JEE Main 8 April 2026 Shift 2: Questions and Solutions

Answer: Option (1)

Key idea: Relations: Counting elements and checking equivalence relation properties (reflexive, symmetric, transitive)

1. The relation $R$ is defined by $(a,b) \in R$ iff $1 + ab > 0$, i.e., $ab > -1$
2. Check all 25 ordered pairs from $\{-2,-1,0,1,2\} \times \{-2,-1,0,1,2\}$ systematically
3. Count pairs satisfying $ab > -1$: For $a=0$, all 5 pairs work. For other values, check each product. Total = 17 elements. Statement I is **true**
4. Check reflexivity: $(a,a) \in R$ requires $a^2 > -1$, which holds for all $a$. ✓
5. Check symmetry: If $ab > -1$, then $ba > -1$ (commutative). ✓
6. Check transitivity: Counterexample: $(-2,0) \in R$ and $(0,1) \in R$, but $(-2,1) \notin R$ since $1+(-2) = -1 \not> 0$. ✗
7. $R$ is not an equivalence relation. Statement II is **false**. Answer: Only I is true

Answer: Option (2)

Key idea: Intersection of circle and ellipse in complex plane

1. Recognize that $|z - (4 + 8i)| = \sqrt{10}$ is a circle with center $(4, 8)$ and radius $\sqrt{10}$
2. The equation $|z - (3 + 5i)| + |z - (5 + 11i)| = 4\sqrt{5}$ represents an ellipse with foci at $A(3, 5)$ and $B(5, 11)$
3. Find the ellipse parameters: distance between foci $= 2\sqrt{10}$, so $c = \sqrt{10}$. Since $2a = 4\sqrt{5}$, we get $a = 2\sqrt{5}$ and $b = \sqrt{a^2 - c^2} = \sqrt{10}$
4. The ellipse center is the midpoint of $AB$: $\left(\frac{3+5}{2}, \frac{5+11}{2}\right) = (4, 8)$, which equals the circle center!
5. Since the circle radius $r = \sqrt{10}$ equals the ellipse semi-minor axis $b = \sqrt{10}$, the circle intersects the ellipse at exactly the two endpoints of the minor axis
6. Verify: The intersection points are $(1, 9)$ and $(7, 7)$, both satisfying both equations

Answer: Option (3)

Key idea: Condition for infinitely many solutions in a system of linear equations

1. For infinitely many solutions, the determinant of coefficient matrix must be zero: $\det(A) = 0$
2. Calculate $\det(A) = 1(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) = \lambda - 6$
3. Setting $\det(A) = 0$ gives $\lambda = 6$
4. For consistency, the third equation must be a linear combination of the first two equations
5. Find $a$ and $b$ such that $a(x+y+z) + b(x+2y+5z) = 2x+3y+6z$, which gives $a=1, b=1$
6. Therefore $\mu = 1(6) + 1(10) = 16$
7. Thus $\lambda + \mu = 6 + 16 = 22$

Answer: Option (3)

Key idea: Determinants, Adjoint Matrix Properties, and det(adj(A)) = [det(A)]^(n-1)

1. Calculate $\det(A) = 15\alpha + 6$ by expanding along the first row
2. Calculate the determinant of the diagonal matrix in $B$: $\det(M) = 10\alpha^2$
3. Use the property: For a $3 \times 3$ matrix, $\det(\text{adj}(A)) = [\det(A)]^2$
4. From $\det(B) = 66$ and the answer options being perfect squares (289, 361, 441), deduce that $\det(A)$ must be 17, 19, or 21
5. Test $\alpha = 1$: This gives $\det(A) = 21$, which is consistent with the problem structure
6. Calculate $\det(\text{adj}(A)) = (21)^2 = 441$

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Key idea: Arithmetic Progressions, Trigonometric Equations, and Trigonometric Identities

Answer: Option (2)

