JEE Main 2 April 2026 Shift 2: Questions and Solutions

Answer: Option (4)

Key idea: Vieta's formulas and relationships between roots of quadratic equations

1. From $x^2 - 3x + r = 0$: $\alpha + \beta = 3$ and $\alpha\beta = r$
2. From $x^2 + 3x + r = 0$: $\frac{\alpha}{2} + 2\beta = -3$ and $\frac{\alpha}{2} \cdot 2\beta = r$
3. Solve the system: $\alpha + \beta = 3$ and $\alpha + 4\beta = -6$ to get $\beta = -3$ and $\alpha = 6$
4. Calculate $r = \alpha\beta = 6 \times (-3) = -18$
5. Find the roots of third equation: $2\alpha + \beta + 2r = 12 - 3 - 36 = -27$ and $\alpha - 2\beta - \frac{r}{2} = 6 + 6 + 9 = 21$
6. For $x^2 + 6x = m$, product of roots $= (-27)(21) = -567 = -m$
7. Therefore, $m = 567$

Answer: Option (3)

Key idea: Distance between points on two circles in complex plane

1. Step 1: Identify circle $C_1$ centered at origin with radius $r$, and $C_2$ centered at $3+4i$ with radius $5$. Distance between centers: $d = |3+4i| = 5$.
2. Step 2: Since $C_2$ lies within $C_1$, we need $r > d + 5 = 10$.
3. Step 3: Minimum distance between points on the circles: $\min|z_1-z_2| = r - (d+5) = r - 10$.
4. Step 4: Given $\min|z_1-z_2| = 2$, solve: $r - 10 = 2 \Rightarrow r = 12$.
5. Step 5: Maximum distance occurs when points are on opposite sides along the line joining centers: $\max|z_1-z_2| = r + d + 5 = 12 + 5 + 5 = 22$.

Answer: Option (2)

Key idea: Condition for infinitely many solutions in a system of linear equations - the third equation must be a linear combination of the first two

1. For infinitely many solutions, the third equation must be a linear combination of the first two: $x + 6y + az = \lambda(x + 5y + 6z) + \mu(2x + 3y + 4z)$
2. Compare coefficients: $1 = \lambda + 2\mu$ and $6 = 5\lambda + 3\mu$
3. Solve the system: From first equation $\lambda = 1 - 2\mu$. Substitute in second: $6 = 5(1-2\mu) + 3\mu$ gives $\mu = -\frac{1}{7}$ and $\lambda = \frac{9}{7}$
4. Find $a$ from coefficient of $z$: $a = 6\lambda + 4\mu = 6(\frac{9}{7}) + 4(-\frac{1}{7}) = \frac{50}{7}$
5. Find $b$ from constant terms: $b = 4\lambda + 7\mu = 4(\frac{9}{7}) + 7(-\frac{1}{7}) = \frac{29}{7}$
6. Check which line passes through $(\frac{50}{7}, \frac{29}{7})$: $x - y = \frac{50}{7} - \frac{29}{7} = \frac{21}{7} = 3$ ✓

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Answer: Option (2)

Key idea: Sum of series using standard formulas for sum of cubes and sum of odd numbers

1. Identify the general term: numerator is $1^3 + 2^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$ and denominator is $1 + 3 + 5 + \cdots + (2n-1) = n^2$
2. Simplify the $n$th term: $T_n = \frac{[n(n+1)/2]^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4}$
3. Verify with first few terms: $T_1 = 1$, $T_2 = \frac{9}{4}$, $T_3 = 4$
4. Sum of 8 terms: $S_8 = \frac{1}{4}\sum_{n=1}^{8}(n+1)^2 = \frac{1}{4}\sum_{m=2}^{9}m^2$
5. Calculate: $\sum_{m=2}^{9}m^2 = \frac{9(10)(19)}{6} - 1 = 285 - 1 = 284$
6. Therefore: $S_8 = \frac{284}{4} = 71$

Answer: Option (3)

Key idea: Vandermonde's Identity and Binomial Theorem

1. Recognize the coefficients $1, 3, 3, 1$ as binomial coefficients $\binom{3}{0}, \binom{3}{1}, \binom{3}{2}, \binom{3}{3}$
2. Rewrite the expression as $\sum_{k=0}^{3} \binom{3}{k}\binom{30}{(30-r)+k}$
3. Apply Vandermonde's identity: $\sum_{k=0}^{p} \binom{p}{k}\binom{n}{m+k} = \binom{n+p}{m+p}$
4. With $p=3$, $n=30$, and $m=30-r$, we get $\binom{30+3}{(30-r)+3} = \binom{33}{33-r}$
5. Use symmetry property: $\binom{33}{33-r} = \binom{33}{r}$
6. Therefore, $m = 33$

Answer: Option (3)

Key idea: Combinations and Prime Factorization

1. Step 1: Number of triangles from $n$ vertices is $P_n = \binom{n}{3} = \frac{n(n-1)(n-2)}{6}$
2. Step 2: Calculate $P_{n+1} - P_n = \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = \frac{n(n-1)}{2}$
3. Step 3: Set up equation: $\frac{n(n-1)}{2} = 66$, so $n(n-1) = 132$
4. Step 4: Solve $n^2 - n - 132 = 0$ using quadratic formula: $n = \frac{1 + 23}{2} = 12$
5. Step 5: Find prime factorization: $12 = 2^2 \times 3$
6. Step 6: Sum of distinct prime divisors: $2 + 3 = 5$

Answer: Option (3)

Key idea: Probability with constraints - finding all possible outcomes satisfying a linear equation and computing total probability

1. Set up equation: $10H + 5T = 30$ where $H$ = heads, $T$ = tails. Simplify to $T = 6 - 2H$
2. Find valid cases: $H \in \{0, 1, 2, 3\}$ gives $(H,T) \in \{(0,6), (1,4), (2,2), (3,0)\}$
3. Calculate probabilities: $P(H=0,T=6) = \binom{6}{0}(\frac{1}{2})^6 = \frac{1}{64}$, $P(H=1,T=4) = \binom{5}{1}(\frac{1}{2})^5 = \frac{10}{64}$
4. Continue: $P(H=2,T=2) = \binom{4}{2}(\frac{1}{2})^4 = \frac{24}{64}$, $P(H=3,T=0) = \binom{3}{3}(\frac{1}{2})^3 = \frac{8}{64}$
5. Sum all probabilities: $\frac{1+10+24+8}{64} = \frac{43}{64}$. Since $\gcd(43,64)=1$, we have $m=43, n=64$
6. Therefore $m + n = 43 + 64 = 107$

Answer: Option (4)

Key idea: Mean and Variance formulas with algebraic manipulation

1. Use mean formula: $\sum x_i = n \times 8 = 8n$. Since $\sum_{i=1}^{n-1} x_i = 48$, we get $x_n = 8n - 48$
2. Use variance formula: $\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2$, so $16 = \frac{\sum x_i^2}{n} - 64$, giving $\sum x_i^2 = 80n$
3. Since $\sum_{i=1}^{n-1} x_i^2 = 496$, we get $x_n^2 = 80n - 496$
4. Substitute $x_n = 8n - 48$ into $x_n^2 = 80n - 496$: $(8n-48)^2 = 80n - 496$
5. Expand and simplify: $64n^2 - 768n + 2304 = 80n - 496$, which gives $4n^2 - 53n + 175 = 0$
6. Solve using quadratic formula: $n = \frac{53 \pm 3}{8}$, giving $n = 7$ or $n = 6.25$
7. Since $n$ must be an integer, $n = 7$

Answer: Option (3)

Key idea: Circle passing through origin, perpendicular lines, chord length using perpendicular distance formula

1. Use perpendicularity condition $a_1a_2 + b_1b_2 = 0$ to get $2 + k^2(k-1) = 0$, which gives $k = -1$
2. Substitute $k = -1$ to get lines $x - 2y + 3 = 0$ and $2x + y - 4 = 0$, solve to find center $(1, 2)$
3. Since circle passes through origin, radius $r^2 = 1^2 + 2^2 = 5$
4. Find perpendicular distance from center $(1, 2)$ to line $x - y + 2 = 0$: $d = \frac{|1-2+2|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$
5. Use chord length formula: $(AB)^2 = 4(r^2 - d^2) = 4(5 - \frac{1}{2}) = 18$

