \times 60} = \frac{40}{60}$; the resistance per cm cancels.
With a wire of $2r$ per cm the ratio is still $\frac{l}{100 - l}$, so the balance point stays at 40 cm.
Answer: (D). | ||||
| 50 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $g = \frac{4\pi^2L}{T^2}$, so $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$.
$\frac{\Delta L}{L} = \frac{0.2}{20} = 0.01$. For the time, $\frac{\Delta t}{t} = \frac{1}{40}$ (the same fraction for one period).
$\frac{\Delta g}{g} = 0.01 + 2 \times 0.025 = 0.06 = 6\%$. $N = 6$.
Answer: (A). | ||||
| 51 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion At the highest point the velocity is horizontal, $u\cos 45^\circ = \frac{u}{\sqrt{2}}$, at height $H = \frac{u^2\sin^2 45^\circ}{2g} = \frac{u^2}{4g}$.
About the point of projection, $L = m \cdot \frac{u}{\sqrt{2}} \cdot H = \frac{mu^3}{4\sqrt{2}g} = \frac{\sqrt{2}mu^3}{8g}$.
So $X = 8$.
Answer: 8. | ||||
| 52 | Motion of System of Particles and Rigid BodyMoment of inertia | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on moment of inertia, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Each sphere (taken solid) has its centre $0.75$ m from the axis. By the parallel axis theorem:
$I_1 = \frac{2}{5}MR^2 + Md^2 = \frac{2}{5}(2)(0.5)^2 + 2(0.75)^2 = 0.2 + 1.125 = 1.325$ kg m$^2$.
For both: $I = 2.65$ kg m$^2 = \frac{53}{20}$ kg m$^2$, so $x = 53$.
Answer: 53. | ||||
| 53 | 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 Tension in the wire: $T = \frac{2m_1m_2g}{m_1 + m_2} = \frac{2 \times 2 \times 4 \times 10}{6} = \frac{80}{3}$ N.
Cross-section: $A = \pi r^2 = \pi \times 16 \times 10^{-10}$ m$^2$.
Strain $= \frac{T}{AY} = \frac{80/3}{16\pi \times 10^{-10} \times 2 \times 10^{11}} = \frac{80/3}{320\pi} = \frac{80}{960\pi} = \frac{1}{12\pi}$.
So $\alpha = 12$.
Answer: 12. | ||||
| 54 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | NumericalMedium | |
A medium numerical question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves From the figure: two springs $k$ hang side by side from the ceiling and are joined below to a third spring $k$ that goes to the block; a fourth spring $k$ goes straight from the ceiling to the block.
The two side-by-side springs are in parallel: $2k$. In series with $k$: $\frac{2k \cdot k}{3k} = \frac{2k}{3}$.
This branch is in parallel with the single spring: $k_{eq} = \frac{2k}{3} + k = \frac{5k}{3}$.
$T = 2\pi\sqrt{\frac{M}{k_{eq}}} = 2\pi\sqrt{\frac{3M}{5k}} = \pi\sqrt{\frac{12M}{5k}}$, so $\alpha = 12$.
Answer: 12. | ||||
| 55 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance From the figure, along a line through $O$: $+2q$ at $-2l$, $-q$ at $-l$, $+q$ at $+l$, $-2q$ at $+2l$; $OP$ makes $60^\circ$ with the line, measured towards $+q$.
Dipole 1 ($\pm q$): moment $2ql$, pointing from $-q$ to $+q$, i.e. along $+x$.
Dipole 2 ($\pm 2q$): moment $2q \times 4l = 8ql$, pointing from $-2q$ to $+2q$, i.e. along $-x$.
Net moment $p = 6ql$ along $-x$, making $180^\circ - 60^\circ = 120^\circ$ with $OP$.
For $r \gg l$: $V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos 120^\circ}{r^2} = 9 \times 10^9 \times \frac{6ql \times (-\frac{1}{2})}{r^2} = -27\frac{ql}{r^2} \times 10^9$ V.
