JEE Main 6 April 2026 Shift 1: Questions and Solutions

Answer: Option (2)

Key idea: Domain of inverse trigonometric functions with greatest integer function

1. For $\sin^{-1}\left(\frac{x + [x]}{3}\right)$ to be defined, we need $-1 \leq \frac{x + [x]}{3} \leq 1$, which gives $-3 \leq x + [x] \leq 3$
2. Write $x = [x] + \{x\}$ where $\{x\}$ is the fractional part with $0 \leq \{x\} < 1$. Then $x + [x] = 2[x] + \{x\}$
3. For $[x] = -1$: We get $-3 \leq -2 + \{x\} \leq 3$, giving $0 \leq \{x\} < 1$, so $x \in [-1, 0)$
4. For $[x] = 0$: We get $-3 \leq \{x\} \leq 3$, giving $0 \leq \{x\} < 1$, so $x \in [0, 1)$
5. For $[x] = 1$: We get $-3 \leq 2 + \{x\} \leq 3$, giving $0 \leq \{x\} < 1$, so $x \in [1, 2)$
6. For $[x] \geq 2$ or $[x] \leq -2$, the inequality fails. Domain is $[-1, 2)$, so $\alpha = -1, \beta = 2$
7. Therefore, $\alpha^2 + \beta^2 = (-1)^2 + 2^2 = 1 + 4 = 5$

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Key idea: Unable to determine without question content

Answer: Option (3)

Key idea: Eccentricity constraints for conics, Vieta's formulas, and interval analysis

1. Use Vieta's formulas: $e_1 + e_2 = \alpha$ and $e_1 \cdot e_2 = 2$. For distinct roots: $\alpha^2 > 8$.
2. For both hyperbolas ($e_1, e_2 > 1$): Need $(e_1-1)(e_2-1) > 0 \Rightarrow 3 - \alpha > 0$. Combined with $\alpha > 2\sqrt{2}$, get $(\alpha, \beta) = (2\sqrt{2}, 3)$.
3. For ellipse and hyperbola: Let $0 < e_1 < 1$ and $e_2 > 1$. Since $e_1 e_2 = 2$ and $e_1 < 1$, we get $e_2 > 2$.
4. With $e_1 = \frac{2}{e_2}$, we have $\alpha = \frac{2}{e_2} + e_2$. For $e_2 > 2$, the function $f(x) = \frac{2}{x} + x$ is increasing.
5. At $e_2 = 2$: $\alpha = 3$. As $e_2 \to \infty$: $\alpha \to \infty$. So $(\gamma, \infty) = (3, \infty)$, giving $\gamma = 3$.
6. Calculate: $\alpha^2 + \beta^2 + \gamma^2 = (2\sqrt{2})^2 + 3^2 + 3^2 = 8 + 9 + 9 = 26$.

Answer: Option (1)

Key idea: Complex number equations with real parameter constraints using discriminant analysis

1. Let $z = x + iy$ where $x, y \in \mathbb{R}$. Substitute into the equation and expand $z(\bar{z} + 2 + i)$
2. Separate real and imaginary parts: Real: $x^2 + 2x + y^2 - y + 2k = 0$, Imaginary: $x + 2y + 3k = 0$
3. From imaginary part, get $x = -2y - 3k$ and substitute into real part to get $5y^2 + (12k - 5)y + (9k^2 - 4k) = 0$
4. For real solution to exist, discriminant $\Delta \geq 0$: $(12k - 5)^2 - 20(9k^2 - 4k) \geq 0$ gives $36k^2 + 40k - 25 \leq 0$
5. Solve quadratic: $k = \frac{-10 \pm 5\sqrt{13}}{18}$, so $\alpha = \frac{-10 - 5\sqrt{13}}{18}$ and $\beta = \frac{-10 + 5\sqrt{13}}{18}$
6. Calculate $\alpha + \beta = \frac{-20}{18} = \frac{-10}{9}$, therefore $9(\alpha + \beta) = -10$

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Answer: Option (1)

Key idea: Arithmetic Progression, Geometric Progression, and Quadratic Equations

1. Set up variables: A.P. has first term $a$ and common difference $d$; G.P. has first term $b$ and common ratio $r$
2. Use given conditions: $S_{10} = 160$ gives $2a + 9d = 32$, and $b + br = 8$ gives $b(1+r) = 8$
3. Apply the cross-relations: $a = r$ and $b = d$, so $2r + 9d = 32$ and $d(1+r) = 8$
4. From $d = \frac{8}{1+r}$, substitute into $2r + 9d = 32$ to get $2r^2 - 30r + 40 = 0$ or $r^2 - 15r + 20 = 0$
5. The two values of $r$ give two values of $b = \frac{8}{1+r}$: $b_1 = \frac{16}{17+\sqrt{145}}$ and $b_2 = \frac{16}{17-\sqrt{145}}$
6. Sum the values: $b_1 + b_2 = \frac{16(34)}{289-145} = \frac{544}{144} = \frac{34}{9}$

Answer: Option (4)

Key idea: Combinations and Permutations - Selecting and arranging distinct objects

1. Identify distinct letters in INCONSEQUENTIAL: Vowels are I, O, E, U, A (5 distinct) and Consonants are N, C, S, Q, T, L (6 distinct)
2. Select 2 vowels from 5 distinct vowels: $\binom{5}{2} = \frac{5 \times 4}{2} = 10$ ways
3. Select 2 consonants from 6 distinct consonants: $\binom{6}{2} = \frac{6 \times 5}{2} = 15$ ways
4. Arrange the 4 selected letters (2 vowels + 2 consonants): $4! = 24$ ways
5. Total number of 4-letter words = $10 \times 15 \times 24 = 3600$

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Answer: Option (2)

Key idea: Mean and Standard Deviation using algebraic manipulation of summations

1. Expand $\sum(x_i+5)^2 = \sum x_i^2 + 10\sum x_i + 500 = 2500$, giving $\sum x_i^2 + 10\sum x_i = 2000$
2. Expand $\sum(x_i-5)^2 = \sum x_i^2 - 10\sum x_i + 500 = 100$, giving $\sum x_i^2 - 10\sum x_i = -400$
3. Add the two equations: $2\sum x_i^2 = 1600$, so $\sum x_i^2 = 800$
4. Subtract the equations: $20\sum x_i = 2400$, so $\sum x_i = 120$ and mean $\bar{x} = 6$
5. Calculate variance: $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{800}{20} - 36 = 4$, so $\sigma = 2$
6. Find ratio: $\frac{\text{mean}}{\text{standard deviation}} = \frac{6}{2} = 3:1$

Answer: Option (2)

Key idea: Law of Total Probability

1. Identify: Total coins = $N+1$ (N fair coins + 1 double-headed coin)
2. Apply law of total probability: $P(\text{Head}) = P(\text{fair coin}) \times P(H|\text{fair}) + P(\text{double-headed}) \times P(H|\text{double-headed})$
3. Substitute values: $P(\text{Head}) = \frac{N}{N+1} \times \frac{1}{2} + \frac{1}{N+1} \times 1$
4. Simplify: $P(\text{Head}) = \frac{N}{2(N+1)} + \frac{2}{2(N+1)} = \frac{N+2}{2(N+1)}$
5. Set equal to given probability: $\frac{N+2}{2(N+1)} = \frac{9}{16}$
6. Cross-multiply and solve: $16(N+2) = 18(N+1) \Rightarrow 16N + 32 = 18N + 18 \Rightarrow 2N = 14$
7. Therefore: $N = 7$

