JEE Main 6 April 2026 Shift 2: Questions and Solutions

Answer: Option (4)

Key idea: Testing injectivity and surjectivity of rational functions using derivatives and range analysis

1. Verify domain: Check $3x^2 + x + 3 = 0$. Discriminant $= 1 - 36 = -35 < 0$, so denominator is always positive. Function is well-defined on $\mathbb{R}$.
2. Test one-one: Find $f'(x) = \frac{11(x^2-1)}{(3x^2+x+3)^2}$ using quotient rule.
3. Analyze $f'(x)$: Since $f'(x) < 0$ for $x \in (-1,1)$ and $f'(x) > 0$ for $x \in (-\infty,-1) \cup (1,\infty)$, function is not monotonic. Hence NOT one-one.
4. Test onto: Find critical values $f(-1) = \frac{7}{5}$ (local min) and $f(1) = \frac{1}{7}$ (local max).
5. As $x \to \pm\infty$, $f(x) \to \frac{2}{3}$. Range is $[\frac{1}{7}, \frac{7}{5}] \neq \mathbb{R}$. Hence NOT onto.
6. Conclusion: $f$ is neither one-one nor onto.

Answer: Option (3)

Key idea: Quadratic equations (sum and product of roots), optimization using calculus, geometric progression sum formula

1. Step 1: For the quadratic $(n^2-2n+2)x^2 - 3x + (n^2-2n+2)^2 = 0$, let $a = n^2-2n+2$. Product of roots = $a$ and sum of roots = $\frac{3}{a}$
2. Step 2: Find minimum of $a = n^2-2n+2 = (n-1)^2 + 1$. Minimum occurs at $n=1$, giving $\alpha = 1$
3. Step 3: To maximize sum of roots $\frac{3}{a}$, minimize $a$. Maximum sum is $\beta = \frac{3}{1} = 3$
4. Step 4: G.P. has first term $\alpha = 1$ and common ratio $r = \frac{\alpha}{\beta} = \frac{1}{3}$
5. Step 5: Sum of first 6 terms: $S_6 = 1 \cdot \frac{1-(1/3)^6}{1-1/3} = \frac{1-1/729}{2/3} = \frac{728/729}{2/3}$
6. Step 6: Simplify: $S_6 = \frac{728}{729} \times \frac{3}{2} = \frac{2184}{1458} = \frac{364}{243}$

Answer: Option (1)

Key idea: Vieta's formulas and recurrence relations for sum of powers of roots

1. Step 1: From the equation $z^2 + \sqrt{6}iz - 3 = 0$, use Vieta's formulas: $z_1 + z_2 = -\sqrt{6}i$ and $z_1z_2 = -3$
2. Step 2: Define $S_n = z_1^n + z_2^n$. From the equation, derive recurrence: $S_n = -\sqrt{6}i \cdot S_{n-1} + 3S_{n-2}$
3. Step 3: Calculate initial values: $S_0 = 2$, $S_1 = -\sqrt{6}i$, $S_2 = (z_1+z_2)^2 - 2z_1z_2 = -6 + 6 = 0$
4. Step 4: Use recurrence to find: $S_3 = -3\sqrt{6}i$, $S_4 = -18$, $S_5 = 9\sqrt{6}i$, $S_6 = 0$, $S_7 = 27\sqrt{6}i$
5. Step 5: Calculate $S_8 = -\sqrt{6}i \cdot 27\sqrt{6}i + 3(0) = -27 \cdot 6 \cdot i^2 = 162$

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Key idea: Determinant condition for non-trivial solutions of homogeneous linear systems and solving trigonometric equations

Answer: Option (3)

Key idea: Matrix Powers using Binomial Theorem for Nilpotent Matrices

1. Step 1: Write $A = I + N$ where $N = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix}$ is strictly lower triangular
2. Step 2: Compute $N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$ and $N^3 = 0$ (nilpotent)
3. Step 3: Use binomial expansion: $A^n = I + nN + \frac{n(n-1)}{2}N^2$
4. Step 4: For $n=99$: $A^{99} = \begin{bmatrix} 1 & 0 & 0 \\ 297 & 1 & 0 \\ 44550 & 297 & 1 \end{bmatrix}$
5. Step 5: Calculate $B = A^{99} - I = \begin{bmatrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 44550 & 297 & 0 \end{bmatrix}$
6. Step 6: Find $\frac{b_{31} - b_{21}}{b_{32}} = \frac{44550 - 297}{297} = \frac{44253}{297} = 149$

Answer: Option (3)

Key idea: Sum of series using formula for sum of squares

1. Identify the general term: $T_n = \frac{1}{n}(1^2 + 2^2 + ... + n^2)$
2. Use the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ to get $T_n = \frac{(n+1)(2n+1)}{6}$
3. Expand: $(n+1)(2n+1) = 2n^2 + 3n + 1$
4. Write $S_{10} = \frac{1}{6}\sum_{n=1}^{10}(2n^2 + 3n + 1) = \frac{1}{6}[2\sum n^2 + 3\sum n + 10]$
5. Calculate: $\sum_{n=1}^{10} n^2 = 385$, $\sum_{n=1}^{10} n = 55$
6. Substitute: $S_{10} = \frac{1}{6}[2(385) + 3(55) + 10] = \frac{945}{6} = \frac{315}{2}$

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Answer: Option (2)

Key idea: Mean and Variance of data, forming quadratic equations from roots

1. Use mean condition: $\frac{2+4+\alpha+8+\beta+12+14}{7} = 8$ gives $\alpha + \beta = 16$
2. Use variance formula: $\text{Var} = \frac{\sum x_i^2}{n} - (\text{mean})^2 = 16$ gives $\alpha^2 + \beta^2 = 136$
3. From $(\alpha+\beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$: $256 = 136 + 2\alpha\beta$, so $\alpha\beta = 60$
4. Solve $t^2 - 16t + 60 = 0$ to get $\alpha = 6, \beta = 10$ (since $\alpha < \beta$)
5. Calculate roots: $3\alpha + 2 = 20$ and $2\beta + 1 = 21$
6. Form equation: $x^2 - (20+21)x + (20)(21) = 0$, which gives $x^2 - 41x + 420 = 0$

Answer: Option (3)

Key idea: Probability with combinations and arrangements

1. We have 6 blue and 6 green balls. We draw 6 pairs without replacement.
2. **Total arrangements**: Arrange 6 B's and 6 G's in 12 positions where positions form pairs: $(1,2), (3,4), (5,6), (7,8), (9,10), (11,12)$. Total = $\binom{12}{6} = 924$
3. **Favorable arrangements**: Each pair must have exactly 1 blue and 1 green ball.
4. For each of the 6 pairs, choose which position gets blue (other gets green): $2^6 = 64$ ways
5. **Probability** = $\frac{64}{924} = \frac{16}{231}$

Answer: 6√2

Key idea: Circle touching x-axis, y-intercept, and chord length using perpendicular distance formula