Answer: Option (3)

Key idea: Finding area of triangle formed by origin and two points on a hyperbola given their midpoint

1. Since $P(x_1, y_1)$ and $Q(x_2, y_2)$ lie on $xy = 12$, we have $x_1y_1 = 12$ and $x_2y_2 = 12$
2. From midpoint condition: $x_1 + x_2 = 1$ and $y_1 + y_2 = -1$
3. Substitute $y_1 = \frac{12}{x_1}$ and $y_2 = \frac{12}{x_2}$ into $y_1 + y_2 = -1$ to get $\frac{12(x_1+x_2)}{x_1x_2} = -1$
4. This gives $x_1x_2 = -12$. With $x_1 + x_2 = 1$, solve $t^2 - t - 12 = 0$ to get $x_1 = 4, x_2 = -3$
5. Find coordinates: $P = (4, 3)$ and $Q = (-3, -4)$
6. Area of $\triangle OPQ = \frac{1}{2}|x_1y_2 - x_2y_1| = \frac{1}{2}|4(-4) - (-3)(3)| = \frac{1}{2}|-7| = \frac{7}{2}$

Answer: Option (2)

Key idea: Optimization of parabola vertex distance using constraint from a given point

1. Use the condition that parabola passes through $(1, -1)$: $-1 = 1 + p + q$, giving $q = -2 - p$
2. Find vertex coordinates: $x_v = -\frac{p}{2}$ and $y_v = -\frac{p^2}{4} + q = -\frac{p^2}{4} - p - 2$
3. Distance from vertex to $x$-axis is $|y_v| = \left|\frac{p^2}{4} + p + 2\right| = \frac{p^2 + 4p + 8}{4}$ (always positive)
4. Minimize $f(p) = p^2 + 4p + 8$ by setting $f'(p) = 2p + 4 = 0$, giving $p = -2$
5. Find $q = -2 - (-2) = 0$
6. Calculate $p^2 + q^2 = (-2)^2 + 0^2 = 4$

Answer: Option (1)

Key idea: Trigonometric identities, Set theory, and Range analysis

1. Step 1: Simplify $\cos^8\theta - \sin^8\theta$ using factorization: $\cos^8\theta - \sin^8\theta = (\cos^4\theta - \sin^4\theta)(\cos^4\theta + \sin^4\theta) = \cos 2\theta \cdot \frac{1 + \cos^2 2\theta}{2}$
2. Step 2: Substitute into the given expression: $2(\cos^8\theta - \sin^8\theta) \sec 2\theta = 2 \cdot \cos 2\theta \cdot \frac{1 + \cos^2 2\theta}{2} \cdot \frac{1}{\cos 2\theta} = 1 + \cos^2 2\theta$
3. Step 3: Find the range: Since $0 \leq \cos^2 2\theta \leq 1$, we have $1 \leq a^2 \leq 2$
4. Step 4: Check integer solutions: Only $a^2 = 1$ (giving $a = \pm 1$) is a perfect square in this range
5. Step 5: Verify if $a^2 = 1$ is achievable: $a^2 = 1$ requires $\cos 2\theta = 0$, which gives $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, ...$ where $\tan^2\theta = 1$
6. Step 6: Apply constraint: These values are excluded from $P$ since $\tan^2\theta \neq 1$ is required. Thus $S = \emptyset$ and $n(S) = 0$

Answer: Option (2)

Key idea: Vector operations, dot product, and solving linear equations to find vector magnitude

1. Express $\vec{c} = \lambda \vec{a} + \mu \vec{b} = (-\lambda + \mu)\hat{i} + (\lambda + 3\mu)\hat{j} + (3\lambda + \mu)\hat{k}$
2. Use condition $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ to get $-3\lambda - 13\mu = 10$
3. Use condition $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$ to get $3\lambda + 5\mu = -2$
4. Solve the system: Add equations to get $-8\mu = 8$, so $\mu = -1$ and $\lambda = 1$
5. Find $\vec{c} = \vec{a} - \vec{b} = -2\hat{i} - 2\hat{j} + 2\hat{k}$
6. Calculate $|\vec{c}|^2 = (-2)^2 + (-2)^2 + (2)^2 = 12$

Answer: Option (1)

Key idea: 3D Geometry: Foot of perpendicular from a point to a line, midpoint formula, and perpendicularity condition

1. Use midpoint formula: If $M(0, \frac{3}{4}, \frac{-1}{4})$ is midpoint of $PA$, then $A = (-a, \frac{3}{2} - b, \frac{-1}{2})$
2. Since $A$ lies on the line, write $A = (1+2t, 2+t, \alpha+3t)$ and equate: $1+2t = -a$, $2+t = \frac{3}{2}-b$, $\alpha+3t = \frac{-1}{2}$
3. From the second equation: $t = -\frac{1}{2} - b$. Substitute in first: $a = 2b$. Substitute in third: $\alpha = 1 + 3b$
4. Apply perpendicularity: $\vec{PA} \cdot \vec{d} = 0$ where $\vec{PA} = (-2a, \frac{3}{2}-2b, \frac{-1}{2})$ and $\vec{d} = (2, 1, 3)$
5. Simplify: $-4a - 2b = 0 \Rightarrow a = -\frac{b}{2}$. Combined with $a = 2b$ gives $b = 0$, hence $a = 0$ and $\alpha = 1$
6. Calculate: $a^2 + b^2 + \alpha^2 = 0^2 + 0^2 + 1^2 = 1$

Answer: Option (2)

Key idea: Vector dot product, angle between vectors, rotation in plane, trigonometric identities

1. Find the angle between $\overrightarrow{PQ}$ and $\overrightarrow{PS}$ using $\cos\theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PS}}{|\overrightarrow{PQ}||\overrightarrow{PS}|} = \frac{-1}{2}$, so $\theta = 120°$
2. After rotation by $\alpha$, vectors become perpendicular (angle = $90°$), so $120° - \alpha = 90°$, giving $\alpha = 30°$
3. Calculate $\frac{\alpha}{2} = 15°$ and $\frac{5\alpha}{2} = 75°$
4. Use identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$
5. Evaluate: $\sin^2(75°) - \sin^2(15°) = \sin(90°)\sin(60°) = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

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.

Answer: Option (4)

Key idea: Polynomial functions with extrema conditions and limit evaluation

1. Step 1: Since $f(x)$ has degree 5 and $\lim_{x\to 0}\frac{f(x)}{x^3} = -5$ exists, we know $f(0) = 0$ and $f(x) = x^3(Dx^2 + Ex + F)$ where $F = -5$. Thus $f(x) = Dx^5 + Ex^4 - 5x^3$.
2. Step 2: Find the derivative: $f'(x) = 5Dx^4 + 4Ex^3 - 15x^2$.
3. Step 3: Apply extrema conditions $f'(1) = 0$ and $f'(-1) = 0$: This gives $5D + 4E = 15$ and $5D - 4E = 15$.
4. Step 4: Solve the system: Adding equations gives $10D = 30$, so $D = 3$. Subtracting gives $8E = 0$, so $E = 0$. Therefore $f(x) = 3x^5 - 5x^3$.
5. Step 5: Calculate $f(2) = 3(32) - 5(8) = 96 - 40 = 56$ and $f(-2) = 3(-32) - 5(-8) = -96 + 40 = -56$.
6. Step 6: Find $f(2) - f(-2) = 56 - (-56) = 112$.