So $\alpha = 27$.
Answer: 27. | ||||
| 56 | 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 The left and right sides of the square are plain wires, so the network has two terminals: left (L) and right (R).
Between them: the top $2\,\Omega$, the bottom $2\,\Omega$, and the two diagonals, each two $2\,\Omega$ resistors meeting at the centre. By symmetry the centre is at the mid potential, so each diagonal path adds L–centre–R; the four diagonal halves form $1\,\Omega + 1\,\Omega = 2\,\Omega$ between L and R.
Three $2\,\Omega$ paths in parallel: $\frac{2}{3}\,\Omega$.
Total resistance with the internal resistance: $\frac{2}{3} + \frac{2}{3} = \frac{4}{3}\,\Omega$, so $I = \frac{2}{4/3} = 1.5$ A.
Power consumed in the entire circuit (including the cell's internal resistance): $P = EI = 2 \times 1.5 = 3$ W.
Answer: 3. | ||||
| 57 | Magnetic Effects of Current and MagnetismBiot-Savart law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on biot-savart law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Field at the centre of a coil: $B = \frac{\mu_0NI}{2R}$ with $R = \pi$ cm $= 0.01\pi$ m.
$B_P = \frac{4\pi \times 10^{-7} \times 100 \times 1}{2 \times 0.01\pi} = 2 \times 10^{-3}$ T $= 2$ mT; $B_Q = 4$ mT (current 2 A).
The planes are perpendicular, so the fields are perpendicular: $B = \sqrt{2^2 + 4^2} = \sqrt{20}$ mT. $x = 20$.
Answer: 20. | ||||
| 58 | 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 $\varepsilon = -\frac{d\phi}{dt} = -(10t - 36)$. At $t = 2$ s: $\varepsilon = -(20 - 36) = 16$ V.
$I = \frac{16}{8} = 2$ A.
Answer: 2. | ||||
| 59 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Refraction at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$ with $u = -100$ cm, $R = +20$ cm, $\mu_1 = 1$, $\mu_2 = 1.5$.
$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20} = \frac{1}{40} \Rightarrow \frac{1.5}{v} = \frac{1}{40} - \frac{1}{100} = \frac{3}{200} \Rightarrow v = 100$ cm.
The image is 100 cm inside the medium, on the other side of the surface; its distance from the object is $100 + 100 = 200$ cm.
Answer: 200. | ||||
| 60 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | NumericalEasy | |
A easy numerical question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei $R = R_0A^{1/3}$, so the volume $V = \frac{4}{3}\pi R^3 \propto A$.
$\frac{V_2}{V_1} = \frac{A_2}{A_1} = 4$.
Answer: 4. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Some Basic Concepts of ChemistryStoichiometry | Class 11: Some Basic Concepts of Chemistry | Single correctMedium | |
A medium single correct question on stoichiometry, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry On ignition $\mathrm{CaCO_3 \to CaO + CO_2}$ and $\mathrm{MgCO_3 \to MgO + CO_2}$, leaving the oxides.
Let the mixture hold $x$ g $\mathrm{CaCO_3}$ and $(2.21 - x)$ g $\mathrm{MgCO_3}$. Residue: $\frac{56}{100}x + \frac{40}{84}(2.21 - x) = 1.152$.
$0.56x + 1.0524 - 0.47619x = 1.152 \Rightarrow 0.08381x = 0.0996 \Rightarrow x \approx 1.188$ g.
So about 1.187 g $\mathrm{CaCO_3}$ and 1.023 g $\mathrm{MgCO_3}$. Check: $0.56 \times 1.187 + \frac{40}{84}\times 1.023 = 0.665 + 0.487 = 1.152$ g. (Options C and D do not even add up to 2.21 g.)
Answer: (B). | ||||
| 62 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom K: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$. The outermost electron is in 4s: $n = 4$, $l = 0$, $m = 0$, $s = +\frac{1}{2}$.
