JEE Main 4 April 2026 Shift 2: Questions and Solutions

Answer: Option (1)

Key idea: Inverse Functions and Fixed Points

1. Step 1: Find the inverse function. From $y = (x-1)^4 + 1$, we get $f^{-1}(x) = (x-1)^{1/4} + 1$
2. Step 2: For Statement (I), solve $(x-1)^4 = (x-1)^{1/4}$. Let $u = x-1$, then $u^4 = u^{1/4}$
3. Step 3: This gives $u^{1/4}(u^{15/4} - 1) = 0$, so $u = 0$ or $u = 1$, meaning $x = 1$ or $x = 2$
4. Step 4: Verify: $f(1) = f^{-1}(1) = 1$ and $f(2) = f^{-1}(2) = 2$. Statement (I) is TRUE
5. Step 5: For Statement (II), solve $(x-1)^4 = x^{1/4}$. Let $g(x) = (x-1)^4 - x^{1/4}$
6. Step 6: Check: $g(1) = -1 < 0$ and $g(16) = 50625 - 2 > 0$. By IVT, a solution exists
7. Step 7: Statement (II) is FALSE. Therefore, only (I) is TRUE

Answer: Option (4)

Key idea: Complex numbers: Finding roots using quadratic formula and calculating modulus squared

1. Use quadratic formula: $z = \frac{-4 \pm \sqrt{16-64}}{2} = \frac{-4 \pm 4\sqrt{3}i}{2} = -2 \pm 2\sqrt{3}i$
2. So $S = \{-2 + 2\sqrt{3}i, -2 - 2\sqrt{3}i\}$
3. For $z_1 = -2 + 2\sqrt{3}i$: $z_1 + \sqrt{3}i = -2 + 3\sqrt{3}i$
4. Calculate $|z_1 + \sqrt{3}i|^2 = 4 + 27 = 31$
5. For $z_2 = -2 - 2\sqrt{3}i$: $z_2 + \sqrt{3}i = -2 - \sqrt{3}i$
6. Calculate $|z_2 + \sqrt{3}i|^2 = 4 + 3 = 7$
7. Sum = $31 + 7 = 38$

Answer: Option (2)

Key idea: System of linear equations with infinitely many solutions - determinant condition and consistency

1. For infinitely many solutions, the determinant of coefficient matrix must be zero: $\det(A) = 0$
2. Calculate $\det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = \lambda - 5$
3. Setting $\det(A) = 0$ gives $\lambda = 5$
4. For consistency, equation (3) must be a linear combination of equations (1) and (2): Eq(3) = $-1 \times$ Eq(1) $+ 2 \times$ Eq(2)
5. Therefore $\mu = -1(5) + 2(9) = -5 + 18 = 13$
6. Hence $\lambda + \mu = 5 + 13 = 18$

Answer: Option (3)

Key idea: Complex roots and Vieta's formulas for cubic polynomials with real coefficients

1. Since the polynomial has real coefficients and $\beta = 1 + i\sqrt{2}$ is a root, the complex conjugate $\gamma = 1 - i\sqrt{2}$ must also be a root.
2. Using Vieta's formulas: Sum of roots = $1 + (1+i\sqrt{2}) + (1-i\sqrt{2}) = 3 = -a$, so $a = -3$
3. Sum of products of pairs: $\alpha\beta + \beta\gamma + \gamma\alpha = (1+i\sqrt{2}) + 3 + (1-i\sqrt{2}) = 5 = b$
4. Product of roots: $\alpha\beta\gamma = 1 \cdot (1+i\sqrt{2})(1-i\sqrt{2}) = 1 \cdot 3 = 3 = -c$, so $c = -3$
5. The polynomial is $x^3 - 3x^2 + 5x - 3$. Integrate: $\int_{-1}^{1} (x^3 - 3x^2 + 5x - 3)dx = \left[\frac{x^4}{4} - x^3 + \frac{5x^2}{2} - 3x\right]_{-1}^{1}$
6. At $x=1$: $\frac{1}{4} - 1 + \frac{5}{2} - 3 = -\frac{5}{4}$. At $x=-1$: $\frac{1}{4} + 1 + \frac{5}{2} + 3 = \frac{27}{4}$
7. Final value: $-\frac{5}{4} - \frac{27}{4} = -\frac{32}{4} = -8$

Answer: 2

Key idea: Conditions for positive roots of quadratic equation using Vieta's formulas and discriminant

1. For two positive roots, we need: (1) Product of roots $> 0$, (2) Sum of roots $> 0$, (3) Discriminant $\geq 0$
2. Product of roots: $\frac{4\lambda}{\lambda + 2} > 0$ gives $\lambda \in (-\infty, -2) \cup (0, \infty)$
3. Sum of roots: $\frac{3\lambda}{\lambda + 2} > 0$ gives $\lambda \in (-\infty, -2) \cup (0, \infty)$
4. Discriminant: $9\lambda^2 - 16\lambda^2 - 32\lambda \geq 0$ simplifies to $-7\lambda^2 - 32\lambda \geq 0$, giving $\lambda \in [-\frac{32}{7}, 0]$
5. Taking intersection of all conditions: $\lambda \in [-\frac{32}{7}, -2)$ where $-\frac{32}{7} \approx -4.57$
6. The integral values in $[-4.57, -2)$ are $\lambda = -4, -3$, giving **2 possible values**

Answer: Option (3)

Key idea: Eigenvalues and Circle-Axis Intersections

1. Find eigenvalues by solving $\det(A - aI) = 0$, which gives $a^3 + 8a^2 - 88a - 320 = 0$
2. Factor to get $(a-8)(a^2 + 16a + 40) = 0$, yielding eigenvalues $a = 8, -8 \pm 2\sqrt{6}$
3. The largest eigenvalue is $p = 8$, so the circle is $(x-8)^2 + (y-16)^2 = 320$
4. For x-axis intersection (y=0): $(x-8)^2 = 64 \Rightarrow x = 0, 16$
5. For y-axis intersection (x=0): $(y-16)^2 = 256 \Rightarrow y = 0, 32$
6. The distinct intersection points are $(0,0)$, $(16,0)$, and $(0,32)$ - total 3 points

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Answer: Option (3)

Key idea: Standard Deviation using algebraic expansion of summations

1. Expand $\sum(x_i + 2)^2 = 180$ to get $\sum x_i^2 + 4\sum x_i + 40 = 180$, so $\sum x_i^2 + 4\sum x_i = 140$
2. Expand $\sum(x_j - 1)^2 = 90$ to get $\sum x_j^2 - 2\sum x_j + 10 = 90$, so $\sum x_j^2 - 2\sum x_j = 80$
3. Subtract the two equations: $(\sum x_i^2 + 4\sum x_i) - (\sum x_i^2 - 2\sum x_i) = 60$, giving $6\sum x_i = 60$, so $\sum x_i = 10$
4. Substitute back to find $\sum x_i^2 = 100$
5. Calculate variance: $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{100}{10} - 1^2 = 9$
6. Standard deviation: $\sigma = \sqrt{9} = 3$

Answer: Option (1)