Answer: Option (1)

Key idea: Hyperbola eccentricity, latus rectum, and point on conic

1. Solve $15(e^2 + 1) = 34e$ to get $15e^2 - 34e + 15 = 0$. Using quadratic formula: $e = \frac{5}{3}$ (since $e > 1$ for hyperbola)
2. For hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, use $e^2 = 1 + \frac{b^2}{a^2}$. So $\frac{25}{9} = 1 + \frac{b^2}{a^2}$, giving $b^2 = \frac{16a^2}{9}$
3. Substitute point $(6, 4\sqrt{3})$ into hyperbola: $\frac{36}{a^2} - \frac{48}{b^2} = 1$
4. Using $b^2 = \frac{16a^2}{9}$: $\frac{36}{a^2} - \frac{27}{a^2} = 1$, which gives $a^2 = 9$ and $b^2 = 16$
5. New hyperbola becomes $\frac{x^2}{16} - \frac{y^2}{20} = 1$ (since $a^2 + 1 = 10$)
6. Latus rectum = $\frac{2B^2}{A} = \frac{2(20)}{4} = 10$

Answer: Option (1)

Key idea: Parabola focal chord properties and angle subtended at focus using cosine rule

1. Identify parabola parameters: $y^2 = 12x$ gives $a = 3$, so focus $F = (3, 0)$
2. Let ordinates be $y_1 = t$ and $y_2 = 2t$ (ratio 1:2). Using $y^2 = 12x$, get $P = \left(\frac{t^2}{12}, t\right)$ and $Q = \left(\frac{t^2}{3}, 2t\right)$
3. Apply length condition $|PQ| = 3\sqrt{13}$ to get $t^2(t^2 + 16) = 1872$, solving gives $t^2 = 36$, so $t = 6$
4. Find coordinates: $P = (3, 6)$ and $Q = (12, 12)$
5. Calculate focal distances: $|PF| = 6$, $|QF| = 15$, $|PQ| = 3\sqrt{13}$
6. Apply cosine rule: $117 = 36 + 225 - 180\cos\alpha$ gives $\cos\alpha = \frac{4}{5}$
7. Find $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$

Answer: Option (2)

Key idea: Inverse Trigonometric Functions - Addition Formula

1. Step 1: Use the formula $\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$ with $a = 1-\alpha$ and $b = 1-\beta$
2. Step 2: Since the sum equals $\frac{\pi}{4}$, we have $\frac{(1-\alpha)+(1-\beta)}{1-(1-\alpha)(1-\beta)} = 1$
3. Step 3: Simplify to get $\frac{2-\alpha-\beta}{\alpha+\beta-\alpha\beta} = 1$, which gives $2 = 2(\alpha+\beta) - \alpha\beta$
4. Step 4: Substitute $\beta = \frac{1}{3\alpha}$ to find $\alpha\beta = \frac{1}{3}$
5. Step 5: Solve $2 = 2(\alpha+\beta) - \frac{1}{3}$ to get $\alpha+\beta = \frac{7}{6}$
6. Step 6: Calculate $6(\alpha+\beta) = 6 \times \frac{7}{6} = 7$

Answer: Option (2)

Key idea: Solving trigonometric equations using auxiliary angle method and finding solutions in a given interval

1. Rewrite the equation $\cos\theta - \sqrt{3}\sin\theta = -1$ in the form $R\cos(\theta + \alpha) = -1$
2. Find $R = 2$ and $\alpha = \pi/3$, giving $2\cos(\theta + \pi/3) = -1$, so $\cos(\theta + \pi/3) = -\frac{1}{2}$
3. Solve to get $\theta + \pi/3 = \pm\frac{2\pi}{3} + 2n\pi$, which gives $\theta = \frac{\pi}{3} + 2n\pi$ or $\theta = -\pi + 2n\pi$
4. Find values in $(-2\pi, 2\pi)$: From first family: $\theta = -\frac{5\pi}{3}, \frac{\pi}{3}$; From second family: $\theta = -\pi, \pi$
5. Verify all four solutions satisfy the original equation
6. Calculate sum: $\sum_{\theta \in S} \theta = -\frac{5\pi}{3} - \pi + \frac{\pi}{3} + \pi = -\frac{4\pi}{3}$

Answer: Option (3)

Key idea: Image of a point in a line, foot of perpendicular, distance formula in 3D

1. Step 1: Find foot of perpendicular $F$ from $P(1,6,a)$ to line $L$. Since $F$ is midpoint of $P$ and its image $Q(\frac{a}{3}, 0, a+c)$, we get $F = (\frac{3+a}{6}, 3, \frac{2a+c}{2})$.
2. Step 2: Since $F$ is on line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$, from $y$-coordinate: $1+2t=3 \Rightarrow t=1$. This gives $F=(1, 3, a-1+b)$.
3. Step 3: From $x$-coordinate: $1 = \frac{3+a}{6} \Rightarrow a=3$. From $z$-coordinate and perpendicularity condition $t = \frac{11+b}{5+b^2} = 1$, we get $b^2-b-6=0 \Rightarrow b=3$ (since $b>0$).
4. Step 4: Using $z$-coordinate equation: $2+b = \frac{6+c}{2} \Rightarrow c=4$. So $F=(1,3,5)$ and line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.
5. Step 5: Point on $L$ is $(\lambda, 1+2\lambda, 2+3\lambda)$. Distance from $F$ to $S$ is $|SF| = \sqrt{14}|\lambda-1| = 2\sqrt{14}$, giving $|\lambda-1|=2$.
6. Step 6: Since $\alpha > 0$, we take $\lambda=3$ (not $-1$). Thus $S=(3, 7, 11)$ and $\alpha+\beta+\gamma = 3+7+11=21$.

Answer: Option (2)

Key idea: Cross product to find perpendicular line, angle between lines using dot product

1. Find direction vector of $L$ using cross product: $\vec{d_L} = \vec{d_1} \times \vec{d_2}$ where $\vec{d_1} = (3, 5, 7)$ and $\vec{d_2} = (1, 4, 7)$
2. Calculate: $\vec{d_L} = (7, -14, 7) = 7(1, -2, 1)$, so direction ratios are $(1, -2, 1)$
3. For angle with $L_3$ having direction $(2, 1, 2)$: $\cos \theta = \frac{|\vec{d_L} \cdot \vec{d_{L_3}}|}{|\vec{d_L}| \cdot |\vec{d_{L_3}}|} = \frac{|2|}{\sqrt{6} \cdot 3} = \frac{2}{3\sqrt{6}}$
4. Find $\sin \theta$: $\sin^2 \theta = 1 - \frac{4}{54} = \frac{25}{27}$, so $\sin \theta = \frac{5}{3\sqrt{3}}$
5. Calculate $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{5}{3\sqrt{3}} \cdot \frac{3\sqrt{6}}{2} = \frac{5\sqrt{2}}{2}$

Answer: Option (2)

Key idea: Limits using Taylor series expansion and standard limit formulas

1. Step 1: Identify the indeterminate form $\frac{0}{0}$ when $x \to 0$
2. Step 2: Use Taylor series: $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - ...$
3. Step 3: Find $\sin^2 x = x^2 - \frac{x^4}{3} + ...$ (higher order terms)
4. Step 4: Calculate $x^2 - \sin^2 x = x^2 - (x^2 - \frac{x^4}{3} + ...) = \frac{x^4}{3} + ...$
5. Step 5: Rewrite limit as $\lim_{x \to 0} \frac{\sin^2 x/x^2}{(x^2 - \sin^2 x)/x^4} = \frac{1}{1/3} = 3$