Key idea: Bayes' Theorem and Law of Total Probability

1. Identify given probabilities: $P(B) = \frac{2}{5}$, $P(S) = \frac{1}{5}$, $P(C) = \frac{2}{5}$ and $P(L|B) = \frac{1}{5}$, $P(L|S) = \frac{1}{3}$, $P(L|C) = \frac{1}{4}$
2. Apply law of total probability: $P(L) = P(L|B) \cdot P(B) + P(L|S) \cdot P(S) + P(L|C) \cdot P(C)$
3. Calculate: $P(L) = \frac{1}{5} \cdot \frac{2}{5} + \frac{1}{3} \cdot \frac{1}{5} + \frac{1}{4} \cdot \frac{2}{5} = \frac{2}{25} + \frac{1}{15} + \frac{1}{10}$
4. Convert to common denominator (150): $P(L) = \frac{12}{150} + \frac{10}{150} + \frac{15}{150} = \frac{37}{150}$
5. Apply Bayes' theorem: $P(B|L) = \frac{P(L|B) \cdot P(B)}{P(L)} = \frac{\frac{2}{25}}{\frac{37}{150}} = \frac{2}{25} \times \frac{150}{37} = \frac{12}{37}$

Answer: Option (3)

Key idea: Combined Variance Formula

1. Calculate combined mean: $\mu = \frac{n_1\mu_1 + n_2\mu_2}{n_1 + n_2} = \frac{4(1) + 6(2)}{10} = 1.6$
2. Find deviations: $d_1 = \mu_1 - \mu = 1 - 1.6 = -0.6$ and $d_2 = \mu_2 - \mu = 2 - 1.6 = 0.4$
3. Square the deviations: $d_1^2 = 0.36$ and $d_2^2 = 0.16$
4. Apply combined variance formula: $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$
5. Substitute values: $\sigma^2 = \frac{4(13 + 0.36) + 6(1 + 0.16)}{10} = \frac{53.44 + 6.96}{10} = 6.04$

Answer: Option (3)

Key idea: Binomial Theorem with Integration

1. Recognize the general term as $\frac{2^{2k+1}}{2k+1}\binom{12}{2k}$ for $k=1,2,...,6$ and use $\frac{2^{2k+1}}{2k+1} = \int_0^2 t^{2k} dt$
2. Convert the sum to integral: $\sum_{k=1}^{6} \frac{2^{2k+1}}{2k+1}\binom{12}{2k} = \int_0^2 \sum_{k=1}^{6} \binom{12}{2k} t^{2k} dt$
3. Use binomial theorem: $(1+t)^{12} + (1-t)^{12} = 2\sum_{k=0}^{6} \binom{12}{2k} t^{2k}$, so $\sum_{k=1}^{6} \binom{12}{2k} t^{2k} = \frac{(1+t)^{12} + (1-t)^{12}}{2} - 1$
4. Evaluate $\int_0^2 (1+t)^{12} dt = \frac{3^{13} - 1}{13}$ and $\int_0^2 (1-t)^{12} dt = \frac{2}{13}$
5. Compute: $\frac{1}{2}\left(\frac{3^{13}-1}{13} + \frac{2}{13}\right) - 2 = \frac{3^{13}+1}{26} - 2 = \frac{3^{13}-51}{26}$
6. Multiply by 26: $26 \cdot \frac{3^{13}-51}{26} = 3^{13} - 51$, therefore $\alpha = 51$

Answer: Option (2)

Key idea: Distribution of distinct objects into distinct boxes with restriction (Stirling numbers of second kind)

1. Since 4 books go into 3 bags with no bag empty, the distribution must be $(2,1,1)$ - one bag gets 2 books, other two get 1 each
2. Choose which 2 books go together: $\binom{4}{2} = 6$ ways
3. Now we have 3 distinct groups: one pair of books and two single books
4. Assign these 3 groups to 3 different bags: $3! = 6$ ways
5. Total number of ways = $\binom{4}{2} \times 3! = 6 \times 6 = 36$

Answer: Option (2)

Key idea: Locus of midpoint of intercepts of a line passing through a fixed point

1. Find the intersection point $A$ of lines $4x + 3y - 1 = 0$ and $3x + 4y - 1 = 0$. Solving gives $A = (\frac{1}{7}, \frac{1}{7})$.
2. Write the line through $A$ in intercept form: $\frac{x}{a} + \frac{y}{b} = 1$, where $P(a,0)$ and $Q(0,b)$ are the intercepts.
3. Since the line passes through $A(\frac{1}{7}, \frac{1}{7})$: $\frac{1}{7a} + \frac{1}{7b} = 1$, which gives $\frac{1}{a} + \frac{1}{b} = 7$.
4. Let midpoint of $PQ$ be $M(h,k)$. Then $h = \frac{a}{2}$ and $k = \frac{b}{2}$, so $a = 2h$ and $b = 2k$.
5. Substitute in the constraint: $\frac{1}{2h} + \frac{1}{2k} = 7 \Rightarrow \frac{1}{h} + \frac{1}{k} = 14 \Rightarrow h + k = 14hk$.
6. Replace $(h,k)$ with $(x,y)$ to get the locus: $x + y - 14xy = 0$.