1. Since the circle touches the x-axis at $(3, 0)$ and center is in first quadrant, center is at $(3, r)$ where $r$ is the radius
2. For y-axis intercept, substitute $x = 0$ in $(x-3)^2 + (y-r)^2 = r^2$ to get intercept length $= 2\sqrt{r^2 - 9}$
3. Given intercept length $= 6\sqrt{3}$, so $2\sqrt{r^2 - 9} = 6\sqrt{3}$ which gives $r^2 = 36$, thus $r = 6$
4. Center is $(3, 6)$ with radius $6$. Find perpendicular distance from center to line $x - y - 3 = 0$: $d = \frac{|3-6-3|}{\sqrt{2}} = 3\sqrt{2}$
5. Apply chord length formula: Length $= 2\sqrt{r^2 - d^2} = 2\sqrt{36 - 18} = 2\sqrt{18} = 6\sqrt{2}$

Answer: Option (4)

Key idea: Ellipse and Hyperbola Properties: Eccentricity, Directrix, Latus Rectum, and Foci

1. Step 1: For ellipse $E$, use directrix formula $\frac{A}{e} = \frac{4\sqrt{6}}{3}$ with $e = \frac{\sqrt{3}}{2}$ to get semi-major axis $A = 2\sqrt{2}$
2. Step 2: Use $e^2 = 1 - \frac{B^2}{A^2}$ to find semi-minor axis $B = \sqrt{2}$, so minor axis length $= 2\sqrt{2}$
3. Step 3: For hyperbola $H$, eccentricity $e_H = 2\sqrt{2}$ (equals $A$ of ellipse). Use $e_H^2 = 1 + \frac{b^2}{a^2}$ to get $\frac{b^2}{a^2} = 7$
4. Step 4: Latus rectum of $H$ equals minor axis of $E$: $\frac{2b^2}{a} = 2\sqrt{2}$, giving $b^2 = a\sqrt{2}$
5. Step 5: Solve $b^2 = 7a^2$ and $b^2 = a\sqrt{2}$ simultaneously to get $a = \frac{\sqrt{2}}{7}$
6. Step 6: Calculate $c = ae_H = \frac{\sqrt{2}}{7} \times 2\sqrt{2} = \frac{4}{7}$
7. Step 7: Distance between foci $= 2c = \frac{8}{7}$

Answer: Option (1)

Key idea: Locus of midpoint of focal chord of ellipse

1. Find semi-major axis: Using directrix $x = \frac{a}{e} = 9$ and $e = \frac{1}{3}$, we get $a = 3$
2. Find focus location: $\alpha = ae = 3 \times \frac{1}{3} = 1$, so $P(1, 0)$
3. Find $b^2$: Using $b^2 = a^2(1-e^2) = 9(1-\frac{1}{9}) = 8$
4. Ellipse equation: $\frac{x^2}{9} + \frac{y^2}{8} = 1$
5. For chord with midpoint $(h,k)$ passing through $P(1,0)$: $\frac{h}{9} = \frac{h^2}{9} + \frac{k^2}{8}$
6. Simplify: $8h = 8h^2 + 9k^2 \Rightarrow 9k^2 = 8h(1-h)$
7. Locus: $9y^2 = 8x(1-x)$

Answer: Option (1)

Key idea: Inverse Trigonometric Functions - Simplification and Equation Solving

1. Let $\theta = \tan^{-1}(x\sqrt{2})$, so $\tan\theta = x\sqrt{2}$. Using a right triangle, $\sin(\tan^{-1}(x\sqrt{2})) = \frac{x\sqrt{2}}{\sqrt{2x^2+1}}$
2. Let $\phi = \sin^{-1}\sqrt{1-x^2}$, so $\sin\phi = \sqrt{1-x^2}$ and $\cos\phi = x$. Thus $\cot(\sin^{-1}\sqrt{1-x^2}) = \frac{x}{\sqrt{1-x^2}}$
3. Equate both sides: $\frac{x\sqrt{2}}{\sqrt{2x^2+1}} = \frac{x}{\sqrt{1-x^2}}$. Divide by $x$ (since $x > 0$): $\frac{\sqrt{2}}{\sqrt{2x^2+1}} = \frac{1}{\sqrt{1-x^2}}$
4. Cross-multiply: $\sqrt{2} \cdot \sqrt{1-x^2} = \sqrt{2x^2+1}$
5. Square both sides: $2(1-x^2) = 2x^2 + 1$, which simplifies to $2 - 2x^2 = 2x^2 + 1$
6. Solve: $1 = 4x^2$, so $x^2 = \frac{1}{4}$, giving $x = \frac{1}{2}$

Answer: Option (3)

Key idea: Shortest distance between two skew lines using the formula d = |((P₂ - P₁) · (b₁ × b₂))| / |b₁ × b₂|

1. Identify points and direction vectors: $P_1 = (4, 3, 2)$, $\vec{b_1} = (1, 2, -3)$ and $P_2 = (-2, 6, 5)$, $\vec{b_2} = (2, 4, -5)$
2. Calculate $P_2 - P_1 = (-6, 3, 3)$
3. Find cross product: $\vec{b_1} \times \vec{b_2} = (2, -1, 0)$
4. Calculate magnitude: $|\vec{b_1} \times \vec{b_2}| = \sqrt{4 + 1 + 0} = \sqrt{5}$
5. Find dot product: $(P_2 - P_1) \cdot (\vec{b_1} \times \vec{b_2}) = (-6)(2) + (3)(-1) + (3)(0) = -15$
6. Apply formula: $d = \frac{|-15|}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$

Answer: 1800

Key idea: Area of triangle using cross product of vectors

1. Step 1: Calculate $2\vec{a} + 3\vec{b} = 4\hat{i} + 6\hat{j} + 6\hat{k} + 18\hat{i} + 9\hat{j} + 9\hat{k} = 22\hat{i} + 15\hat{j} + 15\hat{k}$
2. Step 2: Calculate $\vec{a} - \vec{b} = (2-6)\hat{i} + (3-3)\hat{j} + (3-3)\hat{k} = -4\hat{i}$
3. Step 3: Find cross product $(2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 22 & 15 & 15 \\ -4 & 0 & 0 \end{vmatrix} = -60\hat{j} + 60\hat{k}$
4. Step 4: Calculate magnitude $|\vec{u} \times \vec{v}| = \sqrt{60^2 + 60^2} = \sqrt{7200} = 60\sqrt{2}$
5. Step 5: Area of triangle $= \frac{1}{2}|\vec{u} \times \vec{v}| = \frac{1}{2} \cdot 60\sqrt{2} = 30\sqrt{2}$
6. Step 6: Square of area $= (30\sqrt{2})^2 = 900 \times 2 = 1800$

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Answer: Option (2)

Key idea: Singular Matrix (Determinant = 0), Integration, and Optimization on Closed Interval

1. Find $\alpha$ using $\det(A) = 0$: Expand the determinant to get $\det(A) = 3 - \alpha = 0$, so $\alpha = 3$
2. Integrate to find $f(x) = \int_0^x (t^2 + 2t + 3)dt = \frac{x^3}{3} + x^2 + 3x$
3. Check for critical points: $f'(x) = x^2 + 2x + 3 > 0$ for all $x$ (discriminant $< 0$), so $f$ is strictly increasing
4. Since $f$ is increasing on $[1, 3]$: minimum $m = f(1) = \frac{13}{3}$ and maximum $M = f(3) = 27$
5. Calculate $3(M - m) = 3\left(27 - \frac{13}{3}\right) = 3 \cdot \frac{68}{3} = 68$