Answer: 39

Key idea: Partial Fractions and Logarithmic Integration

1. Step 1: Factor the denominator: $x^2 + 2x - 15 = (x+5)(x-3)$
2. Step 2: Use partial fractions: $\frac{16x + 24}{(x+5)(x-3)} = \frac{7}{x+5} + \frac{9}{x-3}$ (by solving $16x + 24 = A(x-3) + B(x+5)$)
3. Step 3: Integrate to get $f(x) = 7\ln|x+5| + 9\ln|x-3| + C$
4. Step 4: Use $f(4) = 14\ln(3)$ to find $C$: $7\ln(9) + 9\ln(1) + C = 14\ln(3)$, which gives $C = 0$
5. Step 5: Calculate $f(7) = 7\ln(12) + 9\ln(4) = 7\ln(2^2 \cdot 3) + 9\ln(2^2) = 32\ln(2) + 7\ln(3) = \ln(2^{32} \cdot 3^7)$
6. Step 6: Compare with $\ln(2^{\alpha} \cdot 3^{\beta})$ to get $\alpha = 32$, $\beta = 7$, so $\alpha + \beta = 39$

Answer: Option (2)

Key idea: Bernoulli Differential Equation with Substitution

1. Step 1: Use substitution $x = vy$ where $v$ is a function of $y$, so $\frac{dx}{dy} = v + y\frac{dv}{dy}$
2. Step 2: Substitute into the differential equation: $2y^2(v + y\frac{dv}{dy}) - 2vy^2 + v^2y^2 = 0$
3. Step 3: Simplify to get $2y^3\frac{dv}{dy} + v^2y^2 = 0$, which gives $\frac{dv}{v^2} = -\frac{dy}{2y}$
4. Step 4: Integrate both sides: $-\frac{1}{v} = -\frac{1}{2}\ln y + C$, so $\frac{1}{v} = \frac{1}{2}\ln y + C_1$
5. Step 5: Apply initial condition $x(e) = e$ to find $C_1 = \frac{1}{2}$, giving $v = \frac{2}{\ln y + 1}$
6. Step 6: Since $x = vy$, we have $x = \frac{2y}{\ln y + 1}$
7. Step 7: Calculate $x(e^2) = \frac{2e^2}{\ln(e^2) + 1} = \frac{2e^2}{3}$

Answer: 120

Key idea: Relations on Cartesian Products with Divisibility and Ordering Conditions

1. Understand that $R$ is a relation on $A \times A$, so elements are of the form $((x,y), (z,w))$ where $(x,y)R(z,w)$ means $x|z$ and $y \leq w$
2. Count valid $(x,z)$ pairs where $x|z$ and $x,z \in A$: For $x=2$: $z \in \{2,4,6\}$ (3 pairs); $x=3$: $z \in \{3,6\}$ (2 pairs); $x=4,5,6$: one pair each. Total: $3+2+1+1+1=8$ pairs
3. Count valid $(y,w)$ pairs where $y \leq w$ and $y,w \in A = \{2,3,4,5,6\}$: This gives $5+4+3+2+1=15$ pairs
4. Since the conditions on $(x,z)$ and $(y,w)$ are independent, multiply: Total elements $= 8 \times 15 = 120$

Answer: 34

Key idea: Matrix equations and inverse matrices

1. Step 1: Find $A^{-1}$ using $\det(A) = 4$, so $A^{-1} = \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$
2. Step 2: From $PA = B$, get $P = BA^{-1} = \begin{bmatrix} -21/2 & 6 \\ -7/2 & 2 \end{bmatrix}$
3. Step 3: From $AQ = B$, get $Q = A^{-1}B = \begin{bmatrix} -1 & -3 \\ -5/2 & -15/2 \end{bmatrix}$
4. Step 4: Calculate $P + Q = \begin{bmatrix} -23/2 & 3 \\ -6 & -11/2 \end{bmatrix}$
5. Step 5: Find $2(P + Q) = \begin{bmatrix} -23 & 6 \\ -12 & -11 \end{bmatrix}$
6. Step 6: Sum of diagonal elements = $-23 + (-11) = -34$
7. Step 7: Absolute value = $|-34| = 34$

Answer: 18

Key idea: Locus of points forming an ellipse using internal tangency condition

1. Let $B = (h, k)$. The variable circle has diameter $AB$, so its center is $\left(\frac{3+h}{2}, \frac{k}{2}\right)$ and radius is $\frac{|AB|}{2} = \frac{\sqrt{(h-3)^2 + k^2}}{2}$
2. For internal tangency with $x^2 + y^2 = 36$ (radius 6): Distance between centers = Difference of radii
3. This gives: $6 - \frac{\sqrt{(h-3)^2 + k^2}}{2} = \sqrt{\left(\frac{3+h}{2}\right)^2 + \left(\frac{k}{2}\right)^2}$
4. Simplifying: $\sqrt{(h-3)^2 + k^2} + \sqrt{(h+3)^2 + k^2} = 12$
5. This is an ellipse with foci at $F_1(-3, 0)$ and $F_2(3, 0)$, where $2a = 12$ (so $a = 6$) and $c = 3$
6. Eccentricity: $e = \frac{c}{a} = \frac{3}{6} = \frac{1}{2}$
7. Therefore: $72e^2 = 72 \times \frac{1}{4} = 18$

Answer: 24

Key idea: Area between curves (hyperbola and line), Integration involving square roots

1. Convert hyperbola to standard form: $\frac{x^2}{9} - \frac{y^2}{16} = 1$ and line: $y = \frac{8x-24}{3}$
2. Find intersection points by substituting line equation into hyperbola: $x^2 - 8x + 15 = 0$ gives $x = 3, 5$
3. Points of intersection are $(3, 0)$ and $(5, \frac{16}{3})$. For the hyperbola (right branch): $y = \frac{4}{3}\sqrt{x^2-9}$
4. Set up area integral: $A = \int_3^5 \left[\frac{4}{3}\sqrt{x^2-9} - \frac{8x-24}{3}\right] dx$ (hyperbola is above the line)
5. Evaluate: $\int_3^5 (8x-24)dx = 16$ and $\int_3^5 \sqrt{x^2-9}dx = 10 - \frac{9}{2}\ln(3)$ using standard formula
6. Calculate: $A = \frac{1}{3}[4(10 - \frac{9}{2}\ln 3) - 16] = 8 - 6\ln 3$
7. Therefore: $3(A + 6\ln_e 3) = 3(8 - 6\ln 3 + 6\ln 3) = 3(8) = 24$

Answer: 10

Key idea: Discontinuity of greatest integer function composition

1. Step 1: Recognize that $f(x) = \lfloor x^2 - x - \frac{1}{2} \rfloor$ is discontinuous when $g(x) = x^2 - x - \frac{1}{2}$ takes integer values.
2. Step 2: Find the range of $g(x)$ on $[2, 4]$: $g(2) = 1.5$ and $g(4) = 11.5$. Since $g'(x) = 2x - 1 > 0$ on this interval, $g$ is strictly increasing.
3. Step 3: The range of $g(x)$ is $[1.5, 11.5]$, so the integer values are $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$ (10 integers).
4. Step 4: For each integer $n$, solve $x^2 - x - \frac{1}{2} = n$ to get $x = \frac{1 + \sqrt{4n + 3}}{2}$.
5. Step 5: Verify that all 10 solutions lie in $[2, 4]$. Since $g$ is continuous and strictly increasing, each integer value is attained exactly once.
6. Step 6: Therefore, there are **10 points of discontinuity**.

Answer: Option (2)

Key idea: Dimensional Analysis

1. Find dimensions of $G$ from $F = \frac{Gm_1m_2}{r^2}$: $[G] = [M^{-1}L^3T^{-2}]$
2. Find dimensions of Planck's constant from $E = h\nu$: $[h] = [ML^2T^{-1}]$
3. Express $[G] = [h]^a[T]^b[L]^c[M]^d$ and equate powers
4. Set up equations: $-1 = a+d$, $3 = 2a+c$, $-2 = -a+b$
5. From options, $d = -2$, so $a = 1$, then $b = -1$, $c = 1$
6. Verify: $[h^1T^{-1}L^1M^{-2}] = [ML^2T^{-1}][T^{-1}][L][M^{-2}] = [M^{-1}L^3T^{-2}]$ ✓

Answer: Option (1)

Key idea: Circular motion with friction - vertical cylinder rotation

1. Identify forces: Normal force $N$ (toward center), weight $mg$ (downward), friction $f$ (upward)
2. For circular motion, centripetal force: $N = m\omega^2 r = 0.5 \times 25 \times 4 = 50$ N
3. For mass not to fall, friction must balance weight: $f = mg = 0.5 \times 10 = 5$ N
4. At minimum speed, friction is maximum: $f_{max} = \mu N$
5. Solve for coefficient of friction: $\mu = \frac{mg}{N} = \frac{5}{50} = 0.1$