Answer: Option (2)

Key idea: Locus of midpoint of chord, perpendicular chords from vertex of parabola, parametric form of parabola

1. Step 1: Use parametric form: Let $P = (t_1^2, 2t_1)$ and $Q = (t_2^2, 2t_2)$ on parabola $y^2 = 4x$
2. Step 2: Apply perpendicularity: Slopes of OP and OQ are $\frac{2}{t_1}$ and $\frac{2}{t_2}$. Since perpendicular: $\frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \Rightarrow t_1 t_2 = -4$
3. Step 3: Find midpoint: If $(h,k)$ is midpoint of PQ, then $h = \frac{t_1^2 + t_2^2}{2}$ and $k = t_1 + t_2$
4. Step 4: Derive locus: From $k^2 = t_1^2 + t_2^2 + 2t_1t_2 = t_1^2 + t_2^2 - 8$, we get $2h = k^2 + 8$, so $y^2 = 2(x-4)$
5. Step 5: Identify parameters: Comparing $y^2 = 2(x-4)$ with $y^2 = 4a(x-h)$, we get $4a = 2$, so $a = \frac{1}{2}$
6. Step 6: Calculate latus rectum: Length of latus rectum $= 4a = 4 \times \frac{1}{2} = 2$

Answer: Option (1)

Key idea: Range analysis of inverse trigonometric functions and their multiples

1. Find range of $\alpha = 3\sin^{-1}(6/11)$: Since $\sin^{-1}(6/11) \approx 0.576$ rad, we get $\alpha \approx 1.728$ rad
2. Since $\pi/2 \approx 1.571 < \alpha < \pi \approx 3.142$, we have $\alpha \in (\pi/2, \pi)$, so $\cos(\alpha) < 0$ (Statement II is TRUE)
3. Find range of $\beta = 3\cos^{-1}(4/9)$: Since $\cos^{-1}(4/9) \approx 1.107$ rad, we get $\beta \approx 3.321$ rad
4. Since $\pi \approx 3.142 < \beta < 3\pi/2 \approx 4.712$, we have $\beta \in (\pi, 3\pi/2)$
5. Calculate $\alpha + \beta \approx 1.728 + 3.321 = 5.049$ rad, which lies in $(3\pi/2, 2\pi)$
6. Since $\alpha + \beta \in (3\pi/2, 2\pi)$, we have $\cos(\alpha + \beta) > 0$ (Statement I is TRUE)
7. Both statements are true, so the answer is option (1)

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: Differentiability of functions involving absolute values and checking monotonicity using derivatives

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: $\frac{\sqrt{146}}{7}$

Key idea: Distance between parallel lines in 3D, foot of perpendicular from a point to a line

1. Find the foot of perpendicular $(1, \mu, 2)$ on the line by substituting in parametric form: $t = -3$ gives $\mu = 3$
2. Use perpendicularity condition: vector from $(1, 3, 2)$ to $(\lambda, 2, 3)$ is perpendicular to direction $(1, 2, 1)$, giving $(\lambda-1) - 2 + 1 = 0$, so $\lambda = 2$
3. Identify the two lines with $\lambda = 2, \mu = 3$. Both have direction vector $(2, 3, 6)$, so they are parallel
4. For parallel lines, use distance formula: $d = \frac{|\vec{P_1P_2} \times \vec{d}|}{|\vec{d}|}$ where $\vec{P_1P_2} = (1, 1, -1)$
5. Calculate cross product: $(1, 1, -1) \times (2, 3, 6) = (9, -8, 1)$ with magnitude $\sqrt{146}$
6. Distance = $\frac{\sqrt{146}}{7}$

Answer: Option (3)

Key idea: Integration using substitution and algebraic manipulation with factorization of polynomials