Answer: Option (3)

Key idea: Functional equations and first-order linear differential equations

1. Step 1: From $f(xy) = f(x)f(y)$ and $f(0) \neq 0$, set $x=y=0$ to get $f(0) = [f(0)]^2$, so $f(0) = 1$. Then $f(0) = f(x \cdot 0) = f(x)f(0)$ gives $f(x) = 1$ for all $x$.
2. Step 2: Substitute $f(t) = 1$ into the equation: $x^2 g(x) = \int_1^x (t^2 - tg(t)) dt$
3. Step 3: Differentiate both sides using Leibniz rule: $2xg(x) + x^2g'(x) = x^2 - xg(x)$
4. Step 4: Simplify to get the linear ODE: $g'(x) + \frac{3}{x}g(x) = 1$
5. Step 5: Solve using integrating factor $\mu(x) = x^3$: $\frac{d}{dx}[x^3g(x)] = x^3$, so $g(x) = \frac{x}{4} + \frac{C}{x^3}$
6. Step 6: Use initial condition $g(1) = 0$ to find $C = -\frac{1}{4}$, giving $g(x) = \frac{x}{4} - \frac{1}{4x^3}$
7. Step 7: Calculate $g(2) = \frac{2}{4} - \frac{1}{4 \cdot 8} = \frac{1}{2} - \frac{1}{32} = \frac{15}{32}$

Answer: Option (1)

Key idea: Area between curves using definite integration

1. Find intersection points by solving $x^2 - 8x = -x$, which gives $x^2 - 7x = 0$, so $x = 0$ or $x = 7$
2. Identify that the region has $y = x^2 - 8x$ as the lower boundary and $y = -x$ as the upper boundary
3. Set up the area integral: $A = \int_0^7 [(-x) - (x^2 - 8x)] dx = \int_0^7 (7x - x^2) dx$
4. Integrate: $A = \left[\frac{7x^2}{2} - \frac{x^3}{3}\right]_0^7$
5. Evaluate at $x = 7$: $A = \frac{343}{2} - \frac{343}{3} = \frac{1029 - 686}{6} = \frac{343}{6}$

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Answer: 276

Key idea: Transitive Relations and Ordered Pairs

1. Step 1: From $\log_5(x + y) \leq 2$, we get $x + y \leq 25$. So $R = \{(x, y) \in \mathbb{N} \times \mathbb{N} : x + y \leq 25\}$.
2. Step 2: For transitivity, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c)$ must be in $R$. This fails when $a + b \leq 25$, $b + c \leq 25$, but $a + c > 25$.
3. Step 3: A pair $(a, c)$ with $a + c > 25$ needs to be added if there exists $b \in \mathbb{N}$ such that $a + b \leq 25$ and $b + c \leq 25$.
4. Step 4: This requires $b \leq \min(25-a, 25-c)$ and $b \geq 1$, which is possible when $\max(a, c) \leq 24$.
5. Step 5: Count pairs $(a, c)$ where $a, c \in \{1, 2, ..., 24\}$ and $a + c > 25$. For each $a \in \{2, ..., 24\}$, we have $c \in \{26-a, ..., 24\}$.
6. Step 6: For $a = 2$: 1 pair, $a = 3$: 2 pairs, ..., $a = 24$: 23 pairs. Total = $1 + 2 + ... + 23 = \frac{23 \times 24}{2} = 276$.

Answer: 30

Key idea: Binomial Theorem with substitution and coefficient matching

1. Substitute $y = 1-x$ to get $(1-x^3)^{10} = y^{10}(3-3y+y^2)^{10}$
2. Rewrite the given sum as $\sum_{r=0}^{10} a_r (1-y)^r y^{30-2r}$
3. Divide both sides by $y^{10}$ to get $(3-3y+y^2)^{10} = \sum_{r=0}^{10} a_r (1-y)^r y^{20-2r}$
4. Write $(3-3y+y^2)^{10} = (3(1-y)+y^2)^{10}$ and expand using binomial theorem
5. For $a_{10}$: coefficient of $(1-y)^{10}$ is $\binom{10}{10}3^{10} = 3^{10}$
6. For $a_9$: coefficient of $(1-y)^9 y^2$ is $\binom{10}{9}3^9 = 10 \cdot 3^9$
7. Calculate $\frac{9a_9}{a_{10}} = \frac{9 \cdot 10 \cdot 3^9}{3^{10}} = \frac{90}{3} = 30$

Answer: 18

Key idea: Perpendicular bisector of a chord, intersection of line and circle

1. Circle $C$ has center $O(4, -3)$ and radius $3$. Line $x - y = 4$ intersects it at $Q$ and $R$.
2. Since $PQ = PR$, point $P$ lies on the perpendicular bisector of chord $QR$.
3. The perpendicular bisector of any chord passes through the center. Line $QR$ has slope $1$, so the perpendicular bisector has slope $-1$.
4. Perpendicular bisector through $O(4, -3)$ with slope $-1$: $y + 3 = -1(x - 4)$, which gives $x + y = 1$.
5. Point $P(\alpha, \beta)$ satisfies both $(x-4)^2 + (y+3)^2 = 9$ and $x + y = 1$. Substituting $y = 1 - x$: $(x-4)^2 + (4-x)^2 = 9$, giving $2(x-4)^2 = 9$.
6. Solving: $x = 4 \pm \frac{3\sqrt{2}}{2}$ and $y = -3 \mp \frac{3\sqrt{2}}{2}$.
7. For both solutions: $6\alpha + 8\beta = \pm 3\sqrt{2}$, so $(6\alpha + 8\beta)^2 = 18$.

Answer: 162

Key idea: Image of a point in a line, Area of parallelogram using cross product

1. Find image $R$ of $P(0,-5,0)$ in line $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$: Foot of perpendicular $M=(-1,-1,1)$, so $R=2M-P=(-2,3,2)$
2. Find image $S$ of $Q(0,-\frac{1}{2},0)$ in line $\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$: Foot of perpendicular $N=(-1,-1,1)$, so $S=2N-Q=(-2,-\frac{3}{2},2)$
3. Verify $PQRS$ is a parallelogram: Check $\vec{PQ}=\vec{SR}=(0,\frac{9}{2},0)$ and $\vec{PS}=\vec{QR}=(-2,\frac{7}{2},2)$
4. Calculate area using cross product: $\vec{PQ}\times\vec{PS}=(9,0,9)$
5. Find magnitude: $|\vec{PQ}\times\vec{PS}|=\sqrt{81+0+81}=9\sqrt{2}$
6. Square of area: $(9\sqrt{2})^2=81\times 2=162$