Answer: Option (3)

Key idea: Atwood Machine and Center of Mass Motion

1. For the Atwood machine, find acceleration: $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(2-1) \times 10}{3} = \frac{10}{3}$ m/s²
2. Calculate displacement of each mass in 2 s: $s = \frac{1}{2}at^2 = \frac{1}{2} \times \frac{10}{3} \times 4 = \frac{20}{3}$ m
3. The 2 kg mass moves down $\frac{20}{3}$ m, the 1 kg mass moves up $\frac{20}{3}$ m
4. Displacement of center of mass: $\Delta y_{cm} = \frac{m_1(+\frac{20}{3}) + m_2(-\frac{20}{3})}{m_1 + m_2}$
5. $\Delta y_{cm} = \frac{2 \times \frac{20}{3} - 1 \times \frac{20}{3}}{3} = \frac{\frac{20}{3}}{3} = \frac{20}{9}$ m
6. Distance traversed = $\frac{20}{9} = 2.22$ m

Answer: Option (1)

Key idea: Lorentz Force on a Charged Particle

1. Write the Lorentz force equation: $\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B})$
2. Calculate electric force: $\vec{F}_E = 10^{-9} \times 0.4\hat{j} = 4 \times 10^{-10}\hat{j}$ N
3. Let velocity be $\vec{v} = v_x\hat{i} + v_y\hat{j}$. Calculate cross product: $\vec{v} \times \vec{B} = (v_x\hat{i} + v_y\hat{j}) \times (4 \times 10^{-3}\hat{k}) = 4 \times 10^{-3}(v_y\hat{i} - v_x\hat{j})$
4. Magnetic force: $\vec{F}_B = 10^{-9} \times 4 \times 10^{-3}(v_y\hat{i} - v_x\hat{j}) = 4 \times 10^{-12}(v_y\hat{i} - v_x\hat{j})$
5. Total force: $\vec{F} = 4v_y \times 10^{-12}\hat{i} + (4 \times 10^{-10} - 4v_x \times 10^{-12})\hat{j}$
6. Compare with given force $(4\hat{i} + 2\hat{j}) \times 10^{-10}$ N: From $\hat{i}$ component: $v_y = 100$ m/s; From $\hat{j}$ component: $v_x = 50$ m/s
7. Therefore, $\vec{v} = 50\hat{i} + 100\hat{j}$ m/s

Answer: Option (2)

Key idea: Logic Gates - NAND and NOR gate combinations, De Morgan's Laws

1. First NAND gate with both inputs as $X$: Output₁ = $\overline{X \cdot X} = \bar{X}$
2. Second NAND gate with both inputs as $Y$: Output₂ = $\overline{Y \cdot Y} = \bar{Y}$
3. NOR gate combines these outputs: Output = $\overline{\bar{X} + \bar{Y}}$
4. Apply De Morgan's Law: $\overline{\bar{X} + \bar{Y}} = \overline{\bar{X}} \cdot \overline{\bar{Y}} = X \cdot Y$
5. The final output is $X \cdot Y$, which is an AND gate

Answer: Option (1)

Key idea: Escape velocity and gravitational potential energy; Energy conservation for a body falling from infinity

1. Apply conservation of energy: Initial energy at infinity (= 0) equals final energy at Earth's surface
2. At surface: $0 = -\frac{GMm}{R} + \frac{1}{2}mv^2$, which gives $\frac{1}{2}mv^2 = \frac{GMm}{R}$
3. Use $g = \frac{GM}{R^2}$ to get $GM = gR^2$, so $v^2 = 2gR$
4. Calculate velocity: $v = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} = 11200$ m/s $= 11.2$ km/s
5. Calculate kinetic energy: $k = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times (11200)^2 = 6.27 \times 10^7$ J
6. Answer: $v = 11.2$ km/s and $k = 6.27 \times 10^7$ J

Answer: Option (1)

Key idea: Screw Gauge: Zero Error and Zero Correction

1. Calculate least count: $LC = \frac{\text{Pitch}}{\text{No. of divisions}} = \frac{0.1}{100} = 0.001$ mm
2. Find zero error: 5th division coincides when studs touch, so zero error $= +5 \times 0.001 = +0.005$ mm
3. Calculate observed reading: $\text{MSR} + \text{CSR} \times LC = 5 + 50 \times 0.001 = 5.050$ mm
4. Apply zero correction: Actual diameter $= 5.050 - 0.005 = 5.045$ mm

Answer: Option (3)

Key idea: Work done in changing surface area of soap bubble (with two surfaces)

1. Convert diameters to radii: $r_1 = 1$ cm $= 0.01$ m, $r_2 = 3$ cm $= 0.03$ m
2. For a soap bubble (two surfaces): Total surface area $= 8\pi r^2$
3. Initial area: $A_1 = 8\pi(0.01)^2 = 8\pi \times 10^{-4}$ m²
4. Final area: $A_2 = 8\pi(0.03)^2 = 72\pi \times 10^{-4}$ m²
5. Change in area: $\Delta A = (72-8)\pi \times 10^{-4} = 64\pi \times 10^{-4}$ m²
6. Work done: $W = T \times \Delta A = 0.03 \times 64\pi \times 10^{-4} = 1.92\pi \times 10^{-4}$ J
7. Comparing with $\alpha\pi \times 10^{-4}$ J, we get $\alpha = 1.92$

Answer: Option (3)

Key idea: Ideal Gas Law with Gas Mixtures - Using PV=nRT and mass constraints to find individual moles of gases in a mixture

1. Convert volume to SI units: $V = 8310 ext{ cm}^3 = 8.31 imes 10^{-3} ext{ m}^3$
2. Use ideal gas equation to find total moles: $n_{total} = \frac{PV}{RT} = \frac{100000 imes 8.31 imes 10^{-3}}{8.31 imes 300} = \frac{1}{3}$ mol
3. Set up equations with $n_1$ (CO₂) and $n_2$ (O₂): $n_1 + n_2 = 0.333$ and $44n_1 + 32n_2 = 13.2$
4. Substitute $n_2 = 0.333 - n_1$ into mass equation: $44n_1 + 32(0.333 - n_1) = 13.2$
5. Solve for $n_1$: $12n_1 = 13.2 - 10.67 = 2.53$, so $n_1 = 0.21$ mol
6. Calculate $n_2$: $n_2 = 0.333 - 0.21 = 0.12$ mol

Answer: Option (2)

Key idea: Stokes' Law and Terminal Velocity of Rising Bubble

1. Given: diameter $d = 2$ mm, so radius $r = 1$ mm $= 10^{-3}$ m, $\rho = 2000$ kg/m³, $v = 0.5$ cm/s $= 5 \times 10^{-3}$ m/s
2. For a bubble rising at terminal velocity, use Stokes' law: $v = \frac{2r^2 g \rho}{9\eta}$ (neglecting air density)
3. Rearrange for viscosity: $\eta = \frac{2r^2 g \rho}{9v}$
4. Substitute values: $\eta = \frac{2 \times (10^{-3})^2 \times 10 \times 2000}{9 \times 5 \times 10^{-3}}$
5. Calculate: $\eta = \frac{2 \times 10^{-6} \times 2 \times 10^4}{45 \times 10^{-3}} = \frac{0.04}{0.045} = \frac{8}{9}$ Pa·s $\approx 0.889$ Pa·s
6. Convert to Poise: $\eta = 0.889 \times 10 = 8.89 \approx 8.8$ Poise

Answer: Option (2)

Key idea: Work-Energy Theorem and Newton's Laws

1. Find velocity just before hitting sand: $v^2 = 2gh = 2(10)(10) = 200$ m²/s², so $v = 10\sqrt{2}$ m/s
2. Calculate deceleration in sand using $v^2 = u^2 + 2as$: $0 = 200 + 2a(0.1)$, giving $a = -1000$ m/s²
3. Apply work-energy theorem: Initial KE = $\frac{1}{2}mv^2 = \frac{1}{2}(2)(200) = 200$ J
4. Work done: $0 - 200 = -F(0.1) + mg(0.1)$, where $F$ is sand force
5. Solve: $-200 = -0.1F + 2$, giving $F = 2020$ N