1. Factor the denominator: $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$
2. Simplify the integrand to get $\int_{0}^{2} \frac{\sqrt{x}}{(\sqrt{x} + 1)\sqrt{x^2 - x + 1}} dx$
3. Substitute $u = \sqrt{x}$, so $dx = 2u\,du$ and limits become $0$ to $\sqrt{2}$
4. Apply the substitution $t = u + \frac{1}{u}$ to transform the integral
5. Evaluate the resulting logarithmic integral using the limits
6. Simplify to get $\frac{2}{3}\log_e(3 + 2\sqrt{2})$

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

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

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

Answer: Option (4)

Key idea: Continuity and definite integration with absolute value

1. Use continuity at $x=\frac{\pi}{2}$: $\lim_{x\to\frac{\pi}{2}^+} \frac{b(1-\sin x)}{(\pi-2x)^2} = \frac{1}{3}$
2. Substitute $t = x - \frac{\pi}{2}$ and use $1-\cos t \approx \frac{t^2}{2}$ to get $\frac{b}{8} = \frac{1}{3}$, so $b = \frac{8}{3}$
3. Calculate $3b - 6 = 3 \cdot \frac{8}{3} - 6 = 2$
4. Factor $x^2 + 2x - 3 = (x+3)(x-1)$, which is negative on $[0,1)$ and positive on $(1,2]$
5. Evaluate $\int_0^2 |x^2+2x-3|dx = \int_0^1 -(x^2+2x-3)dx + \int_1^2 (x^2+2x-3)dx = \frac{5}{3} + \frac{7}{3} = 4$

Answer: Option (2)

Key idea: Ellipse with major axis condition and properties of strictly decreasing functions

1. For ellipse $\frac{x^2}{A} + \frac{y^2}{B} = 1$ with major axis along $y$-axis, we need $B > A$, i.e., $f(3a+15) > f(a^2+7a+3)$
2. Since $f$ is strictly decreasing, $f(3a+15) > f(a^2+7a+3)$ implies $3a+15 < a^2+7a+3$
3. Rearranging: $a^2 + 4a - 12 > 0$
4. Solving $a^2 + 4a - 12 = 0$: $a = \frac{-4 \pm 8}{2}$, giving $a = 2$ or $a = -6$
5. Since $(a-2)(a+6) > 0$, we get $a \in (-\infty, -6) \cup (2, \infty) = \mathbb{R} - [-6, 2]$
6. Therefore $\alpha = -6$ and $\beta = 2$, so $\alpha^2 + \beta^2 = 36 + 4 = 40$

Answer: 2

Key idea: Logarithmic equations with variable bases and change of base formula

1. Factor the arguments: $2x^2+5x+3 = (2x+3)(x+1)$ and $x^2+2x+1 = (x+1)^2$
2. Apply logarithm properties to get: $\log_{(x+1)}(2x+3) + 1 = 4 - 2\log_{(2x+3)}(x+1)$
3. Let $t = \log_{(x+1)}(2x+3)$, then $\log_{(2x+3)}(x+1) = \frac{1}{t}$, giving $t = 3 - \frac{2}{t}$
4. Solve $t^2 - 3t + 2 = 0$ to get $t = 1$ or $t = 2$
5. For $t=1$: $x+1 = 2x+3 \Rightarrow x = -2$ (invalid: base $x+1 = -1 < 0$)
6. For $t=2$: $(x+1)^2 = 2x+3 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2}$. Only $x = \sqrt{2}$ is valid (check bases are positive and $\neq 1$)
7. Sum of squares: $(\sqrt{2})^2 = 2$

Answer: 12

Key idea: Definite Integration with Trigonometric Identities and Substitution

1. Step 1: Simplify $\cot(x - \frac{\pi}{3})\cot(x + \frac{\pi}{3}) + 1$ using product formulas to get $\frac{2}{2\cos(2x) + 1}$
2. Step 2: Use $\cos(2x) = 1 - 2\sin^2(x)$ to rewrite as $\frac{2\sec^2(x)}{3 - \tan^2(x)}$
3. Step 3: Substitute $u = \tan(x)$, so $du = \sec^2(x)dx$. Limits: $u = \frac{1}{\sqrt{3}}$ to $u = 1$
4. Step 4: Use partial fractions: $\frac{2}{3-u^2} = \frac{1}{\sqrt{3}}\left[\frac{1}{\sqrt{3}-u} + \frac{1}{\sqrt{3}+u}\right]$
5. Step 5: Integrate to get $\frac{1}{\sqrt{3}}\ln\left|\frac{\sqrt{3}+u}{\sqrt{3}-u}\right|$ and evaluate from $\frac{1}{\sqrt{3}}$ to $1$
6. Step 6: Simplify using $\sqrt{3}+1 = \frac{2}{\sqrt{3}-1}$ to get $I = -\frac{2}{\sqrt{3}}\ln(\sqrt{3}-1) = -\frac{2\sqrt{3}}{3}\ln(\sqrt{3}-1)$
7. Step 7: Compare with $\alpha\log_e(\sqrt{3}-1)$ to find $\alpha = -\frac{2\sqrt{3}}{3}$, so $9\alpha^2 = 9 \cdot \frac{4 \cdot 3}{9} = 12$

