JEE Main 5 April 2026 Shift 2: Questions and Solutions

Answer: Option (3)

Key idea: Geometric Progression and Vieta's Formulas

1. Since $\alpha, \beta, \gamma, \delta$ are in G.P., let them be $a, ar, ar^2, ar^3$
2. Using Vieta's formulas: $\alpha + \beta = a(1+r) = 1$ and $\gamma + \delta = ar^2(1+r) = 4$
3. Dividing these equations: $r^2 = 4$, so $r = \pm 2$
4. For $r = 2$: $a = 1/3$ gives non-integer $p$. For $r = -2$: $a = -1$ works
5. With $a = -1, r = -2$: roots are $\alpha = -1, \beta = 2, \gamma = -4, \delta = 8$
6. Calculate $p = \alpha\beta = (-1)(2) = -2$ and $q = \gamma\delta = (-4)(8) = -32$
7. Therefore $|p + q| = |-2 - 32| = 34$

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Key idea: Unable to determine - question content not provided

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Key idea: Determinant evaluation and telescoping sums

Answer: Option (2)

Key idea: Matrix construction from column vectors and solving linear systems

1. Recognize that multiplying $M$ by standard basis vectors gives the columns of $M$
2. Construct matrix $M = \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix}$
3. Set up the system: $x - z = 1$, $2x + y + z = 7$, $3x + 2y + z = 11$
4. From first equation: $x = 1 + z$. Substitute to get $y + 3z = 5$ and $y + 2z = 4$
5. Solve to get $z = 1$, $y = 2$, $x = 2$
6. Calculate $x + y + z = 2 + 2 + 1 = 5$

Answer: Option (3)

Key idea: Telescoping Series and Partial Fractions

1. Factor the denominator: $1 + 4k^4 = (2k^2 - 2k + 1)(2k^2 + 2k + 1)$
2. Use partial fractions to get: $\frac{k}{1+4k^4} = \frac{1}{4}\left(\frac{1}{2k^2 - 2k + 1} - \frac{1}{2k^2 + 2k + 1}\right)$
3. Observe that $2(k+1)^2 - 2(k+1) + 1 = 2k^2 + 2k + 1$, so the series telescopes
4. Sum telescopes to: $S_{10} = \frac{1}{4}\left(1 - \frac{1}{221}\right) = \frac{1}{4} \cdot \frac{220}{221} = \frac{55}{221}$
5. Verify $\gcd(55, 221) = 1$ since $55 = 5 \times 11$ and $221 = 13 \times 17$
6. Therefore $m + n = 55 + 221 = 276$

Answer: Option (4)

Key idea: Arithmetic Progression - Finding arithmetic means and calculating their average

1. The sequence is: $59, A_1, A_2, ..., A_{39}, 159$ with total $41$ terms
2. Find common difference: $d = \frac{159-59}{40} = \frac{100}{40} = 2.5$
3. Calculate the required terms: $A_{25} = 59 + 25(2.5) = 121.5$, $A_{28} = 129$, $A_{31} = 136.5$, $A_{36} = 149$
4. Find the mean: $\text{Mean} = \frac{121.5 + 129 + 136.5 + 149}{4} = \frac{536}{4} = 134$

Answer: Option (2)

Key idea: Binomial Theorem - Finding coefficient of specific term in expansion

1. Write the general term using binomial theorem: $T_k = \binom{10}{k} (2x^2)^{10-k} \left(\frac{1}{x}\right)^k$
2. Simplify the general term: $T_k = \binom{10}{k} \cdot 2^{10-k} \cdot x^{20-3k}$
3. For coefficient of $x^2$, set $20-3k = 2$, which gives $k = 6$
4. Calculate $\binom{10}{6} = \frac{10!}{6! \cdot 4!} = 210$
5. Find coefficient: $210 \cdot 2^4 = 210 \cdot 16 = 3360$

Answer: Option (2)

Key idea: Law of Total Probability

1. Identify given probabilities: $P(A) = 0.6$, $P(B) = 0.4$, $P(W|A) = 0.8$, $P(W|B) = 0.7$
2. Apply Law of Total Probability: $P(W) = P(W|A) \cdot P(A) + P(W|B) \cdot P(B)$
3. Calculate first term: $P(W|A) \cdot P(A) = 0.8 \times 0.6 = 0.48$
4. Calculate second term: $P(W|B) \cdot P(B) = 0.7 \times 0.4 = 0.28$
5. Add both terms: $P(W) = 0.48 + 0.28 = 0.76$

Answer: Option (1)

Key idea: Combinations with constraints - distributing objects with minimum requirements

1. Identify constraints: Select 8 balls with at least 2 of each color from 5 blue, 6 yellow, 4 red balls
2. Set up: Let $(b, y, r)$ be the number of blue, yellow, red balls. Need $b+y+r=8$ with $b \geq 2, y \geq 2, r \geq 2$ and $b \leq 5, y \leq 6, r \leq 4$
3. Find all valid distributions: $(2,2,4), (2,3,3), (2,4,2), (3,2,3), (3,3,2), (4,2,2)$
4. Calculate ways for each case using $\binom{n}{r}$: $150 + 800 + 900 + 600 + 1200 + 450$
5. Sum all cases: $4100$

Answer: Option (1)

Key idea: Mean and Median of Discrete Distribution with Binomial Coefficients

1. Identify that the $k$-th value is $k(k-1)$ with frequency $^nC_k$
2. Calculate mean using $\sum k(k-1) \cdot ^nC_k = n(n-1) \cdot 2^{n-2}$ and total frequency $2^n$
3. Set up equation: $\frac{n(n-1)}{4} = 60$, giving $n(n-1) = 240$
4. Solve $n^2 - n - 240 = 0$ to get $n = 16$
5. Total frequency is $2^{16} = 65536$. By symmetry, $\sum_{k=0}^{7} ^{16}C_k = 32768$
6. The median position is at $32768.5$, between values at $k=7$ (value $42$) and $k=8$ (value $56$)
7. The median is $56$ (the value at the $(n/2 + 1)$-th position)

Answer: Option (2)

Key idea: Parabola vertex, circle geometry, chord properties, and optimization

1. Find vertex of parabola $y = x^2 - 6x + 12$ by completing square: $y = (x-3)^2 + 3$, so $P = (3, 3)$
2. Rewrite circle $x^2 + y^2 - 2x - 4y + 3 = 0$ in standard form: $(x-1)^2 + (y-2)^2 = 2$, center $C = (1, 2)$, radius $r = \sqrt{2}$
3. Calculate distance $PC = \sqrt{(3-1)^2 + (3-2)^2} = \sqrt{5}$. Since $\sqrt{5} > \sqrt{2}$, point $P$ is outside the circle
4. For chord $RS$ with midpoint $M$, we have $PR + PS = 2PM$ where $PM = \sqrt{PC^2 - CM^2} = \sqrt{5 - CM^2}$
5. To maximize $PR + PS$, minimize $CM$. Minimum occurs when line passes through center $C$, giving $CM = 0$
6. Maximum value: $PR + PS = 2\sqrt{5}$, so $(PR + PS)^2 = 4 \times 5 = 20$