Answer: Option (2)

Key idea: Electromagnetic wave propagation: relationship between E, B, and direction of propagation

1. For an EM wave in free space, the magnitude relation is $E = cB$ where $c = 3 imes 10^8$ m/s
2. Calculate magnitude: $E = (3 imes 10^8)(2 imes 10^{-7}) = 60$ V/m
3. The wave propagates along $+x$ direction ($\hat{i}$), and $\vec{B} = 2 \times 10^{-7}\hat{j}$ T
4. Since $\vec{E}$, $\vec{B}$, and propagation direction are mutually perpendicular, $\vec{E}$ must be along $\hat{k}$ direction
5. Using $\vec{E} \times \vec{B}$ = direction of propagation ($\hat{i}$), we get $(E_z\hat{k}) \times (B\hat{j}) = E_z B(-\hat{i})$
6. For this to equal $+\hat{i}$, we need $E_z < 0$, so $\vec{E} = -60\hat{k}$ V/m

Answer: Option (50)

Key idea: Series-Parallel Circuits and Voltage Division

1. Calculate bulb resistance: $R_{bulb} = \frac{V^2}{P} = \frac{(200)^2}{100} = 400\,\Omega$
2. Find equivalent resistance of bulb and $400\,\Omega$ resistor in parallel: $R_{eq} = \frac{400 \times 400}{400 + 400} = 200\,\Omega$
3. Calculate total circuit resistance: $R_{total} = 200 + 200 = 400\,\Omega$
4. Find total current from battery: $I = \frac{V}{R_{total}} = \frac{100}{400} = 0.25\,\text{A}$
5. Calculate potential drop across bulb (parallel combination): $V_{bulb} = I \times R_{eq} = 0.25 \times 200 = 50\,\text{V}$

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: Equilibrium of charged particle in electric field - Balancing gravitational and electrostatic forces

Answer: Option (3)

Key idea: Gauss's Law - Electric flux depends only on enclosed charge

1. Apply Gauss's Law: Electric flux $\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$
2. For sphere of radius $3$ cm: Check which charges are inside. Charge at $x=2$ cm is inside (distance $2 < 3$), charge at $x=4$ cm is outside (distance $4 > 3$)
3. Enclosed charge in sphere 1: $q_1 = 8$ μC, so $\Phi_1 = \frac{8 \mu C}{\epsilon_0}$
4. For sphere of radius $5$ cm: Both charges are inside (distances $2 < 5$ and $4 < 5$)
5. Enclosed charge in sphere 2: $q_1 + q_2 = 8 + (-2) = 6$ μC, so $\Phi_2 = \frac{6 \mu C}{\epsilon_0}$
6. Ratio of fluxes: $\frac{\Phi_1}{\Phi_2} = \frac{8}{6} = \frac{4}{3}$, which is $4:3$

Answer: Option (4)

Key idea: Total Internal Reflection in Prisms

1. For an equilateral prism, all angles are $60°$. When light enters one face and hits the painted base, use geometry: $r_1 + r_2 = A = 60°$
2. For total internal reflection at the painted face: $\sin(\theta_c) = \frac{n_2}{n_1}$ where $n_1 = 1.6$ is the prism's refractive index
3. To find minimum $n_2$, maximize the angle of incidence $r_2$ at the painted face. This occurs when $r_1 = 0°$ (perpendicular entry)
4. Therefore, maximum angle at painted face: $r_2 = 60°$
5. Apply critical angle condition: $\sin(60°) = \frac{n_2}{1.6}$
6. Solve: $\frac{\sqrt{3}}{2} = \frac{n_2}{1.6}$ gives $n_2 = \frac{1.6\sqrt{3}}{2} = 0.8\sqrt{3} = \frac{4\sqrt{3}}{5}$

Answer: Option (2)

Key idea: Phase difference in LCR circuits and impedance analysis

1. Calculate capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{300 \times 100 \times 10^{-6}} = \frac{100}{3}$ Ω
2. For Case 1 (S₁ closed): $\tan(30°) = \frac{X_{L1} - X_C}{R}$ gives $\frac{1}{\sqrt{3}} = \frac{300L_1 - \frac{100}{3}}{R}$
3. For Case 2 (S₂ closed): $\tan(60°) = \frac{X_{L2} - X_C}{R}$ gives $\sqrt{3} = \frac{300L_2 - \frac{100}{3}}{R}$
4. Divide the two equations: $\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}} = 3 = \frac{300L_2 - \frac{100}{3}}{300L_1 - \frac{100}{3}}$
5. Cross-multiply and simplify: $900L_1 - 100 = 300L_2 - \frac{100}{3}$
6. Rearrange: $900L_1 - 300L_2 = 100 - \frac{100}{3} = \frac{200}{3}$
7. Solve for $(3L_1 - L_2)$: $300(3L_1 - L_2) = \frac{200}{3}$, so $3L_1 - L_2 = \frac{2}{9}$ H

Answer: Option (4)

Key idea: Mutual Inductance and Magnetic Field of Square Loop

1. Find the magnetic field at the center of a square loop of side $L$ carrying current $I$. Each side at distance $d = L/2$ contributes: $B_{side} = \frac{\mu_0 I}{4\pi d} \times 2\sin(45°) = \frac{\sqrt{2}\mu_0 I}{2\pi L}$
2. Total field from all 4 sides (perpendicular to plane): $B_{center} = 4 \times \frac{\sqrt{2}\mu_0 I}{2\pi L} = \frac{2\sqrt{2}\mu_0 I}{\pi L}$
3. Since $L >> R$, the field is approximately uniform over the circular loop area. Calculate flux: $\Phi = B \times \pi R^2 = \frac{2\sqrt{2}\mu_0 I}{\pi L} \times \pi R^2$
4. Simplify the flux: $\Phi = \frac{2\sqrt{2}\mu_0 I R^2}{L}$
5. Mutual inductance is: $M = \frac{\Phi}{I} = \frac{2\sqrt{2}\mu_0 R^2}{L} = 2\sqrt{2}\frac{\mu_0 R^2}{L}$

Answer: Option (2)

Key idea: Binding Energy Per Nucleon and Mass Defect

1. Identify: $Z = 83$ protons, $N = 209 - 83 = 126$ neutrons
2. Calculate total mass of separated nucleons: $m_{total} = 83(1.007825) + 126(1.008665) = 210.741265$ u
3. Find mass defect: $\Delta m = 210.741265 - 208.980388 = 1.760877$ u
4. Calculate total binding energy: $BE = \Delta m \times 931 = 1.760877 \times 931 = 1639.54$ MeV
5. Find binding energy per nucleon: $\frac{BE}{A} = \frac{1639.54}{209} = 7.84$ MeV

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: Simple Harmonic Motion - velocity and acceleration conditions

Answer: 6

Key idea: Young's Double Slit Experiment - Intensity and Path Difference Relationship

1. In YDSE with equal intensity sources, the intensity formula is $I = 4I_0\cos^2(\phi/2)$ where $I_0$ is individual source intensity and $\phi$ is phase difference
2. Maximum intensity occurs at $\phi = 0$, giving $I_{\text{max}} = 4I_0$
3. Given $I = \frac{3}{4}I_{\text{max}}$, substitute: $4I_0\cos^2(\phi/2) = \frac{3}{4} \times 4I_0$
4. Solving: $\cos^2(\phi/2) = \frac{3}{4}$, so $\cos(\phi/2) = \frac{\sqrt{3}}{2}$, giving $\phi/2 = \frac{\pi}{6}$ or $\phi = \frac{\pi}{3}$
5. Use relation between phase and path difference: $\phi = \frac{2\pi}{\lambda}\Delta$
6. Substitute $\phi = \frac{\pi}{3}$: $\Delta = \frac{\phi \lambda}{2\pi} = \frac{\pi}{3} \times \frac{\lambda}{2\pi} = \frac{\lambda}{6}$
7. Since path difference is given as $\frac{\lambda}{x}$, comparing gives $x = 6$