Answer: 4

Key idea: 3D Geometry: Line perpendicular to two lines, intersection of lines, distance formula

1. Find direction vectors: $\vec{b_2} = (1, 2, 2)$ for $L_2$ and $\vec{b_3} = (2, 2, 1)$ for $L_3$
2. Direction of $L_1$ perpendicular to both: $\vec{b_1} = \vec{b_2} \times \vec{b_3} = (-2, 3, -2)$
3. Since $L_1$ passes through origin: $L_1: \vec{r} = \lambda(-2, 3, -2)$
4. Find intersection of $L_1$ and $L_2$ by solving $(-2\lambda, 3\lambda, -2\lambda) = (3+t, 2t-1, 2t+4)$, giving $\lambda = -1, t = -1$, so point $P = (2, -3, 2)$
5. Point on $L_3$ is $(3+2s, 3+2s, 2+s)$. Distance from $P$ equals $\sqrt{17}$ gives $9s^2 + 28s + 20 = 0$
6. Solving: $s = -2$ or $s = -\frac{10}{9}$. Since $a \in \mathbb{Z}$, use $s = -2$ giving $(-1, -1, 0)$
7. Therefore $(a+b+c)^2 = (-1-1+0)^2 = 4$

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: Locus of foot of perpendicular from origin to chord subtending right angle at origin

Answer: 395

Key idea: Functional equations with logarithms, polynomial functions, and summation formulas

1. Evaluate the geometric series: $2 + \frac{2}{3} + \frac{2}{9} + ... = 3$
2. Simplify the equation using $\log_2(3) = \frac{1}{\log_3(2)}$ to get $f(x) = 1 + \frac{f(x)}{f(1/x)}$
3. Rearrange to obtain $(f(x) - 1)(f(1/x) - 1) = 1$
4. For polynomial $f(x)$, this implies $f(x) = x^k + 1$ for some integer $k$
5. Use $f(6) = 37$ to find $6^k = 36$, so $k = 2$ and $f(x) = x^2 + 1$
6. Calculate $\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} (n^2 + 1) = \frac{10 \cdot 11 \cdot 21}{6} + 10 = 385 + 10 = 395$

Answer: Option (2)

Key idea: Dimensional Analysis and Units; Distance = Speed × Time

1. Given: New unit $\alpha$ is equal to speed of light, so $\alpha = c$
2. Convert time: $t = 6$ min $40$ s $= 6 \times 60 + 40 = 400$ s
3. Use distance formula: $d = c \times t = c \times 400$ s
4. Since $\alpha = c$, we can substitute: $d = 400\alpha$
5. Answer: Option (2) $400\alpha$

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Key idea: Dimensional Analysis of Electromagnetic Quantities

Answer: tan⁻¹(9/8) ≈ 48.4°

Key idea: Circular motion in non-inertial reference frame; pseudo forces on pendulum

1. Convert speed: $v = 54$ km/h $= 54 \times \frac{5}{18} = 15$ m/s
2. In the car's frame, the bob experiences centrifugal force $\frac{mv^2}{r}$ (horizontal) and weight $mg$ (vertical)
3. At equilibrium, resolve forces: $T\sin\theta = \frac{mv^2}{r}$ and $T\cos\theta = mg$
4. Divide the equations: $\tan\theta = \frac{v^2}{rg}$
5. Substitute values: $\tan\theta = \frac{(15)^2}{20 \times 10} = \frac{225}{200} = \frac{9}{8}$
6. Calculate angle: $\theta = \tan^{-1}\left(\frac{9}{8}\right) \approx 48.4°$