Answer: Option (2)

Key idea: Parabola properties, focal chord, parametric representation, and coordinate geometry

1. Find directrix $x = -2$ from $y^2 = 8x$ (where $a=2$), so $A = (-2, 0)$
2. Use slope condition $\frac{\beta}{\alpha+2} = \frac{3}{5}$ with $\beta^2 = 8\alpha$ to get $3\beta^2 - 40\beta + 48 = 0$
3. Solve to get $\beta = 12$ (valid) or $\beta = \frac{4}{3}$ (invalid as $\alpha \not> 1$), giving $B = (18, 12)$
4. Convert to parametric form: $B = (2t_1^2, 4t_1)$ gives $t_1 = 3$. For focal chord, $t_1 t_2 = -1$, so $t_2 = -\frac{1}{3}$
5. Find $C = (2t_2^2, 4t_2) = (\frac{2}{9}, -\frac{4}{3})$
6. Calculate area using formula: Area$(\triangle ABC) = \frac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| = \frac{80}{3}$
7. Six times the area = $6 \times \frac{80}{3} = 160$

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Answer: Option (3)

Key idea: Finding points on a line at given distance from a point, and calculating centroid of a triangle in 3D

1. Write the parametric form of the line: $(8t-3, 2t+4, 2t-1)$ where $t$ is the parameter
2. Use distance formula: Points at distance 6 from $R(1,2,3)$ satisfy $(8t-4)^2 + (2t+2)^2 + (2t-4)^2 = 36$
3. Simplify to get $72t^2 - 72t + 36 = 36$, which gives $2t(t-1) = 0$, so $t = 0$ or $t = 1$
4. Find coordinates: $P(-3, 4, -1)$ at $t=0$ and $Q(5, 6, 1)$ at $t=1$
5. Calculate centroid: $(\alpha, \beta, \gamma) = \left(\frac{-3+5+1}{3}, \frac{4+6+2}{3}, \frac{-1+1+3}{3}\right) = (1, 4, 1)$
6. Therefore $\alpha + \beta + \gamma = 1 + 4 + 1 = 6$

Answer: Option (3)

Key idea: Image of a point with respect to a line in 3D space

1. Find the foot of perpendicular $M$ from point $(1, 2, 7)$ to the line $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ by setting $\vec{PM} \perp$ direction vector $(1, 1, 2)$
2. Parametrize the line as $(t, 1+t, 2+2t)$ and solve $(t-1, t-1, 2t-5) \cdot (1, 1, 2) = 0$ to get $t = 2$, so $M = (2, 3, 6)$
3. Find the image $P'$ using $M$ as midpoint: $P' = 2M - P = (3, 4, 5)$
4. Calculate distance from $(a, 2, 5)$ to $P'(3, 4, 5)$: $\sqrt{(a-3)^2 + 4} = 4$
5. Solve $(a-3)^2 = 12$ to get $a = 3 \pm 2\sqrt{3}$
6. Sum of all possible values: $(3 + 2\sqrt{3}) + (3 - 2\sqrt{3}) = 6$

Answer: Option (1)

Key idea: Position vectors and vector addition using triangle law

1. Find $\overrightarrow{OR}$: Since $\overrightarrow{OP} = 5\overrightarrow{OR}$, we get $\overrightarrow{OR} = \frac{\vec{a}}{5}$
2. Find $\overrightarrow{RM}$: Since $\overrightarrow{OQ} = 5\overrightarrow{RM}$, we get $\overrightarrow{RM} = \frac{\vec{b}}{5}$
3. Find $\overrightarrow{OM}$: Using $\overrightarrow{OM} = \overrightarrow{OR} + \overrightarrow{RM}$, we get $\overrightarrow{OM} = \frac{\vec{a} + \vec{b}}{5}$
4. Find $\overrightarrow{PM}$: Using $\overrightarrow{PM} = \overrightarrow{OM} - \overrightarrow{OP}$
5. Substitute: $\overrightarrow{PM} = \frac{\vec{a} + \vec{b}}{5} - \vec{a} = \frac{\vec{b} - 4\vec{a}}{5}$

Answer: Option (3)

Key idea: Limits using Taylor series expansion and analyzing transcendental equations

1. Step 1: Find $f(x)$ by substituting $t = xy$ and using Taylor series: $1 - \cos t = \frac{t^2}{2} + O(t^4)$ and $\tan t = t + O(t^3)$
2. Step 2: Expand $(1 - \cos t)\tan t = \frac{t^3}{2} + O(t^5)$, so $f(x) = \lim_{t \to 0} \frac{x^3 \cdot \frac{t^3}{2}}{t^3} = \frac{x^3}{2}$
3. Step 3: Solve $\frac{x^3}{2} = \sin x$ or equivalently $x^3 = 2\sin x$. Define $g(x) = x^3 - 2\sin x$
4. Step 4: Check $x = 0$: $g(0) = 0$, so $x = 0$ is a solution
5. Step 5: Analyze $g'(x) = 3x^2 - 2\cos x$. For $x > 0$, $g$ decreases from 0, reaches a negative minimum, then increases to $+\infty$, giving one positive root
6. Step 6: By odd symmetry $g(-x) = -g(x)$, there is also one negative root
7. Step 7: Total solutions: $x = 0$, one positive, one negative = **3 solutions**

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Answer: Option (1)

Key idea: Solving functional equations involving integrals using differentiation and first-order linear differential equations

1. Substitute $x=1$ in the given equation: $f(1) = 0 + 0 + e = e$
2. Differentiate both sides with respect to $x$ to get: $f'(x) = f(x) - \log_e x + \frac{1}{x}$
3. Rearrange to form a linear ODE: $f'(x) - f(x) = -\log_e x + \frac{1}{x}$
4. Solve using integrating factor $e^{-x}$ and integration by parts to obtain: $f(x) = e^x + \log_e x$
5. Verify: $f(1) = e^1 + \log_e 1 = e$ ✓
6. Calculate $f(f(1)) = f(e) = e^e + \log_e e = e^e + 1 = 1 + e^e$

Answer: Option (2)