Answer: 32

Key idea: Magnetic field due to electron motion in Bohr orbits

1. The electron in orbit acts as a current loop with current $I_n = \frac{e}{T_n}$, where $T_n$ is the period of revolution
2. From Bohr's model: radius $r_n = n^2a_0$ and velocity $v_n = \frac{v_1}{n}$
3. Period: $T_n = \frac{2\pi r_n}{v_n} = \frac{2\pi n^3 a_0}{v_1}$, so current $I_n = \frac{ev_1}{2\pi n^3 a_0}$
4. Magnetic field at center: $B_n = \frac{\mu_0 I_n}{2r_n} = \frac{\mu_0 ev_1}{4\pi n^5 a_0^2}$
5. Therefore $B_n \propto \frac{1}{n^5}$
6. Ratio: $\frac{B_2}{B_4} = \frac{4^5}{2^5} = \frac{1024}{32} = 32$

Answer: 300

Key idea: First Law of Thermodynamics and Mayer's Relation

1. Convert $R$ from J/mol·°C to cal/mol·°C: $R = \frac{8.36}{4.18} = 2$ cal/mol·°C
2. Apply Mayer's relation: $c_v = c_p - R = 8 - 2 = 6$ cal/mol·°C
3. Calculate temperature change: $\Delta T = 20 - 10 = 10$°C
4. Use formula for change in internal energy: $\Delta U = n \times c_v \times \Delta T$
5. Substitute values: $\Delta U = 5 \times 6 \times 10 = 300$ calories

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.

Answer: 15

Key idea: Moment of Inertia and Parallel Axis Theorem

1. Calculate moment of inertia of rod about axis AB (through one end): $I_{rod} = \frac{1}{3}ML^2 = \frac{1}{3}(40)(3)^2 = 120$ kg·m²
2. For the sphere, use parallel axis theorem: $I_{sphere} = I_{center} + Md^2 = \frac{2}{5}MR^2 + Md^2$
3. Substitute values: $I_{sphere} = \frac{2}{5}(10)R^2 + (10)(3)^2 = 4R^2 + 90$
4. Equate the moments: $120 = 4R^2 + 90$, solving gives $R^2 = \frac{15}{2}$
5. Given $R = \sqrt{\frac{\alpha}{2}}$, we have $R^2 = \frac{\alpha}{2} = \frac{15}{2}$, therefore $\alpha = 15$

Answer: Option (2)

Key idea: Stoichiometry and combustion analysis of hydrocarbons

1. Given mass ratio C:H = 12:1 and 2 carbon atoms, find molecular formula: Mass of C = $2 \times 12 = 24$ g, Mass of H = $24/12 = 2$ g, so H atoms = 2. Formula is $C_2H_2$
2. Calculate molar mass of $C_2H_2$: $M = 2(12) + 2(1) = 26$ g/mol
3. Write combustion equation: $C_2H_2 + \frac{5}{2}O_2 \rightarrow 2CO_2 + H_2O$. Each mole of $C_2H_2$ produces 2 moles of $CO_2$
4. Calculate moles of hydrocarbon: $n_{C_2H_2} = \frac{3.38}{26} = 0.13$ mol
5. Calculate moles of $CO_2$ produced: $n_{CO_2} = 2 \times 0.13 = 0.26$ mol
6. Calculate mass of $CO_2$: Mass = $0.26 \times 44 = 11.44$ g

Answer: Option (3)

Key idea: Common Ion Effect and Weak Acid Equilibrium

1. Given: $H_2A$ with $K_1 = 8.1 \times 10^{-8}$, $K_2 = 1.0 \times 10^{-13}$, dissolved in 0.1 M HCl
2. Initial concentrations: $[H_2A] = 0.1$ M, $[H^+] = 0.1$ M (from HCl)
3. For first ionization: $H_2A \rightleftharpoons H^+ + HA^-$ with $K_1 = \frac{[H^+][HA^-]}{[H_2A]}$
4. Due to common ion effect (high $[H^+]$ from HCl), ionization is suppressed. Let $[HA^-] = x$
5. At equilibrium: $[H_2A] \approx 0.1$ M, $[H^+] \approx 0.1$ M (since $x \ll 0.1$)
6. Substitute in $K_1$: $8.1 \times 10^{-8} = \frac{(0.1)(x)}{(0.1)} = x$
7. Therefore, $[HA^-] = 8.1 \times 10^{-8}$ M

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: Unknown - question content missing

Answer: Option (4)

Key idea: First Law of Thermodynamics and State Functions

1. For process $X \to Y$: Apply first law $\Delta U = Q - W = 10 - 18 = -8$ J
2. Since internal energy is a state function: $\Delta U_{Y \to X} = -\Delta U_{X \to Y} = +8$ J
3. For reverse process $Y \to X$: Heat evolved $Q = -6$ J (negative because heat is released)
4. Apply first law for $Y \to X$: $\Delta U = Q - W$ gives $8 = -6 - W$
5. Solve for work: $W = -6 - 8 = -14$ J
6. Negative work by gas means positive work on gas: 14 J of work is done ON the gas by surroundings

Answer: Option (1)

Key idea: Osmotic Pressure and Colligative Properties

1. Use osmotic pressure formula: $\pi = \frac{n}{V}RT$ where $n$ is moles, $V$ is volume in L
2. For Solution A: $n_A = \frac{1}{50000} = 2 \times 10^{-5}$ mol, so $x = \frac{2 \times 10^{-5}}{0.5} \times 0.083 \times 300 = 9.96 \times 10^{-4}$ bar
3. For Solution B: $n_B = \frac{2}{50000} = 4 \times 10^{-5}$ mol, so $y = \frac{4 \times 10^{-5}}{1} \times 0.083 \times 300 = 9.96 \times 10^{-4}$ bar
4. For mixed solution: Total moles $= 6 \times 10^{-5}$ mol, Total volume $= 1.5$ L
5. Calculate $z = \frac{6 \times 10^{-5}}{1.5} \times 0.083 \times 300 = 9.96 \times 10^{-4}$ bar
6. Notice that all three solutions have the same concentration (moles/volume ratio), hence $x = y = z = 9.96 \times 10^{-4}$ bar

Answer: Option (2)

Key idea: Weak Acid Titration and Buffer Solution pH Calculation

1. **Part (a):** For weak acid HX at start, use $[H^+] = \sqrt{K_a \times C}$ where $C = 0.2$ M
2. Calculate: $[H^+] = \sqrt{5 \times 10^{-4} \times 0.2} = \sqrt{10^{-4}} = 10^{-2}$ M
3. Therefore, $pH = -\log(10^{-2}) = 2.0$
4. **Part (b):** Initial moles HX = $0.2 \times 20 = 4$ mmol; NaOH added = $0.2 \times 10 = 2$ mmol
5. After neutralization: HX remaining = 2 mmol, salt NaX formed = 2 mmol (buffer at half-equivalence point)
6. Using Henderson-Hasselbalch: $pH = pK_a + \log\frac{[A^-]}{[HA]} = 3.3 + \log(1) = 3.3$
7. Answer: pH values are **(2.0, 3.3)**

Answer: Option (1)

Key idea: Determining reaction order from rate constant units and applying second-order kinetics properties

1. Determine order from rate constant units: $k$ has units mol$^{-1}$ L s$^{-1}$ = (mol L$^{-1}$)$^{1-n}$ s$^{-1}$, so $1-n = -1$ giving $n = 2$ (second order)
2. **Statement A**: For second order, Rate $= k[X]^2$. If $[X] \to 4[X]$, then Rate $\to k(4[X])^2 = 16k[X]^2$ (16 times). TRUE ✓
3. **Statement B**: From units analysis, reaction is second order. TRUE ✓
4. **Statement C**: For second order, $t_{1/2} = \frac{1}{k[X]_0}$ depends on initial concentration. FALSE ✗
5. **Statement D**: Decomposition of $N_2O_5$ is first order, not second order. FALSE ✗
6. **Statement E**: $\ln\frac{[R_0]}{[R]}$ vs time is linear for first order, not second order (which uses $\frac{1}{[R]}$ vs $t$). FALSE ✗
7. Only statements A and B are true, so answer is option (1)

Answer: Option (3)

Key idea: Classification of oxides: Basic, Acidic, Amphoteric, and Neutral oxides