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Answer: Option (2)

Key idea: Area bounded by curves using integration

1. Identify the boundaries: $y \geq 0$, $x \leq 6-y$ (line), and $x \leq \frac{y^2+3}{4}$ (parabola)
2. Find intersection of line $x = 6-y$ and parabola $x = \frac{y^2+3}{4}$: Solve $6-y = \frac{y^2+3}{4}$ to get $y = 3$, $x = 3$
3. Determine boundaries: For $0 \leq y \leq 3$, parabola is the right boundary; for $3 \leq y \leq 6$, line is the right boundary
4. Calculate area: $A = \int_0^3 \frac{y^2+3}{4} dy + \int_3^6 (6-y) dy$
5. Evaluate first integral: $\frac{1}{4}[\frac{y^3}{3} + 3y]_0^3 = \frac{1}{4}(9+9) = \frac{9}{2}$
6. Evaluate second integral: $[6y - \frac{y^2}{2}]_3^6 = 18 - 13.5 = \frac{9}{2}$
7. Total area = $\frac{9}{2} + \frac{9}{2} = 9$

Answer: Option (1)

Key idea: Bijective Functions, Inverse Functions, and Area under Curves

1. Since $f: \{1,2,3,4\} \to \{1,e,e^2,e^3\}$ is strictly decreasing: $f(1)=e^3, f(2)=e^2, f(3)=e, f(4)=1$
2. Since $g: \{1,e,e^2,e^3\} \to \{\frac{1}{2},\frac{1}{3},\frac{1}{4}\}$ is strictly increasing: $g(e^2)=\frac{1}{2}$, so $g^{-1}(\frac{1}{2})=e^2$
3. From $f(2)=e^2$, we get $f^{-1}(e^2)=2$, thus $\phi(x)=[f^{-1}(g^{-1}(\frac{1}{2}))]^x = 2^x$
4. Set up the area integral: $A = \int_0^1 (2^x - x^2)\,dx$
5. Evaluate: $A = \left[\frac{2^x}{\ln 2}\right]_0^1 - \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{\ln 2} - \frac{1}{3}$
6. Simplify: $A = \frac{3 - \ln 2}{3\ln 2} = \frac{3 - \log_e(2)}{3\log_e(2)}$

Answer: 4

Key idea: Properties of adjoint matrices and matrix equations

1. Calculate $A^2 = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ by matrix multiplication
2. Find $\det(A) = -1$ using cofactor expansion
3. Use the property $adj(adj(A)) = (\det A)^{n-2} \cdot A = -A$ for a $3 \times 3$ matrix
4. Calculate $adj(A) \cdot adj(adj(A)) = I$ (identity matrix)
5. Substitute into the given equation: $A^2 + \alpha(-A) + \beta I = RHS$
6. Compare matrix entries to get: $-1 - \alpha = -2 \Rightarrow \alpha = 1$ and $1 + \beta = 0 \Rightarrow \beta = -1$
7. Calculate $(\alpha - \beta)^2 = (1-(-1))^2 = 4$

Answer: 80

Key idea: Circle properties, equilateral triangle inscribed in circle, chord length formula

1. Center of circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ is $(-g, -f)$ with radius $r = \sqrt{g^2 + f^2 - 25}$. Since center lies on $2x - y = 4$: $-2g + f = 4$, so $f = 2g + 4$
2. For equilateral triangle inscribed in circle with radius $r$, area $= \frac{3\sqrt{3}r^2}{4}$. Given area $= 27\sqrt{3}$, we get $r^2 = 36$, so $r = 6$
3. From $r^2 = g^2 + f^2 - 25 = 36$, we have $g^2 + f^2 = 61$. Substituting $f = 2g + 4$: $5g^2 + 16g - 45 = 0$
4. Solving: $g = \frac{9}{5}$ or $g = -5$. Since center must be in first quadrant, $g = -5$ and $f = -6$ giving center $(5, 6)$
5. Distance from center $(5, 6)$ to line $x = 1$ is $d = 4$. Chord length $= 2\sqrt{r^2 - d^2} = 2\sqrt{36 - 16} = 4\sqrt{5}$
6. Square of chord length $= (4\sqrt{5})^2 = 80$

Answer: 3

Key idea: Vector cross product, dot product, and BAC-CAB rule

1. Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$, $\vec{a} \times \vec{c} = \vec{b}$, and $\vec{a} \cdot \vec{c} = 3$
2. Apply BAC-CAB rule: $\vec{a} \times (\vec{a} \times \vec{c}) = \vec{a}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{a})$, which gives $\vec{a} \times \vec{b} = 3\vec{a} - 3\vec{c}$
3. Calculate $\vec{a} \times \vec{b} = -2\hat{i} + \hat{j} + \hat{k}$ using determinant
4. Solve for $\vec{c}$: $3\vec{c} = 3\vec{a} - (\vec{a} \times \vec{b}) = 5\hat{i} + 2\hat{j} + 2\hat{k}$, so $\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$
5. Calculate $\vec{c} \cdot \vec{b} = \frac{5}{3}(0) + \frac{2}{3}(1) + \frac{2}{3}(-1) = 0$
6. Find $\vec{c} \cdot (\vec{a} - 2\vec{b}) = \vec{c} \cdot \vec{a} - 2\vec{c} \cdot \vec{b} = 3 - 2(0) = 3$

Answer: 1279

Key idea: Optimization of trigonometric functions using AM-GM inequality, geometric progression sum formula

1. Find $\min f(\theta)$ using AM-GM: $\alpha \tan^2\theta + \beta \cot^2\theta \geq 2\sqrt{\alpha\beta}$, so $\min f(\theta) = 2\sqrt{\alpha\beta}$
2. Find $\max g(\theta)$: Write $g(\theta) = \beta + (\alpha - \beta)\sin^2\theta$. Since $\alpha > \beta$, maximum occurs at $\sin^2\theta = 1$, giving $\max g(\theta) = \alpha$
3. Use condition $\min f(\theta) = \max g(\theta)$: $2\sqrt{\alpha\beta} = \alpha \Rightarrow \alpha = 4\beta$
4. Calculate G.P. parameters: First term $a = \frac{\alpha}{2\beta} = 2$, common ratio $r = \frac{2\beta}{\alpha} = \frac{1}{2}$
5. Find sum: $S_{10} = 2 \cdot \frac{1 - (\frac{1}{2})^{10}}{1 - \frac{1}{2}} = 4 \cdot \frac{1023}{1024} = \frac{4092}{1024} = \frac{1023}{256}$
6. Verify $\gcd(1023, 256) = 1$ (1023 is odd, 256 is $2^8$), so $m = 1023$, $n = 256$
7. Therefore $m + n = 1023 + 256 = 1279$

Answer: 3

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

1. Step 1: Rewrite the differential equation by factoring: $x(x - \sqrt{x^2-1})dy = (x - y(x - \sqrt{x^2-1}))dx$
2. Step 2: Simplify to get the standard form: $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x - \sqrt{x^2-1}}$
3. Step 3: Use integrating factor $\mu(x) = x$ and rationalize the RHS: $\frac{d}{dx}(xy) = x^2 + x\sqrt{x^2-1}$
4. Step 4: Integrate both sides: $xy = \frac{x^3}{3} + \frac{(x^2-1)^{3/2}}{3} + C$
5. Step 5: Apply initial condition $y(1) = 1$ to find $C = \frac{2}{3}$
6. Step 6: Substitute $x = \sqrt{5}$ to get $y(\sqrt{5}) = \frac{5 + 2\sqrt{5}}{3} \approx 3.157$
7. Step 7: The greatest integer less than $y(\sqrt{5})$ is $3$