Key idea: Integration of differential equations with boundary conditions

1. Given $f''(x) = g''(x)$, integrate to get $f'(x) - g'(x) = C_1$
2. Use $f'(1) = 4$ and $g'(1) = 2$ to find $C_1 = f'(1) - g'(1) = 2$
3. Integrate again: $f(x) - g(x) = 2x + C_2$
4. Use $f(2) = 3$ and $g(2) = 9$ to find $C_2$: $3 - 9 = 4 + C_2$, so $C_2 = -10$
5. Therefore $f(x) - g(x) = 2x - 10$
6. Calculate $f(25) - g(25) = 2(25) - 10 = 40$

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Answer: 7

Key idea: Reflection of lines about a mirror line, angle between lines, coordinate geometry

1. Find the two incident rays from $(-1, -1)$ making $45°$ with $x+y=0$. Since $x+y=0$ has slope $-1$ (making $135°$ with x-axis), the rays at $45°$ to it are horizontal ($y=-1$) and vertical ($x=-1$).
2. For the horizontal ray $y=-1$, find its reflection about mirror $x+2y=1$. Use the reflection formula: from point $(x_0, -1)$, find the perpendicular to the mirror, locate the foot of perpendicular, and find the image point.
3. The reflected ray from $y=-1$ passes through $(3/5, 11/5)$ and $(18/5, -9/5)$, giving slope $-4/3$. The equation is $4x+3y=9$.
4. For the vertical ray $x=-1$, similarly find its reflection. The reflected ray passes through $(3/5, 11/5)$ and $(-17/5, -4/5)$, giving slope $3/4$. The equation is $-3x+4y=7$.
5. Identify coefficients: $ax+by=9$ gives $a=4, b=3$ and $cx+dy=7$ gives $c=-3, d=4$.
6. Calculate $ad+bc = 4(4) + 3(-3) = 16-9 = 7$.

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Key idea: Trigonometric equations using product-to-sum formulas

Answer: 16

Key idea: Functional equations, optimization of trigonometric functions, and area between curves

1. Solve the functional equation by substituting $x \to \frac{\pi}{2} - x$ to get $f\left(\frac{\pi}{2} - x\right) + 3f(x) = \cos x$, then solve the system to find $f(x) = \frac{3\cos x - \sin x}{8}$
2. Find maximum of $f(x)$ using $3\cos x - \sin x = \sqrt{10}\cos(x + \theta)$, giving $\alpha = \frac{\sqrt{10}}{8}$
3. Find intersection points of $x^2 = \beta x^3$: $x = 0$ and $x = \frac{1}{\beta}$
4. Calculate area: $A = \int_0^{1/\beta} (x^2 - \beta x^3)dx = \frac{1}{12\beta^3}$
5. Set $\frac{1}{12\beta^3} = \alpha^2 = \frac{10}{64} = \frac{5}{32}$ to get $\beta^3 = \frac{8}{15}$
6. Calculate $30\beta^3 = 30 \times \frac{8}{15} = 16$

Answer: 48

Key idea: First-order linear differential equation with integrating factor method

1. Rewrite the equation as $\frac{dy}{dx} + (\tan x) y = \frac{\sec^3 x}{(\tan x)^{1/2}}$
2. Find integrating factor: $I.F. = e^{\int \tan x dx} = \sec x$
3. Multiply by I.F. and integrate: $y \sec x = \int \frac{\sec^4 x}{(\tan x)^{1/2}} dx$
4. Use substitution $u = \tan x$ to get $\int (u^{-1/2} + u^{3/2}) du = 2u^{1/2} + \frac{2u^{5/2}}{5} + C$
5. Apply initial condition $y(\pi/4) = \frac{6\sqrt{2}}{5}$ to find $C = 0$
6. Evaluate $y(\pi/3) = 3^{1/4} \cdot \frac{8}{5} = \frac{4\alpha}{5}$ to get $\alpha = 2 \cdot 3^{1/4}$
7. Calculate $\alpha^4 = 16 \times 3 = 48$

Answer: Option (2)

Key idea: Dimensional Analysis and Planck Units

1. Write dimensions: $\hbar = [ML^2T^{-1}]$, $G = [M^{-1}L^3T^{-2}]$, $c = [LT^{-1}]$
2. For I: $\sqrt{\frac{\hbar c}{G}} = \sqrt{\frac{[ML^3T^{-2}]}{[M^{-1}L^3T^{-2}]}} = \sqrt{[M^2]} = [M]$ matches C (Kilogram)
3. For II: $\sqrt{\frac{G\hbar}{c^5}} = \sqrt{\frac{[L^5T^{-3}]}{[L^5T^{-5}]}} = \sqrt{[T^2]} = [T]$ matches B (Second)
4. For IV: $\sqrt{\frac{G\hbar}{c^3}} = \sqrt{\frac{[L^5T^{-3}]}{[L^3T^{-3}]}} = \sqrt{[L^2]} = [L]$ matches A (Meter)
5. For III: By elimination and dimensional construction, $\sqrt{\frac{K^2L^2c^3}{G\hbar}}$ gives $[K]$ matches D (Kelvin)
6. Therefore: A-IV, B-II, C-I, D-III

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Key idea: Unknown - question content missing

Answer: Option (1)

Key idea: Work done by friction on an object moving with an inclined plane assembly

1. Given: mass $m = 1$ kg on inclined plane ($\theta = 30°$), assembly moves up with constant velocity $v = 4$ m/s for $t = 2$ s
2. Since mass is at rest on the plane and assembly moves with constant velocity, friction balances the weight component: $f = mg\sin(30°) = 1 \times 10 \times 0.5 = 5$ N (up the plane)
3. Vertical displacement of mass: $s = vt = 4 \times 2 = 8$ m
4. Component of displacement along the inclined plane: $s_{\parallel} = s\sin(30°) = 8 \times 0.5 = 4$ m
5. Work done by friction: $W = f \times s_{\parallel} = 5 \times 4 = 20$ J

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Key idea: Velocity-Time Graph Analysis: Distance vs Displacement

Answer: Option (2)

Key idea: Rotational Kinematics - Equations of motion for uniform angular acceleration

1. Given: Initial angular velocity $\omega_0 = 0$, uniform angular acceleration $\alpha$
2. For first 2 s: Using $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$, we get $\theta_1 = \frac{1}{2}\alpha(2)^2 = 2\alpha$
3. For total 4 s: $\theta_{total} = \frac{1}{2}\alpha(4)^2 = 8\alpha$
4. For next 2 s (from t=2s to t=4s): $\theta_2 = \theta_{total} - \theta_1 = 8\alpha - 2\alpha = 6\alpha$
5. Calculate ratio: $\frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3$

Answer: Option (4)