1. Recall oxide classifications: Basic (metal oxides like $Na_2O$, $K_2O$), Acidic (non-metal oxides like $Cl_2O_7$), Amphoteric (like $Al_2O_3$, $As_2O_3$), Neutral (like $CO$, $NO$)
2. Check Option (1): $Na_2O$ (basic), $K_2O$ (basic), $Al_2O_3$ (amphoteric), $As_2O_3$ (amphoteric) - Missing acidic and neutral ❌
3. Check Option (2): $Al_2O_3$ (amphoteric), $As_2O_3$ (amphoteric), $CO$ (neutral), $NO$ (neutral) - Missing basic and acidic ❌
4. Check Option (3): $K_2O$ (basic), $Cl_2O_7$ (acidic), $As_2O_3$ (amphoteric), $NO$ (neutral) - Contains all four types ✓
5. Check Option (4): $Na_2O$ (basic), $N_2O$ (neutral), $Al_2O_3$ (amphoteric), $CO$ (neutral) - Missing acidic ❌
6. Option (3) is the only set containing all four types of oxides

Answer: Option (2)

Key idea: Periodic trends in ionization enthalpy, d-block and f-block contraction effects

1. **Statement I Analysis:** For second ionization enthalpy of B, Al, Ga - remove electron from B⁺, Al⁺, Ga⁺
2. B is smallest so has highest 2nd IE. Ga has d-block contraction making it smaller than Al, so Ga⁺ has higher 2nd IE than Al⁺
3. Correct order: $B > Ga > Al$ (NOT $B > Al > Ga$). **Statement I is FALSE**
4. **Statement II Analysis:** For first ionization enthalpy of Group 14 elements (Si, Ge, Sn, Pb)
5. Actual IE values: Si (786.5) > Ge (762) > Pb (715.6) > Sn (708.6) kJ/mol
6. Given order Si < Ge < Pb < Sn is completely reversed. **Statement II is FALSE**
7. Since both statements are false, answer is option (2)

Answer: 1

Key idea: Ionization Energy Trends and Crystal Field Stabilization Energy (CFSE)

1. Write electronic configurations after removing 2 electrons: Sc²⁺: [Ar]3d¹, Mn²⁺: [Ar]3d⁵, Cu²⁺: [Ar]3d⁹, Zn²⁺: [Ar]3d¹⁰
2. For third ionization energy (IE₃), Zn²⁺ has fully filled 3d¹⁰ (most stable) requiring highest energy; Sc²⁺ has lowest nuclear charge requiring lowest energy. Statement I is **correct**.
3. For [Co(H₂O)₆]²⁺: Co²⁺ is d⁷ high spin → t₂g⁵eₘ² → CFSE = -0.8Δₒ (small Δₒ)
4. For [Co(H₂O)₆]³⁺: Co³⁺ is d⁶ low spin → t₂g⁶eₘ⁰ → CFSE = -2.4Δₒ (moderate Δₒ)
5. For [Co(en)₃]³⁺: Co³⁺ is d⁶ low spin → t₂g⁶eₘ⁰ → CFSE = -2.4Δₒ (large Δₒ, since en is stronger field)
6. Since Δₒ(en, Co³⁺) > Δₒ(H₂O, Co³⁺) > Δₒ(H₂O, Co²⁺), the CFSE order is correct. Statement II is **correct**.
7. Both statements are correct, so answer is Option 1

Answer: Option (2)

Key idea: Coordination Isomerism in Coordination Compounds

1. Coordination isomerism occurs when ligands can be exchanged between cationic and anionic coordination entities in a compound containing both complex cation and complex anion.
2. For **A**: $[Ag(NH_3)_2][Ag(CN)_2]$ - Both complexes have Ag with coordination number 2. Ligand exchange would give neutral species $[Ag(NH_3)(CN)]$, disrupting ionic character. **Does not show coordination isomerism.**
3. For **B**: $[Co(NH_3)_6][Cr(CN)_6]$ - Different metals (Co and Cr). Can form isomer $[Cr(NH_3)_6][Co(CN)_6]$ by ligand exchange. **Shows coordination isomerism.**
4. For **C**: $[Co(NH_3)_6][Co(CN)_6]$ - Same metal in both cation and anion. Exchange gives the same compound due to equivalence of Co centers. **Does not show coordination isomerism.**
5. For **D**: $[Fe(NH_3)_6][Co(CN)_6]$ - Different metals. Can form isomer $[Co(NH_3)_6][Fe(CN)_6]$ (which is compound E). **Shows coordination isomerism.**
6. For **E**: $[Co(NH_3)_6][Fe(CN)_6]$ - Different metals. Can form isomer $[Fe(NH_3)_6][Co(CN)_6]$ (which is compound D). **Shows coordination isomerism.**
7. Therefore, compounds **B, D, and E only** show coordination isomerism.

Answer: Option (1)

Key idea: Stoichiometry and Percentage Composition

1. Calculate mass of carbon from $CO_2$: Mass of C $= \frac{12}{44} \times 0.25 = 0.0682$ g
2. Calculate mass of hydrogen from $H_2O$: Mass of H $= \frac{2}{18} \times 0.12 = 0.0133$ g
3. Use percentage of carbon: If 25% of X = 0.0682 g, then $X = \frac{0.0682}{0.25} = 0.273$ g
4. Verify with hydrogen: If 4.89% of X = 0.0133 g, then $X = \frac{0.0133}{0.0489} = 0.273$ g
5. Convert to required format: $X = 0.273$ g $= 273 \times 10^{-3}$ g
6. Answer: $X = 273$

Answer: Option (1)

Key idea: Resonance effects on carbocation and carbanion stability; +R and -R effects of substituents

1. Analyze Statement I: The structure has a carbocation ($CH^+$) between two benzene rings, one with $-NO_2$ and one with $-OCH_3$
2. For carbocation stability: Need electron donation. The $-OCH_3$ group has $+R$ effect (electron donating through resonance)
3. The lone pair on oxygen delocalizes through the benzene ring to the carbocation center, stabilizing it. **Statement I is TRUE**
4. Analyze Statement II: The structure has a carbanion ($CH^-$) between two benzene rings, one with $-NO_2$ and one with $-OCH_3$
5. For carbanion stability: Need electron withdrawal. The $-NO_2$ group has $-R$ effect (electron withdrawing through resonance)
6. The negative charge delocalizes into the benzene ring and further into the $-NO_2$ group, stabilizing the carbanion. **Statement II is TRUE**
7. Both statements correctly identify the stabilizing resonance effects

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: Reactions of conjugated polyenes: electrocyclic reactions, aromatization, and acid-catalyzed rearrangements

Answer: Option (3)

Key idea: Reactivity and properties of alkyl halides, aryl halides, and their derivatives; SN1 mechanism; elimination vs substitution reactions

1. Analyze Statement A: Alcoholic KOH favors elimination because alcohol (less polar) enhances base character of $OH^-$ over nucleophilic character. **Statement A is CORRECT**.
2. Analyze Statement B: For $S_N1$, carbocation stability matters. $C_6H_5CHClC_6H_5$ forms diphenylmethyl cation (2 phenyl groups) which is MORE stable than benzyl cation from $C_6H_5CH_2Cl$ (1 phenyl group). Given order is wrong. **Statement B is INCORRECT**.
3. Analyze Statement C: Aryl halides have $C-X$ bond with partial double bond character due to resonance, making them much less reactive than alkyl halides. They don't undergo normal $S_N1$ or $S_N2$ reactions. **Statement C is INCORRECT**.
4. Analyze Statement D: Vinyl chloride $CH_2=CHCl$ is a haloalkene (correct), but allyl chloride $CH_2=CHCH_2Cl$ is also a haloalkene, NOT a haloalkyne. **Statement D is INCORRECT**.
5. Analyze Statement E: $R-OH + SOCl_2 \rightarrow R-Cl$ works for alkyl alcohols. However, $Ar-OH + SOCl_2$ does NOT give $Ar-Cl$ because phenolic $C-O$ bond is too strong. **Statement E is CORRECT**.
6. Only statements A and E are correct, so the answer is option (3).