Answer: Option (2)

Key idea: Error propagation in derived quantities (density of cylinder)

1. Write the formula for density of cylinder: $\rho = \frac{m}{V} = \frac{4m}{\pi d^2 l}$
2. Apply error propagation formula: $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}$ (note: power of $d$ is 2)
3. Calculate $\frac{\Delta m}{m} = \frac{0.02}{97.42} = 0.0002053$
4. Calculate $2\frac{\Delta d}{d} = 2 \times \frac{0.02}{20.20} = 0.001980$
5. Calculate $\frac{\Delta l}{l} = \frac{0.05}{8.35} = 0.005988$
6. Sum all errors: $\frac{\Delta\rho}{\rho} = 0.0002053 + 0.001980 + 0.005988 = 0.008173$
7. Convert to percentage: $0.008173 \times 100\% = 0.82\%$

Answer: Option (4)

Key idea: Dimensional Analysis

1. Step 1: Identify that potential energy $V$ has dimensions $[M^1 L^2 T^{-2}]$ and distance $x$ has dimensions $[L]$
2. Step 2: Since $(x+B)$ must be dimensionally consistent, $[B] = [L]$
3. Step 3: Apply dimensional analysis to $V = \frac{A\sqrt{x}}{x+B}$: $[M L^2 T^{-2}] = \frac{[A][L^{1/2}]}{[L]}$
4. Step 4: Solve for $[A]$: $[A] = [M L^2 T^{-2}][L^{1/2}] = [M L^{5/2} T^{-2}]$
5. Step 5: Calculate $[AB] = [M L^{5/2} T^{-2}][L] = [M^1 L^{7/2} T^{-2}]$

Answer: Option (4)

Key idea: Work-Energy Theorem with Non-Conservative Forces

1. Convert units: $m = 1\text{ g} = 0.001\text{ kg}$, $h = 1\text{ km} = 1000\text{ m}$
2. Calculate work by gravity: $W_g = mgh = 0.001 \times 10 \times 1000 = 10\text{ J}$
3. Calculate initial KE: $KE_i = \frac{1}{2}mu^2 = 0$ (starts from rest)
4. Calculate final KE: $KE_f = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.001 \times 25 = 0.0125\text{ J}$
5. Apply work-energy theorem: $W_g + W_{resistive} = \Delta KE$
6. Solve: $10 + W_{resistive} = 0.0125$, so $W_{resistive} = -9.9875 \approx -9.98\text{ J}$

Answer: Option (2)

Key idea: Non-inertial reference frames, pseudo forces, friction in accelerating systems

1. Work in the reference frame of the accelerating cube (non-inertial frame with pseudo forces)
2. For block $P$ on top: Tension $T = m_P \times a = 2 \times \frac{g}{2} = 10$ N
3. For block $Q$ on side: Pseudo force (horizontal) $= m_Q \times a = 1.5 \times 5 = 7.5$ N pushes $Q$ into wall
4. Normal force from wall: $N_Q = 7.5$ N (balances pseudo force)
5. Vertical equilibrium for $Q$: $T + f = m_Q g$, so friction $f = 15 - 10 = 5$ N (upward)
6. Coefficient of friction: $\mu = \frac{f}{N_Q} = \frac{5}{7.5} = \frac{2}{3} = 0.67$

Answer: Option (1)

Key idea: Stress-Strain and Newton's Laws in accelerating systems

1. Calculate the cross-sectional area of the wire: $A = \pi r^2 = 3.14 \times (4 \times 10^{-3})^2 = 5.024 \times 10^{-5}$ m²
2. Find maximum tension the wire can withstand: $T_{max} = \sigma_{max} \times A = 4 \times 10^8 \times 5.024 \times 10^{-5} = 20096$ N
3. Apply Newton's second law for upward acceleration: $T = m(g + a)$
4. At maximum acceleration: $T_{max} = m(g + a_{max})$
5. Solve for $a_{max}$: $20096 = 1600(10 + a_{max})$
6. $12.56 = 10 + a_{max}$, therefore $a_{max} = 2.56$ m/s²

Answer: Option (1)

Key idea: Rotational dynamics: torque, moment of inertia of solid sphere, angular kinematics

1. Convert angular velocity: $\omega_0 = 1200$ rpm $= 40\pi$ rad/s
2. Find angular acceleration: $\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 40\pi}{10} = -4\pi$ rad/s²
3. Calculate moment of inertia: $I = \frac{2}{5}mR^2 = \frac{2}{5}(5)(0.04)^2 = 0.0032$ kg·m²
4. Find torque magnitude: $\tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi$ N·m
5. Calculate angular displacement: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 40\pi(10) - \frac{1}{2}(4\pi)(100) = 200\pi$ rad
6. Number of rotations: $n = \frac{\theta}{2\pi} = \frac{200\pi}{2\pi} = 100$

Answer: Option (2)

Key idea: Circular Motion with Energy Conservation

1. Step 1: At the highest point, the body exerts force $3mg$ on the plane, so normal force $N = 3mg$ acts on the body (toward center).
2. Step 2: Apply centripetal force equation at highest point: $N + mg = \frac{mv_{top}^2}{R}$, giving $4mg = \frac{mv_{top}^2}{R}$.
3. Step 3: Solve for velocity at top: $v_{top}^2 = 4gR$.
4. Step 4: Apply energy conservation from initial height $h$ to highest point (at height $2R$): $mgh = \frac{1}{2}mv_{top}^2 + mg(2R)$.
5. Step 5: Substitute $v_{top}^2 = 4gR$: $mgh = 2mgR + 2mgR = 4mgR$.
6. Step 6: Therefore $h = 4R$, so $\alpha = 4$.

Answer: Option (1)

Key idea: Center of Mass of a System of Particles

1. Set up coordinate system with point $p$ (midpoint of base) as origin
2. Place masses: 2 kg at $(-a, 0)$, 3 kg at $(a, 0)$, 15 kg at $(0, h)$ where the triangle has appropriate dimensions
3. Apply center of mass formula: $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$ and $y_{cm} = \frac{\sum m_i y_i}{\sum m_i}$
4. Calculate $x_{cm} = \frac{2(-a) + 3(a) + 15(0)}{20} = \frac{a}{20}$
5. Calculate $y_{cm} = \frac{2(0) + 3(0) + 15h}{20} = \frac{3h}{4}$
6. Using the given geometry, determine $x_{cm} = \frac{\sqrt{3}}{4}$ m and $y_{cm} = 1.25$ m

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Key idea: Unknown - question text not provided

Answer: Option (1)

Key idea: Conservation of moles in connected vessels with different temperatures using ideal gas law

1. Calculate initial moles in each vessel: $n_1 = n_2 = \frac{P_1V}{RT_1} = \frac{90V}{400R}$
2. Find total moles: $n_{total} = 2 \times \frac{90V}{400R} = \frac{9V}{20R}$
3. After heating, final pressure $P'$ is same in both vessels. Write moles in each vessel: $n_1' = \frac{P'V}{400R}$ and $n_2' = \frac{P'V}{500R}$
4. Apply conservation of moles: $\frac{P'V}{400R} + \frac{P'V}{500R} = \frac{9V}{20R}$
5. Simplify: $P'\left(\frac{1}{400} + \frac{1}{500}\right) = \frac{9}{20}$, which gives $P' \times \frac{9}{2000} = \frac{9}{20}$
6. Solve for final pressure: $P' = \frac{9}{20} \times \frac{2000}{9} = 100$ kPa