Answer: (C). | ||||
| 63 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure By Fajans' rules, covalent character grows with the polarising power of the cation.
$\mathrm{K^+}$ and $\mathrm{Ba^{2+}}$ are large ions with a noble-gas core and polarise little: $\mathrm{KCl}$ and $\mathrm{BaCl_2}$ are largely ionic.
$\mathrm{Ag^+}$ has a pseudo-noble-gas ($4d^{10}$) configuration; the poorly shielding d-electrons make it strongly polarising, so $\mathrm{AgCl}$ has marked covalent character (it is insoluble in water, unlike the other chlorides).
$\mathrm{AgCl}$ is taken as the least ionic.
Answer: (B). | ||||
| 64 | EquilibriumChemical equilibrium | Class 11: Equilibrium | Single correctMedium | |
A medium single correct question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Start with 1 mol A. At equilibrium: A $= 1 - \alpha$, B $= \alpha$, C $= \frac{\alpha}{2}$; total $= 1 + \frac{\alpha}{2} = \frac{2 + \alpha}{2}$.
Partial pressures: $p_A = \frac{2(1 - \alpha)}{2 + \alpha}P$, $p_B = \frac{2\alpha}{2 + \alpha}P$, $p_C = \frac{\alpha}{2 + \alpha}P$.
$K_P = \frac{p_B\,p_C^{1/2}}{p_A} = \frac{\frac{2\alpha P}{2 + \alpha}\left(\frac{\alpha P}{2 + \alpha}\right)^{1/2}}{\frac{2(1 - \alpha)P}{2 + \alpha}} = \frac{\alpha^{3/2}P^{1/2}}{(2 + \alpha)^{1/2}(1 - \alpha)}$.
Answer: (A). | ||||
| 65 | Redox ReactionsOxidation number | Class 11: Redox Reactions | StatementsMedium | |
A medium statements question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions Statement I: $\mathrm{S_8 + 12OH^- \to 4S^{2-} + 2S_2O_3^{2-} + 6H_2O}$. Sulphur goes from 0 to $-2$ and to $+2$: a disproportionation. Correct.
Statement II: in $\mathrm{ClO_4^-}$ chlorine is already in its highest oxidation state, $+7$; it cannot be oxidised further, so it cannot disproportionate. Incorrect.
Answer: (C). | ||||
| 66 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsMedium | |
A medium statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties A' and B' are in one period (B' to the right); C' and D' are directly below them in the next period.
Across a period size and metallic character decrease; down a group they increase. So B' < A', D' < C', and each lower element is larger than the one above it (A' < C', B' < D').
A. B' < A' < D' < C': true (taking the increase down a group to outweigh the decrease across one step, as usual).
B. Metallic character follows the same order: true.
C. D' < C' < B' < A' puts the upper elements as larger: false.
D. Cations of the same charge follow the same trend as the atoms: B'$^+$ < A'$^+$ < D'$^+$ < C'$^+$: true.
So A, B and D only.
Answer: (A). | ||||
| 67 | p-Block ElementsGroup 13 elements | No matching NCERT chapter | StatementsEasy | |
A easy statements question on group 13 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: the trihalides of group 13 are covalent and electron-deficient, and are readily hydrolysed by water. True.
Statement II: in acidified aqueous solution $\mathrm{AlCl_3}$ gives the octahedral $\mathrm{[Al(H_2O)_6]^{3+}}$ ion (Al uses its vacant d-orbitals, $sp^3d^2$). True.
Answer: (A). | ||||
| 68 | p-Block ElementsGroup 16 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 16 elements, from p-Block Elements. NCERT: No matching NCERT chapter A. The group 16 elements form $\mathrm{EO_2}$ and $\mathrm{EO_3}$ oxides, and both types are described as acidic. Taken as true.
B. The reducing property of the dioxides falls from $\mathrm{SO_2}$ to $\mathrm{TeO_2}$: $\mathrm{SO_2}$ is reducing, $\mathrm{TeO_2}$ is an oxidising agent. True.