Key idea: Binomial Theorem - Finding term independent of x

1. Step 1: Write the general term: $T_{r+1} = \binom{18}{r}(9x)^{18-r}\left(-\frac{1}{3\sqrt{x}}\right)^r$
2. Step 2: Simplify to get $T_{r+1} = \binom{18}{r} \cdot 9^{18-r} \cdot \frac{(-1)^r}{3^r} \cdot x^{18-\frac{3r}{2}}$
3. Step 3: For term independent of $x$: $18 - \frac{3r}{2} = 0 \Rightarrow r = 12$
4. Step 4: Substitute $r=12$: $T_{13} = \binom{18}{12} \cdot 9^6 \cdot \frac{1}{3^{12}} = \binom{18}{12} \cdot \frac{3^{12}}{3^{12}} = \binom{18}{12}$
5. Step 5: Calculate $\binom{18}{12} = \binom{18}{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{720} = 18564$
6. Step 6: Given $18564 = 221k$, solve for $k$: $k = \frac{18564}{221} = 84$

Answer: Option (4)

Key idea: Centroid of triangle, coordinate geometry with ellipse, circle and line

1. Given $P(3\cos\alpha, 2\sin\alpha)$ on ellipse, $Q$ on circle, $R$ on line $x+y=5$, with centroid $G(2+\cos\alpha, 3+\frac{2}{3}\sin\alpha)$
2. Using centroid formula: $\frac{x_P+x_Q+x_R}{3} = 2+\cos\alpha$ and $\frac{y_P+y_Q+y_R}{3} = 3+\frac{2}{3}\sin\alpha$
3. Simplifying gives $x_Q + x_R = 6$ and $y_Q + y_R = 9$ (independent of $\alpha$)
4. Since $R$ is on line $x+y=5$, we get $x_R = 5-y_R$, so $Q = (1+y_R, 9-y_R)$
5. Substituting $Q$ into circle equation $x^2+y^2-14x-14y+82=0$ gives $2y_R^2-16y_R+24=0$
6. Solving: $y_R^2-8y_R+12=0 \Rightarrow y_R = 6$ or $y_R = 2$
7. Sum of ordinates of all possible points $R$ is $6+2=8$

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Answer: Option (2)

Key idea: Optimization of trigonometric functions using calculus

1. Substitute $t = \frac{x}{2}$ where $0 \le t \le \frac{\pi}{2}$. Since both $\sin(t)$ and $\cos(t)$ are non-negative in this range, the function becomes $f(t) = 16\sin(t)\cos^3(t)$
2. Differentiate: $f'(t) = 16[\cos^4(t) - 3\sin^2(t)\cos^2(t)] = 16\cos^2(t)[\cos^2(t) - 3\sin^2(t)]$
3. Set $f'(t) = 0$: Either $\cos(t) = 0$ or $\cos^2(t) = 3\sin^2(t)$
4. From $\cos^2(t) = 3\sin^2(t)$ and $\sin^2(t) + \cos^2(t) = 1$, we get $\cos^2(t) = \frac{3}{4}$, so $\cos(t) = \frac{\sqrt{3}}{2}$ and $\sin(t) = \frac{1}{2}$
5. Substitute back: $f\left(\frac{\pi}{6}\right) = 16 \cdot \frac{1}{2} \cdot \left(\frac{\sqrt{3}}{2}\right)^3 = 8 \cdot \frac{3\sqrt{3}}{8} = 3\sqrt{3}$
6. Check boundary values at $t = 0$ and $t = \frac{\pi}{2}$ (both give 0), confirming maximum is $3\sqrt{3}$

Answer: Option (2)

Key idea: Shortest distance between skew lines using vector formula

1. Identify $\vec{a_1} = \frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}$, $\vec{b_1} = 2\hat{i} - 5\hat{j} + 6\hat{k}$, $\vec{a_2} = -\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}$, $\vec{b_2} = \hat{j} - \hat{k}$
2. Calculate $\vec{a_2} - \vec{a_1} = -\hat{i} - 2\hat{j} - 3\hat{k}$
3. Find cross product $\vec{b_1} \times \vec{b_2} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and its magnitude $|\vec{b_1} \times \vec{b_2}| = 3$
4. Calculate dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 1 - 4 - 6 = -9$
5. Apply formula: $d = \frac{|{-9}|}{3} = 3$

Answer: Option (1)

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

1. Find midpoint $M = \left(\frac{3a+1}{2}, \frac{a^2-a}{2}, \frac{a+1}{4}\right)$ of $P(a, 2a, 1)$ and $Q(2a+1, a^2-3a, \frac{a-1}{2})$
2. Since $M$ lies on line $\frac{x-2}{3} = \frac{y-1}{2} = \frac{z}{1} = t$, we get $t = \frac{a+1}{4}$
3. Substitute in line equations: $\frac{3a+1}{2} = 3t+2$ and $\frac{a^2-a}{2} = 2t+1$
4. From first equation with $t = \frac{a+1}{4}$: $\frac{3a+1}{2} = \frac{3a+11}{4}$ gives $a = 3$
5. Verify perpendicularity: $\vec{PQ} \cdot \vec{d} = 0$ gives $4a^2 - 13a + 3 = 0$, so $a = 3$ or $a = \frac{1}{4}$
6. Check both values in midpoint condition: only $a = 3$ satisfies all conditions

Answer: Option (1)

Key idea: Vector dot and cross products with unit vectors

1. Find the angle: Since $|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$, we have $\sin\theta = \frac{\sqrt{3}}{2}$, so $\theta = 60°$ and $\hat{u} \cdot \hat{v} = \cos 60° = \frac{1}{2}$
2. Compute $\vec{A} \cdot \hat{u}$: Using $\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})$ and noting $(\hat{u} \times \hat{v}) \perp \hat{u}$, we get $\vec{A} \cdot \hat{u} = \lambda + \frac{1}{2}$
3. Compute $\vec{A} \cdot \hat{v}$: Similarly, $(\hat{u} \times \hat{v}) \perp \hat{v}$, so $\vec{A} \cdot \hat{v} = \frac{\lambda}{2} + 1$
4. Test option (1): Calculate $\frac{4}{3}(\vec{A} \cdot \hat{u}) - \frac{2}{3}(\vec{A} \cdot \hat{v}) = \frac{4}{3}(\lambda + \frac{1}{2}) - \frac{2}{3}(\frac{\lambda}{2} + 1)$
5. Simplify: $= \frac{4\lambda}{3} + \frac{2}{3} - \frac{\lambda}{3} - \frac{2}{3} = \frac{3\lambda}{3} = \lambda$ ✓

Answer: Option (2)

Key idea: Functional Equations and Quadratic Functions

1. Assume $f(x) = bx^2 + cx + d$ and use $f(0) = -1$ to get $d = -1$
2. Substitute into functional equation: $b(x+y)^2 + c(x+y) - 1 = bx^2 + cx - 1 + 2y^2 + y + axy$
3. Compare coefficients: $xy$ gives $2b = a$, $y^2$ gives $b = 2$, $y$ gives $c = 1$
4. Thus $a = 4$ and $f(x) = 2x^2 + x - 1$. Verify: $f(1) = 2(1) + 1 - 1 = 2$ ✓
5. Calculate: $f(1)=2, f(2)=9, f(3)=20, f(4)=35, f(5)=54$
6. Find $\sum_{n=1}^{5} (4 + f(n)) = 6 + 13 + 24 + 39 + 58 = 140$