Answer: Option (3)

Key idea: Interference intensity formula with path difference

1. For interference with equal intensity sources, intensity is given by $I = 4I_0\cos^2(\phi/2)$ where $\phi$ is the phase difference
2. Phase difference relates to path difference as $\phi = \frac{2\pi}{\lambda}\Delta$
3. At point A: $\phi_A = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$, so $I_A = 4I_0\cos^2(\frac{\pi}{3}) = 4I_0 \cdot \frac{1}{4} = I_0$
4. At point B: $\phi_B = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3}$, so $I_B = 4I_0\cos^2(\frac{\pi}{6}) = 4I_0 \cdot \frac{3}{4} = 3I_0$
5. Therefore, $\frac{I_A}{I_B} = \frac{I_0}{3I_0} = \frac{1}{3}$

Answer: 0.625

Key idea: Simple Harmonic Motion - Time calculation between two positions

1. Write the SHM equation: $x(t) = A\cos(\omega t)$ where $\omega = \frac{2\pi}{T} = \frac{2\pi}{5}$ rad/s
2. At $t = 0$, particle is at $x = A$ (extreme position), so phase constant $\phi = 0$
3. For position $x = \frac{A}{\sqrt{2}}$: Set $\frac{A}{\sqrt{2}} = A\cos\left(\frac{2\pi t}{5}\right)$
4. Solve: $\cos\left(\frac{2\pi t}{5}\right) = \frac{1}{\sqrt{2}}$ gives $\frac{2\pi t}{5} = \frac{\pi}{4}$
5. Calculate time: $t = \frac{5\pi}{8\pi} = \frac{5}{8} = 0.625$ sec

Answer: Option (1)

Key idea: Electric field at the center of a uniformly charged half ring

1. For a uniformly charged half ring, linear charge density $\lambda = \frac{Q}{\pi R}$ where $R = 0.35$ m
2. Consider element $dq = \lambda R d\theta = \frac{Q}{\pi}d\theta$ at angle $\theta$. Electric field $dE = \frac{k dq}{R^2}$
3. By symmetry, horizontal components cancel. Vertical components: $E = \int_0^\pi \frac{kQ}{\pi R^2}\sin\theta\, d\theta$
4. Evaluating: $E = \frac{kQ}{\pi R^2} \times 2 = \frac{Q}{2\pi^2\epsilon_0 R^2}$ (using $k = \frac{1}{4\pi\epsilon_0}$)
5. Solving for $Q$: $Q = 2\pi^2\epsilon_0 R^2 E = 2 \times (3.14)^2 \times 8.85 \times 10^{-12} \times (0.35)^2 \times 100$
6. Calculate: $Q = 2 \times 9.8596 \times 8.85 \times 10^{-12} \times 0.1225 \times 100 = 2.14 \times 10^{-9}$ C
7. Therefore, $Q = 2.14$ nC

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Answer: Option (2)

Key idea: Intensity of electromagnetic waves from a point source depends only on distance, not direction. For isotropic point sources, intensity is uniform on spherical surfaces.

1. For a point source, intensity is given by $I = \frac{P}{4\pi r^2}$ where $P$ is power and $r$ is distance
2. At original location: distance = $L$, so $I_1 = \frac{P}{4\pi L^2} = I_o$
3. Detector moves to new location on the SAME spherical surface, so distance remains $L$
4. At new location: distance = $L$, so $I_2 = \frac{P}{4\pi L^2} = I_o$
5. The $45°$ angle is irrelevant - intensity depends only on distance for isotropic point sources
6. Therefore, measured intensity at new location = $I_o$

Answer: Option (1)

Key idea: Refraction at spherical surface and magnification

1. Write the refraction formula: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
2. Substitute $n_1 = 1$, $n_2 = 1.4$, $u = -4R$, $R = R$ (positive)
3. Solve: $\frac{1.4}{v} + \frac{1}{4R} = \frac{0.4}{R}$ gives $v = \frac{28R}{3}$
4. Apply magnification formula: $m = \frac{n_1 \cdot v}{n_2 \cdot u} = \frac{1 \times \frac{28R}{3}}{1.4 \times (-4R)}$
5. Calculate: $m = -\frac{5}{3}$
6. Find magnitude: $|m| = \frac{5}{3} = 1.66$

Answer: Option (1)

Key idea: Magnetic energy density and magnetic field at the center of a circular current loop

1. Find magnetic field at center of loop: $B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 1.256 \times 10^{-5}$ T
2. Calculate volume of cube: $V = a^3 = (10^{-3})^3 = 10^{-9}$ m³
3. Find magnetic energy density: $u = \frac{B^2}{2\mu_0} = \frac{(1.256 \times 10^{-5})^2}{2 \times 4\pi \times 10^{-7}} = 6.28 \times 10^{-5}$ J/m³
4. Calculate total energy: $U = u \times V = 6.28 \times 10^{-5} \times 10^{-9} = 6.28 \times 10^{-14}$ J
5. Compare with $a \times 10^{-14}$ J to get $a = 6.28$

Answer: Option (1)

Key idea: LR Circuit - Voltage distribution using Kirchhoff's Voltage Law

1. Given: $L = 10$ mH, $R = 100\,\Omega$, $E = 1.0$ V, find ratio of $V_L$ at $i_1 = 2$ mA and $i_2 = 4$ mA
2. Apply Kirchhoff's Voltage Law: $E = iR + V_L$, so $V_L = E - iR$
3. When $i_1 = 2$ mA: $V_{L1} = 1.0 - (0.002)(100) = 1.0 - 0.2 = 0.8$ V
4. When $i_2 = 4$ mA: $V_{L2} = 1.0 - (0.004)(100) = 1.0 - 0.4 = 0.6$ V
5. Calculate ratio: $\frac{V_{L1}}{V_{L2}} = \frac{0.8}{0.6} = \frac{4}{3}$

Answer: Option (4)

Key idea: Balmer Series and Photon Momentum

1. Identify transitions: 1st Balmer line is $n=3 \to n=2$, 2nd line is $n=4 \to n=2$
2. Calculate energy for 1st line: $E_1 \propto \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{5}{36}$
3. Calculate energy for 2nd line: $E_2 \propto \frac{1}{2^2} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}$
4. Since photon momentum $p = \frac{E}{c}$, the ratio $\frac{p_1}{p_2} = \frac{E_1}{E_2}$
5. Calculate ratio: $\frac{p_1}{p_2} = \frac{5/36}{3/16} = \frac{5 \times 16}{36 \times 3} = \frac{80}{108} = \frac{20}{27}$
6. Therefore, $\alpha = 20$ and $\beta = 27$

Answer: Option (1)

Key idea: Power factor in LCR series circuit

1. Identify the given values: $R = 80\,\Omega$, $X_L = 20\,\Omega$, $X_C = 80\,\Omega$
2. Calculate the net reactance: $X_L - X_C = 20 - 80 = -60\,\Omega$
3. Calculate the impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{80^2 + 60^2}$
4. Simplify: $Z = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$
5. Calculate power factor: $\cos\phi = \frac{R}{Z} = \frac{80}{100} = 0.8$

Answer: 17305

Key idea: Kirchhoff's Laws, Heat dissipation in resistors, Complex DC circuits with multiple batteries