Key idea: Energy Conservation in Rolling Motion

1. Write energy conservation: Initial KE (translational + rotational) = Final PE at maximum height
2. Express total initial energy: $\frac{1}{2}mv_o^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv_o^2(1+k)$ where $I = kmR^2$
3. Use rolling condition $v_o = R\omega$ and energy conservation: $\frac{1}{2}mv_o^2(1+k) = mgh$
4. Solve for height: $h = \frac{v_o^2(1+k)}{2g}$
5. Equate with given height: $\frac{v_o^2(1+k)}{2g} = \frac{7v_o^2}{10g}$, which gives $k = \frac{2}{5}$
6. Identify object: For solid sphere, $I = \frac{2}{5}mR^2$, so $k = \frac{2}{5}$ ✓

Answer: Option (4)

Key idea: Gravitational Potential Energy and its variation with height

1. Identify positions: Initial at surface ($r_1 = R_e$), Final at height $2R_e$ ($r_2 = 3R_e$)
2. Use gravitational PE formula: $U(r) = -\frac{GMm}{r}$ where $GM = gR_e^2$
3. Calculate initial PE: $U_1 = -\frac{GMm}{R_e} = -mgR_e$
4. Calculate final PE: $U_2 = -\frac{GMm}{3R_e} = -\frac{mgR_e}{3}$
5. Find change in PE: $\Delta U = U_2 - U_1 = -\frac{mgR_e}{3} + mgR_e = \frac{2}{3}mgR_e$

Answer: Option (2)

Key idea: Surface Energy and Coalescence of Drops

1. Use volume conservation: $8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$, which gives $R^3 = 8r^3$, so $R = 2r$
2. Calculate initial surface area: $A_i = 8 \times 4\pi r^2 = 32\pi r^2$
3. Calculate final surface area: $A_f = 4\pi R^2 = 4\pi(2r)^2 = 16\pi r^2$
4. Initial surface energy: $E_i = S \times 32\pi r^2$
5. Final surface energy: $E_f = S \times 16\pi r^2$
6. Energy released: $\Delta E = E_i - E_f = 32\pi r^2 S - 16\pi r^2 S = 16\pi r^2 S$

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Key idea: Coefficient of volume expansion for ideal gas in a polytropic process

Answer: Option (1)

Key idea: Periodic Motion and Simple Harmonic Motion (SHM) - Identifying periodicity and SHM characteristics using trigonometric identities

1. For **A**: Use $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$. Period is $T = \frac{\pi}{\omega}$. About mean position $\frac{1}{2}$, it satisfies SHM equation $\frac{d^2y}{dt^2} = -4\omega^2 y$ → **III**
2. For **B**: Use $\sin^3(2\omega t) = \frac{3\sin(2\omega t) - \sin(6\omega t)}{4}$. Period is LCM of $\frac{\pi}{\omega}$ and $\frac{\pi}{3\omega}$ which is $\frac{\pi}{\omega}$. Sum of different frequencies → NOT SHM → **I**
3. For **C**: $\sin(\omega t)$ has period $\frac{2\pi}{\omega}$ and $\cos(\pi\omega t)$ has period $\frac{2}{\omega}$. Ratio of periods is $\pi$ (irrational) → **Non-periodic** → **IV**
4. For **D**: $\cos(\omega t)$ has period $\frac{2\pi}{\omega}$ and $\cos(2\omega t)$ has period $\frac{\pi}{\omega}$. LCM gives $T = \frac{2\pi}{\omega}$. Sum of different frequencies → NOT SHM → **II**
5. Match: A-III, B-I, C-IV, D-II

Answer: Option (1)

Key idea: Motional EMF in rotating rod with non-uniform magnetic field; Integration by parts

1. For a small element $dr$ at distance $r$ from origin, velocity is $v = \omega r$
2. Induced emf in element: $d\varepsilon = B(r) \cdot v \cdot dr = B_o e^{-\lambda r} \cdot \omega r \cdot dr$
3. Total emf: $\varepsilon = B_o \omega \int_0^L r e^{-\lambda r} dr$
4. Use integration by parts with $u = r$, $dv = e^{-\lambda r}dr$
5. $\int r e^{-\lambda r} dr = -\frac{r e^{-\lambda r}}{\lambda} - \frac{e^{-\lambda r}}{\lambda^2}$
6. Evaluate limits from $0$ to $L$: $\left[-e^{-\lambda r}\left(\frac{r}{\lambda} + \frac{1}{\lambda^2}\right)\right]_0^L = \frac{1}{\lambda^2} - e^{-\lambda L}\left(\frac{1}{\lambda^2} + \frac{L}{\lambda}\right)$
7. Final answer: $\varepsilon = B_o \omega \left[\frac{1}{\lambda^2} - e^{-\lambda L}\left(\frac{1}{\lambda^2} + \frac{L}{\lambda}\right)\right]$

Answer: Option (2)

Key idea: Capacitor in steady state acts as open circuit; no current flows through capacitor branch, so entire branch voltage appears across capacitor

1. In steady state, capacitor is fully charged and acts as **open circuit** (no current through that branch)
2. Current flows only through: $2\Omega$ resistor → $6\Omega$ resistor path
3. Total resistance: $R_{total} = 2\Omega + 6\Omega = 8\Omega$
4. Current: $I = \frac{V}{R} = \frac{2V}{8\Omega} = 0.25A$
5. Voltage across parallel combination (including capacitor branch): $V = I \times 6\Omega = 0.25 \times 6 = 1.5V$
6. Since no current through $4\Omega$ resistor, its voltage drop = 0V, so **entire 1.5V appears across capacitor**

Answer: Option (1)

Key idea: Work-Energy Theorem with Electromagnetic Forces

1. Apply work-energy theorem: $\Delta KE = W_{total}$
2. Note that magnetic force does no work since $\vec{F}_B \perp \vec{v}$ always
3. Only electric force contributes: $W = \int_0^L q\vec{E} \cdot d\vec{r}$
4. Since $\vec{E} = E_0 e^{-\lambda x}\hat{x}$, we get $W = qE_0 \int_0^L e^{-\lambda x} dx$
5. Evaluate integral: $W = qE_0 \left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^L$
6. Simplify: $\Delta KE = \frac{qE_0}{\lambda}\left[1 - e^{-\lambda L}\right]$

Answer: Option (2)

Key idea: Radiation pressure and momentum of electromagnetic waves

1. **Assertion (A)**: EM waves exert pressure on surfaces - This is **TRUE**. Radiation pressure $P = I/c$ (absorption) or $P = 2I/c$ (reflection)
2. **Reason (R)**: EM waves have no mass - This is **TRUE**. Photons have zero rest mass
3. **Key Question**: Does (R) explain (A)?
4. EM waves exert pressure because they carry **momentum** $p = E/c = h/\lambda$, not because they lack mass
5. The momentum transfer (despite zero mass) causes the pressure, so (R) does NOT correctly explain (A)
6. **Conclusion**: Both statements are true, but (R) is not the correct explanation of (A)