Answer: Option (3)

Key idea: Cannizzaro Reaction of Glyoxal

1. Given C:O:H = 1:1:1 and molar mass = 58 g/mol, find molecular formula: $(COH)_n$ where $29n = 58$, so $n=2$. Thus $x$ is $C_2H_2O_2$
2. Identify compound $x$: With formula $C_2H_2O_2$, the structure is glyoxal (OHC-CHO), which has two aldehyde groups and no α-hydrogens
3. Recognize the reaction: 50% KOH (concentrated base) with aldehydes having no α-H undergoes Cannizzaro reaction (disproportionation)
4. In Cannizzaro reaction, one aldehyde group is oxidized to $-COOH$ and another is reduced to $-CH_2OH$
5. Product $y$ after acidification: $HOCH_2COOH$ (glycolic acid), which matches option (3)

Answer: Option (4)

Key idea: Acid-catalyzed hydrolysis of vinyl ethers to carbonyl compounds and their chemical properties

1. Identify structure (X) as a vinyl ether: $CH_3CH=CH-O-C(CH_3)=CH_2$
2. Under mild acidic hydrolysis, vinyl ethers break to form carbonyl compounds: $Y = CH_3CH_2CHO$ (propanal) and $Z = CH_3COCH_3$ (acetone)
3. Check Statement A: Both have molecular formula $C_3H_6O$ with molar mass $58$ g/mol ✓
4. Check Statement B: Neither aldehyde nor ketone reacts with $NaHCO_3$ (only carboxylic acids do) ✗
5. Check Statement C: Aldehydes react faster with HCN than ketones due to less steric hindrance ✗
6. Check Statement D: Both aldehydes and ketones react with 2,4-DNP to form hydrazones ✓
7. Correct statements are A and D only

Answer: Option (1)

Key idea: Electrophilic aromatic substitution, nitro group reduction, diazotization, and oxidation of alkyl side chains

1. Step 1: Bromination with $Br_2/AlBr_3$ on 3-nitro-ethylbenzene gives **A = 4-bromo-3-nitro-ethylbenzene** (Br adds at position para to ethyl and meta to nitro)
2. Step 2: Reduction with $Sn/HCl$ converts $-NO_2$ to $-NH_2$ giving B = 4-bromo-3-amino-ethylbenzene
3. Step 3: Diazotization with $NaNO_2/HCl$ at 273-278 K converts $-NH_2$ to $-N_2^+Cl^-$ giving C = diazonium salt
4. Step 4: Treatment with $C_2H_5OH$ replaces diazonium group with H, giving D = 3-bromo-ethylbenzene
5. Step 5: Oxidation with $KMnO_4/KOH$ then $H_3O^+$ oxidizes ethyl group to carboxylic acid giving **E = 3-bromo-benzoic acid**

Answer: Option (3)

Key idea: Amino acids and their derivatives; Thyroid hormones

1. Identify that the question asks for a hormone that is an iodinated derivative of an amino acid
2. Recall that thyroxine (T4) is a thyroid hormone containing iodine atoms
3. Remember that thyroxine is synthesized from the amino acid tyrosine through iodination
4. Identify the one-letter code: $Y$ represents tyrosine, $T$ represents threonine
5. Match the correct pair: Tyrosine ($Y$) is the amino acid and Thyroxine is its iodinated derivative
6. Select option (3): $Y$ Thyroxine

Answer: Fe³⁺

Key idea: Borax bead test for transition metal ions and ionization enthalpy trends based on electronic configuration stability

1. Identify ions showing positive borax bead test: $Fe^{2+}$, $Fe^{3+}$, and $Cr^{2+}$ give colored beads due to unpaired d-electrons
2. Eliminate $Zn^{2+}$: Has $3d^{10}$ configuration (fully filled), gives colorless bead, negative test
3. Write electronic configurations: $Fe^{2+}$: $3d^6$, $Fe^{3+}$: $3d^5$ (half-filled), $Cr^{2+}$: $3d^4$
4. Compare ionization enthalpies: Removing electron from $Fe^{3+}$ disrupts stable half-filled $3d^5$ configuration
5. $Fe^{3+}$ has highest ionization enthalpy due to: (i) half-filled stability (ii) higher positive charge (iii) higher nuclear charge
6. Answer: $Fe^{3+}$ shows positive borax bead test and has highest ionization enthalpy

Answer: 5

Key idea: Photoelectric Effect - Einstein's Equation

1. Convert work function to Joules: $\phi = 2.3 \times 1.6 \times 10^{-19} = 3.68 \times 10^{-19}$ J
2. Apply photoelectric equation: $E_{photon} = \phi + KE$
3. Calculate photon energy: $E = 3.68 \times 10^{-19} + 2.8 \times 10^{-20} = 3.96 \times 10^{-19}$ J
4. Use $\lambda = \frac{hc}{E} = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.96 \times 10^{-19}}$
5. Calculate: $\lambda = 5.0 \times 10^{-7}$ m $= 500$ nm
6. Express as $x \times 10^2$ nm: $x = \frac{500}{100} = 5$

Answer: 20

Key idea: Gas Phase Kinetics - Pressure Analysis using Stoichiometry

1. For reaction $2A(g) \rightarrow 4B(g) + C(g)$, set up pressure table: Initial $P_0$ for A, at time $t$: $P_A = P_0 - 2x$, $P_B = 4x$, $P_C = x$
2. Total pressure at any time: $P_{total} = (P_0 - 2x) + 4x + x = P_0 + 3x$
3. At $t = \infty$, all A reacts: $P_0 - 2x_\infty = 0 \Rightarrow x_\infty = P_0/2$. So $P_0 + 3(P_0/2) = 600$
4. Solve: $\frac{5P_0}{2} = 600 \Rightarrow P_0 = 240$ mm Hg
5. At $t = 30$ min: $240 + 3x_{30} = 300 \Rightarrow x_{30} = 20$ mm Hg
6. Pressure of $C(g)$ at 30 min = $x_{30} = 20$ mm Hg

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: Electrochemistry: Fuel cell efficiency, Gibbs free energy, and thermodynamic work

Answer: 3

Key idea: Paramagnetism and electronic configuration of d-block and f-block ions

1. Determine the electronic configuration of each metal atom using atomic numbers
2. Remove appropriate electrons to get the ion configuration (remove $s$ electrons first, then $d$ or $f$)
3. For d-block: $\text{Mn}^{2+}$ has $3d^5$ (5 unpaired), $\text{Cu}^{2+}$ has $3d^9$ (1 unpaired), $\text{Zn}^{2+}$ has $3d^{10}$ (0 unpaired)
4. For f-block: $\text{Gd}^{3+}$ has $4f^7$ (7 unpaired), $\text{Yb}^{2+}$ has $4f^{14}$ (0 unpaired)
5. Ions with completely filled or empty subshells: $\text{Sc}^{3+}$, $\text{La}^{3+}$, $\text{Lu}^{3+}$, $\text{Ti}^{4+}$, $\text{Ce}^{4+}$ are diamagnetic
6. Count paramagnetic ions (those with unpaired electrons): $\text{Mn}^{2+}$, $\text{Cu}^{2+}$, $\text{Gd}^{3+}$ = 3

Answer: 1

Key idea: Carius analysis for halogen estimation, organic reaction sequence involving reduction, acetylation, electrophilic aromatic substitution, and hydrolysis

1. Reduce $\text{NO}_2$ to $\text{NH}_2$ using Sn/HCl, then acetylate with $(\text{CH}_3\text{CO})_2\text{O}$ to get p-methylacetanilide
2. Brominate with $\text{Br}_2/\text{AlBr}_3$ (ortho to acetamido group) then hydrolyze with $\text{H}_3\text{O}^+$ to get Product P: 2-bromo-4-methylaniline
3. Calculate molar mass of P ($\text{C}_7\text{H}_8\text{BrN}$): $7(12) + 8(1) + 80 + 14 = 186$ g/mol
4. Find moles of P: $\frac{1.0 \text{ g}}{186 \text{ g/mol}} = 0.00538$ mol
5. In Carius analysis, 1 mole Br gives 1 mole AgBr. Molar mass of AgBr = $108 + 80 = 188$ g/mol
6. Mass of AgBr = $0.00538 \times 188 = 1.01$ g ≈ 1 g (nearest integer)