Answer: 3

Key idea: Composition of piecewise functions and continuity

1. Find $f(x)$ values: For $x < 0$, $f(x) = x^3 + 8$; for $x \ge 0$, $f(x) = x^2 - 4$
2. Determine when $f(x) < 0$ and $f(x) \ge 0$ in each region to apply correct formula for $g$
3. For $x < -2$: $f(x) < 0$, so $g(f(x)) = (x^3)^{1/3} = x$. For $-2 \le x < 0$: $f(x) \ge 0$, so $g(f(x)) = (x^3 + 12)^{1/2}$
4. For $0 \le x < 2$: $f(x) < 0$, so $g(f(x)) = (x^2 - 12)^{1/3}$. For $x \ge 2$: $f(x) \ge 0$, so $g(f(x)) = x$
5. Check continuity at transition points $x = -2, 0, 2$ by comparing left and right limits
6. At $x = -2$: LHL $= -2$, RHL $= 2$ (discontinuous). At $x = 0$: LHL $= 2\sqrt{3}$, RHL $= -\sqrt[3]{12}$ (discontinuous). At $x = 2$: LHL $= -2$, RHL $= 2$ (discontinuous)
7. Total number of discontinuities is $3$

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Key idea: Unknown - question text not visible

Answer: Option (3)

Key idea: Dimensional Analysis of Physical Constants

1. **Boltzmann constant:** From $PV = Nk_BT$, we get $k_B = \frac{PV}{NT} = \frac{[ML^{-1}T^{-2}][L^3]}{[K]} = [ML^2T^{-2}K^{-1}]$ → **III**
2. **Stefan's constant:** From $P = \sigma AT^4$, we get $\sigma = \frac{P}{AT^4} = \frac{[ML^2T^{-3}]}{[L^2][K^4]} = [ML^0T^{-3}K^{-4}]$ → **IV**
3. **Planck's constant:** From $E = h\nu$, we get $h = \frac{E}{\nu} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$ → **II**
4. **Gravitational constant:** From $F = \frac{Gm_1m_2}{r^2}$, we get $G = \frac{Fr^2}{m_1m_2} = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$ → **I**
5. Matching: A-III, B-IV, C-II, D-I

Answer: Option (1)

Key idea: Rolling motion with tangential force; moment of inertia of sphere and shell

1. Write equations for rolling motion: Linear equation $F + f = Ma$ and rotational equation $FR - fR = I\alpha$
2. Use rolling constraint $a = \alpha R$ to get $F - f = \frac{Ia}{R^2}$
3. Solve the two equations to get $a = \frac{2FR^2}{MR^2 + I}$
4. For solid sphere A: $I_A = \frac{2}{5}(5m)R^2 = 2mR^2$, so $a_A = \frac{2FR^2}{7mR^2} = \frac{2F}{7m}$
5. For spherical shell B: $I_B = \frac{2}{3}mR^2$, so $a_B = \frac{2FR^2}{\frac{5mR^2}{3}} = \frac{6F}{5m}$
6. Calculate ratio: $\frac{a_A}{a_B} = \frac{2F/7m}{6F/5m} = \frac{5}{21}$

Answer: Option (3)

Key idea: Work-Energy Theorem

1. Given: mass $m = 1$ kg, velocity $v = 2x^2$, displacement from $x = 0$ to $x = 5$ m
2. Apply Work-Energy Theorem: $W = \Delta KE = KE_f - KE_i$
3. Find initial velocity at $x = 0$: $v_0 = 2(0)^2 = 0$ m/s
4. Find final velocity at $x = 5$: $v_5 = 2(5)^2 = 50$ m/s
5. Calculate $KE_i = \frac{1}{2}(1)(0)^2 = 0$ J and $KE_f = \frac{1}{2}(1)(50)^2 = 1250$ J
6. Work done $W = 1250 - 0 = 1250$ J

Answer: Option (2)

Key idea: Adiabatic process for ideal gas with moving piston

1. Identify that the right compartment undergoes adiabatic compression (adiabatic walls, no heating). The piston ensures equal pressure on both sides.
2. Write the adiabatic equation for the right side: $PV^\gamma = P_fV_f^\gamma$ where $\gamma = 3/2$
3. Substitute values: $P \times 9^{3/2} = \frac{27P}{8} \times V_f^{3/2}$
4. Calculate $9^{3/2} = (\sqrt{9})^3 = 3^3 = 27$, so: $27 = \frac{27}{8} \times V_f^{3/2}$
5. Solve for $V_f^{3/2}$: $V_f^{3/2} = 27 \times \frac{8}{27} = 8$
6. Find $V_f = 8^{2/3} = (2^3)^{2/3} = 2^2 = 4$ litres

Answer: Option (1)

Key idea: Electromagnetic wave relations: magnetic field in terms of electric field and propagation vector

1. For an EM wave in vacuum, $\vec{E}$, $\vec{B}$, and $\vec{k}$ are mutually perpendicular and form a right-handed system
2. The magnitudes are related by $B = \frac{E}{c}$ where $c$ is the speed of light
3. For a wave, $k = \frac{\omega}{c}$, so $c = \frac{\omega}{k}$
4. The magnetic field must be perpendicular to both $\vec{E}$ and $\vec{k}$, suggesting a cross product
5. Using the relation $\vec{B} = \frac{1}{\omega}(\vec{k} \times \vec{E})$ or equivalently $\vec{B} = \frac{1}{\omega}(\vec{E} \times \vec{k})$ with appropriate sign
6. The answer is $\vec{B} = \frac{\vec{E} \times \vec{k}}{\omega}$

Answer: Option (3)

Key idea: Center of Mass Motion - When acceleration is parallel to initial velocity, the path is a straight line

1. Find initial velocity of COM: $\vec{v}_{CM} = \frac{m\vec{v}_1 + m\vec{v}_2}{2m} = \frac{4\hat{i} + 4\hat{j}}{2} = 2\hat{i} + 2\hat{j}$ m/s
2. Find acceleration of COM: $\vec{a}_{CM} = \frac{m\vec{a}_1 + m\vec{a}_2}{2m} = \frac{6\hat{i} + 6\hat{j}}{2} = 3\hat{i} + 3\hat{j}$ m/s²
3. Check if $\vec{a}_{CM}$ is parallel to $\vec{v}_{CM}$: $\vec{v}_{CM} = 2(\hat{i} + \hat{j})$ and $\vec{a}_{CM} = 3(\hat{i} + \hat{j})$
4. Since $\vec{a}_{CM} = \frac{3}{2}\vec{v}_{CM}$, acceleration is parallel to initial velocity
5. When acceleration is parallel to velocity, the motion is along a straight line

Answer: Option (1)

Key idea: Young's Modulus and Stress-Strain Relationship

1. Read from graph: At $W = 60$ N, $\Delta l = 0.6 \times 10^{-4}$ m
2. Young's modulus formula: $Y = \frac{F \cdot L}{A \cdot \Delta l}$
3. Given: $L = 1$ m, $A = 10^{-5}$ m², $F = 60$ N, $\Delta l = 0.6 \times 10^{-4}$ m
4. Substitute: $Y = \frac{60 \times 1}{10^{-5} \times 0.6 \times 10^{-4}}$
5. Calculate denominator: $10^{-5} \times 0.6 \times 10^{-4} = 0.6 \times 10^{-9}$ m³
6. Simplify: $Y = \frac{60}{0.6 \times 10^{-9}} = \frac{100}{10^{-9}} = 10^{11}$ N/m²
7. Therefore, $Y = 1.0 \times 10^{11}$ N/m²

Answer: Option (2)