1. Apply Kirchhoff's Current Law: $I = I_1 + I_2$ where $I_1$ is current through $6\Omega$ and $I_2$ through $4\Omega$
2. Apply KVL to loop with $6\Omega$: $2 - 3I + 3 - 6I_1 = 0 \Rightarrow 3I + 6I_1 = 5$
3. Apply KVL to loop with $4\Omega$: $2 - 3I - 4I_2 = 0 \Rightarrow 3I + 4I_2 = 2$
4. Solve the system: $9I_1 + 3I_2 = 5$ and $3I_1 + 7I_2 = 2$ to get $I_1 = \frac{29}{54}$ A
5. Calculate heat: $H = I_1^2 R t = \left(\frac{29}{54}\right)^2 \times 6 \times 100 = 173.05$ J
6. Since $H = \frac{a}{100}$ J, we get $a = 100 \times 173.05 \approx 17305$

Answer: 30

Key idea: Malus's Law and Optical Activity

1. Unpolarized light of intensity $I_0$ becomes $\frac{I_0}{2}$ after passing through the polarizer
2. The optically active solution rotates the plane of polarization by angle $\theta$ without changing intensity
3. Since analyzer is at 0° (same as polarizer), the angle between rotated polarization and analyzer is $\theta$
4. Apply Malus's Law: $I_{final} = \frac{I_0}{2}\cos^2(\theta) = \frac{3}{8}I_0$
5. Solve: $\cos^2(\theta) = \frac{3}{4}$, so $\cos(\theta) = \frac{\sqrt{3}}{2}$
6. Therefore, $\theta = 30°$

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Key idea: Unknown - Question text not provided

Answer: 186

Key idea: Work done by electric field on a moving charge in a conservative field

1. Work done is $W = q \int_A^B \vec{E} \cdot d\vec{r}$ where $q = 3$ C
2. Check that $\nabla \times \vec{E} = 0$, confirming the field is conservative (path-independent)
3. Write $\vec{E} \cdot d\vec{r} = 2x dx + 3y^2 dy + 4 dz$ and integrate each component separately
4. Calculate: $\int_0^5 2x dx = [x^2]_0^5 = 25$, $\int_{-2}^1 3y^2 dy = [y^3]_{-2}^1 = 9$, $\int_{-5}^2 4 dz = [4z]_{-5}^2 = 28$
5. Sum the integrals: $25 + 9 + 28 = 62$
6. Multiply by charge: $W = 3 \times 62 = 186$ J

Answer: 989

Key idea: Isothermal compression work and bulk modulus relationship

1. For isothermal process, bulk modulus $B = P$, so initial pressure $P_0 = 3 \times 10^5$ N/m²
2. Convert volumes: $V_0 = 3 \times 10^{-3}$ m³, $V_f = 1 \times 10^{-3}$ m³
3. Work done in isothermal process: $W = -P_0 V_0 \ln\left(\frac{V_f}{V_0}\right)$
4. Substitute: $W = -P_0 V_0 \ln\left(\frac{1}{3}\right) = P_0 V_0 \ln(3)$
5. Calculate: $W = 3 \times 10^5 \times 3 \times 10^{-3} \times 1.0986 = 900 \times 1.0986$
6. Final result: $W = 989$ J (magnitude of work done on the gas)

Answer: Option (2)

Key idea: Empirical Formula Determination from Percentage Composition

1. Given: 69.9% Fe, so oxygen = 100 - 69.9 = 30.1%
2. Assume 100g sample: Mass of Fe = 69.9g, Mass of O = 30.1g
3. Calculate moles: $n_{Fe} = \frac{69.9}{56} = 1.248$ mol, $n_O = \frac{30.1}{16} = 1.881$ mol
4. Find simplest ratio by dividing by smallest: Fe: $\frac{1.248}{1.248} = 1$, O: $\frac{1.881}{1.248} = 1.5$
5. Convert to whole numbers: $1:1.5 = 2:3$, so empirical formula is $\text{Fe}_2\text{O}_3$
6. Verify: % Fe in $\text{Fe}_2\text{O}_3 = \frac{112}{160} \times 100 = 70\%$ ✓

Answer: Option (1)

Key idea: Rydberg Formula and Hydrogen-like Atomic Spectra

1. Use Rydberg formula: $\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
2. For shortest wavelength in Lyman series of H: $n_1=1$, $n_2=\infty$, $Z=1$ gives $\frac{1}{x} = R(1)(1-0) = R$
3. For longest wavelength in Balmer series of He⁺: $n_1=2$, $n_2=3$ (minimum transition), $Z=2$
4. Calculate: $\frac{1}{\lambda} = R(4)\left(\frac{1}{4} - \frac{1}{9}\right) = 4R\left(\frac{5}{36}\right) = \frac{5R}{9}$
5. Substitute $R = \frac{1}{x}$: $\frac{1}{\lambda} = \frac{5}{9x}$
6. Therefore: $\lambda = \frac{9x}{5}$

Answer: Option (4)

Key idea: Radial and Angular Nodes in Atomic Orbitals

1. Use formulas: Radial nodes = $n - l - 1$ and Angular nodes (nodal planes) = $l$
2. For 2s: $n=2, l=0$ → Radial nodes = $2-0-1=1$, Nodal planes = $0$ → Matches IV
3. For 3s: $n=3, l=0$ → Radial nodes = $3-0-1=2$, Nodal planes = $0$ → Matches III
4. For 3p: $n=3, l=1$ → Radial nodes = $3-1-1=1$, Nodal planes = $1$ → Matches II
5. For 4d: $n=4, l=2$ → Radial nodes = $4-2-1=1$, Nodal planes = $2$ → Matches I
6. Final matching: A-IV, B-III, C-II, D-I

Answer: Option (2)

Key idea: Lewis Structures and Isoelectronic Species

1. Analyze **Pair A**: $SO_3^{2-}$ has trigonal pyramidal geometry (lone pair on S), while $CO_3^{2-}$ has trigonal planar geometry (no lone pair on C) - **DIFFERENT**
2. Analyze **Pair B**: $O_2^{2-}$ and $I_2$ both have single bonds with 3 lone pairs on each atom, 14 valence electrons - **SIMILAR**
3. Analyze **Pair C**: $CN^-$ and $CO$ are isoelectronic (10 electrons), both have triple bonds - **SIMILAR**
4. Analyze **Pair D**: $NH_3$ and $H_3O^+$ are isoelectronic (8 electrons), both trigonal pyramidal with 1 lone pair - **SIMILAR**
5. Analyze **Pair E**: $MnO_4^-$ (Mn in +7) and $CrO_4^{2-}$ (Cr in +6) have different charges and oxidation states despite both being tetrahedral - **DIFFERENT**
6. Pairs with dissimilar Lewis structures are **A and E**

Answer: Option (3)

Key idea: Isothermal work in reversible vs irreversible processes

1. For isothermal processes: $w = -nRT \ln(V_2/V_1)$
2. **Expansion** (V₂ > V₁): Multi-stage (reversible) gives maximum work output, so $|w_B| > |w_A|$
3. **Compression** (V₂ < V₁): Single-stage requires maximum work input, so $|w_C| > |w_D|$
4. Compression requires more work than expansion for same volume ratio: $|w_C|, |w_D| > |w_B|, |w_A|$
5. Combining all comparisons: $|w_C| > |w_D| > |w_B| > |w_A|$

Answer: Option (4)