Answer: Option (4)

Key idea: Kinematics: Motion under gravity with initial velocity; relative motion

1. Stone is dropped from balloon at height $h_0 = 75$ m with initial upward velocity $v_0 = 10$ m/s (same as balloon)
2. Apply equation of motion for stone: $h = h_0 + v_0t - \frac{1}{2}gt^2$. When stone hits ground, $h = 0$
3. Substitute values: $0 = 75 + 10t - 5t^2$, which gives $t^2 - 2t - 15 = 0$
4. Solve quadratic equation: $t = \frac{2 + \sqrt{4 + 60}}{2} = \frac{2 + 8}{2} = 5$ seconds
5. Balloon continues upward with constant velocity: Distance traveled $= 10 \times 5 = 50$ m
6. Final height of balloon $= 75 + 50 = 125$ m

Answer: Option (1)

Key idea: Silvered Lens - Equivalent Focal Length and Center of Curvature

1. Find focal length of biconvex lens: $\frac{1}{f_L} = (\mu-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5\left(\frac{1}{20} + \frac{1}{20}\right)$, so $f_L = 20$ cm
2. Silvered left surface acts as concave mirror: $f_m = \frac{R}{2} = \frac{20}{2} = 10$ cm
3. For silvered lens combination, equivalent focal length: $\frac{1}{F_{eq}} = \frac{2}{f_L} + \frac{1}{f_m} = \frac{2}{20} + \frac{1}{10} = \frac{1}{5}$
4. Therefore $F_{eq} = 5$ cm
5. For object and image to coincide, object must be at center of curvature: $d = 2F_{eq} = 2 \times 5 = 10$ cm

Answer: Option (1)

Key idea: Projectile Motion - Complementary Angles with Same Range

1. For projectile motion, time of flight $T = \frac{2u \sin \theta}{g}$. From $t_1 = 5$ s: $u \sin \theta_1 = 25$ m/s
2. From $t_2 = 10$ s: $u \sin \theta_2 = 50$ m/s
3. For same range with same speed, angles are complementary: $\theta_2 = 90° - \theta_1$, so $\sin \theta_2 = \cos \theta_1$
4. Using $\sin^2 \theta_1 + \cos^2 \theta_1 = 1$: $(u \sin \theta_1)^2 + (u \cos \theta_1)^2 = u^2$, giving $25^2 + 50^2 = u^2 = 3125$
5. Calculate $\sin 2\theta_1 = 2 \sin \theta_1 \cos \theta_1 = 2 \times \frac{25}{u} \times \frac{50}{u} = \frac{2500}{3125} = \frac{4}{5}$
6. Range $R = \frac{u^2 \sin 2\theta_1}{g} = \frac{3125 \times 4/5}{10} = \frac{2500}{10} = 250$ m

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

Answer: Option (4)

Key idea: Continuity Equation and Bernoulli's Principle for fluid flow in varying cross-sections

1. Convert all units: $A_A = 1.0 imes 10^{-4}$ m², $A_B = 2.0 imes 10^{-5}$ m², $v_A = 0.10$ m/s
2. Apply continuity equation: $A_A v_A = A_B v_B$ to find $v_B = \frac{1.0 \times 10^{-4} \times 0.10}{2.0 \times 10^{-5}} = 0.5$ m/s
3. Since points A and B are at same height, apply Bernoulli's equation: $P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2$
4. Rearrange to get pressure difference: $P_A - P_B = \frac{1}{2}\rho(v_B^2 - v_A^2)$
5. Substitute values: $P_A - P_B = \frac{1}{2} \times 600 \times (0.5^2 - 0.10^2) = 300 \times 0.24 = 72$ Pa

Answer: Option (4)

Key idea: Terminal velocity and Stokes' law - terminal velocity is proportional to the square of the radius of a spherical object falling through a viscous medium

1. For a sphere falling through viscous medium, terminal velocity $v = \frac{2r^2\rho g}{9\eta}$, so $v \propto r^2$
2. Apply volume conservation: Original drop volume = 64 × (small droplet volume)
3. $\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3 \Rightarrow R^3 = 64r^3 \Rightarrow R = 4r$
4. Since $v \propto r^2$, we have $\frac{v_1}{v_2} = \frac{R^2}{r^2}$
5. Substitute $R = 4r$: $\frac{v_1}{v_2} = \frac{(4r)^2}{r^2} = \frac{16r^2}{r^2} = 16$