Answer: Option (1)

Key idea: Combination of thin lenses in contact - Power and focal length addition

1. For two thin lenses in contact: $\frac{1}{f_{combined}} = \frac{1}{f_{convex}} + \frac{1}{f_{concave}}$
2. Let $f_{convex} > 0$ (positive) and $f_{concave} < 0$ (negative), so $\frac{1}{f_{combined}} = \frac{1}{f_{convex}} - \frac{1}{|f_{concave}|}$
3. Rewrite: $\frac{1}{f_{combined}} = \frac{|f_{concave}| - f_{convex}}{f_{convex} \cdot |f_{concave}|}$
4. If $|f_{convex}| > |f_{concave}|$, then numerator is negative, so $f_{combined} < 0$ (concave lens behavior)
5. If $|f_{convex}| < |f_{concave}|$, then numerator is positive, so $f_{combined} > 0$ (convex lens behavior)
6. Since addition is commutative, interchanging lens positions doesn't change $f_{combined}$
7. Therefore, option (1) is correct

Answer: Option (2)

Key idea: Two lens system - successive image formation using lens formula

1. For lens P: Use lens formula with $u_P = -15$ cm, $f_P = 10$ cm to get $v_P = 30$ cm (real image 30 cm right of P)
2. Calculate magnification by P: $m_P = v_P/u_P = 30/(-15) = -2$
3. Image from P acts as virtual object for Q at distance $u_Q = +(30-15) = +15$ cm (positive sign for virtual object)
4. For lens Q: Use lens formula with $u_Q = +15$ cm, $f_Q = 15$ cm to get $1/v_Q = 1/15 + 1/15 = 2/15$, so $v_Q = 7.5$ cm
5. Calculate magnification by Q: $m_Q = v_Q/u_Q = 7.5/15 = 1/2$
6. Total magnification: $m_{total} = m_P \times m_Q = (-2) \times (1/2) = -1$, so size is same as AB
7. Final image is real (positive $v_Q$), at 7.5 cm right of Q, same size as AB

Answer: Option (2)

Key idea: Young's Double Slit Experiment - Intensity at arbitrary point using path difference and phase difference

1. Given: $d = 5\lambda$, $D = 10d = 50\lambda$. Point P is directly opposite to slit S₁.
2. Calculate distances: $r_1 = D = 50\lambda$ and $r_2 = \sqrt{D^2 + d^2} = \sqrt{(50\lambda)^2 + (5\lambda)^2} = 5\lambda\sqrt{101}$
3. Find path difference: $\Delta = r_2 - r_1 = 5\lambda(\sqrt{101} - 10) \approx 5\lambda(0.0499) \approx \frac{\lambda}{4}$
4. Calculate phase difference: $\phi = \frac{2\pi}{\lambda} \times \Delta = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$
5. Use intensity formula: $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$ where $I_1 = I_2 = \frac{I_0}{4}$
6. Substitute: $I = \frac{I_0}{4} + \frac{I_0}{4} + 2\left(\frac{I_0}{4}\right)\cos\left(\frac{\pi}{2}\right) = \frac{I_0}{2} + 0 = \frac{I_0}{2}$

Answer: Option (3)

Key idea: de Broglie wavelength and its inverse relationship with velocity

1. Write the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$
2. In free space: $\lambda_0 = \frac{h}{mv}$
3. Velocity in medium is reduced by 20%: $v_{medium} = 0.8v$
4. In medium: $\lambda_{medium} = \frac{h}{m(0.8v)} = \frac{h}{0.8mv}$
5. Find ratio: $\alpha = \frac{\lambda_{medium}}{\lambda_0} = \frac{h/(0.8mv)}{h/(mv)} = \frac{1}{0.8} = 1.25$

Answer: Option (2)

Key idea: Mass Defect and Nuclear Binding Energy

1. Identify that $^{12}_6$C has 6 protons and 6 neutrons (A - Z = 12 - 6 = 6)
2. Calculate total mass of separated nucleons: $6 \times 1.00727 + 6 \times 1.00866 = 12.09558$ u
3. Find mass defect: $\Delta m = 12.09558 - 12 = 0.09558$ u
4. Convert to MeV/c²: $0.09558 \times 931.5 = 89.03$ MeV/c²

Answer: Option (2)

Key idea: Depletion region width in p-n junction with unequal doping

1. Given: $N_a = 10^{15}$ atoms/cm³ (p-side), $N_d = 10^{18}$ atoms/cm³ (n-side)
2. Apply charge neutrality condition at junction: $N_a \times x_p = N_d \times x_n$
3. Calculate width ratio: $\frac{x_p}{x_n} = \frac{N_d}{N_a} = \frac{10^{18}}{10^{15}} = 1000$
4. Therefore: $x_p = 1000 \times x_n$
5. The depletion region is much wider on the lightly doped p-side

Answer: 33

Key idea: Elastic Potential Energy in Stretched Wire

1. Given: $L = 3$ m, $\Delta L = 3 \times 10^{-3}$ m, $V = 600 \times 10^{-6}$ m³, $Y = 1.1 \times 10^{11}$ N/m²
2. Find cross-sectional area: $A = \frac{V}{L} = \frac{600 \times 10^{-6}}{3} = 200 \times 10^{-6}$ m²
3. Calculate strain: $\epsilon = \frac{\Delta L}{L} = \frac{3 \times 10^{-3}}{3} = 10^{-3}$
4. Use elastic potential energy formula: $U = \frac{1}{2} Y \epsilon^2 V$
5. Substitute values: $U = \frac{1}{2} \times 1.1 \times 10^{11} \times (10^{-3})^2 \times 600 \times 10^{-6}$
6. Calculate: $U = \frac{1}{2} \times 1.1 \times 10^{11} \times 10^{-6} \times 6 \times 10^{-10} = 33$ J

Answer: 922

Key idea: Heat Transfer and Phase Change - Energy Conservation

1. Heat extracted from $x$ g water cooling from 50°C to 0°C: $Q_1 = \frac{x}{1000} \times 4200 \times 50 = 210x$ J
2. Heat needed to raise $(1000-x)$ g water from 50°C to 100°C: $Q_{2a} = \frac{1000-x}{1000} \times 4200 \times 50 = 210(1000-x)$ J
3. Heat needed to evaporate $(1000-x)$ g water at 100°C: $Q_{2b} = \frac{1000-x}{1000} \times 2256000 = 2256(1000-x)$ J
4. Total heat required: $Q_2 = 210(1000-x) + 2256(1000-x) = 2466(1000-x)$ J
5. Apply energy conservation: $210x = 2466(1000-x)$
6. Solve: $2676x = 2466000 \Rightarrow x = 921.52$
7. Closest integer value: $x = 922$