C. Down the group the H–E bond weakens, so the reducing property increases from $\mathrm{H_2S}$ to $\mathrm{H_2Te}$. False.
D. In $\mathrm{O_3}$ the central O has one lone pair, one terminal O has two and the other three: six lone pairs in all. False.
So A and B only.
Answer: (B). | ||||
| 69 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | StatementsMedium | |
A medium statements question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements A. $\mathrm{Mn_2O_7}$ is a green oil at room temperature. True.
B. $\mathrm{V_2O_4}$ (V in +4) dissolves in acids to give the vanadyl ion, $\mathrm{VO^{2+}}$. Taken as true (see note).
C. CrO, with Cr in its lowest oxidation state +2, is basic. True.
D. $\mathrm{V_2O_5}$ is amphoteric: it reacts with acids (giving $\mathrm{VO_2^+}$ salts) as well as with alkalis. False.
So A, B and C only.
Answer: (B). | ||||
| 70 | Coordination CompoundsValence bond theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on valence bond theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{[Ni(CO)_4]}$: Ni(0), $3d^84s^2$; the strong-field CO pairs all electrons into $3d^{10}$, $sp^3$ tetrahedral. No unpaired electrons: diamagnetic.
$\mathrm{[NiCl_4]^{2-}}$: $\mathrm{Ni^{2+}}$, $3d^8$; weak-field $\mathrm{Cl^-}$ does not pair the electrons, $sp^3$ tetrahedral with 2 unpaired electrons: paramagnetic.
Answer: (B). | ||||
| 71 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Matching listEasy | |
A easy matching list question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Water is a weak-field ligand, so the high-spin arrangement applies where there is a choice.
A. $\mathrm{Cr^{3+}}$, $d^3$: $t_{2g}^3e_g^0$ (II).
B. $\mathrm{Fe^{3+}}$, $d^5$ high spin: $t_{2g}^3e_g^2$ (III).
C. $\mathrm{Ni^{2+}}$, $d^8$: $t_{2g}^6e_g^2$ (IV).
D. $\mathrm{V^{3+}}$, $d^2$: $t_{2g}^2e_g^0$ (I).
A→II, B→III, C→IV, D→I.
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 Substances that are steam volatile and immiscible with water are separated by steam distillation: the mixture boils when the sum of the two vapour pressures reaches atmospheric pressure, i.e. below the oil's own boiling point, so the oil distils without decomposing.
Steam distillation.
Answer: (D). | ||||
| 73 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{-CH_3}$ (B) activates the ring (+I and hyperconjugation): B > A (benzene).
$\mathrm{-Cl}$ (C) deactivates mildly (−I outweighs +R): A > C.
$\mathrm{-NO_2}$ (D) deactivates strongly (−I and −R): C > D.
Order: B > A > C > D.
Answer: (D). | ||||
| 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 Pentane chain C1–C5 with a phenyl on C1 and Br on C2 and C3: $\mathrm{C_6H_5-CH_2-CHBr-CHBr-CH_2-CH_3}$.
(A) has the bromines on C3 and C4 (3,4-dibromo-1-phenylpentane); (B) on C1 and C2; (D) on C1 and C3.
(C) has $\mathrm{C_6H_5CH_2-}$, then CHBr, CHBr and an ethyl end: the required compound.
Answer: (C). | ||||
| 75 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 1-Methylcyclopentene adds D–Cl by Markovnikov's rule: $\mathrm{D^+}$ adds to the ring CH carbon, giving the more stable tertiary carbocation at the carbon bearing $\mathrm{CH_3}$, which then takes $\mathrm{Cl^-}$.
Product: 1-chloro-2-deuterio-1-methylcyclopentane (Cl and $\mathrm{CH_3}$ on one carbon, D and H on the next). Options (C) and (D) put Cl on the wrong carbon.