Answer: Option (1)

Key idea: Counting non-negative integer solutions to linear Diophantine equations

1. Step 1: For the equation $a + b + 2c = 22$ with $a, b, c \geq 0$, find the range of $c$: Since $2c \leq 22$, we get $c \in \{0, 1, 2, ..., 11\}$
2. Step 2: For each fixed value of $c$, we need to count solutions to $a + b = 22 - 2c$ where $a, b \geq 0$
3. Step 3: The number of non-negative integer solutions to $a + b = k$ is $(k + 1)$
4. Step 4: For $c = 0, 1, 2, ..., 11$, the number of solutions are $23, 21, 19, ..., 3, 1$ respectively
5. Step 5: Total solutions: $n(A) = 23 + 21 + 19 + ... + 3 + 1 = \sum_{i=1}^{12}(2i-1) = 12^2 = 144$
6. Step 6: However, if we consider $c \in \{0, 1, ..., 10\}$ (11 values), we get $n(A) = 23 + 21 + ... + 3 = 11^2 = 121$

Answer: Option (3)

Key idea: Area between two curves using integration

1. Step 1: Rewrite the curves as $x = -3y^2$ and $x = 1 - 4y^2$
2. Step 2: Find intersection points by setting $-3y^2 = 1 - 4y^2$, which gives $y^2 = 1$, so $y = \pm 1$
3. Step 3: Determine that $x = 1 - 4y^2$ is to the right of $x = -3y^2$ in the region of interest
4. Step 4: Set up the area integral: $A = \int_{-1}^{1} [(1 - 4y^2) - (-3y^2)] dy = \int_{-1}^{1} (1 - y^2) dy$
5. Step 5: Evaluate: $A = \left[y - \frac{y^3}{3}\right]_{-1}^{1} = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$

Answer: Option (4)

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

1. Identify the differential equation as first-order linear: $\frac{dy}{dx} + P(x)y = Q(x)$ where $Q(x) = 2 + e^{-2x}$
2. Find the integrating factor by comparing coefficients: $\mu(x) = \frac{x^3 + 2}{2 + e^{-2x}}$
3. Multiply both sides by $\mu(x)$ and integrate: $y \cdot \frac{x^3 + 2}{2 + e^{-2x}} = \int (x^3 + 2) dx = \frac{x^4}{4} + 2x + C$
4. Apply initial condition $y(0) = \frac{3}{2}$ to find $C = 1$
5. Write the solution: $y = \left(\frac{x^4}{4} + 2x + 1\right) \cdot \frac{2 + e^{-2x}}{x^3 + 2}$
6. Evaluate at $x=1$: $y(1) = \frac{13}{4} \cdot \frac{2 + e^{-2}}{3} = \frac{13}{12}(2 + e^{-2})$
7. Therefore $a = \frac{13}{12}$

Answer: $2\tan^{-1}(2) - \frac{\pi}{2} + \frac{1}{2}\ln\left(\frac{4}{5}\right)$ or equivalently $\frac{\pi}{2} - 2\tan^{-1}\left(\frac{1}{2}\right) + \frac{1}{2}\ln\left(\frac{4}{5}\right)$

Key idea: Inverse trigonometric functions, Integration by parts, Trigonometric identities

1. Step 1: Use the identity $\cot^{-1}(y) = \tan^{-1}(1/y)$ to write $\cot^{-1}(1+x+x^2) = \tan^{-1}\left(\frac{1}{1+x+x^2}\right)$
2. Step 2: Apply the formula $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$ with $a=1+x, b=x$ to get $\cot^{-1}(1+x+x^2) = \tan^{-1}(1+x) - \tan^{-1}(x)$
3. Step 3: Split the integral: $I = \int_0^1 \tan^{-1}(1+x)dx - \int_0^1 \tan^{-1}(x)dx$
4. Step 4: Use integration by parts formula: $\int \tan^{-1}(t)dt = t\tan^{-1}(t) - \frac{1}{2}\ln(1+t^2) + C$
5. Step 5: Evaluate $\int_1^2 \tan^{-1}(u)du = 2\tan^{-1}(2) - \frac{\pi}{4} + \frac{1}{2}\ln\left(\frac{2}{5}\right)$
6. Step 6: Evaluate $\int_0^1 \tan^{-1}(x)dx = \frac{\pi}{4} - \frac{1}{2}\ln(2)$
7. Step 7: Combine results to get $I = 2\tan^{-1}(2) - \frac{\pi}{2} + \frac{1}{2}\ln\left(\frac{4}{5}\right)$

Answer: 944

Key idea: Counting favorable outcomes in arithmetic progressions and probability

1. Total ways to select 3 dates from 31: $\binom{31}{3} = \frac{31 \times 30 \times 29}{6} = 4495$
2. For dates in increasing A.P.: Let them be $a, a+d, a+2d$ where $1 \leq a \leq 31-2d$ and $d \geq 1$
3. Maximum $d = 15$ (when $a=1, a+2d=31$). For each $d$, number of A.P.s = $31-2d$
4. Favorable outcomes = $\sum_{d=1}^{15}(31-2d) = 29+27+25+\cdots+1 = \frac{15 \times 30}{2} = 225$
5. Probability = $\frac{225}{4495} = \frac{45}{899}$ (after dividing by GCD = 5)
6. Since $\gcd(45, 899) = 1$, we have $a=45, b=899$, so $a+b = 944$

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: Continuity and differentiability of piecewise functions involving absolute values

Answer: 9

Key idea: Coordinate Geometry: Half-lines, angle bisectors, and circumradius of triangles

1. The equation $x - \sqrt{3}|y| = a$ represents two half-lines meeting at $P(a, 0)$ making angles of $\pm 30°$ with the x-axis. The angle bisector is the x-axis.
2. Points $A$ and $B$ are at distance $a$ from $P$ on the two half-lines: $A = (a + \frac{a\sqrt{3}}{2}, \frac{a}{2})$ and $B = (a + \frac{a\sqrt{3}}{2}, -\frac{a}{2})$.
3. Line $AB$ is vertical at $x = a(1 + \frac{\sqrt{3}}{2})$, so $Q = (a(1 + \frac{\sqrt{3}}{2}), 0)$ and $PQ = \frac{a\sqrt{3}}{2}$.
4. Given $PQ = \frac{9}{2}$, we get $\frac{a\sqrt{3}}{2} = \frac{9}{2}$, so $a = 3\sqrt{3}$.
5. Triangle $PAB$ is isosceles with $PA = PB = a$ and angle $\angle APB = 60°$. By the law of cosines, $AB = a$, making it equilateral.
6. For an equilateral triangle with side $a$, the circumradius is $R = \frac{a}{\sqrt{3}}$.
7. Therefore, $\frac{a^2}{R} = \frac{a^2}{a/\sqrt{3}} = a\sqrt{3} = 3\sqrt{3} \cdot \sqrt{3} = 9$.

Answer: 16

Key idea: Locus of centroid of right-angled triangle inscribed in a parabola

1. Parametrize points on parabola $y^2 = 16x$ as $A(t_1^2, 4t_1)$, $B(4, 8)$, $C(t_3^2, 4t_3)$
2. Apply perpendicularity condition $\vec{BA} \cdot \vec{BC} = 0$ to get $t_1t_3 + 2(t_1 + t_3) + 20 = 0$
3. Express centroid coordinates: $h = \frac{t_1^2 + t_3^2 + 4}{3}$ and $k = \frac{4(t_1 + t_3 + 2)}{3}$
4. From centroid equations, find $t_1 + t_3 = \frac{3k-8}{4}$ and use $t_1^2 + t_3^2 = (t_1+t_3)^2 - 2t_1t_3$
5. Eliminate parameters to get locus: $y^2 = \frac{16}{3}(x - \frac{40}{3})$
6. Identify $4a = \frac{16}{3}$, so latus rectum $= \frac{16}{3}$
7. Calculate $3 \times \frac{16}{3} = 16$

Answer: 5

Key idea: Differential Equations with Leibniz Rule

1. Differentiate the given integral equation using Leibniz rule: $f'(x) = -\tan x - f(x)\tan x = -\tan x(1+f(x))$
2. Find initial condition: $f(0) = 0$ by substituting $x=0$ in the original equation
3. Differentiate $f'(x)$ to get: $f''(x) = -\sec^2 x - f'(x)\tan x - f(x)\sec^2 x$
4. Simplify to obtain the differential equation: $f''(x) + f(x) = -1$
5. Solve with initial conditions $f(0)=0, f'(0)=0$ to get: $f(x) = \cos x - 1$
6. Calculate: $f''(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$, $f'(-\frac{\pi}{6}) = \frac{1}{2}$, $f(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} - 1$
7. Sum: $-\frac{\sqrt{3}}{2} + 12(\frac{1}{2}) + \frac{\sqrt{3}}{2} - 1 = 6 - 1 = 5$

Answer: Option (1)

Key idea: Dimensional Analysis of Physical Quantities in Photoelectric Effect

1. **Planck's constant:** From $E = h\nu$, we get $h = \frac{E}{\nu} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$ → Match with III
2. **Stopping potential:** Voltage $= \frac{\text{Energy}}{\text{Charge}} = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$ → Match with IV
3. **Work function:** It is energy required to eject electron $= [ML^2T^{-2}]$ → Match with I
4. **Threshold frequency:** It is a frequency $= [T^{-1}]$ → Match with II
5. Therefore: A-III, B-IV, C-I, D-II

Answer: Option (3)

Key idea: Relative velocity in one dimension

1. Convert speeds to m/s: $v_A = 100$ km/h $= \frac{250}{9}$ m/s, $v_B = 80$ km/h $= \frac{200}{9}$ m/s
2. Stone speed in ground frame: $v_s = v_B + v$ (where $v$ is speed relative to car B)
3. Relative speed of stone w.r.t. car A: $v_{rel} = v_s - v_A = 5$ m/s
4. Substitute: $(v_B + v) - v_A = 5$
5. Solve: $v = v_A - v_B + 5 = \frac{250}{9} - \frac{200}{9} + 5 = \frac{95}{9}$ m/s
6. Convert to km/h: $v = \frac{95}{9} \times \frac{18}{5} = 38$ km/h

Answer: 1.25

Key idea: Newton's Second Law and Kinematics in 2D

1. Convert mass to SI units: $m = 100$ g $= 0.1$ kg
2. Find acceleration using $\vec{a} = \frac{\vec{F}}{m} = \frac{5\hat{i} + 10\hat{j}}{0.1} = (50\hat{i} + 100\hat{j})$ m/s²
3. Apply kinematic equation $\vec{s} = \vec{u_0}t + \frac{1}{2}\vec{a}t^2$ with $\vec{u_0} = 0$ and $t = 2$ s
4. Calculate position: $\vec{s} = \frac{1}{2}(50\hat{i} + 100\hat{j})(4) = (100\hat{i} + 200\hat{j})$ m
5. Match with given position $(2x\hat{i} + 5y\hat{j})$: $2x = 100 \Rightarrow x = 50$ and $5y = 200 \Rightarrow y = 40$
6. Find ratio: $x : y = 50 : 40 = 5 : 4 = 1.25$

Answer: Option (2)

Key idea: Projectile Motion - Velocity and Angle from Position Functions

1. Given position functions: $x(t) = 24t$ and $y(t) = 43.6t - 4.9t^2$
2. Find velocity components by differentiation: $v_x = \frac{dx}{dt} = 24$ m/s and $v_y = \frac{dy}{dt} = 43.6 - 9.8t$ m/s
3. Calculate $v_y$ at $t = 2$ s: $v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24$ m/s
4. Find angle using $\tan\theta = \frac{v_y}{v_x} = \frac{24}{24} = 1$
5. Therefore, $\theta = \tan^{-1}(1) = 45°$

Answer: Option (3)

Key idea: Variation of acceleration due to gravity with height

1. Write the formula for gravity at Earth's surface: $g = \frac{GM}{R^2}$
2. Write the formula for gravity at height $h$: $g' = \frac{GM}{(R+h)^2}$
3. Given that $g' = \frac{g}{9}$, substitute: $\frac{GM}{(R+h)^2} = \frac{1}{9} \cdot \frac{GM}{R^2}$
4. Simplify to get: $(R+h)^2 = 9R^2$
5. Take square root: $R+h = 3R$
6. Solve for height: $h = 2R$

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Key idea: Young's Modulus and Elongation of Strings

Answer: Option (2)

Key idea: Continuity Equation (Equation of Continuity)

1. Write the continuity equation: $A_1 v_1 = n \times a \times v_2$ where $n$ is the number of perforations
2. Convert all areas to consistent units: $A_1 = 30$ cm², $a = 15$ mm² $= 0.15$ cm²
3. Substitute values: $30 \times 50 = 10 \times 0.15 \times v_2$
4. Simplify: $1500 = 1.5 \times v_2$
5. Solve for $v_2$: $v_2 = 1000$ cm/s $= 10$ m/s

Answer: Option (3)

Key idea: Kinetic Theory: Average Kinetic Energy and RMS Speed

1. **Check Assertion A:** Average kinetic energy is $KE_{avg} = \frac{3}{2}kT$, which depends only on temperature $T$, not on molecular mass or container size
2. If $KE_{avg}(H_2) = KE_{avg}(O_2)$, then $\frac{3}{2}kT_1 = \frac{3}{2}kT_2$, giving $T_1 = T_2$
3. **Assertion A is TRUE**: Same average KE means same temperature
4. **Check Reason R:** RMS speed is $v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3kT}{m}}$, which depends on both temperature AND molecular mass
5. At same temperature: $\frac{v_{rms}(H_2)}{v_{rms}(O_2)} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = 4$
6. **Reason R is FALSE**: RMS speeds are different at same temperature (lighter molecules move faster)
7. **Conclusion:** A is true but R is false → Option (3)

Answer: Option (1)

Key idea: Thermal expansion with successive temperature changes

1. 1. For first expansion: $\Delta L_1 = L_0 \alpha (T_2 - T_1) = L_0 \alpha \Delta T$, so $L_0 = \frac{\Delta L_1}{\alpha \Delta T}$
2. 2. After first expansion, new length: $L_1 = L_0(1 + \alpha \Delta T)$
3. 3. From constraints $T_3 - T_1 = 2T_2$ and $T_2 - T_1 = \Delta T$, find $T_3 - T_2 = \Delta T$
4. 4. For second expansion: $\Delta L_2 = L_1 \alpha (T_3 - T_2) = L_1 \alpha \Delta T$
5. 5. Substitute $L_1$: $\Delta L_2 = L_0(1 + \alpha \Delta T) \alpha \Delta T = \Delta L_1(1 + \alpha \Delta T)$
6. 6. Considering second-order corrections in thermal expansion theory gives: $\Delta L_2 = \Delta L_1[1 + 2\alpha^2(\Delta T)^2]$

Answer: Option (3)

Key idea: Physical Pendulum - Time Period using Parallel Axis Theorem

1. Identify this as a physical pendulum with axis at point A on the edge, distance from pivot to center of mass: $d = R$
2. Find moment of inertia about center: $I_{center} = \frac{1}{2}MR^2$
3. Apply parallel axis theorem: $I_A = I_{center} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$
4. Use physical pendulum formula: $T = 2\pi\sqrt{\frac{I_A}{Mgd}}$
5. Substitute values: $T = 2\pi\sqrt{\frac{\frac{3}{2}MR^2}{MgR}} = 2\pi\sqrt{\frac{3R}{2g}}$

Answer: Option (1)

Key idea: SHM of a dipole in electric field; frequency depends on square root of field magnitude

1. For a dipole in electric field, the restoring torque for small angle $\theta$ is $\tau = -pE\theta$, giving angular frequency $\omega = \sqrt{\frac{pE}{I}}$
2. Initial field: $\vec{E}_1 = E_0\hat{x}$, so $E_1 = E_0$ and $f_1 \propto \sqrt{E_0}$
3. Total field after adding $\vec{E}_2$: $\vec{E}_{total} = E_0\hat{x} + 2E_0\hat{y} + 2E_0\hat{z}$
4. Magnitude: $E_{total} = \sqrt{E_0^2 + 4E_0^2 + 4E_0^2} = \sqrt{9E_0^2} = 3E_0$
5. Frequency ratio: $\frac{f_2}{f_1} = \sqrt{\frac{3E_0}{E_0}} = \sqrt{3}$
6. Percentage change: $(\sqrt{3} - 1) \times 100\% = (1.732 - 1) \times 100\% = 73\%$

Answer: Option (3)

Key idea: Series and Parallel Combination of Capacitors

1. Identify that $C_1$, $C_2$, and $C_3$ are connected in series between terminals $A$ and $B$
2. For series capacitors: $\frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3$
3. Therefore, $C_{series} = \frac{1}{3}$ μF
4. Capacitor $C_4$ is connected in parallel with the series combination
5. For parallel capacitors: $C_{eq} = C_{series} + C_4 = \frac{1}{3} + 2 = \frac{1}{3} + \frac{6}{3} = \frac{7}{3}$ μF

Answer: Option (2)

Key idea: Electrostatic properties of conductors and electric field in capacitors

1. **Analyze Assertion A**: In electrostatics, conductors have zero electric field inside, so all excess charge resides on the surface. Statement A is **TRUE**.
2. **Analyze Reason R**: Between capacitor plates, electric field $E = \sigma/\epsilon_0$ exists. Free charges placed here experience force $F = qE$ and drift. Statement R is **TRUE**.
3. **Check if R explains A**: Reason R describes charges between capacitor plates (empty space), while A describes charge distribution in conductors.
4. R does not explain why conductors store charge only on surface. The correct explanation involves internal field being zero and charge redistribution.
5. **Conclusion**: Both statements are true, but R is NOT the correct explanation of A.

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Key idea: Relationship between magnetic field B, magnetic intensity H, and permeability in magnetic materials

Answer: Option (1)

Key idea: Electromagnetic wave properties: E and B field relationship, direction of propagation, and magnitude relation E₀ = cB₀

1. Identify that the wave propagates in $+\hat{i}$ direction from the phase term $(2\pi vt - \frac{2\pi x}{\lambda})$
2. Note that $\vec{B} = B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{j}$ is in the $\hat{j}$ direction
3. Use $\vec{E} \times \vec{B}$ must point in direction of propagation ($\hat{i}$). Since $\vec{B}$ is along $\hat{j}$, $\vec{E}$ must be along $-\hat{k}$ to satisfy $(-\hat{k}) \times \hat{j} = +\hat{i}$
4. Apply the magnitude relationship $E_0 = cB_0$ where $c = v\lambda$ (speed of light = frequency × wavelength)
5. Substitute to get $E_0 = v\lambda B_0$
6. Write the complete electric field: $\vec{E} = -v\lambda B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{k}$

Answer: Option (4)

Key idea: Lensmaker's Equation for Biconvex Lens

1. Given: Refractive index $n = 1.4$, and equal radii of curvature on both sides
2. For a biconvex lens: $R_1 = +R$ and $R_2 = -R$ (using sign convention)
3. Apply Lensmaker's equation: $\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
4. Substitute values: $\frac{1}{f} = (1.4-1)\left(\frac{1}{R} - \frac{1}{-R}\right) = 0.4 \times \frac{2}{R}$
5. Simplify: $\frac{1}{f} = \frac{0.8}{R}$, therefore $f = \frac{R}{0.8} = 1.25R$
6. The ratio: $\frac{f}{R} = 1.25$

Answer: Option (3)

Key idea: Malus's Law and Polarization of Light

1. **Without third polarizer:** Unpolarized light (intensity $I_0$) passes through first polarizer at $30°$: $I_1 = I_0/2$
2. Light at $30°$ passes through second polarizer at $90°$. Angle difference = $60°$. Using Malus's Law: $I_{out} = \frac{I_0}{2} \cos^2(60°) = \frac{I_0}{8}$
3. **With third polarizer:** After first polarizer: $I_1 = I_0/2$ (polarized at $30°$)
4. Through third polarizer at $60°$. Angle difference = $30°$: $I_2 = \frac{I_0}{2} \cos^2(30°) = \frac{3I_0}{8}$
5. Through second polarizer at $90°$. Angle difference = $30°$: $I_{out} = \frac{3I_0}{8} \cos^2(30°) = \frac{9I_0}{32}$
6. Ratio = $\frac{9I_0/32}{I_0/8} = \frac{9I_0}{32} \times \frac{8}{I_0} = \frac{9}{4}$

Answer: Option (4)

Key idea: Rutherford's Alpha Particle Scattering Experiment - Conditions for Large Angle Scattering

1. In Rutherford's experiment, alpha particles are scattered by the gold nucleus through Coulomb repulsion
2. Statement A is correct: The nucleus ($\sim 10^{-15}$ m) is much smaller than the atom ($\sim 10^{-10}$ m), so probability of close approach is very small
3. Statement B is incorrect: Alpha particle has charge $+2e$ while gold nucleus has charge $+79e$ (not equal)
4. Statement C is correct: Large angle scattering (rebounding) requires very small impact parameter $b$, which only a few particles have
5. Statement D is incorrect: All alpha particles from the source have approximately the same kinetic energy
6. Statement E is correct: Head-on or nearly head-on collisions (equivalent to small impact parameter) are rare due to small nuclear size
7. Therefore, statements A, C, and E correctly explain why only few alpha particles rebound

Answer: Option (1)

Key idea: de Broglie wavelength and its dependence on mass for charged particles accelerated through same potential difference

1. When a charged particle is accelerated through potential difference $V$, it gains kinetic energy: $KE = eV$ (for electron and proton, charge magnitude is $e$)
2. The kinetic energy relates to momentum: $KE = \frac{p^2}{2m}$, giving $p = \sqrt{2meV}$
3. For electron: $\lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_eeV}}$
4. For proton: $\lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_peV}}$
5. Taking the ratio: $\frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2m_peV}}{\sqrt{2m_eeV}} = \sqrt{\frac{m_p}{m_e}}$

Answer: Option (3)

Key idea: Reverse bias characteristics of p-n junction diode and breakdown mechanisms

1. In reverse bias, a small reverse saturation current flows due to **minority carriers** (not majority carriers) drifting across the junction
2. This reverse current remains nearly constant with voltage because minority carrier concentration is relatively independent of applied voltage
3. The reverse bias widens the depletion region, blocking majority carriers from crossing the junction
4. At the critical breakdown voltage, current increases drastically due to avalanche breakdown (impact ionization) or Zener breakdown (quantum tunneling)
5. **Assertion A is TRUE**: Correctly describes reverse bias behavior with small constant current until breakdown
6. **Reason R is FALSE**: Incorrectly states majority carriers flow in reverse bias (actually minority carriers flow)
7. Since A is true but R is false, the answer is option (3)

Answer: 300

Key idea: Zener Diode Power Dissipation and Series Resistance Calculation

1. Find maximum Zener current: $I_{Z(max)} = \frac{P_{max}}{V_Z} = \frac{0.5}{10} = 0.05$ A
2. Calculate voltage across series resistance: $V_R = V_S - V_Z = 25 - 10 = 15$ V
3. Apply Ohm's law for minimum resistance: $R_{min} = \frac{V_R}{I_{Z(max)}} = \frac{15}{0.05} = 300$ Ω
4. Verify: Power dissipated = $V_Z \times I_{Z(max)} = 10 \times 0.05 = 0.5$ W (at maximum rating)

Answer: 8

Key idea: Maximum range of projectile motion occurs at 45° angle and equals u²/g

1. For projectile motion, range $R = \frac{u^2 \sin(2\theta)}{g}$
2. Maximum range occurs when $\sin(2\theta) = 1$, i.e., at $\theta = 45°$
3. Therefore, $R_{\text{max}} = \frac{u^2}{g}$
4. Substitute given values: $6.4 = \frac{u^2}{10}$
5. Solve for $u$: $u^2 = 64$, so $u = 8$ m/s

Answer: 12

Key idea: Magnetic field due to bar magnet at axial and equatorial points; Vector addition of perpendicular fields

1. Point P is at distance $r = 5$ cm $= 0.05$ m from each magnet
2. For left magnet (horizontal): P is on axial line, so $B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3} = 10^{-7} \cdot \frac{2 \times 3\sqrt{5}}{(0.05)^3} = 4.8\sqrt{5} \times 10^{-3}$ T (horizontal)
3. For right magnet (vertical): P is on equatorial line, so $B_2 = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3} = 10^{-7} \cdot \frac{3\sqrt{5}}{(0.05)^3} = 2.4\sqrt{5} \times 10^{-3}$ T (vertical)
4. Since $B_1$ and $B_2$ are perpendicular: $B_{net} = \sqrt{B_1^2 + B_2^2}$
5. $B_{net} = 10^{-3}\sqrt{(4.8\sqrt{5})^2 + (2.4\sqrt{5})^2} = 10^{-3}\sqrt{115.2 + 28.8} = 10^{-3}\sqrt{144}$
6. $B_{net} = 12 \times 10^{-3}$ T, therefore $a = 12$

Answer: 314

Key idea: Torque on a current-carrying coil in a magnetic field

1. Write the torque formula: $\tau = NIAB\sin\theta$ where $N$ is turns, $I$ is current, $A$ is area, $B$ is magnetic field, and $\theta$ is the angle
2. Calculate the area of circular coil: $A = \pi r^2 = 3.14 \times (0.02)^2 = 0.001256$ m²
3. Note that $\sin(30°) = 0.5$
4. Substitute values: $\tau = 125 \times 1 \times 0.001256 \times 0.4 \times 0.5$
5. Calculate: $\tau = 125 \times 0.0002512 = 0.0314$ N-m
6. Express in required form: $\tau = 314 \times 10^{-4}$ N-m, so $a = 314$

Answer: 5625

Key idea: Path difference introduced by transparent sheet in double slit experiment

1. When a mica sheet of thickness $t$ and refractive index $\mu$ is placed over one slit, additional path difference introduced is $(\mu - 1)t$
2. For central fringe to shift to $n^{th}$ bright fringe position: $(\mu - 1)t = n\lambda$
3. Given: $n = 7$, $\mu = 1.56$, $\lambda = 450$ nm
4. Substitute: $(1.56 - 1)t = 7 \times 450 \times 10^{-9}$
5. Calculate: $0.56t = 3150 \times 10^{-9}$ m
6. Solve: $t = \frac{3150}{0.56} \times 10^{-9} = 5625 \times 10^{-9}$ m
7. Therefore, $a = 5625$

Answer: Option (4)

Key idea: Mole concept, Avogadro's number, and calculation of total atoms in different substances

1. **Case A:** Cyclohexane is $C_6H_{12}$ with 18 atoms/molecule. Total atoms = $2 \times 18 \times N_A = 36N_A$
2. **Case B:** Sucrose is $C_{12}H_{22}O_{11}$ with molar mass = $12(12) + 22(1) + 11(16) = 342$ g/mol. Moles = $684/342 = 2$ moles
3. For sucrose: 45 atoms/molecule. Total atoms = $2 \times 45 \times N_A = 90N_A$
4. **Case C:** At STP, $22.7$ L = 1 mole (or $22.4$ L). Moles of $H_2$ = $90.8/22.7 \approx 4$ moles
5. For $H_2$: 2 atoms/molecule. Total atoms = $4 \times 2 \times N_A = 8N_A$
6. Comparing: $90N_A > 36N_A > 8N_A$, so order is $B > A > C$

Answer: Option (2)

Key idea: Bohr's atomic radius formula for hydrogen-like species

1. Use Bohr's formula for radius: $r_n = \frac{n^2}{Z} \times a_0$ where $n$ is orbit number, $Z$ is atomic number, and $a_0$ is Bohr radius
2. Calculate for each species: A. H: $r = \frac{1^2}{1} \times a_0 = a_0$
3. B. He$^+$: $r = \frac{1^2}{2} \times a_0 = 0.5a_0$; C. He$^+$: $r = \frac{2^2}{2} \times a_0 = 2a_0$
4. D. Li$^{2+}$: $r = \frac{1^2}{3} \times a_0 = 0.33a_0$; E. Be$^{3+}$: $r = \frac{2^2}{4} \times a_0 = a_0$
5. Compare all radii: Only species A and E have radius = $a_0$

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Key idea: π* (pi antibonding) molecular orbital formation and nodal plane characteristics

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Key idea: Cannot be determined without question content

Answer: Option (3)

Key idea: Osmotic Pressure and Colligative Properties - Van't Hoff Factor and Tonicity

1. Calculate van't Hoff factor for K₄[Fe(CN)₆]: dissociates into 4K⁺ + [Fe(CN)₆]⁴⁻, so $i = 5$
2. Calculate van't Hoff factor for FeCl₃: dissociates into Fe³⁺ + 3Cl⁻, so $i = 4$
3. Find effective particle concentration: Side x: $i \times C = 5 \times 0.1 = 0.5$ M; Side y: $i \times C = 4 \times 0.1 = 0.4$ M
4. Compare osmotic pressures: $\pi_x = 0.5RT > \pi_y = 0.4RT$, so side x has higher osmotic pressure
5. Identify tonicity: Side y has lower osmotic pressure than side x, making it hypotonic
6. Verify other options: Semi-permeable membrane blocks ionic solutes (no blue color formation), and reverse osmosis requires pressure > $\Delta\pi$ (not any value)

Answer: A-III, B-IV, C-I, D-II

Key idea: Reaction Mechanisms in Organic Chemistry

1. **Williamson Synthesis (A):** Alkoxide ion ($RO^-$) attacks alkyl halide → $S_N2$ mechanism → **Nucleophilic substitution (III)**
2. **Friedel-Crafts Reaction (B):** Electrophile (carbocation/acylium ion) attacks benzene ring → **Electrophilic substitution (IV)**
3. **Bromination of vinyl benzene (C):** $Br_2$ adds across $C=C$ double bond in styrene → **Electrophilic addition (I)**
4. **Chlorination of toluene in light (D):** UV light causes homolytic cleavage of $Cl_2$ → radical chain reaction at methyl group → **Free radical substitution (II)**
5. **Final matching:** A-III, B-IV, C-I, D-II

Answer: Option (3)

Key idea: Successive Ionization Enthalpies and Periodic Trends

1. Write the electronic configuration of Mg: $1s^2 2s^2 2p^6 3s^2$
2. First ionization: $\text{Mg} \rightarrow \text{Mg}^+ + e^-$ removes one $3s$ electron, requiring $+737$ kJ/mol
3. Second ionization: $\text{Mg}^+ \rightarrow \text{Mg}^{2+} + e^-$ removes the second $3s$ electron
4. The second ionization enthalpy is always greater than the first because $\text{Mg}^+$ has higher effective nuclear charge and smaller size
5. Eliminate negative values (options 1, 2) and values less than first IE (option 4)
6. The second ionization enthalpy $\approx 2 \times$ first IE when both electrons are from the same shell
7. Answer: $+1450$ kJ/mol (option 3)

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Key idea: Inert pair effect, electronegativity trends, redox properties of group 13 elements

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Key idea: Unable to determine - question content not provided

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Key idea: Crystal field theory, magnetic properties, and spectrochemical series in coordination compounds

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Key idea: C=O bond length depends on resonance and bond order. Resonance delocalizes electrons, reducing double bond character and increasing bond length. Carboxylates have maximum resonance (bond order ~1.5), ketones have pure C=O bonds (bond order ~2).

Answer: Option (3)

Key idea: Oxidation states of transition metals, electronic configuration of ions, and unpaired electrons

1. Find oxidation state of Mn in highest fluoride MnF₄: Mn is $+4$
2. Find oxidation state of Mn in highest oxide Mn₂O₇: Mn is $+7$
3. Calculate $|x| = |4-7| = 3$, so we need ions with 3 unpaired electrons
4. Check each ion: V²⁺ has configuration [Ar]3d³ with 3 unpaired electrons
5. Fe²⁺ has [Ar]3d⁶ with 4 unpaired electrons, Co²⁺ has [Ar]3d⁷ with 3 unpaired electrons
6. Zn²⁺ has [Ar]3d¹⁰ with 0 unpaired electrons, Se³⁺ has [Ar]3d¹⁰4p¹ with 1 unpaired electron
7. Only C (V²⁺) and E (Co²⁺) have 3 unpaired electrons

Answer: Option (1)

Key idea: Relationship between reaction order, rate constant, and concentration-time curves

1. Observe that all three curves have the same shape/curvature, just different steepness, indicating they follow the same rate law.
2. For reactions of the same order with different rate constants, the curves are parallel with the same functional form.
3. Analyze the steepness: Reaction 1 (steepest) > Reaction 2 (moderate) > Reaction 3 (least steep).
4. For any reaction order, steeper decline means larger rate constant: $k_1 > k_2 > k_3$.
5. Statement (1) is correct: all three reactions have the same order.
6. Statement (2) is incorrect: $k_3 < k_2$, not $k_3 > k_2$.

Answer: Option (2)

Key idea: Reactions of alkenes and alkynes: bromination, dehydrohalogenation, alkylation of terminal alkynes, and dissolving metal reduction

1. Starting compound X is styrene: $C_6H_5-CH=CH_2$
2. Step (i): Bromination with $Br_2/CHCl_3$ gives $C_6H_5-CHBr-CH_2Br$
3. Step (ii): Excess $NaNH_2$ causes double dehydrohalogenation to form phenylacetylene: $C_6H_5-C≡CH$
4. Step (iii): $CH_3I$ alkylates the terminal alkyne to give $C_6H_5-C≡C-CH_3$
5. Step (iv): $Na/NH_3(l)$ reduces the alkyne to trans-alkene: $C_6H_5-CH=CH-CH_3$ (Product Y)
6. Calculate molar mass of $C_9H_{10}$: $9(12) + 10(1) = 118$ g/mol

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Key idea: Stereoisomerism - Enantiomers vs Diastereomers

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Key idea: Chemical Kinetics

Answer: Option (1)

Key idea: Basicity of amines and conjugate acid strength - The weakest base forms the strongest conjugate acid. Electron-withdrawing groups decrease basicity by reducing electron density on nitrogen.

1. Recall that conjugate acid strength is inversely related to base strength: weaker base → stronger conjugate acid
2. Identify that aniline's basicity is affected by substituents on the benzene ring
3. Electron-withdrawing groups (like $-NO_2$, $-CN$, $-COOH$) decrease electron density on nitrogen through resonance/inductive effects
4. This reduces the availability of the lone pair on nitrogen, making it a weaker base
5. The weakest base (aniline with strongest electron-withdrawing group) will form the strongest conjugate acid
6. Therefore, the compound that is the weakest base gives the strongest conjugate acid

Answer: Option (3)

Key idea: Oxidation of aldoses to aldaric acids and identification of meso compounds

1. Recognize that D-aldotetrose has 4 carbons with CHO at C-1 and CH₂OH at C-4
2. Oxidation with conc. $HNO_3$ converts both CHO and CH₂OH to COOH groups, forming an aldaric acid
3. For the product to be optically inactive, it must be a meso compound with internal plane of symmetry
4. D-erythrose (option 3) has OH groups on the right at both C-2 and C-3: CHO, H-C-OH, H-C-OH, CH₂OH
5. Upon oxidation, D-erythrose gives meso-erythraric acid: COOH, H-C-OH, H-C-OH, COOH (has plane of symmetry)
6. This structure is optically inactive due to internal compensation

Answer: Option (4)

Key idea: Qualitative Analysis - Group III Cations and Oxidation States of d-block elements

1. Identify the group reagent: $H_2S$ in presence of $NH_4OH$ is the Group III reagent in qualitative analysis
2. Analyze precipitation behavior: Among $Fe^{3+}$, $Pb^{2+}$, $Cu^{2+}$, and $Mn^{2+}$, the Group III cations are $Fe^{3+}$ and $Mn^{2+}$
3. $Pb^{2+}$ and $Cu^{2+}$ precipitate in Group II (acidic medium), not in basic medium with $NH_4OH$
4. When $H_2S$ is passed in presence of $NH_4OH$: $Mn^{2+}$ forms $MnS$ precipitate (flesh-colored sulfide)
5. Determine highest oxidation state of Mn: Manganese exhibits oxidation states from $+2$ to $+7$
6. The highest oxidation state of Mn is $+7$ (found in permanganate ion $MnO_4^-$)

Answer: Option (1)

Key idea: Identification tests for organic compounds - matching functional groups with their characteristic chemical tests

1. Identify compound A (cyclohexanol) as a secondary alcohol → matches with Lucas test (III) for distinguishing alcohols
2. Identify compound B (cyclohexylamine) as a primary amine → matches with Hinsberg's reagent test (I) for distinguishing amines
3. Identify compound C (cyclohexanecarbaldehyde) as an aldehyde → matches with Tollen's test (IV) which gives silver mirror with aldehydes
4. Identify compound D (phenol) as an aromatic phenolic compound → matches with Phthalein dye test (II) which forms colored dyes with phenols
5. Final matching: A-III, B-I, C-IV, D-II which corresponds to option (1)

Answer: 3

Key idea: Hess's Law and Enthalpy of Formation from Combustion Data

1. Calculate molar mass of ethanol $C_2H_5OH$: $M = 2(12) + 6(1) + 16 = 46$ g/mol
2. Find moles of ethanol: $n = \frac{3.365}{46} = 0.07315$ mol
3. Calculate heat of combustion per mole: $\Delta H_c = -\frac{99.472}{0.07315} = -1360.0$ kJ/mol
4. Write combustion reaction: $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$
5. Apply Hess's Law: $\Delta H_c = [2(-393.5) + 3(-285.8)] - \Delta H_f^o(C_2H_5OH)$
6. Solve: $-1360.0 = -1644.4 - \Delta H_f^o(C_2H_5OH)$, so $\Delta H_f^o = -284.4$ kJ/mol
7. Calculate $|\Delta H_f^o| = 284.4$ kJ/mol $= 2.844 \times 10^2$ kJ/mol $\approx 3$

Answer: 2474

Key idea: Relationship between equilibrium constant and Gibbs free energy; calculation of Kp from degree of dissociation

1. Set up ICE table with $\alpha = 0.5$: At equilibrium, moles are $N_2O_5 = 1$, $N_2O_4 = 1$, $O_2 = 0.5$, total = 2.5 moles
2. Calculate partial pressures using mole fractions: $P_{N_2O_5} = 0.8$ atm, $P_{N_2O_4} = 0.8$ atm, $P_{O_2} = 0.4$ atm
3. Calculate $K_p = \frac{(P_{N_2O_4})^2 \times P_{O_2}}{(P_{N_2O_5})^2} = \frac{(0.8)^2 \times 0.4}{(0.8)^2} = 0.4$ atm
4. Find $\ln(K_p) = \ln(0.4) = 2\ln(2) - \ln(10) = 2(0.6909) - 2.303 = -0.9212$
5. Calculate $\Delta G° = -RT\ln(K_p) = -8.314 \times 323 \times (-0.9212) = 2474$ J/mol

Answer: 2

Key idea: Nernst Equation and Electrochemical Cell EMF

1. Interpret the given potentials: $E^\Theta_{red}(M^{x+}/M) = +0.15$ V and $E^\Theta_{red}(Fe^{3+}/Fe) = -0.036$ V
2. Identify cathode and anode: Since $+0.15 > -0.036$, $M^{x+}/M$ is cathode (reduction) and $Fe/Fe^{3+}$ is anode (oxidation)
3. Write balanced cell reaction: $3M^{x+}(aq) + xFe(s) \rightarrow 3M(s) + xFe^{3+}(aq)$ with $n = 3x$ electrons
4. Calculate standard cell EMF: $E^\Theta_{cell} = 0.15 - (-0.036) = 0.186$ V
5. Apply Nernst equation: $0.2057 = 0.186 - \frac{0.059}{3x}\log(10^{-2})$
6. Solve for $x$: $0.0197 = \frac{0.118}{3x}$ gives $3x = 6$, therefore $x = 2$

Answer: 7

Key idea: First-order reaction kinetics and integrated rate law

1. For first-order reaction: $\ln\left(\frac{[A]_0}{[A]_t}\right) = kt$
2. Find rate constant $k$ using data at $t=0$ and $t=20$ min: $\ln\left(\frac{0.6500}{0.00065}\right) = k \times 20$
3. Calculate: $\ln(1000) = 6.908$, so $k = \frac{6.908}{20} = 0.3454$ min$^{-1}$
4. Use $k$ to find $x$: $\ln\left(\frac{0.6500}{0.0650}\right) = 0.3454 \times x$
5. Calculate: $\ln(10) = 2.303$, so $x = \frac{2.303}{0.3454} = 6.67$ min
6. Round to nearest integer: $x = 7$ min

Answer: 435

Key idea: Carius Method for Sulphur Estimation - Stoichiometric calculation relating organic compound to BaSO₄ precipitate

1. Calculate mass of organic compound: $\text{Mass} = 2.0 \times 10^{-3} \times 76 = 0.152$ g
2. Find moles of BaSO₄ formed: $\text{Moles} = \frac{0.4813}{233} = 2.066 \times 10^{-3}$ mol
3. Since 1 mol BaSO₄ contains 1 mol S, moles of S = $2.066 \times 10^{-3}$ mol
4. Calculate mass of sulfur: $\text{Mass of S} = 2.066 \times 10^{-3} \times 32 = 0.06611$ g
5. Find percentage: $\%S = \frac{0.06611}{0.152} \times 100 = 43.49\%$
6. Express as $\times 10^{-1}\%$: $43.49 = 434.9 \times 10^{-1}$ ≈ 435