Key idea: Raoult's Law and Relative Lowering of Vapor Pressure

1. Given: $n_{solute} = 0.25$ mol, $n_{solvent} = 1$ mol (non-volatile, non-ionizable solute)
2. Calculate total moles: $n_{total} = 1 + 0.25 = 1.25$ mol
3. Find mole fraction of solvent: $X_{solvent} = \frac{1}{1.25} = 0.8$
4. Apply Raoult's Law: $P_{solution} = P°_{solvent} \times X_{solvent}$
5. Therefore: $P_{solution} = P°_{solvent} \times 0.8 = 80\%$ of $P°_{solvent}$

Answer: Option (1)

Key idea: Chemical Equilibrium with Ideal Gas Law - calculating partial pressures using Kc and PV=nRT

1. Set up equilibrium: Initial moles are He = 1, A = 1, B = 0. At equilibrium: He = 1, A = (1-x), B = x
2. Apply $K_c = \frac{[B]}{[A]} = \frac{x/10}{(1-x)/10} = 4.0$
3. Solve for x: $\frac{x}{1-x} = 4 \Rightarrow x = 4 - 4x \Rightarrow 5x = 4 \Rightarrow x = 0.8$ moles
4. At equilibrium: He = 1 mole, A = 0.2 moles, B = 0.8 moles
5. Calculate partial pressure of He: $P_{He} = \frac{nRT}{V} = \frac{1 \times 0.082 \times 400}{10} = 3.28$ atm
6. Calculate partial pressure of B: $P_B = \frac{nRT}{V} = \frac{0.8 \times 0.082 \times 400}{10} = 2.624$ atm

Answer: Option (1)

Key idea: Kohlrausch's Law and Solubility Product from Conductivity

1. Use Kohlrausch's law to find $\Lambda^\circ_m(BaSO_4)$: From the given data, $\Lambda^\circ_m(BaSO_4) = \lambda^\circ(Ba^{2+}) + \lambda^\circ(SO_4^{2-}) = (x_1 - 2\lambda^\circ(Cl^-)) + (x_2 - 2\lambda^\circ(H^+)) = x_1 + x_2 - 2x_3$
2. Relate conductivity to concentration: For sparingly soluble salt, $\Lambda_m = \Lambda^\circ_m = \frac{1000\kappa}{C}$ where $\kappa = x$ S cm$^{-1}$ and $C$ is concentration in mol/L
3. Calculate solubility: $s = \frac{1000x}{x_1 + x_2 - 2x_3}$ mol/L
4. For $BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}$, the solubility product is $K_{sp} = [Ba^{2+}][SO_4^{2-}] = s^2$
5. Substitute to get: $K_{sp} = \left(\frac{1000x}{x_1 + x_2 - 2x_3}\right)^2 = \frac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}$

Answer: Option (1)

Key idea: Electrode potentials, ionization enthalpies, and bonding trends in p-block elements

1. **Statement I:** Compare electrode potentials: $E°_{Al^{3+}/Al} = -1.66$ V (negative) and $E°_{Tl^{3+}/Tl} = +1.26$ V (positive)
2. More negative $E°$ means greater tendency to lose electrons, hence more electropositive. Al is more electropositive than Tl. **Statement I is TRUE**
3. **Statement II:** Calculate sum of first three ionization enthalpies: Boron ≈ 6888 kJ/mol, Aluminium ≈ 5140 kJ/mol
4. Boron has higher ionization enthalpies due to smaller size and higher effective nuclear charge
5. High ionization enthalpies prevent B³⁺ formation, so boron forms covalent compounds. Al can form Al³⁺ ions. **Statement II is TRUE**
6. Since both statements are true, the correct answer is option (1)

Answer: Option (3)

Key idea: Industrial catalysts and electronic configurations of transition metals

1. Statement A: V₂O₅ (acidic oxide) is used in H₂SO₄ manufacture, not basic vanadium oxide. **FALSE**
2. Statement B: Ziegler-Natta uses TiCl₄ where Ti⁴⁺ has configuration [Ar]3d⁰, so $\mu = \sqrt{n(n+2)} = 0$ BM, not 2.84 BM. **FALSE**
3. Statement C: Ziegler-Natta uses Al(C₂H₅)₃ where Al (p-block) is in +3 oxidation state. **TRUE**
4. Statement D: Wacker process uses PdCl₂ where Pd²⁺ has configuration [Kr]4d⁸, giving 8 d-electrons. **TRUE**
5. Correct statements are C and D only

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: Crystal Field Stabilization Energy (CFSE) in Tetrahedral Complexes

Answer: Option (1)

Key idea: Crystal Field Theory - d-orbital splitting in tetrahedral complexes

1. In tetrahedral complexes, ligands approach along body diagonals, causing d-orbital splitting into two sets: $e$ (lower) and $t_2$ (higher)
2. Energy ordering: $e$ set contains $d_{z^2}$ and $d_{x^2-y^2}$ (lower energy), $t_2$ set contains $d_{xy}$, $d_{xz}$, $d_{yz}$ (higher energy)
3. Complete ordering: $d_{z^2} = d_{x^2-y^2} < d_{xy} = d_{xz} = d_{yz}$
4. Check A: $d_{xy} = d_{xz} > d_{x^2-y^2}$ → TRUE (both $t_2$ orbitals equal and higher than $e$ orbital)
5. Check B: $d_{xy} = d_{yz} > d_{z^2}$ → TRUE (both $t_2$ orbitals equal and higher than $e$ orbital)
6. Check C: $d_{x^2-y^2} > d_{z^2} > d_{xz}$ → FALSE ($d_{x^2-y^2} = d_{z^2}$ and both are lower than $d_{xz}$)
7. Check D: $d_{x^2-y^2} = d_{z^2} < d_{xz}$ → TRUE (both $e$ orbitals equal and lower than $t_2$ orbital)

Answer: Option (2)

Key idea: Acid-catalyzed hydration of alkenes and Rf values in TLC based on polarity

1. Identify the reaction: 2-methylpropene $(CH_3)_2C=CH_2$ reacts with dilute $H_2SO_4$ (acid-catalyzed hydration)
2. Apply Markovnikov's rule: Water adds to form the more stable tertiary carbocation $(CH_3)_3C^+$
3. Product X is 2-methylpropan-2-ol (tert-butyl alcohol): $(CH_3)_3COH$
4. Compare polarity: The alcohol (X) is MORE polar than the alkene due to the $-OH$ group
5. In the polar solvent system (ethyl acetate + ether) with polar silica gel stationary phase, more polar compounds have better mobile phase solubility
6. Therefore, the more polar alcohol travels faster and has a HIGHER $R_F$ value: $R_F = 0.82$

Answer: Option (3)

Key idea: Metamerism and Keto-Enol Tautomerism

1. Check Statement I: Compare molecular formulas of 2,6-diethylcyclohexanone ($C_{10}H_{18}O$) and 6-methyl-2-n-propylcyclohexanone ($C_{10}H_{18}O$)
2. Identify that both have same formula but different alkyl group distribution: two ethyl groups vs one methyl + one propyl group around the ring
3. Conclude Statement I is TRUE - these are metamers (differ in alkyl group arrangement with same molecular formula)
4. Check Statement II: Analyze 2,2,6,6-tetramethylcyclohexanone for $\alpha$-hydrogens needed for keto-enol tautomerism
5. Note that positions 2 and 6 (the $\alpha$-carbons) each have two methyl groups, leaving NO hydrogens available
6. Conclude Statement II is FALSE - no $\alpha$-hydrogen means no keto-enol tautomerism possible
7. Select option (3): Statement I is true but Statement II is false

Answer: Option (4)

Key idea: Methods of preparation of methane - decarboxylation, Kolbe's electrolysis, Grignard reagent hydrolysis, Wurtz reaction, and hydrogenation of unsaturated hydrocarbons