(A) and (B) differ only in stereochemistry. Hydrogen-halide addition to such cycloalkenes goes mainly anti (D and Cl on opposite faces): in (A) D is up and Cl down, so (A) is the major product.
Answer: (A). | ||||
| 76 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Single correctMedium | |
A medium single correct question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Step (i)/(ii) with NaOH then HCl replaces Br by OH (nucleophilic substitution, then acidification of the phenoxide). An aryl C–Br bond is replaced this way only when electron-withdrawing nitro groups at the ortho/para positions activate it; with three of them (2,4,6) the replacement is easy.
(A) and (B) keep the Br in B and add OH elsewhere, which is not what NaOH does; (D) loses Br in the nitration step, which cannot happen.
(C) is the consistent sequence: nitration gives 2,4,6-trinitrobromobenzene (A), and NaOH then HCl gives 2,4,6-trinitrophenol, picric acid (B).
Answer: (C). | ||||
| 77 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons This is Friedel–Crafts acylation: anhydrous $\mathrm{AlCl_3}$ takes $\mathrm{Cl^-}$ from benzoyl chloride to give the acylium ion $\mathrm{C_6H_5CO^+}$, which substitutes into benzene.
$\mathrm{C_6H_6 + C_6H_5COCl \xrightarrow{AlCl_3} C_6H_5COC_6H_5 + HCl}$.
P is benzophenone.
Answer: (B). | ||||
| 78 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Benzene treated with carbon monoxide and HCl in the presence of anhydrous $\mathrm{AlCl_3}$/CuCl gives benzaldehyde: the Gattermann–Koch reaction.
(Etard: toluene with chromyl chloride; Stephen: nitrile with $\mathrm{SnCl_2}$/HCl; Rosenmund: acid chloride with $\mathrm{H_2}$/Pd–$\mathrm{BaSO_4}$.)
Answer: (C). | ||||
| 79 | AminesClassification and identification of amines | Class 12: Amines | StatementsMedium | |
A medium statements question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Statement I: sulphonation of aniline and heating at 453–473 K gives sulphanilic acid (p-aminobenzenesulphonic acid). It contains both N and S, so in Lassaigne's test it gives NaSCN, which with $\mathrm{Fe^{3+}}$ gives a blood-red colour. True.
Statement II: aniline, a Lewis base, forms a salt with the Lewis acid $\mathrm{AlCl_3}$; the nitrogen becomes positively charged and strongly deactivates the ring, so aniline does not undergo Friedel–Crafts reactions. True.
Answer: (A). | ||||
| 80 | AminesDiazonium salts | Class 12: Amines | Single correctMedium | |
A medium single correct question on diazonium salts, from Amines. NCERT: Class 12: Amines Sulphanilic acid is diazotised by $\mathrm{NaNO_2}$/acetic acid: X is the diazonium salt $\mathrm{HO_3S-C_6H_4-N_2^+}$.
It couples with 1-naphthylamine at the position para to $\mathrm{-NH_2}$ (C-4, in the same ring), the most activated free position.
Y is 4-(4-amino-1-naphthylazo)benzenesulphonic acid: the sulphonated benzene ring–N=N–naphthalene with $\mathrm{NH_2}$ para to the azo link in the same ring, option (A). (This red dye is the basis of the test for nitrite.)
Answer: (A). | ||||
| 81 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $\mu = q \times d$, so $q = \frac{1.2 \times 10^{-18}\text{ esu cm}}{1 \times 10^{-8}\text{ cm}} = 1.2 \times 10^{-10}$ esu.
As a fraction of the electronic charge ($4.8 \times 10^{-10}$ esu): $\frac{1.2 \times 10^{-10}}{4.8 \times 10^{-10}} = 0.25 = 2.5 \times 10^{-1}$.
The stem does not ask for a rounded value, so the exact 2.5 is kept; either nearest integer, 2 or 3, is also accepted.
Answer: 2.5. | ||||
| 82 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics For a reversible isothermal expansion $w = -2.303\,nRT\log\frac{V_2}{V_1}$.