Answer: 200

Key idea: Resonance in LCR circuits - At resonance, inductive reactance equals capacitive reactance

1. Given: $R = 20\,\Omega$, $L = 1.6\,\text{H}$, $C = 40\,\mu\text{F} = 40 \times 10^{-6}\,\text{F}$
2. At resonance, the angular frequency is $\omega_0 = \frac{1}{\sqrt{LC}}$
3. Calculate: $\omega_0 = \frac{1}{\sqrt{1.6 \times 40 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = 125\,\text{rad/s}$
4. Inductive reactance: $X_L = \omega_0 L = 125 \times 1.6 = 200\,\Omega$
5. Verify: At resonance $X_L = X_C$, so $X_C = \frac{1}{\omega_0 C} = \frac{1}{125 \times 40 \times 10^{-6}} = 200\,\Omega$ ✓

Answer: 1

Key idea: EMF and internal resistance of a cell using Ohm's law

1. For a cell with internal resistance $r$ and EMF $E$: $E = I(R + r)$
2. **Case 1:** With $R_1 = 5\Omega$ and $I_1 = 0.25$ A: $E = 0.25(5 + r) = 1.25 + 0.25r$
3. **Case 2:** With $R_2 = 2\Omega$ and $I_2 = 0.5$ A: $E = 0.5(2 + r) = 1 + 0.5r$
4. Equate both expressions: $1.25 + 0.25r = 1 + 0.5r$
5. Solve: $0.25 = 0.25r$, therefore $r = 1\Omega$
6. Verify: $E = 1.5$ V gives correct currents in both cases

Answer: Option (3)

Key idea: Stoichiometry, Limiting Reagent, Gas Volume at STP

1. Calculate mass of $H_2SO_4$ solution: $50 \text{ mL} \times 1.3 \text{ g/mL} = 65$ g
2. Find mass of pure $H_2SO_4$: $65 \times 0.5 = 32.5$ g (50% purity)
3. Calculate moles: $H_2SO_4 = \frac{32.5}{98} = 0.3316$ mol and $Zn = \frac{20}{65} = 0.3077$ mol
4. Identify limiting reagent: $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$ (1:1:1 ratio), Zn is limiting
5. Calculate $H_2$ volume at STP: $0.3077 \text{ mol} \times 22.4 \text{ L/mol} = 6.892$ L

Answer: Option (2)

Key idea: Aufbau Principle, Orbital Energy, Orbital Size, and Radial Nodes

1. **Statement A**: According to Aufbau principle (n+l rule), when two orbitals have same $(n+l)$ value, the one with lower $n$ has lower energy. TRUE ✓
2. **Statement B**: As atomic number increases, effective nuclear charge increases, making orbital energies MORE NEGATIVE (lower, not higher). FALSE ✗
3. **Statement C**: For same orbital type, size increases with principal quantum number $n$. Since $n=3 > n=2$, size of $3p_z > 2p_z$. TRUE ✓
4. **Statement D**: Calculate radial nodes using formula: $n - l - 1$. For $5d$: $5 - 2 - 1 = 2$ radial nodes. So the statement 'none have 2 radial nodes' is FALSE ✗
5. Only statements A and C are true

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Key idea: Covalent radius and bond length in diatomic molecules

Answer: Option (2)

Key idea: Gibbs Free Energy calculation using ΔG° = ΔH° - TΔS°

1. Calculate $\Delta_r H^\circ = \Sigma \Delta_f H^\circ_{\text{products}} - \Sigma \Delta_f H^\circ_{\text{reactants}} = 2(42) - (8 + 80) = -4$ kJ/mol
2. Calculate $\Delta_r S^\circ = \Sigma S^\circ_{\text{products}} - \Sigma S^\circ_{\text{reactants}} = 2(200) - (140 + 250) = 10$ J/(mol·K)
3. Convert $\Delta_r S^\circ$ to kJ/(mol·K): $\Delta_r S^\circ = 10$ J/(mol·K) $= 0.010$ kJ/(mol·K)
4. Apply Gibbs equation: $\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ$
5. Substitute values: $\Delta_r G^\circ = -4 - (600)(0.010) = -4 - 6 = -10$ kJ/mol

Answer: Option (2)

Key idea: Molar heat capacities of different states of matter and their dependence on degrees of freedom

1. Identify the state of each substance: He(g) is monoatomic gas, Cu(s) is solid metal, Br₂(l) is liquid
2. For monoatomic ideal gas (He): $C_p = \frac{5}{2}R \approx 20.8$ J/(mol·K)
3. For solid metal (Cu): Use Dulong-Petit law, $C_p \approx 3R \approx 24.4$ J/(mol·K)
4. For liquid (Br₂): Molecular liquid has highest heat capacity due to multiple degrees of freedom, $C_p \approx 75.7$ J/(mol·K)
5. Compare values: Br₂(l) > Cu(s) > He(g)
6. Answer: Option (2)

Answer: Option (2)

Key idea: Nernst Equation and Cell Potential

1. Identify cell reaction: Zn acts as anode (oxidation) and Ag as cathode (reduction). Overall: $Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$
2. Calculate standard cell potential: $E^o_{cell} = E^o_{Ag^+/Ag} - E^o_{Zn^{2+}/Zn} = 0.80 - (-0.76) = 1.56$ V
3. Apply Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Ag^+]^2}$ where $n=2$
4. Substitute values: $1.60 = 1.56 - \frac{0.059}{2} \log \frac{1}{x^2}$ where $x = [Ag^+]$
5. Simplify: $0.04 = \frac{0.059}{2} \times 2\log x = 0.059 \log x$
6. Solve for $\log x$: $\log x = \frac{0.04}{0.059} = \frac{4}{5.9}$

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Nature of oxides (acidic, basic, amphoteric, neutral) and their classification based on metal oxidation states

Answer: Option (4)