Key idea: Torricelli's Theorem and Projectile Motion

1. Find height of vessel: $V = \pi r^2 h \Rightarrow h = \frac{528 \times 10^{-3}}{\pi (0.4)^2} = 1.05$ m = 105 cm
2. Hole is 70 cm below water surface. Height of hole above ground = (105 + 105) - 70 = 140 cm
3. Velocity of efflux using Torricelli's theorem: $v = \sqrt{2gh_1} = \sqrt{2 \times 10 \times 0.7} = \sqrt{14}$ m/s
4. Time to fall from height $h_2 = 140$ cm: $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 \times 1.4}{10}} = \sqrt{0.28} = \frac{\sqrt{7}}{5}$ s
5. Horizontal range: $R = vt = \sqrt{14} \times \frac{\sqrt{7}}{5} = \frac{\sqrt{98}}{5} = \frac{7\sqrt{2}}{5}$ m = $140\sqrt{2}$ cm

Answer: Option (4)

Key idea: Conservation of Internal Energy in Gas Mixtures

1. For monoatomic gas, $C_v = \frac{3}{2}R$ and internal energy $U = nC_vT$
2. Calculate initial internal energy of gas 1: $U_1 = 2 \times \frac{3}{2}R \times T = 3RT$
3. Calculate initial internal energy of gas 2: $U_2 = 6 \times \frac{3}{2}R \times 2T = 18RT$
4. Total initial energy: $U_{initial} = 3RT + 18RT = 21RT$
5. After mixing, total moles = 8, so $U_{final} = 8 \times \frac{3}{2}R \times T_f = 12RT_f$
6. By energy conservation: $21RT = 12RT_f$, giving $T_f = \frac{7T}{4}$

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 (1)

Key idea: Relation between Electric Field and Electric Potential

1. Given: $V = 5(x^2 - y^2)$ V. Use the relation $\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j}\right)$
2. Calculate partial derivatives: $\frac{\partial V}{\partial x} = 5(2x) = 10x$ and $\frac{\partial V}{\partial y} = 5(-2y) = -10y$
3. Write electric field: $\vec{E} = -10x\hat{i} + 10y\hat{j}$ V/m
4. Substitute point $(2, 3)$: $\vec{E} = -10(2)\hat{i} + 10(3)\hat{j}$
5. Final result: $\vec{E} = (-20\hat{i} + 30\hat{j})$ V/m

Answer: Option (300)

Key idea: Magnetic field due to parallel current-carrying wires; superposition principle

1. Distance from each wire to midpoint: $r = \frac{8}{2} = 4$ cm $= 0.04$ m
2. Magnetic field due to one wire: $B_1 = \frac{\mu_0}{4\pi} \times \frac{2I}{r} = 10^{-7} \times \frac{2 \times 30}{0.04}$
3. Calculate: $B_1 = 10^{-7} \times 1500 = 1.5 \times 10^{-4}$ T $= 150$ μT
4. Since currents are opposite, both fields at midpoint point in the same direction (use right-hand rule)
5. Total magnetic field: $B_{net} = B_1 + B_2 = 150 + 150 = 300$ μT

Answer: Option (4)

Key idea: Faraday's Law of Electromagnetic Induction

1. Calculate area of square loop: $A = (0.02)^2 = 4 \times 10^{-4}$ m²
2. Find magnetic flux: $\Phi = BA\cos(60°) = 0.4\sin(300t) \times 4 \times 10^{-4} \times 0.5$
3. Simplify flux: $\Phi = 8 \times 10^{-5}\sin(300t)$ Wb
4. Apply Faraday's law: $\varepsilon = -\frac{d\Phi}{dt} = -8 \times 10^{-5} \times 300 \times \cos(300t)$
5. Calculate: $\varepsilon = -24\cos(300t)$ mV
6. Maximum induced emf occurs when $|\cos(300t)| = 1$, giving $\varepsilon_{max} = 24$ mV

Answer: Option (2)

Key idea: Energy loss in charge redistribution between capacitors

1. Calculate initial charge: $Q_1 = C_1V_1 = 100 \times 10^{-12} \times 100 = 10^{-8}$ C
2. Calculate initial energy: $U_i = \frac{1}{2}C_1V_1^2 = \frac{1}{2} \times 100 \times 10^{-12} \times 10^4 = 5 \times 10^{-7}$ J
3. When spheres touch, charge distributes over combined capacitance: $V_f = \frac{Q_{total}}{C_{combined}} = \frac{10^{-8}}{200 \times 10^{-12}} = 50$ V
4. Calculate final energy: $U_f = \frac{1}{2}C_{combined}V_f^2 = \frac{1}{2} \times 200 \times 10^{-12} \times 2500 = 2.5 \times 10^{-7}$ J
5. Change in energy: $|\Delta U| = |U_f - U_i| = |2.5 - 5| \times 10^{-7} = 2.5 \times 10^{-7}$ J
6. Therefore, $\alpha = 2.5 = \frac{5}{2}$

Answer: Option (2)

Key idea: Nuclear Binding Energy and Mass-Energy Equivalence

1. Identify the reaction: $2(^2_1H) \to ^4_2He$
2. Calculate total BE of reactants: $2 \times 2 \times 1.1 = 4.4$ MeV
3. Calculate total BE of product: $4 \times 7.2 = 28.8$ MeV
4. Energy released $= BE_{product} - BE_{reactants} = 28.8 - 4.4 = 24.4$ MeV

Answer: Option (3)

Key idea: Prism formula relating refractive index with angle of prism and minimum deviation

1. For an equilateral prism, angle of prism $A = 60°$
2. Given: minimum deviation $\delta_m = \frac{A}{2} = 30°$
3. Apply prism formula: $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
4. Substitute values: $\mu = \frac{\sin(45°)}{\sin(30°)} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}$
5. Simplify: $\mu = \frac{2}{\sqrt{2}} = \sqrt{2}$

Answer: Option (2)

Key idea: Logic Gates - NAND and NOR gate operations

1. Identify the logic gates in the circuit: NAND and NOR gates with inputs A and B
2. For **Case 1** $(A=1, B=1)$: Calculate NAND$(1,1) = \overline{1 \cdot 1} = \overline{1} = 0$
3. For **Case 1**: Calculate NOR$(1,1) = \overline{1 + 1} = \overline{1} = 0$
4. For **Case 2** $(A=0, B=1)$: Calculate NAND$(0,1) = \overline{0 \cdot 1} = \overline{0} = 1$
5. For **Case 2**: Calculate NOR$(0,1) = \overline{0 + 1} = \overline{1} = 0$
6. Determine output Y based on circuit configuration: Y follows NAND operation giving outputs $(0, 1)$ for the two input cases
7. Match with options: Output is $0$ for $(A=1,B=1)$ and $1$ for $(A=0,B=1)$ → Option (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 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: Interference of light waves and intensity distribution in Young's double slit experiment

Answer: 75

Key idea: Motion on inclined plane with and without friction, kinematic equations

1. For motion with friction: acceleration $a_1 = g\sin 45° - \mu g\cos 45° = \frac{g}{\sqrt{2}}(1-\mu)$
2. Distance traveled: $s = \frac{1}{2}a_1 t^2 = \frac{1}{2} \cdot \frac{g}{\sqrt{2}}(1-\mu) \cdot t^2$
3. For frictionless motion: acceleration $a_2 = g\sin 45° = \frac{g}{\sqrt{2}}$
4. Distance traveled: $s = \frac{1}{2}a_2 \left(\frac{t}{2}\right)^2 = \frac{1}{2} \cdot \frac{g}{\sqrt{2}} \cdot \frac{t^2}{4}$
5. Equating distances: $\frac{1}{2}(1-\mu) = \frac{1}{8}$, which gives $\mu = \frac{3}{4}$
6. Since $\mu = \frac{\alpha}{100}$, we get $\alpha = 75$

Answer: 4

Key idea: de Broglie wavelength and its dependence on accelerating potential

1. For an electron accelerated through potential $V$, de Broglie wavelength $\lambda = \frac{h}{\sqrt{2meV}}$, so $\lambda \propto \frac{1}{\sqrt{V}}$
2. Given that wavelength increases by 50%, so $\lambda_2 = 1.5\lambda_1$
3. Using the relation $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$, substitute $\lambda_2 = 1.5\lambda_1$
4. This gives $\frac{\lambda_1}{1.5\lambda_1} = \frac{2}{3} = \sqrt{\frac{V_2}{V_1}}$
5. Squaring both sides: $\frac{4}{9} = \frac{V_2}{V_1}$, so $\frac{V_1}{V_2} = \frac{9}{4}$
6. Comparing with $\frac{V_1}{V_2} = \frac{9}{a}$, we get $a = 4$

Answer: 50

Key idea: Galvanometer with shunt and series resistance; current division and Ohm's law

1. Let $G$ be the galvanometer resistance and $I_g$ be the full-scale deflection current.
2. **Shunt case:** When shunted with $2\,\Omega$, total current is $0.5\,\text{A}$. Using current division: $I_g = \frac{1}{2+G}$
3. **Series case:** With $470\,\Omega$ in series and $10\,\text{V}$ applied: $I_g = \frac{10}{G+470}$
4. Equate both expressions for $I_g$: $\frac{1}{2+G} = \frac{10}{G+470}$
5. Cross-multiply: $G + 470 = 10(2+G) = 20 + 10G$
6. Solve: $450 = 9G \Rightarrow G = 50\,\Omega$

Answer: 3

Key idea: Cells in parallel with same and opposite polarities, equivalent EMF and internal resistance

1. For parallel cells with same polarity: $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{1/2 + 2/1}{1/2 + 1} = \frac{5/3}{3/2} = \frac{5}{3}$ V
2. Equivalent internal resistance: $r_{eq} = \frac{1}{1/r_1 + 1/r_2} = \frac{1}{3/2} = \frac{2}{3}$ Ω
3. Using $I = \frac{E_{eq}}{r_{eq} + R}$: $1 = \frac{5/3}{2/3 + R}$ gives $R = 1$ Ω
4. When polarity reversed: $E_{eq} = \frac{E_2/r_2 - E_1/r_1}{1/r_1 + 1/r_2} = \frac{2 - 1/2}{3/2} = 1$ V
5. New current: $I = \frac{1}{2/3 + 1} = \frac{1}{5/3} = \frac{3}{5}$ A
6. Comparing with $\frac{\alpha}{5}$ A: $\alpha = 3$

Answer: 10

Key idea: Mirror equation and magnification for concave mirrors with two object positions giving same magnitude of magnification

1. Given: Concave mirror with $f = -10$ cm, magnification $|m| = 2$
2. **Case 1:** Real image with $m = -2$, so $v = 2u$. Using $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$: $\frac{-1}{10} = \frac{1}{2u} + \frac{1}{u} = \frac{3}{2u}$, giving $u_1 = -15$ cm
3. **Case 2:** Virtual image with $m = +2$, so $v = -2u$. Using mirror equation: $\frac{-1}{10} = \frac{-1}{2u} + \frac{1}{u} = \frac{1}{2u}$, giving $u_2 = -5$ cm
4. Verify: For $u_1 = -15$ cm, $v = -30$ cm, $m = -2$ ✓; For $u_2 = -5$ cm, $v = +10$ cm, $m = +2$ ✓
5. Distance between two positions = $|u_1 - u_2| = |-15 - (-5)| = 10$ cm

Answer: Option (4)

Key idea: Mole concept, Avogadro's number, and calculation of number of atoms from mass, volume at STP, and moles

1. Calculate moles for each option: A: $n = \frac{2}{32} = 0.0625$ mol O₂; B: $n = \frac{4}{64} = 0.0625$ mol SO₂; C: $n = \frac{1400}{22400} = 0.0625$ mol O₂; D: $n = \frac{50}{22400} = 0.00223$ mol He; E: $n = 0.0625$ mol H₂
2. For A: Number of atoms = $0.0625 \times N_A \times 2 = 0.125 N_A$ (O₂ has 2 atoms per molecule)
3. For B: Number of atoms = $0.0625 \times N_A \times 3 = 0.1875 N_A$ (SO₂ has 3 atoms per molecule)
4. For C: Number of atoms = $0.0625 \times N_A \times 2 = 0.125 N_A$ (O₂ has 2 atoms per molecule)
5. For D: Number of atoms = $0.00223 \times N_A \times 1 = 0.00223 N_A$ (He is monoatomic)
6. For E: Number of atoms = $0.0625 \times N_A \times 2 = 0.125 N_A$ (H₂ has 2 atoms per molecule)
7. Compare: A, C, and E all have $0.125 N_A$ atoms, so answer is option (4)

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: Bohr's Model - Radius of hydrogen-like species

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: Octet Rule, VSEPR Theory, and Lone Pair Determination

Answer: Option (3)

Key idea: Isothermal processes and thermodynamic expressions for different types of expansions/compressions

1. **A. Reversible expansion (Isothermal):** For isothermal reversible process, $q = nRT \ln \frac{V_2}{V_1}$ → Matches with **II**
2. **B. Free expansion (Isothermal):** $P_{ext} = 0$, so $w = 0$. Since $\Delta U = 0$ (isothermal), $q = 0$ → Matches with **I**
3. **C. Irreversible Compression (Isothermal):** Work done is $w = -P_{ext}(V_2 - V_1) = -P_{ext}(V_1 - V_2)$ for compression → Matches with **III**
4. **D. Cyclic reversible (Isothermal):** For a complete cycle, $\Delta S = 0$, so $\frac{q_{rev}}{T} = 0$ → Matches with **IV**
5. Therefore: A-II, B-I, C-III, D-IV

Answer: Option (4)

Key idea: Osmotic Pressure and Colligative Properties

1. Calculate molarity of Chamber 1: $M_1 = \frac{18/180}{0.1} = 1$ M
2. Calculate molarity of Chamber 2: $M_2 = \frac{30/180}{0.25} = 0.668$ M
3. Since $M_1 > M_2$, water moves from Chamber 2 to Chamber 1, so Statement I is **false**
4. For Statement II, calculate moles of K₂SO₄: $n = \frac{0.050}{174} = 2.874 \times 10^{-4}$ mol
5. K₂SO₄ dissociates into 3 ions, so van't Hoff factor $i = 3$
6. Calculate osmotic pressure: $\pi = iCRT = 3 \times \frac{2.874 \times 10^{-4}}{2} \times 0.083 \times 300 = 0.0107$ bar
7. Statement II is **true**, so answer is option (4)

Answer: Option (2)

Key idea: Degree of dissociation of weak electrolyte using equilibrium constant with small alpha approximation

1. Write dissociation: $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$
2. Set up equilibrium concentrations: $[A_xB_y] = c(1-\alpha)$, $[A^{y+}] = xc\alpha$, $[B^{x-}] = yc\alpha$
3. Write $K = \frac{(xc\alpha)^x(yc\alpha)^y}{c(1-\alpha)} = \frac{x^x y^y c^{x+y} \alpha^{x+y}}{c(1-\alpha)}$
4. For concentrated solution, $\alpha \ll 1$, so $(1-\alpha) \approx 1$
5. Simplify: $K = x^x y^y c^{x+y-1} \alpha^{x+y}$
6. Solve for $\alpha$: $\alpha = \left(\frac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}$

Answer: Option (4)

Key idea: Nernst Equation and Electrochemical Work

1. Write the balanced overall reaction by multiplying anode by $n_2$ and cathode by $n_1$ to balance electrons: $n_2\text{Red}_1 + n_1\text{Ox}_2 \to n_2\text{Ox}_1^{n_1^+} + n_1\text{Red}_2^{n_2^-}$ ✓
2. Verify statement (2): Electrons cancel in overall reaction since electrons produced at anode equal electrons consumed at cathode ✓
3. Check statement (3) using Nernst equation: $E = E° - \frac{2.303RT}{nF}\log_{10}Q$, so $\frac{E-E°}{RT/F} = -\frac{2.303}{n}\log_{10}Q$ with slope $\propto \frac{1}{n}$ ✓
4. Analyze statement (4): For reversible electrochemical work, $W_{\text{elec}} = nFE = Q \times E$ (charge × potential)
5. Statement (4) incorrectly states work equals ratio $Q/E$ instead of product $Q \times E$
6. Therefore, statement (4) is INCORRECT

Answer: Option (1)

Key idea: Periodic trends: Ionization enthalpy and electron gain enthalpy across a period

1. Identify extreme left elements in a period: **Alkali metals** (Group 1)
2. Alkali metals have largest atomic size and lowest effective nuclear charge, so they have **lowest first ionization enthalpy** in the period
3. Identify extreme right elements (except noble gases): **Halogens** (Group 17)
4. Halogens have smallest atomic size (except noble gases) and highest effective nuclear charge, with strong tendency to gain electrons
5. Halogens have **most negative (lowest) electron gain enthalpy** as they release maximum energy when gaining an electron
6. Therefore: First ionization enthalpy = **lowest**, Electron gain enthalpy = **lowest** (most negative)

Answer: Option (3)

Key idea: Bond angles in bent molecules and ionic vs covalent character in fluorides

1. **Statement I Analysis:** Compare bond angles in $F_2O$, $H_2O$, and $Cl_2O$ (all bent with 2 lone pairs on O)
2. In $F_2O$: F is more electronegative than O, pulls bonding electrons away, increases lone pair repulsion → bond angle $\approx 103°$
3. In $H_2O$: Intermediate electronegativity → bond angle $\approx 104.5°$
4. In $Cl_2O$: Cl is less electronegative than O, bonding electrons closer to O, reduces lone pair repulsion; Cl is larger → bond angle $\approx 110.9°$
5. Therefore $F_2O < H_2O < Cl_2O$ is correct. **Statement I is TRUE**
6. **Statement II Analysis:** $SiF_4$, $SnF_4$, and $PbF_4$ all have high oxidation state (+4) and form covalent bonds due to high charge density and polarization. **Statement II is FALSE**
7. Answer: Statement I is true but Statement II is false → Option (3)

Answer: Option (2)

Key idea: Ionization enthalpy trends in d-block elements based on electronic configuration stability

1. Write electronic configurations: Cr = [Ar] $3d^5 4s^1$, Mn = [Ar] $3d^5 4s^2$
2. For first ionization: Mn has higher nuclear charge (Z=25) than Cr (Z=24), so $\Delta_i H_1$: Mn > Cr (Statement C is correct)
3. After first ionization: Cr$^+$ = [Ar] $3d^5$ (half-filled, very stable), Mn$^+$ = [Ar] $3d^5 4s^1$
4. For second ionization: Cr$^+$ must lose electron from stable $d^5$ (difficult), while Mn$^+$ loses from $4s$ to achieve stable $d^5$ (easier)
5. Therefore $\Delta_i H_2$: Cr > Mn (Statement B is correct)
6. Statements B and C are correct, so answer is option (2)

Answer: Option (4)

Key idea: Hybridization, geometry, and magnetic properties of coordination compounds based on ligand field strength

1. For **A. $[NiCl_4]^{2-}$**: $Ni^{2+}$ has $3d^8$ configuration. $Cl^-$ is weak field, no pairing occurs. Tetrahedral with $sp^3$ hybridization and 2 unpaired electrons → **paramagnetic** ✓
2. For **B. $[Ni(NH_3)_6]^{2+}$**: $Ni^{2+}$ with $3d^8$. $NH_3$ is moderate field ligand. Octahedral with $sp^3d^2$ hybridization and 2 unpaired electrons → **paramagnetic** ✓
3. For **C. $[Ni(CO)_4]$**: $Ni^0$ has $3d^{10}$ after electron rearrangement. $CO$ is strong field causing complete pairing. Tetrahedral with $sp^3$ but **diamagnetic** (0 unpaired electrons) ✗
4. For **D. $[Ni(CN)_4]^{2-}$**: $Ni^{2+}$ with $3d^8$. $CN^-$ is strong field causing pairing. Square planar with $dsp^2$ hybridization and 0 unpaired electrons → **diamagnetic** ✓
5. Statement C is incorrect because $[Ni(CO)_4]$ is diamagnetic, not paramagnetic
6. Therefore, A, B, and D are correct sequences

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: Separation techniques based on differential solubility and steam distillation

Answer: α < β < γ

Key idea: Bond lengths in esters: resonance effects and hybridization

1. Identify the three bonds: $\alpha$ is C=O (carbonyl), $\beta$ is C-O (ester linkage), $\gamma$ is O-CH₃
2. The $\alpha$ bond is a C=O double bond, which is the shortest (~1.20 Å)
3. The $\beta$ bond has partial double bond character due to resonance (lone pair on O delocalizes into carbonyl), making it shorter than a normal single bond (~1.35 Å)
4. The $\gamma$ bond is a regular C-O single bond with no resonance effects, making it the longest (~1.43 Å)
5. Apply the rule: double bonds < resonance-stabilized single bonds < regular single bonds
6. Therefore, the correct order is: $\alpha < \beta < \gamma$

Answer: Option (2)

Key idea: Aldol condensation of ketones; Hydration of alkynes; Grignard reaction with nitriles

1. Step 1: Hydration of prop-1-yne ($HC≡C-CH_3$) with $HgSO_4/H^+$ gives propanone: $HC≡C-CH_3 + H_2O → CH_3COCH_3$ (compound $x$)
2. Step 2: Reaction of ethane nitrile ($CH_3CN$) with $CH_3MgBr$ followed by hydrolysis gives propanone: $CH_3CN + CH_3MgBr → CH_3COCH_3$ (compound $y$)
3. Step 3: Both $x$ and $y$ are acetone ($CH_3COCH_3$). In presence of $Ba(OH)_2$, aldol addition occurs between two acetone molecules to form $CH_3COCH_2C(OH)(CH_3)_2$
4. Step 4: Upon heating, dehydration occurs to give the α,β-unsaturated ketone: $CH_3CO-CH=C(CH_3)_2$
5. Step 5: Numbering from carbonyl end: $CH_3-CO-CH=C(CH_3)-CH_3$ gives 4-Methylpent-3-en-2-one

Answer: Option (1)

Key idea: Identification of functional groups through chemical tests; Haloform test for methyl ketones

1. Optically active $C_4H_9Br$ (2-bromobutane) undergoes E2 elimination with ethanolic KOH to give but-2-ene [A]: $CH_3CH=CHCH_3$
2. [A] reacts with $Br_2$ to form 2,3-dibromobutane [B]: $CH_3CHBr-CHBrCH_3$
3. [B] undergoes double elimination with ethanolic KOH and sodamide to form but-2-yne [C]: $CH_3C≡CCH_3$
4. [C] undergoes hydration with $HgSO_4/H_2SO_4$ at 333 K to form butan-2-one [D]: $CH_3COCH_2CH_3$ (a methyl ketone)
5. Methyl ketones give positive **Haloform test** with $I_2/NaOH$, producing yellow $CHI_3$ precipitate

Answer: Option (4)

Key idea: Ether cleavage with HI - regioselectivity based on carbocation stability

1. Benzyl phenyl ether reacts with HI: $C_6H_5-O-CH_2-C_6H_5 + HI$
2. Oxygen gets protonated: $C_6H_5-\overset{+}{O}H-CH_2-C_6H_5$
3. The $O-CH_2$ bond cleaves (not $C_6H_5-O$) because benzyl cation $(^+CH_2C_6H_5)$ is stabilized by resonance, while phenyl cation is extremely unstable
4. Products formed: Phenol $(C_6H_5OH)$ and Benzyl iodide $(C_6H_5CH_2I)$
5. Statement I is FALSE (claims benzyl alcohol + iodobenzene, but actual products are phenol + benzyl iodide)
6. Statement II is TRUE (correctly identifies $O-CH_2$ bond cleavage)
7. Answer: Statement I is false but Statement II is true

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: Hofmann Bromamide Degradation and Diazotization of Primary Aliphatic Amines

Answer: Option (4)

Key idea: Reduction of carbohydrates, catalytic aromatization, Friedel-Crafts acylation, and counting π electrons and lone pairs in benzophenone

1. Step 1: Treat glucose with $\text{HI}/\Delta$ to reduce all $\text{-OH}$ and $\text{-CHO}$ groups, forming n-hexane ($\text{C}_6\text{H}_{14}$)
2. Step 2: Catalytic dehydrogenation with $\text{V}_2\text{O}_5$ at high temperature converts n-hexane to benzene through cyclization and aromatization
3. Step 3: Friedel-Crafts acylation with benzoyl chloride and $\text{AlCl}_3$ gives benzophenone: $\text{C}_6\text{H}_5\text{-CO-C}_6\text{H}_5$
4. Count π electrons: First benzene ring has 3 π bonds = 6 electrons, second benzene ring has 3 π bonds = 6 electrons
5. Count carbonyl: $\text{C=O}$ has 1 π bond = 2 electrons
6. Count lone pairs: Oxygen has 2 lone pairs = 4 electrons
7. Total electrons = $6 + 6 + 2 + 4 = 18$ electrons

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: Brown Ring Test for Nitrates - Formation of Nitrosyl Complex

Answer: 5

Key idea: Coordination Chemistry - Ionizable vs Non-ionizable Ligands

1. Write the formula of tetraaquadichloridochromium(III) chloride: $[Cr(H_2O)_4Cl_2]Cl$
2. Identify ionizable chlorides: Only the $Cl^-$ outside the coordination sphere (counter ion) precipitates with $AgNO_3$. The 2 $Cl^-$ inside are non-ionizable.
3. Calculate moles of complex: $n = M \times V = 0.05 \times 0.1 = 5 \times 10^{-3}$ mol
4. Since 1 mole of complex gives 1 mole of ionizable $Cl^-$: Moles of $AgCl$ = $5 \times 10^{-3}$ mol
5. Express in required format: $5 \times 10^{-3}$ mol, so answer = 5

Answer: 4

Key idea: Alkane combustion stoichiometry and structural symmetry in organic compounds

1. Use combustion formula: $C_nH_{2n+2} + \frac{3n+1}{2}O_2 \rightarrow nCO_2 + (n+1)H_2O$. Given 8 moles $O_2$, so $\frac{3n+1}{2} = 8$, giving $n = 5$.
2. The alkane is $C_5H_{12}$ (pentane isomer).
3. Since only one monochlorinated product forms, all H atoms must be equivalent. This indicates high symmetry.
4. Among $C_5H_{12}$ isomers, only neopentane (2,2-dimethylpropane) has this symmetry: $(CH_3)_3C-CH_3$.
5. Neopentane has 1 central quaternary carbon and 4 primary carbons (the four $-CH_3$ groups).
6. Total number of primary carbon atoms = 4

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Key idea: Redox Titration and Stoichiometry

Answer: 8

Key idea: Chemical Equilibrium, Gibbs Free Energy, Equilibrium Constant

1. Set up ICE table: Initial 1 mol $X_2Y_4$, at equilibrium with 75% dissociation gives 0.25 mol $X_2Y_4$ and 1.5 mol $XY_2$, total = 1.75 mol
2. Calculate partial pressures: $P_{X_2Y_4} = \frac{0.25}{1.75} \times 1 = \frac{1}{7}$ atm, $P_{XY_2} = \frac{1.5}{1.75} \times 1 = \frac{6}{7}$ atm
3. Calculate $K_p = \frac{(P_{XY_2})^2}{P_{X_2Y_4}} = \frac{(6/7)^2}{1/7} = \frac{36}{7}$
4. Find $\ln K_p = \ln 36 - \ln 7 = (2\ln 2 + 2\ln 3) - \ln 7 = 2(0.69) + 2(1.104) - 1.932 = 1.656$
5. Calculate $\Delta_rG^\circ = -RT \ln K_p = -8.3 \times 600 \times 1.656 = -8247$ J/mol $= -8.247$ kJ/mol
6. Magnitude of $\Delta_rG^\circ = 8$ kJ/mol (nearest integer)

Answer: 232

Key idea: Arrhenius Equation and Activation Energy

1. Write the Arrhenius equation: $k = A e^{-E_a/RT}$
2. Compare with given equation: $k = (5.5 \times 10^{11}) e^{-28000/T}$
3. Match the exponents: $\frac{-28000}{T} = \frac{-E_a}{RT}$
4. Simplify to get: $28000 = \frac{E_a}{R}$
5. Calculate: $E_a = 28000 \times 8.3 = 232400$ J mol$^{-1}$
6. Convert to kJ mol$^{-1}$: $E_a = 232.4 \approx 232$ kJ mol$^{-1}$