$w = -2.303 \times 5 \times 8.314 \times 300 \times \log 10 = -2.303 \times 12471 = -28720.7$ J.
So $x \approx 28721$.
Answer: 28721. | ||||
| 83 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Mass of 1 L of solution $= 1000 \times 1.54 = 1540$ g; mass of $\mathrm{H_3PO_4}$ $= 0.70 \times 1540 = 1078$ g.
Moles $= \frac{1078}{98} = 11$; in 1 L this is 11 M.
Answer: 11. | ||||
| 84 | ElectrochemistryConductance | Class 12: Electrochemistry | NumericalMedium | |
A medium numerical question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry These are the values of the standard table of conductivities at 298.15 K:
$2.1 \times 10^3$ sodium (metallic conductor); $1.0 \times 10^{-16}$ glass (insulator); $1.2 \times 10$ graphite (conductor); 3.91 0.1 M HCl (electrolytic conductor); $1.5 \times 10^{-2}$ silicon (semiconductor); $1 \times 10^{-7}$ CuO (semiconductor); $1.0 \times 10^3$ iron (metallic conductor).
Conductors: sodium, graphite, 0.1 M HCl and iron, i.e. 4.
Answer: 4. | ||||
| 85 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalEasy | |
A easy numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The reaction is first order, so $t_{1/2} = 120$ min and $k = \frac{0.693}{120}$ min$^{-1}$.
$t_{90\%} = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303 \times 120}{0.693} \approx 398.8$ min (exactly $120 \times \frac{\ln 10}{\ln 2} = 398.6$ min).
To the nearest integer: 399 minutes.
Answer: 399. | ||||
| 86 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalEasy | |
A easy numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Half-reactions: $\mathrm{MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O}$ and $\mathrm{C_2O_4^{2-} \to 2CO_2 + 2e^-}$.
Overall: $\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O}$.
Per mole of $\mathrm{MnO_4^-}$: 8 mol $\mathrm{H^+}$.
Answer: 8. | ||||
| 87 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons 2-Methylbutane, $\mathrm{(CH_3)_2CH-CH_2-CH_3}$, has four kinds of H, giving four structural products:
1-chloro-2-methylbutane (chiral: 2 enantiomers); 2-chloro-2-methylbutane (achiral); 2-chloro-3-methylbutane (chiral: 2 enantiomers); 1-chloro-3-methylbutane (achiral).
Counting stereoisomers: $2 + 1 + 2 + 1 = 6$.
Answer: 6. | ||||
| 88 | AminesClassification and identification of amines | Class 12: Amines | NumericalEasy | |
A easy numerical question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Each acetylation replaces an H of $\mathrm{-NH_2}$ by $\mathrm{-COCH_3}$, adding $43 - 1 = 42$ g mol$^{-1}$.
Increase $= 192 - 108 = 84 = 2 \times 42$, so two amino groups are acetylated (e.g. phenylenediamine, $\mathrm{C_6H_4(NH_2)_2}$, M = 108).
Answer: 2. | ||||
| 89 | BiomoleculesNucleic acids and vitamins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on nucleic acids and vitamins, from Biomolecules. NCERT: Class 12: Biomolecules The fat-soluble vitamins A, D, E and K are stored in the liver and adipose tissue.
The water-soluble vitamins (B group and C) are excreted and must be taken regularly, except vitamin $\mathrm{B_{12}}$, which is stored in the body.
Stored: A, D, E, K and $\mathrm{B_{12}}$, i.e. 5.
Answer: 5. | ||||
| 90 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | NumericalEasy | |
A easy numerical question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter This is the chromyl chloride test: $\mathrm{K_2Cr_2O_7 + 4KCl + 6H_2SO_4 \to 2CrO_2Cl_2 + 6KHSO_4 + 3H_2O}$.
In $\mathrm{CrO_2Cl_2}$: $x + 2(-2) + 2(-1) = 0$, so $x = +6$.
Answer: 6. | ||||
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