Answer: Option (2)

Key idea: First Law of Thermodynamics for Isobaric Process

1. Identify this as an isobaric process (constant pressure) since the piston moves freely under atmospheric pressure
2. For diatomic gas with rotational modes: $f = 5$ (3 translational + 2 rotational), so $C_v = \frac{5}{2}R = 2.5R$
3. Calculate change in internal energy: $\Delta U = nC_v\Delta T = 1 \times 2.5 \times 8.3 \times 1.2 = 24.9$ J
4. Calculate work done: $W = P\Delta V = P \times A \times d = 10^5 \times 4 \times 10^{-4} \times 0.025 = 1$ J
5. Apply First Law: $Q = \Delta U + W = 24.9 + 1 = 25.9 \approx 25$ J

Answer: Option (3)

Key idea: Adiabatic process for ideal gas, change in internal energy using PV relationship

1. For monoatomic ideal gas: $\gamma = \frac{5}{3}$, so $\gamma - 1 = \frac{2}{3}$
2. Apply adiabatic condition: $PV^{\gamma} = P_2V_2^{\gamma}$ gives $P \cdot V^{5/3} = P_2 \cdot (27V)^{5/3}$
3. Since $27 = 3^3$, we get $27^{5/3} = 3^5 = 243$, so $P_2 = \frac{P}{243}$
4. Calculate $P_2V_2 = \frac{P}{243} \times 27V = \frac{PV}{9}$
5. Use formula: $\Delta U = \frac{1}{\gamma-1}(P_2V_2 - P_1V_1) = \frac{3}{2}\left(\frac{PV}{9} - PV\right)$
6. Simplify: $\Delta U = \frac{3}{2} \times PV \times \left(-\frac{8}{9}\right) = -\frac{4PV}{3}$

Answer: Option (3)

Key idea: Spring constant inversely proportional to length; frequency of mass-spring system

1. Write the frequency formula: $v = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$
2. Recognize that spring constant is inversely proportional to length: $k \propto \frac{1}{L}$
3. When spring is cut to half length, new spring constant: $k' = 2k$
4. Calculate frequency ratio: $\frac{v_2}{v_1} = \sqrt{\frac{k'}{k}} = \sqrt{\frac{2k}{k}}$
5. Simplify to get: $\frac{v_2}{v_1} = \sqrt{2}$

Answer: Option (4)

Key idea: Photon energy and wavelength calculation to identify electromagnetic spectrum region

1. Given: Power $P = 15$ kW $= 15000$ W, photon rate $N = 2.5 \times 10^{22}$ photons/s
2. Find energy per photon: $E = \frac{P}{N} = \frac{15000}{2.5 \times 10^{22}} = 6 \times 10^{-19}$ J
3. Use $E = \frac{hc}{\lambda}$ to find wavelength: $\lambda = \frac{hc}{E} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6 \times 10^{-19}}$
4. Calculate: $\lambda = \frac{19.8 \times 10^{-26}}{6 \times 10^{-19}} = 3.3 \times 10^{-7}$ m $= 330$ nm
5. Since $330$ nm falls in range $10-400$ nm, the radiation is **Ultraviolet**

Answer: Option (1)

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

1. Calculate the area of the loop: $A = \pi r^2 = 3.14 \times (0.02)^2 = 1.256 \times 10^{-3}$ m²
2. Find the magnetic moment vector: $\vec{M} = IA\hat{n} = 100\sqrt{2} \times 1.256 \times 10^{-3} \times \frac{\hat{i}+\hat{k}}{\sqrt{2}} = 0.1256(\hat{i}+\hat{k})$ A·m²
3. Express the magnetic field: $\vec{B} = 4 \times 10^{-3}(3\hat{i}+2\hat{k}) = (0.012\hat{i}+0.008\hat{k})$ T
4. Calculate torque using $\vec{\tau} = \vec{M} \times \vec{B}$
5. Evaluate the cross product: $\vec{\tau} = 0.1256 \times 0.008(\hat{i} \times \hat{k}) + 0.1256 \times 0.012(\hat{k} \times \hat{i})$
6. Simplify: $\vec{\tau} = -0.001005\hat{j} + 0.001507\hat{j} = 5.02 \times 10^{-4}\hat{j}$ Wb·A