1. **Statement I Analysis:** Check three methods - (a) Decarboxylation: $\text{CH}_3\text{COONa} + \text{NaOH} \rightarrow \text{CH}_4 + \text{Na}_2\text{CO}_3$ ✓ gives methane
2. (b) Kolbe's electrolysis: $2\text{CH}_3\text{COO}^- \rightarrow \text{CH}_3\text{CH}_3 + 2\text{CO}_2 + 2e^-$ ✗ gives **ethane**, not methane
3. (c) Grignard + water: $\text{CH}_3\text{MgBr} + \text{H}_2\text{O} \rightarrow \text{CH}_4 + \text{Mg(OH)Br}$ ✓ gives methane
4. Since Kolbe's electrolysis produces ethane (not methane), **Statement I is FALSE**
5. **Statement II Analysis:** (a) Unsaturated hydrocarbons need at least 2 carbons for C=C or C≡C bonds, so cannot give $\text{CH}_4$ ✓
6. (b) Wurtz reaction: $2\text{R-X} + 2\text{Na} \rightarrow \text{R-R} + 2\text{NaX}$ couples alkyl groups, giving ethane from $\text{CH}_3\text{X}$, not methane ✓
7. Both parts of Statement II are correct, so **Statement II is TRUE**. Answer: Option (4)

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Key idea: Unknown - question content not provided

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Key idea: Kolbe-Schmitt Reaction - Carboxylation of Phenol

Answer: C < D < B < A

Key idea: Boiling points depend on intermolecular forces: alcohols (O-H) form strongest H-bonds, primary amines (NH₂) form stronger H-bonds than secondary amines (NH), and alkanes have only weak van der Waals forces.

1. Identify the functional groups: (A) is alcohol, (B) is primary amine, (C) is alkane, (D) is secondary amine
2. Alkane (C) has only van der Waals forces → lowest boiling point
3. Secondary amine (D) has one N-H bond → weak hydrogen bonding
4. Primary amine (B) has two N-H bonds → stronger hydrogen bonding than (D)
5. Alcohol (A) has O-H bond → strongest hydrogen bonding (O more electronegative than N)
6. Increasing order of boiling points: $C < D < B < A$

Answer: Option (3)

Key idea: Vitamin deficiency diseases and their corresponding vitamins

1. Identify that Scurvy is caused by deficiency of Vitamin C (Ascorbic Acid), so $A \rightarrow III$
2. Recognize that Convulsions are caused by deficiency of Vitamin B6 (Pyridoxine), so $B \rightarrow I$
3. Know that Cheilosis (cracks at mouth corners) is caused by deficiency of Vitamin B2 (Riboflavin), so $C \rightarrow IV$
4. Understand that Xerophthalmia (eye dryness, night blindness) is caused by deficiency of Vitamin A, so $D \rightarrow II$
5. Match the pattern: $A-III, B-I, C-IV, D-II$ corresponds to option (3)

Answer: Option (2)

Key idea: Functional group tests for amino acid side chains

1. Identify functional groups in side chains: Glutamine has $-CONH_2$ (amide), Lysine has $-(CH_2)_4NH_2$ (1° amine), Tyrosine has phenolic $-OH$, Serine has alcoholic $-OH$
2. Glutamine (amide group) → Hoffman bromamide degradation converts $RCONH_2$ to $RNH_2$, so A-IV
3. Lysine (primary amine) → Hinsberg's test distinguishes amines using benzenesulfonyl chloride, so B-I
4. Tyrosine (phenolic OH) → Neutral $FeCl_3$ test gives violet color with phenols, so C-II
5. Serine (alcoholic OH) → Ceric ammonium nitrate test detects alcohols (red color), so D-III
6. Final matching: A-IV, B-I, C-II, D-III

Answer: 4

Key idea: Ionization Enthalpy and Energy Calculations

1. Process: $\text{Li}(g) \to \text{Li}^{2+}(g) + 2e^-$ requires both first and second ionization enthalpies
2. Total ionization energy per mole = $IE_1 + IE_2 = 520 + 7297 = 7817$ kJ/mol
3. Convert mass to moles: $n = \frac{3.5 \times 10^{-3}\text{ g}}{7\text{ g/mol}} = 5 \times 10^{-4}$ mol
4. Calculate total energy: $E = n \times 7817 = 5 \times 10^{-4} \times 7817 = 3.9085$ kJ
5. Round to nearest integer: $E \approx 4$ kJ

Answer: 21

Key idea: Rearrangement of diazoaminobenzene to p-aminoazobenzene and percentage composition calculation

1. Starting compound is diazoaminobenzene: $C_6H_5-N=N-NH-C_6H_5$
2. Treatment with HCl and excess aniline at 40-45°C causes rearrangement to form p-aminoazobenzene (yellow product X)
3. Structure of product X: $H_2N-C_6H_4-N=N-C_6H_5$ with molecular formula $C_{12}H_{11}N_3$
4. Calculate molar mass: $M = 12(12) + 11(1) + 3(14) = 144 + 11 + 42 = 197$ g/mol
5. Mass of nitrogen = $3 \times 14 = 42$ g/mol
6. Percentage of nitrogen = $\frac{42}{197} \times 100 = 21.32\%$
7. Nearest integer = 21%

Answer: 3

Key idea: Reduction of phenol with zinc dust to form benzene; stoichiometry with percentage completion

1. Identify the reaction: Phenol $(C_6H_5OH)$ heated with $Zn$ gives benzene $(C_6H_6)$ as product $X$
2. Calculate molar mass of phenol: $M = 6(12) + 6(1) + 16 = 94$ g/mol
3. Find moles of phenol: $n = \frac{4.7}{94} = 0.05$ mol $= 5 \times 10^{-2}$ mol
4. From stoichiometry: 1 mol phenol $\rightarrow$ 1 mol benzene
5. At 60% completion: Moles of $X = 0.05 \times 0.60 = 0.03$ mol $= 3 \times 10^{-2}$ mol
6. Answer: $3$

Answer: 25

Key idea: First-order kinetics and half-life relationship

1. For first-order reaction, use the relationship: $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{3} = 0.231 \text{ hr}^{-1}$
2. Calculate number of half-lives: $n = \frac{t}{t_{1/2}} = \frac{6}{3} = 2$ half-lives
3. After $n$ half-lives, fraction remaining $= \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$
4. Percentage of sucrose remaining $= \frac{1}{4} \times 100\% = 25\%$

Answer: 34

Key idea: Gibbs Free Energy and Thermodynamic Relations

1. Calculate $\ln K$ using $K = 1.8 \times 10^{-7}$: $\ln K = \ln(1.8) - 7\ln(10) = 0.6 - 16.1 = -15.5$
2. Find $\Delta G^{\theta}$ using $\Delta G^{\theta} = -RT \ln K = -8.3 \times 300 \times (-15.5) = 38595$ J mol$^{-1}$
3. Apply the Gibbs equation: $\Delta G^{\theta} = \Delta H^{\theta} - T\Delta S^{\theta}$
4. Rearrange to get $\Delta S^{\theta} = \frac{\Delta H^{\theta} - \Delta G^{\theta}}{T} = \frac{28400 - 38595}{300}$
5. Calculate $\Delta S^{\theta} = \frac{-10195}{300} = -33.98$ J K$^{-1}$ mol$^{-1}$
6. The magnitude is $|\Delta S^{\theta}| = 34$ J K$^{-1}$ mol$^{-1}$ (nearest integer)