Key idea: VSEPR Theory and Coordination Chemistry

1. **Statement I Analysis:** Al reacts with NaOH to form $[Al(OH)_4]^-$ (tetrahedral), NOT $[Al(OH)_6]^{3-}$. Statement I is **FALSE**.
2. **For $ICl_4^-$:** I has 7 valence e⁻, 4 bonds, 1 negative charge → Total = 12 e⁻ = 6 pairs. With 4 bonding pairs and 2 lone pairs, geometry is **square planar** ✓
3. **For $ClO_3^-$:** Cl has 7 valence e⁻, 3 bonds, 1 negative charge → 8 e⁻ total. With 3 bonding pairs and 1 lone pair (steric number = 4), geometry is **pyramidal** ✓
4. **For $IBr_2^-$:** I has 7 valence e⁻, 2 bonds, 1 negative charge → 10 e⁻ = 5 pairs. With 2 bonding pairs and 3 lone pairs, geometry is **linear** ✓
5. **Statement II Analysis:** All three geometries are correctly stated. Statement II is **TRUE**.
6. **Conclusion:** Statement I is false, Statement II is true → Answer is option **(4)**

Answer: Option (1)

Key idea: Crystal Field Theory - d-orbital splitting in octahedral and tetrahedral complexes, and relationship between unpaired electrons and enthalpy of atomization in transition metals

1. **Statement I:** More unpaired d-electrons in transition metals → more electrons participate in metallic bonding → stronger metal-metal bonds → higher enthalpy of atomization. Statement I is **correct**.
2. **Statement II - Part 1:** $[Fe(H_2O)_6]^{3+}$ is octahedral (weak field ligand H₂O). In octahedral field: $t_{2g}$ (lower) and $e_g$ (higher).
3. For octahedral: $d_{xy} = d_{xz} = d_{yz} < d_{x^2-y^2} = d_{z^2}$. This matches the statement. ✓
4. **Statement II - Part 2:** $[Ni(Cl)_4]^{2-}$ is tetrahedral (4 ligands, weak field Cl⁻). In tetrahedral field: $e$ (lower) and $t_2$ (higher).
5. For tetrahedral: $d_{x^2-y^2} = d_{z^2} < d_{xy} = d_{xz} = d_{yz}$. This matches the statement. ✓
6. Both Statement I and Statement II are correct.

Answer: Option (3)

Key idea: Stability of coordination complexes, chelate effect, hybridization, linkage isomerism, and ambidentate ligands

1. **Statement A**: $[Fe(C_2O_4)_3]^{3-}$ has oxalate (bidentate chelating ligand) which provides chelate effect, making it more stable than complexes with monodentate ligands OH⁻ and SCN⁻. **CORRECT**
2. **Statement B**: $[Cu(en)_2]^{2+}$ has ethylenediamine (bidentate) which is more stable than $[Cu(NH_3)_4]^{2+}$ with monodentate NH₃ due to chelate effect. **INCORRECT**
3. **Statement C**: In $K_4[Fe(CN)_6]$, Fe is in +2 state with configuration 3d⁶. CN⁻ is strong field, causing pairing. Hybridization is $d^2sp^3$ (inner orbital complex). **CORRECT**
4. **Statement D**: NO₂⁻ is ambidentate (can bind through N or O), so $[Fe(NO_2)_3Cl_3]^{3-}$ exhibits linkage isomerism (nitro vs nitrito). **CORRECT**
5. **Statement E**: Both NO₂⁻ (N or O binding) and SCN⁻ (S or N binding) ARE ambidentate ligands. **INCORRECT**
6. Correct statements are A, C, and D only

Answer: Option (2)

Key idea: Purification techniques in organic chemistry - matching distillation methods with their applications

1. **Simple distillation (A)**: Used to separate two liquids with **large difference** in boiling points (>25°C). Match: $A \rightarrow II$
2. **Fractional distillation (B)**: Used to separate two liquids with **close boiling points** using a fractionating column. Match: $B \rightarrow IV$
3. **Steam distillation (C)**: Used to separate **steam volatile compounds** that are immiscible with water and heat-sensitive. Match: $C \rightarrow I$
4. **Distillation under reduced pressure (D)**: Used for liquids that **decompose at their boiling point** by lowering the bp through vacuum. Match: $D \rightarrow III$
5. Final matching: $A-II, B-IV, C-I, D-III$ which corresponds to **option (2)**

Answer: Option (1)

Key idea: Alkene stability based on degree of substitution and hyperconjugation

1. Identify the degree of substitution for each alkene by counting alkyl groups attached to the double bond carbons
2. **C (2,3-Dimethylbut-2-ene)**: $(CH_3)_2C=C(CH_3)_2$ - tetrasubstituted (4 alkyl groups)
3. **A (2-Methylbut-2-ene)**: $CH_3-C(CH_3)=CH-CH_3$ - trisubstituted (3 alkyl groups)
4. **B (cis-But-2-ene)**: $CH_3-CH=CH-CH_3$ (cis) - disubstituted (2 alkyl groups)
5. **D (Prop-1-ene)**: $CH_2=CH-CH_3$ - monosubstituted (1 alkyl group)
6. Apply stability rule: More substituted alkenes are more stable due to hyperconjugation and inductive effects
7. Stability order: $C > A > B > D$ (tetra > tri > di > mono)

Answer: Option (3)

Key idea: Primary carbon identification and molecular formula determination from molar mass

1. Determine molecular formula: Molar mass = 72 g/mol. For alkane $C_nH_{2n+2}$: $12n + 2n + 2 = 72$ gives $n = 5$, so $C_5H_{12}$
2. Recall: Primary carbon ($1°$) is bonded to only ONE other carbon atom
3. Check option (1): 1,1-Dimethylcyclopropane is $C_5H_{10}$ (mass = 70 g/mol) ✗
4. Check option (2): 2,2-Dimethylpropane has 4 primary carbons (all four $CH_3$ groups) ✗
5. Check option (3): 2-Methylbutane: $CH_3-CH(CH_3)-CH_2-CH_3$ has exactly 3 primary carbons (three $CH_3$ groups) ✓
6. Check option (4): n-Pentane has only 2 primary carbons (two terminal $CH_3$ groups) ✗
7. Answer: 2-Methylbutane satisfies both conditions (mass = 72 g/mol and 3 primary carbons)

Answer: Option (1)

Key idea: Hell-Volhard-Zelinsky (HVZ) reaction for α-halogenation of carboxylic acids and nitrile hydrolysis

1. Start with propanenitrile ($CH_3CH_2CN$)
2. Step (i) & (ii): Basic hydrolysis followed by acidification converts nitrile to carboxylic acid: $CH_3CH_2CN \xrightarrow{OH^-/H_2O/\Delta} CH_3CH_2COO^- \xrightarrow{H_3O^+} CH_3CH_2COOH$
3. Step (iii): Hell-Volhard-Zelinsky reaction - $Cl_2$/Red P causes α-halogenation of carboxylic acid via acyl chloride intermediate
4. Red P + $Cl_2$ forms $PCl_3$, which converts $COOH$ to $COCl$ and enables enolization for α-chlorination
5. Step (iv): Hydrolysis of acyl chloride: $CH_3CHClCOCl + H_2O \rightarrow CH_3CHClCOOH$
6. Final product P is 2-chloropropanoic acid (chlorine at α-carbon, carbon-2)

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Condensation reactions of carbonyl compounds with semicarbazide and hydrolysis of hemiacetals

Answer: Option (4)

Key idea: Hofmann Bromamide Degradation and Electrophilic Aromatic Substitution in Aniline

1. **Statement I Analysis:** Hofmann bromamide degradation of benzamide: $\text{C}_6\text{H}_5\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{NH}_2$ (aniline)
2. The product is **aniline** ($\text{C}_6\text{H}_5\text{NH}_2$), NOT benzylamine ($\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$)
3. Therefore, **Statement I is FALSE**
4. **Statement II Analysis:** In acidic medium ($\text{HNO}_3/\text{H}_2\text{SO}_4$), aniline gets protonated to anilinium ion ($\text{C}_6\text{H}_5\text{NH}_3^+$)
5. The $-\text{NH}_3^+$ group is **meta-directing** (electron-withdrawing), so $m$-nitroaniline is the major product
6. Therefore, **Statement II is TRUE**
7. **Answer:** Statement I is false but Statement II is true → Option (4)

Answer: Option (3)

Key idea: Tertiary Structure of Proteins and Denaturation

1. Analyze statement (1): Proteins can indeed be fibrous or globular - this is **correct**
2. Analyze statement (2): Tertiary structure is stabilized by H-bonds, disulfide links, van der Waals forces, and electrostatic interactions - this is **correct**
3. Analyze statement (3): Tertiary structure is **highly sensitive** to pH changes, which can cause denaturation by disrupting ionic interactions and hydrogen bonds
4. Statement (3) claims structure remains intact with pH changes - this is **INCORRECT** (proteins denature with pH changes)
5. Analyze statement (4): The hierarchy Primary → Secondary → Tertiary structure is correct
6. Therefore, option **(3)** is the incorrect statement

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Anomers of glucose and stereochemistry of monosaccharides

Answer: Option (4)

Key idea: Redox reactions of transition metal compounds; Reduction of dichromate by SO₂

1. Identify that $SO_2$ acts as a reducing agent in acidic medium
2. Analyze that the green color indicates formation of $Cr^{3+}$ ions
3. Write the reduction reaction: $Cr_2O_7^{2-}$ (orange) $+ 3SO_2 + 2H^+ \rightarrow 2Cr^{3+}$ (green) $+ 3SO_4^{2-} + H_2O$
4. Recognize this as a characteristic test for potassium dichromate
5. Therefore, compound X is $K_2Cr_2O_7$

Answer: 15

Key idea: Crystal Field Theory - Electron Configuration and Unpaired Electrons in Octahedral Complexes

1. In octahedral complexes, d-orbitals split into $t_{2g}$ (3 orbitals, lower energy) and $e_g$ (2 orbitals, higher energy)
2. $d^3$: Configuration is $t_{2g}^3$ with all electrons unpaired = **3 unpaired electrons**
3. $d^4$ (low spin): Strong field causes pairing in $t_{2g}$. Configuration: $t_{2g}^4$ (↑↓ ↑ ↑) = **2 unpaired electrons**
4. $d^5$ (high spin): Weak field, all orbitals filled singly first. Configuration: $t_{2g}^3 e_g^2$ = **5 unpaired electrons**
5. $d^6$ (high spin): Configuration is $t_{2g}^4 e_g^2$ (↑↓ ↑ ↑ | ↑ ↑) = **4 unpaired electrons**
6. $d^7$ (low spin): Strong field causes maximum pairing. Configuration: $t_{2g}^6 e_g^1$ = **1 unpaired electron**
7. Total unpaired electrons = $3 + 2 + 5 + 4 + 1 = 15$

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Grignard reagent hydrolysis, iodoform formation, and Wurtz-type reductive coupling

Answer: 62

Key idea: Osmotic Pressure and Colligative Properties

1. Calculate osmotic pressure from height: $\pi = \rho gh = 1000 \times 10 \times 0.08 = 800$ Pa $= 0.8$ kPa
2. Apply van't Hoff equation: $\pi = \frac{n}{V}RT$ where $n$ is moles of hemoglobin
3. Calculate moles: $n = \frac{\pi V}{RT} = \frac{0.8 \times 1}{8.3 \times 300} = \frac{0.8}{2490} = 3.213 \times 10^{-4}$ mol
4. Convert mass to kg: mass $= 20$ g $= 0.02$ kg
5. Calculate molar mass: $M = \frac{\text{mass}}{n} = \frac{0.02}{3.213 \times 10^{-4}} = 62.24$ kg mol$^{-1}$
6. Round to nearest integer: $M = 62$ kg mol$^{-1}$

Answer: 15

Key idea: Relationship between conductivity, molar conductivity, and concentration; conversion between molarity and weight/weight percentage

1. Calculate conductivity: $\kappa = G \times \text{cell constant} = 1.9 \times 10^{-3} \times 1.3 = 2.47 \times 10^{-3}$ S cm$^{-1}$
2. Use the relation $\Lambda_m = \frac{\kappa \times 1000}{C}$ to find molarity: $C = \frac{2.47 \times 10^{-3} \times 1000}{123.5} = 0.02$ mol L$^{-1}$
3. For $x\%$ (w/w) solution: 100 g solution contains $x$ g of MX with volume = $\frac{100}{1.0} = 100$ mL = 0.1 L
4. Express molarity in terms of $x$: $C = \frac{x}{75 \times 0.1} = \frac{x}{7.5}$ mol L$^{-1}$
5. Solve for $x$: $0.02 = \frac{x}{7.5} \Rightarrow x = 0.15 = 15 \times 10^{-2}$

Answer: 90

Key idea: Zero-order reaction kinetics: half-life is directly proportional to initial concentration

1. Observe that $t_{1/2}$ vs $[A]_0$ is a straight line through origin, indicating **zero-order reaction**
2. For zero-order reaction: $t_{1/2} = \frac{[A]_0}{2k}$, so $t_{1/2} \propto [A]_0$
3. From graph, at $[A]_0 = 4 \times 10^{-3}$ mol L$^{-1}$, $t_{1/2} = 240$ s
4. Calculate slope: $m = \frac{240}{4 \times 10^{-3}} = 60000$ s·L·mol$^{-1}$
5. Find $x$ at $[A]_0 = 1.5 \times 10^{-3}$: $x = 60000 \times 1.5 \times 10^{-3} = 90$ s