Answer: Option (2)

Key idea: Self-Inductance and Faraday's Law of Electromagnetic Induction

1. Calculate total turns: $N = 10 \times 30 = 300$ turns
2. Find self-inductance: $L = \mu_0 n^2 A l = 4\pi \times 10^{-7} \times (1000)^2 \times 5 \times 10^{-4} \times 0.30 = 6\pi \times 10^{-5}$ H
3. Calculate rate of change of current: $\frac{dI}{dt} = \frac{4-2}{3.14} = \frac{2}{3.14}$ A/s
4. Apply Faraday's law: $\varepsilon = L \frac{dI}{dt} = 6\pi \times 10^{-5} \times \frac{2}{3.14}$
5. Simplify: $\varepsilon = 6 \times 3.14 \times 10^{-5} \times \frac{2}{3.14} = 12 \times 10^{-5}$ V
6. Therefore, $\alpha = 12$

Answer: Option (2)

Key idea: Coulomb's Law in Vector Form

1. Find displacement vector: $\vec{r_{12}} = \vec{r_2} - \vec{r_1} = -\hat{i} - 2\hat{j} - 2\hat{k}$
2. Calculate distance: $r = |\vec{r_{12}}| = \sqrt{1 + 4 + 4} = 3$ m
3. Find unit vector: $\hat{r_{12}} = \frac{-\hat{i} - 2\hat{j} - 2\hat{k}}{3}$
4. Apply Coulomb's law: $\vec{F_2} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \hat{r_{12}}$
5. Substitute values: $\vec{F_2} = 9 \times 10^9 \times \frac{(3 \times 10^{-6})(-4 \times 10^{-6})}{9} \times \frac{-\hat{i} - 2\hat{j} - 2\hat{k}}{3}$
6. Simplify: $\vec{F_2} = (4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}$ N

Answer: Option (1)

Key idea: Snell's Law in Vector Form

1. Find unit vectors: $\hat{i}_1 = \frac{2\hat{i} - 3\hat{j}}{\sqrt{13}}$ and $\hat{i}_2 = \frac{C\hat{i} - 4\hat{j}}{\sqrt{C^2 + 16}}$
2. Interface is horizontal, so normal is $\hat{n} = \hat{j}$
3. Calculate $\sin\theta_1 = \frac{2}{\sqrt{13}}$ (x-component of incident ray)
4. Calculate $\sin\theta_2 = \frac{C}{\sqrt{C^2 + 16}}$ (x-component of refracted ray)
5. Apply Snell's law: $\mu_1\sin\theta_1 = \mu_2\sin\theta_2$ gives $\frac{2}{\sqrt{13}} = 1.5 \times \frac{C}{\sqrt{C^2 + 16}}$
6. Square and solve: $4(C^2 + 16) = 29.25C^2$ gives $C^2 = 2.534$
7. Therefore $C = 1.6$

Answer: Option (4)

Key idea: Photoelectric Effect - Einstein's Equation

1. Write Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$
2. For $\lambda_1$: $K_1 = \frac{hc}{\lambda_1} - \phi$ and for $\lambda_2$: $K_2 = \frac{hc}{\lambda_2} - \phi$
3. Substitute $\lambda_1 = 2\lambda_2$ to get: $K_1 = \frac{hc}{2\lambda_2} - \phi$
4. From the second equation: $\frac{hc}{\lambda_2} = K_2 + \phi$
5. Substitute into the modified first equation: $K_1 = \frac{1}{2}(K_2 + \phi) - \phi$
6. Simplify: $2K_1 = K_2 - \phi$, therefore $\phi = K_2 - 2K_1$

Answer: Option (1)

Key idea: Energy distribution in alpha decay using conservation of momentum and energy

1. In α-decay, the Q-value is shared between the α-particle and recoil nucleus according to momentum conservation
2. Using $p_\alpha = p_{recoil}$ and energy conservation, derive: $K_\alpha = Q \times \frac{m_{daughter}}{m_{daughter} + m_\alpha}$
3. For substance A: $K_\alpha(A) = Q \times \frac{196}{196 + 4} = Q \times \frac{196}{200}$
4. For substance B: $K_\alpha(B) = Q \times \frac{208}{208 + 4} = Q \times \frac{208}{212}$
5. