JEE (Main) · 2025
JEE (Main) 2025 — 23 January, 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
22 Q
- Integral Calculus3
- Sets, Relations and Functions2
- Matrices and Determinants2
9 easy · 13 medium · 0 hard
Physics
13 Q
- Optics3
- Electromagnetic Induction and Alternating Currents2
- Electrostatics1
6 easy · 5 medium · 2 hard
Chemistry
24 Q
- Thermodynamics2
- Equilibrium2
- Hydrocarbons2
14 easy · 10 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 FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f(g(x)) = \log_e g(x)$ is defined exactly where $g(x)$ is defined and $g(x) > 0$.
Denominator: $2x^2 - 2x + 1$ has discriminant $4 - 8 < 0$, so it is always positive; $g$ is defined for every real $x$.
Numerator: $(x^2 - x + 1)^2 = x^4 - 2x^3 + 3x^2 - 2x + 1$, so the numerator is $(x^2 - x + 1)^2 + 1 > 0$.
Hence $g(x) > 0$ for all real $x$ and the domain of $f \circ g$ is $\mathbb{R}$.
Answer: (A) $\mathbb{R}$. | ||||
| 2 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Since $1 \sim 2$ and $2 \sim 3$, any equivalence relation containing $R$ puts $1, 2, 3$ in one class; the smallest one has classes $\{1, 2, 3\}$ and $\{4\}$.
It contains $3^2 + 1^2 = 10$ ordered pairs: all 9 pairs from $\{1,2,3\}$ and $(4, 4)$.
$R$ already has 3 of them, so $10 - 3 = 7$ pairs must be added: $(1,1), (2,2), (4,4), (2,1), (3,2), (1,3), (3,1)$.
Answer: (A) $7$. | ||||
| 3 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $\bar{z} = x + iy$. Then $3|\bar{z} - i| = |2\bar{z} + i|$:
$9\left(x^2 + (y-1)^2\right) = 4x^2 + (2y+1)^2 \Rightarrow 5x^2 + 5y^2 - 22y + 8 = 0 \Rightarrow x^2 + y^2 - \frac{22}{5}y + \frac{8}{5} = 0.$
So $\bar{z}$ lies on a circle with centre $\left(0, \frac{11}{5}\right)$, and $z$ (its mirror image in the real axis) on a circle with centre $C = \left(0, -\frac{11}{5}\right)$. Either way C is at distance $\frac{11}{5}$ from the x-axis.
Area of the triangle with vertices $(0,0)$, C, $(\alpha, 0)$ $= \frac{1}{2}|\alpha| \cdot \frac{11}{5} = 11 \Rightarrow |\alpha| = 10$, so $\alpha^2 = 100$.
Answer: (B) $100$. | ||||
| 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 Call the equations $E_1, E_2, E_3$. Replace them by $E_1 - E_2$, $E_2$, $E_3 - E_2$:
$E_1 - E_2:\ -x - 3y + 4z = -2$; $\quad E_3 - E_2:\ x + 3y + (2 - 2\lambda)z = 2$.
Adding these two: $(6 - 2\lambda)z = 0$.
Determinant: $\Delta = (6 - 2\lambda)\begin{vmatrix}\lambda & \lambda - 1\\ 1 & 3\end{vmatrix} = (6 - 2\lambda)(2\lambda + 1)$, which vanishes for $\lambda = 3$ or $\lambda = -\frac{1}{2}$.
$\lambda = 3$: the sum above reads $0 = 0$, and $E_3 - E_2$ is just $-(E_1 - E_2)$, so only two independent consistent equations remain: infinitely many solutions.
$\lambda = -\frac{1}{2}$: the sum gives $7z = 0$, so $z = 0$; but $E_3 + 2E_2$ gives $-6z = 16$, a contradiction: no solution.
So $\lambda = 3$ and $\lambda^2 + \lambda = 12$.
Answer: (D) $12$. | ||||
| 5 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For an invertible $M$, $\mathrm{adj}(M) = |M| M^{-1}$, so $\mathrm{adj}(A^{-1}) = |A^{-1}| A = \frac{A}{|A|}$ and $\mathrm{adj}(B^{-1}) = \frac{B}{|B|}$.
The inverse of $A\,X^{-1}B$ (with $X = \mathrm{adj}(A^{-1}) + \mathrm{adj}(B^{-1})$) is $B^{-1}XA^{-1}$:
$B^{-1}\left(\frac{A}{|A|} + \frac{B}{|B|}\right)A^{-1} = \frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}.$
With $B^{-1} = \frac{\mathrm{adj}B}{|B|}$ and $A^{-1} = \frac{\mathrm{adj}A}{|A|}$ this is $\frac{\mathrm{adj}(B) + \mathrm{adj}(A)}{|A||B|} = \frac{1}{|AB|}(\mathrm{adj}(B) + \mathrm{adj}(A))$.
Answer: (C). | ||||
| 6 | 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 $S_4 = \frac{1}{5}(S_8 - S_4) \Rightarrow 6S_4 = S_8$.
With $a = 3$: $S_4 = 2(6 + 3d) = 12 + 6d$ and $S_8 = 4(6 + 7d) = 24 + 28d$.
$72 + 36d = 24 + 28d \Rightarrow d = -6$.
$S_{20} = 10\,(2\cdot3 + 19\cdot(-6)) = 10(6 - 114) = -1080$.
Answer: (A) $-1080$. | ||||
| 7 | 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 DAUGHTER has 8 distinct letters, of which A, U, E are vowels.
All arrangements: $8! = 40320$. Vowels all together: treat them as one block, $6!\times3! = 720\times6 = 4320$.
Vowels never all together: $40320 - 4320 = 36000$.
Answer: (C) $36000$. | ||||
| 8 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability Count the 36 equally likely face pairs (faces counted with their multiplicities).
Sum 4: (1,3): $2\times2 = 4$; (2,2): $2\times2 = 4$; (3,1): $1\times1 = 1$; total 9.
Sum 5: (1,4): $2\times1 = 2$; (2,3): $2\times2 = 4$; (3,2): $1\times2 = 2$; (4,1): $1\times1 = 1$; total 9.
$P = \frac{9 + 9}{36} = \frac{1}{2}$.
Answer: (B) $\frac{1}{2}$. | ||||
| 9 | 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 PQ is horizontal ($y = 4$) of length 7, so the altitude from R is vertical and passes through the orthocentre: $a = 2$.
Area: $\frac{1}{2}\cdot7\cdot|b - 4| = 35 \Rightarrow b = 14$ or $b = -6$.
The altitude from P is perpendicular to QR $= (a + 2, b - 4) = (4, b - 4)$ and passes through O: $\overrightarrow{PO} = \left(-3, -\frac{6}{5}\right)$, so $-12 - \frac{6}{5}(b - 4) = 0 \Rightarrow b = -6$.
Centroid: $c = \frac{5 - 2 + 2}{3} = \frac{5}{3}$, $d = \frac{4 + 4 - 6}{3} = \frac{2}{3}$.
$c + 2d = \frac{5}{3} + \frac{4}{3} = 3$.
Answer: (D) $3$. | ||||
| 10 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $\alpha = \tan^{-1}\frac{5}{12}$, so $\cos\alpha = \frac{12}{13}$, $\sin\alpha = \frac{5}{13}$, and $0 < \alpha < \frac{\pi}{4}$.
The argument is $\cos x\cos\alpha + \sin x\sin\alpha = \cos(x - \alpha)$.
For $\frac{\pi}{2} \le x \le \frac{3\pi}{4}$, $x - \alpha$ lies between $\frac{\pi}{4}$ and $\frac{3\pi}{4}$, inside $[0, \pi]$, so $\cos^{-1}\cos(x - \alpha) = x - \alpha$.
Answer: (C) $x - \tan^{-1}\frac{5}{12}$. | ||||
| 11 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\cot10^\circ\cot70^\circ - 1 = \frac{\cos10^\circ\cos70^\circ - \sin10^\circ\sin70^\circ}{\sin10^\circ\sin70^\circ} = \frac{\cos80^\circ}{\sin10^\circ\sin70^\circ}$.
Multiplying by $\sin70^\circ$: $\frac{\cos80^\circ}{\sin10^\circ} = 1$, since $\cos80^\circ = \sin10^\circ$.
Answer: (B) $1$. | ||||
| 12 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections From the line, $y = \frac{3x + 12}{2}$; then $2(3x + 12) = 3x^2 \Rightarrow x^2 - 2x - 8 = 0 \Rightarrow x = 4$ or $-2$.
$A(4, 12)$, $B(-2, 3)$; the vertex is $O(0, 0)$.
For $\vec{OA} = (4, 12)$ and $\vec{OB} = (-2, 3)$: dot product $= -8 + 36 = 28$, cross product $= 4\cdot3 - 12\cdot(-2) = 36$.
$\tan\theta = \frac{36}{28} = \frac{9}{7}$ with $\theta$ acute (dot product positive), so $\theta = \tan^{-1}\frac{9}{7}$.
Answer: (C) $\tan^{-1}\left(\frac{9}{7}\right)$. | ||||
| 13 | Vector AlgebraComponents in two and three dimensions | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on components in two and three dimensions, from Vector Algebra. NCERT: Class 12: Vector Algebra Arc lengths are proportional to the angles, so $\angle AOB = 15^\circ$ and $\angle BOC = 75^\circ$.
Take unit radius, $\overrightarrow{OA} = \hat{i}$, $\overrightarrow{OC} = \hat{j}$, $\overrightarrow{OB} = \cos15^\circ\,\hat{i} + \sin15^\circ\,\hat{j}$.
$\hat{j} = \alpha\hat{i} + \beta(\cos15^\circ\hat{i} + \sin15^\circ\hat{j})$ gives $\beta = \frac{1}{\sin15^\circ}$ and $\alpha = -\cot15^\circ = -(2 + \sqrt{3})$.
$\sin15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}$, so $\sqrt{2}(\sqrt{3} - 1)\beta = \sqrt{2}(\sqrt{3} - 1)\cdot\frac{2\sqrt{2}}{\sqrt{3} - 1} = 4$.
$\alpha + 4 = 2 - \sqrt{3}$.
Answer: (C) $2 - \sqrt{3}$. | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line is $(3 + 7t,\ 2 - t,\ -1 - 2t)$. For the foot P, $\overrightarrow{QP}$ is perpendicular to $(7, -1, -2)$:
$\overrightarrow{QP} = (7t - 7,\ 5 - t,\ -2t)$, and $7(7t - 7) - (5 - t) - 2(-2t) = 54t - 54 = 0 \Rightarrow t = 1$, so $P = (10, 1, -3)$.
$\overrightarrow{QP} = (0, 4, -2)$ and $\overrightarrow{QR} = (-7, 1, 2)$: $\overrightarrow{QP}\cdot\overrightarrow{QR} = 0 + 4 - 4 = 0$, so the right angle is at Q.
Area $= \frac{1}{2}|QP|\,|QR| = \frac{1}{2}\sqrt{20}\,\sqrt{54} = \frac{1}{2}\sqrt{1080} = 3\sqrt{30}$.
Answer: (D) $3\sqrt{30}$. | ||||
| 16 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Left limit: $\frac{2}{x}\{\sin(k_1 + 1)x + \sin(k_2 - 1)x\} \to 2\{(k_1 + 1) + (k_2 - 1)\} = 2(k_1 + k_2)$. Setting it equal to 4: $k_1 + k_2 = 2$.
Right limit: $\frac{2}{x}\left[\log_e\left(1 + \frac{k_1x}{2}\right) - \log_e\left(1 + \frac{k_2x}{2}\right)\right] \to 2\left(\frac{k_1}{2} - \frac{k_2}{2}\right) = k_1 - k_2 = 4$.
So $k_1 = 3$, $k_2 = -1$ and $k_1^2 + k_2^2 = 10$.
Answer: (C) $10$. | ||||
| 17 | StatisticsMean, median and mode | Class 11: Statistics | Single correctEasy | |
A easy single correct question on mean, median and mode, from Statistics. NCERT: Class 11: Statistics Median $= l + \frac{\frac{N}{2} - cf}{f}\times h$ with $l = 12$, $h = 6$, $f = 12$, $cf = 18$:
$14 = 12 + \frac{\frac{N}{2} - 18}{12}\times6 \Rightarrow \frac{N}{2} - 18 = 4 \Rightarrow N = 44$.
Answer: (B) $44$. | ||||
| 18 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Write the integrand as $\left(\frac{x - 11}{x + 15}\right)^{-\frac{11}{13}}\frac{1}{(x + 15)^2}$.
Put $t = \frac{x - 11}{x + 15}$: $dt = \frac{26}{(x + 15)^2}dx$.
$I = \frac{1}{26}\int t^{-\frac{11}{13}}dt = \frac{1}{26}\cdot\frac{13}{2}t^{\frac{2}{13}} = \frac{1}{4}\left(\frac{x - 11}{x + 15}\right)^{\frac{2}{13}} + C$.
$I(37) - I(24) = \frac{1}{4}\left[\left(\frac{1}{2}\right)^{\frac{2}{13}} - \left(\frac{1}{3}\right)^{\frac{2}{13}}\right] = \frac{1}{4}\left(\frac{1}{4^{\frac{1}{13}}} - \frac{1}{9^{\frac{1}{13}}}\right)$.
So $b = 4$, $c = 9$ and $3(b + c) = 39$.
Answer: (C) $39$. | ||||
| 19 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t = \log_e x$, $dt = \frac{dx}{x}$; the limits become 2 and 4:
$I = \int_2^4 \frac{e^{1/(t^2 + 1)}}{e^{1/(t^2 + 1)} + e^{1/((6 - t)^2 + 1)}}dt$.
Replacing $t$ by $2 + 4 - t = 6 - t$ swaps the two exponentials, so adding the two forms gives $2I = \int_2^4 1\,dt = 2$.
$I = 1$.
Answer: (A) $1$. | ||||
| 20 | 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 $\frac{dy}{3 + y} = \frac{2e^{2x}}{7 + e^{2x}}dx \Rightarrow \ln(3 + y) = \ln(7 + e^{2x}) + C$.
At $(0, 5)$: $\ln 8 = \ln 8 + C$, so $C = 0$ and $3 + y = 7 + e^{2x}$, i.e. $y = 4 + e^{2x}$.
At $x = \log_e 2$: $k = 4 + 4 = 8$.
Answer: (B) $8$. | ||||
| 21 | 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 The coefficients add to $ab - ac + bc - ab + ac - bc = 0$, so $x = 1$ is a root; with equal roots both roots are 1.
Product of roots: $\frac{c(a - b)}{a(b - c)} = 1 \Rightarrow ca - bc = ab - ac \Rightarrow 2ac = b(a + c)$.
$2ac = \frac{36}{5}\times15 = 108 \Rightarrow ac = 54$.
$a^2 + c^2 = (a + c)^2 - 2ac = 225 - 108 = 117$.
Answer: 117. | ||||
| 22 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersections: $y = x - 1$ meets the circle where $x^2 - x - 12 = 0$, at $(4, 3)$; $y = 1 - x$ meets it at $(-3, 4)$.
The V-shaped graph cuts the disc into an upper part (above the V) and a lower part.
Upper part $= \int_{-3}^{4}\sqrt{25 - x^2}\,dx - \int_{-3}^{4}|x - 1|\,dx$.
$\int_{-3}^{4}\sqrt{25 - x^2}\,dx = \left[\frac{x}{2}\sqrt{25 - x^2} + \frac{25}{2}\sin^{-1}\frac{x}{5}\right]_{-3}^{4} = (6 + 6) + \frac{25}{2}\left(\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{3}{5}\right) = 12 + \frac{25\pi}{4}$.
$\int_{-3}^{4}|x - 1|\,dx = \frac{16}{2} + \frac{9}{2} = \frac{25}{2}$.
Upper part $= \frac{25\pi}{4} - \frac{1}{2}$; lower (larger) part $= 25\pi - \frac{25\pi}{4} + \frac{1}{2} = \frac{1}{4}(75\pi + 2)$.
$b = 75$, $c = 2$, $b + c = 77$.
Answer: 77. | ||||
| 23 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem A general term is $\frac{6!}{p!\,q!\,r!}\,2^{q/3}\,3^{r/2}$ with $p + q + r = 6$; it is rational when $q \in \{0, 3, 6\}$ and $r$ is even.
$q = 0$: $r = 0$: $1$; $r = 2$: $15\cdot3 = 45$; $r = 4$: $15\cdot9 = 135$; $r = 6$: $27$. Sum $208$.
$q = 3$: $r = 0$ ($p = 3$): $20\cdot2 = 40$; $r = 2$ ($p = 1$): $60\cdot2\cdot3 = 360$. Sum $400$.
$q = 6$: $2^2 = 4$.
Total $= 208 + 400 + 4 = 612$.
Answer: 612. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 36 | ElectrostaticsElectric dipole | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctHard | |
A hard single correct question on electric dipole, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Field at P of dipole 1 (P on its axis, at distance r from its centre), using the exact point-charge sum:
$E_1 = kq\left[\frac{1}{(r - a)^2} - \frac{1}{(r + a)^2}\right] = \frac{4kqar}{(r^2 - a^2)^2}$.
Field at P of dipole 2 (P on its perpendicular bisector, at distance r): $E_2 = \frac{2kqa}{(r^2 + a^2)^{3/2}}$.
The two fields point in opposite directions (dipole 1's field at P points away from its nearer $+q$; dipole 2's equatorial field is antiparallel to its moment), so the force on Q vanishes when $E_1 = E_2$:
$\frac{4ar}{(r^2 - a^2)^2} = \frac{2a}{(r^2 + a^2)^{3/2}} \Rightarrow 2r(r^2 + a^2)^{3/2} = (r^2 - a^2)^2$.
With $x = \frac{a}{r}$: $2(1 + x^2)^{3/2} = (1 - x^2)^2$. For $x < 1$ the left side exceeds 2 while the right side is below 1, so there is no small root; solving numerically gives $x \approx 2.98$, i.e. $\frac{a}{r} \sim 3$.
Answer: (C) $\frac{a}{r} \sim 3$. | ||||
| 37 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | StatementsEasy | |
A easy statements question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current A: true. For a solenoid $L = \mu_0\mu_r n^2 A l$: it depends on the number of turns, area and length, i.e. the geometry.
B: false. The same formula shows $L \propto \mu$, the permeability of the core medium.
C: true. By Lenz's law the self-induced e.m.f. $-L\frac{dI}{dt}$ opposes any change in the current.
D: true. Inductance resists change of current as mass resists change of velocity; the energy $\frac{1}{2}LI^2$ parallels $\frac{1}{2}mv^2$.
E: true. While the current is building up, the source does work against the back e.m.f.; it is stored as $\frac{1}{2}LI^2$.
Correct: A, C, D, E.
Answer: (D) A, C, D, E only. | ||||
| 39 | Electromagnetic WavesProperties of electromagnetic waves | Class 12: Electromagnetic Waves | Single correctMedium | |
A medium single correct question on properties of electromagnetic waves, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves The phase $\omega t - \vec{k}\cdot\vec{r}$ shows the wave travels along $\hat{n} = \frac{3\hat{i} + 4\hat{j}}{5}$ (and $\omega/k = 7.5\times10^6/(5\times10^{-3}\times5) = 3\times10^8$ m/s).
$\vec{B} = \frac{1}{c}\hat{n}\times\vec{E}$. Here $\hat{n}\times(4\hat{i} - 3\hat{j}) = \frac{1}{5}(3\hat{i} + 4\hat{j})\times(4\hat{i} - 3\hat{j}) = \frac{1}{5}(-9 - 16)\hat{k} = -5\hat{k}$.
So $\vec{B} = -\frac{57}{3\times10^8}\cos[\ldots](5\hat{k})$; its magnitude $\frac{57\times5}{c}$ is $|\vec{E}|/c$, as it must be, and $\vec{E}\perp\vec{B}\perp\hat{n}$.
Answer: (C). | ||||
| 40 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctHard | |
A hard single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics The image coincides with the object when the light, after refraction at the first surface, strikes the silvered second surface normally, i.e. travels towards the centre of curvature of the second surface. For the biconvex lens that centre lies on the object side, at distance $|R_2|$ from the (thin) lens.
So the first surface must form a (virtual) image of the object at $v = -|R_2|$ inside the glass. Refraction at the first surface (radius $+|R_1|$, object at $u = -x$):
$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{|R_1|} \Rightarrow -\frac{\mu_2}{|R_2|} + \frac{\mu_1}{x} = \frac{\mu_2 - \mu_1}{|R_1|}$.
$\frac{\mu_1}{x} = \frac{(\mu_2 - \mu_1)|R_2| + \mu_2|R_1|}{|R_1||R_2|} \Rightarrow x = \frac{\mu_1|R_1||R_2|}{\mu_2(|R_1| + |R_2|) - \mu_1|R_2|}$.
The rays then retrace their path and the real, inverted image forms on the object itself.
Answer: (D). | ||||
| 41 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Inside the slab the ray travels a length $\frac{h}{\cos r}$ at angle $r$ to the normal. The emergent ray is parallel to the incident one, and the perpendicular distance between them is that path length times the sine of the angle between the refracted ray and the original direction, $(i - r)$:
$d = \frac{h\sin(i - r)}{\cos r}$.
Answer: (D). | ||||
| 42 | OpticsReflection and refraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on reflection and refraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Refraction at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$ with $\mu_1 = 1$, $\mu_2 = 1.5$, $R = +R$ (centre in the glass).
With $PO = PI = x$: $u = -x$, $v = +x$.
$\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R} \Rightarrow x = 5R$.
Answer: (A) 5 R. | ||||
| 43 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda = \frac{h}{mv} = \frac{6.63\times10^{-34}}{10^{-30}\times2.21\times10^6} = 3\times10^{-10}$ m $= 0.3$ nm $= 3$ Å.
A wavelength of a few ångström lies in the X-ray region.
Answer: (C) X-rays. | ||||
| 44 | 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 Let $\lambda_1 = \lambda$, $\lambda_2 = 3\lambda$ and $T = \frac{\ln2}{\lambda}$ the half-life of $n_1$.
$\frac{N_2}{N_1} = \frac{N_0e^{-3\lambda T}}{N_0e^{-\lambda T}} = e^{-2\lambda T} = e^{-2\ln2} = \frac{1}{4}$.
Answer: (B) $1/4$. | ||||
| 45 | Electronic Devicesp-n junction diode | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | StatementsMedium | |
A medium statements question on p-n junction diode, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits The cell's longer plate (positive) is at the top, so current flows through the ammeter to C and down to D; the diode points from D to A and is forward biased (taken as ideal).
Then the two lower 4 Ω resistors (D to bottom, and A to B) are in parallel: 2 Ω, in series with the 4 Ω between C and D.
Total resistance $= 4 + 2 = 6\ \Omega$: A true, E false.
Current $= \frac{6}{6} = 1$ A: B true.
$V_{CD} = 1\times4 = 4$ V: D true. $V_{AB} = 1\times2 = 2$ V: C false.
Answer: (D) A, B and D Only. | ||||
| 46 | KinematicsScalars and vectors | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on scalars and vectors, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Equal magnitudes: $4 + 9n^2 + 4 = 4 + 4 + 16p^2 \Rightarrow 9n^2 = 16p^2 \Rightarrow p = \pm\frac{3n}{4}$.
Perpendicular: $\vec{A}\cdot\vec{B} = 4 - 6n + 8p = 0$.
$p = +\frac{3n}{4}$ gives $4 = 0$, impossible; $p = -\frac{3n}{4}$ gives $4 - 12n = 0$, so $n = \frac{1}{3}$.
$n^{-1} = 3$.
Answer: 3. | ||||
| 47 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power The force is not conservative ($\frac{\partial F_x}{\partial y} = x^2 \ne \frac{\partial F_y}{\partial x} = 0$), so the work depends on the path; take the straight displacement from $(0, 0)$ to $(4, 2)$, along $y = \frac{x}{2}$.
$W = \int F_x\,dx + \int F_y\,dy = \int_0^4 x^2\cdot\frac{x}{2}\,dx + \int_0^2 y^2\,dy = \frac{1}{2}\cdot\frac{4^4}{4} + \frac{8}{3} = 32 + 2.67 = 34.67$ J.
To the nearest integer, 35 J.
Answer: 35. | ||||
| 48 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics A sudden compression is adiabatic: $TV^{\gamma - 1} = $ constant, with $\gamma = \frac{3}{2}$.
$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 273\times4^{1/2} = 546$ K.
Change in temperature $= 546 - 273 = 273$ K.
Answer: 273. | ||||
| 49 | Electromagnetic Induction and Alternating CurrentsSelf and mutual inductance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on self and mutual inductance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Pulling the slider outwards increases R, so the current decreases: $\frac{dI}{dt} = -8$ A/s.
Kirchhoff's loop rule: $E = IR + L\frac{dI}{dt} \Rightarrow 12 = 12I + 3(-8) \Rightarrow 12I = 36 \Rightarrow I = 3$ A.
(Taking the current as increasing would give a negative current, which is impossible.)
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | Single correctEasy | |
A easy single correct question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles removed $= \frac{10^{21}}{6.02\times10^{23}} = 1.661\times10^{-3}$ mol.
Moles initially $= 1.661\times10^{-3} + 2.8\times10^{-3} = 4.461\times10^{-3}$ mol.
Mass $= 4.461\times10^{-3}\times44 = 0.1963$ g $\approx 196.2$ mg.
Answer: (D) 196.2 mg. | ||||
| 52 | Structure of AtomBohr model | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on bohr model, from Structure of Atom. NCERT: Class 11: Structure of Atom $\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.
Paschen, $\infty \to 3$: $\frac{1}{\lambda} = \frac{10^5}{9}\ \mathrm{cm^{-1}}$, so $\lambda = 9\times10^{-5}$ cm $= 900$ nm, exactly the wavelength needed.
(Paschen $5 \to 3$: $\frac{1}{\lambda} = 10^5\left(\frac{1}{9} - \frac{1}{25}\right)$, $\lambda \approx 1406$ nm; Balmer $\infty \to 2$: 400 nm; Lyman $\infty \to 1$: 100 nm.)
Answer: (A) Paschen series, $\infty \to 3$. | ||||
| 53 | 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: $\mathrm{CCl_4}$ and $\mathrm{CO_2}$ have eight electrons round every central atom: IV.
B: in $\mathrm{BCl_3}$ and $\mathrm{AlCl_3}$ the central atom has only six electrons: II.
C: $\mathrm{NO}$ (11 valence electrons) and $\mathrm{NO_2}$ (17) are odd-electron molecules: I.
D: S in $\mathrm{H_2SO_4}$ (12 electrons) and P in $\mathrm{PCl_5}$ (10) have expanded octets: III.
Answer: (D) A-IV, B-II, C-I, D-III. | ||||
| 54 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics Entropy is a state function; follow a reversible path in five steps at constant pressure:
1. heat ice from 268 K to 273 K: $\int\frac{C_{p,m}(\text{ice})}{T}dT$;
2. melt at $T_f = 273$ K: $\frac{\Delta H_{m,fusion}}{T_f}$;
3. heat water from 273 K to 373 K: $\int\frac{C_{p,m}(\text{water})}{T}dT$;
4. vaporise at $T_b = 373$ K: $\frac{\Delta H_{m,vaporisation}}{T_b}$;
5. heat vapour from 373 K to 383 K: $\int\frac{C_{p,m}(\text{vapour})}{T}dT$.
Heating steps contribute $\int\frac{C_p}{T}dT$, not $\int C_p\,dT$ (which is an enthalpy), so (A), (B) and (C) are wrong; (D) is the sum above.
Answer: (D). | ||||
| 55 | Solutionsvan't Hoff factor | Class 12: Solutions | Single correctEasy | |
A easy single correct question on van't hoff factor, from Solutions. NCERT: Class 12: Solutions $\Delta T_f = iK_fm \Rightarrow 0.558 = i\times1.86\times0.1 \Rightarrow i = 3$.
Three ions per formula unit: one complex cation and two $\mathrm{Cl^-}$, i.e. $\mathrm{[Cr(NH_3)_5Cl]^{2+}}$ + $2\mathrm{Cl^-}$ (coordination number 6 = five $\mathrm{NH_3}$ + one Cl).
Answer: (B) $\mathrm{[Cr(NH_3)_5Cl]Cl_2}$. | ||||
| 56 | EquilibriumSolubility product | Class 11: Equilibrium | Single correctEasy | |
A easy single correct question on solubility product, from Equilibrium. NCERT: Class 11: Equilibrium Hydroxide ion concentration needed to start precipitation:
$\mathrm{A(OH)_2}$: $[\mathrm{OH^-}] = \sqrt{\frac{9\times10^{-10}}{1}} = 3\times10^{-5}$ M.
$\mathrm{B(OH)_3}$: $[\mathrm{OH^-}] = \sqrt[3]{\frac{27\times10^{-18}}{1}} = 3\times10^{-6}$ M.
As $\mathrm{NH_4OH}$ is added, $[\mathrm{OH^-}]$ rises and reaches $3\times10^{-6}$ M first, so $\mathrm{B(OH)_3}$ precipitates first.
Answer: (A). | ||||
| 57 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Potentials do not add, free energies do: $\Delta G^\circ = -nFE^\circ$.
$\mathrm{FeO_4^{2-} \to Fe^{3+}}$ (Fe +6 to +3): $n = 3$, $E^\circ = 2.0$ V. $\mathrm{Fe^{3+} \to Fe^{2+}}$: $n = 1$, $E^\circ = 0.8$ V.
$4E^\circ = 3(2.0) + 1(0.8) = 6.8 \Rightarrow E^\circ_{\mathrm{FeO_4^{2-}/Fe^{2+}}} = 1.7$ V.
Answer: (B) 1.7 V. | ||||
| 58 | Classification of Elements and PeriodicityModern periodic table | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on modern periodic table, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Os (76), Ir (77) and Pt (78) are 5d elements of period 6; Pd (46) is a 4d element of period 5.
Answer: (C) Palladium. | ||||
| 59 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Correct: N has no d orbitals and cannot expand its octet, so $\mathrm{NF_5}$ does not exist, while $\mathrm{PF_3}$ (and $\mathrm{PF_5}$) do.
(B) Correct: the lone pair of P in $\mathrm{PH_3}$ is in a larger, more diffuse orbital, so $\mathrm{PH_3}$ is a weaker base than $\mathrm{NH_3}$.
(C) Correct: $\mathrm{NO_2}$ is an odd-electron molecule and dimerises to $\mathrm{N_2O_4}$.
(D) Incorrect: S in $\mathrm{SO_2}$ is +4 and can go up (to +6, e.g. decolourising acidified $\mathrm{KMnO_4}$) or down (to 0, with $\mathrm{H_2S}$), so it is both a reducing and an oxidising agent; it acts mainly as a reducing agent.
Answer: (D). | ||||
| 60 | d- and f-Block ElementsTransition element properties | Class 12: The d- and f-Block Elements | Single correctMedium | |
A medium single correct question on transition element properties, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Colours of the aquated ions: $\mathrm{V^{2+}}$ violet, $\mathrm{Cr^{3+}}$ violet, $\mathrm{Mn^{3+}}$ violet: all the same.
(A) $\mathrm{Sc^{3+}}$ colourless, $\mathrm{Ti^{3+}}$ purple, $\mathrm{Cr^{2+}}$ blue. (C) $\mathrm{Ti^{4+}}$ colourless, $\mathrm{V^{4+}}$ blue, $\mathrm{Mn^{2+}}$ pink. (D) $\mathrm{Zn^{2+}}$ colourless, $\mathrm{V^{3+}}$ green, $\mathrm{Fe^{3+}}$ yellow.
Answer: (B). | ||||
| 61 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Co(II) is $d^7$. $\mu = \sqrt{n(n + 2)}$: 3 unpaired electrons give 3.87 BM, close to 3.95 BM, so the complex is high spin.
Octahedral high-spin $d^7$: $t_{2g}^5e_g^2$ (3 unpaired). Low spin $t_{2g}^6e_g^1$ has only 1 unpaired; $e^4t_2^3$ is tetrahedral notation; $t_{2g}^3e_g^0$ is $d^3$.
Answer: (A) $t_{2g}^5e_g^2$. | ||||
| 63 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Of the dichloropropanes, only 1,2-dichloropropane, $\mathrm{CH_2Cl{-}CHCl{-}CH_3}$, is chiral (C-2 carries H, Cl, $\mathrm{CH_3}$ and $\mathrm{CH_2Cl}$), so x is 1,2-dichloropropane.
Replacing one more H by Cl:
on C-1: $\mathrm{CHCl_2{-}CHCl{-}CH_3}$ (1,1,2-trichloropropane);
on C-2: $\mathrm{CH_2Cl{-}CCl_2{-}CH_3}$ (1,2,2-trichloropropane);
on C-3: $\mathrm{CH_2Cl{-}CHCl{-}CH_2Cl}$ (1,2,3-trichloropropane).
Three structural isomers.
Answer: (A) $3$. | ||||
| 64 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons q, the cyclopentadienyl anion, is planar, cyclic and fully conjugated with $6\pi$ electrons ($4n + 2$, $n = 1$): aromatic, most stable.
r, cyclooctatetraene, would be antiaromatic ($8\pi$) if planar, so it puckers into a tub and is non-aromatic: ordinary stability.
p, the cyclopropenyl anion, is planar and conjugated with $4\pi$ electrons ($4n$): antiaromatic, least stable.
Order: $q > r > p$.
Answer: (C). | ||||
| 65 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Matching listEasy | |
A easy matching list question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes A. Swarts: $\mathrm{C_2H_5Br + AgF \to C_2H_5F}$, ethyl fluoride: IV.
B. Sandmeyer: $\mathrm{C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN}$, cyanobenzene: III.
C. Wurtz–Fittig: $\mathrm{C_6H_5Br + C_2H_5Br \xrightarrow{Na,\ ether} C_6H_5C_2H_5}$, ethyl benzene: I.
D. Finkelstein: $\mathrm{C_2H_5Cl + NaI \xrightarrow{acetone} C_2H_5I}$, ethyl iodide: II.
Answer: (C) A-IV, B-III, C-I, D-II. | ||||
| 66 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers $\mathrm{C_6H_5OH + 3Br_2 \to C_6H_2Br_3OH + 3HBr}$.
Phenol: $M = 94$; $\frac{2}{94} = 0.02128$ mol, which needs $3\times0.02128 = 0.06383$ mol $\mathrm{Br_2}$.
Mass $= 0.06383\times160 = 10.21$ g $\approx 10.22$ g.
Answer: (C) 10.22 g. | ||||
| 67 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Propanal has two α-hydrogens; HCHO has none and is the better electrophile.
Two crossed aldol additions put a $\mathrm{CH_2OH}$ on the α-carbon each time: $\mathrm{CH_3CH(CH_2OH)CHO}$, then $\mathrm{CH_3C(CH_2OH)_2CHO}$ (no α-H left).
With excess HCHO and alkali, a crossed Cannizzaro reaction follows: HCHO is oxidised to formate and the aldehyde is reduced to $\mathrm{CH_3C(CH_2OH)_3}$ (1,1,1-tris(hydroxymethyl)ethane).
Answer: (A). | ||||
| 68 | AminesClassification and identification of amines | Class 12: Amines | StatementsEasy | |
A easy statements question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Hinsberg's reagent (benzenesulphonyl chloride) reacts with amines that have an N–H bond: primary and secondary amines.
A aniline (1°), C methylamine (1°) and E diphenylamine (2°) react.
B N,N-dimethylaniline and D trimethylamine are tertiary and do not react.
Answer: (A) A, C and E Only. | ||||
| 69 | BiomoleculesCarbohydrates | Class 12: Biomolecules | StatementsMedium | |
A medium statements question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Statement I: true. Fructose is a ketose, yet it reduces Tollens' reagent.
Statement II: true. In the alkaline medium of Tollens' reagent, fructose isomerises through an enediol intermediate (Lobry de Bruyn–van Ekenstein rearrangement) to glucose and mannose, which carry –CHO groups; this is why it reduces the reagent.
Answer: (A). | ||||
| 70 | Practical Organic ChemistryDetection of elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on detection of elements, from Practical Organic Chemistry. NCERT: No matching NCERT chapter Statement I: true. Fusion with sodium converts covalently bound N, S and halogens into ionic NaCN, $\mathrm{Na_2S}$, NaSCN and NaX, which can be tested in water.
Statement II: false. When both N and S are present, sodium thiocyanate forms; with iron(III) it gives a blood-red colour ($\mathrm{[Fe(SCN)]^{2+}}$), not Prussian blue. Also, the Prussian blue test is done with $\mathrm{FeSO_4}$ followed by acidification ($\mathrm{H_2SO_4}$); sodium ferrocyanide itself is not the reagent.
Answer: (C). | ||||
| 71 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ with $T = 298.15$ K:
$\Delta G^\circ = 55000 - 298.15\times175.0 = 55000 - 52176.25 = 2823.75\ \mathrm{J\ mol^{-1}}$.
Nearest integer: 2824.
With T = 298 K: ΔG = 55000 − 298 × 175 = 2850 J mol⁻¹ (2824 with 298.15 K).
Answer: 2850 (2824 with T = 298.15 K). | ||||
| 72 | EquilibriumIonic equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on ionic equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium pH = 9, so pOH = 5 and $[\mathrm{OH^-}] = 10^{-5}$ M (at 298 K).
$K_b = \frac{[\mathrm{OH^-}]^2}{C} = \frac{(10^{-5})^2}{10^{-3}} = 10^{-7}$.
$x = 7$.
Answer: 7. | ||||
| 73 | Chemical KineticsIntegrated rate laws and half-life | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on integrated rate laws and half-life, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The unit $\mathrm{s^{-1}}$ makes the reaction first order: $\ln\frac{p_0}{p} = kt = 4.606\times10^{-2}\times100 = 4.606 = 2\times2.303$, so $\frac{p_0}{p} = 100$.
Pressure of $\mathrm{N_2O_5}$ left: $p = \frac{0.6}{100} = 0.006$ atm; decomposed: $0.594$ atm.
Every 2 mol of $\mathrm{N_2O_5}$ give 3 mol of gas, so the total pressure rises by half the amount decomposed:
$p_{total} = 0.6 + \frac{0.594}{2} = 0.897$ atm $= 897\times10^{-3}$ atm.
Answer: 897. | ||||
| 74 | AminesDiazonium salts | Class 12: Amines | NumericalMedium | |
A medium numerical question on diazonium salts, from Amines. NCERT: Class 12: Amines i) $\mathrm{Br_2/Fe}$ brominates the ring of 3-nitrotoluene, mainly where the activating $\mathrm{CH_3}$ directs (ortho/para to it).
ii) Sn/HCl reduces $\mathrm{NO_2}$ to $\mathrm{NH_2}$.
iii) $\mathrm{NaNO_2}$/HCl at 273 K diazotises the $\mathrm{NH_2}$ to $\mathrm{N_2^+Cl^-}$.
iv) $\mathrm{H_3PO_2}$/$\mathrm{H_2O}$ replaces the diazonium group by H.
So (A) is a bromotoluene, $\mathrm{C_7H_7Br}$, whichever ring position the bromine took: $M = 7(12) + 7(1) + 80 = 171\ \mathrm{g\ mol^{-1}}$.
Answer: 171. | ||||
| 75 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques $M(\mathrm{BaSO_4}) = 137 + 32 + 64 = 233$, so 466 mg is 2 mmol of $\mathrm{BaSO_4}$, containing 2 mmol = 64 mg of S.
% S $= \frac{64}{160}\times100 = 40\%$.
Answer: 40. | ||||
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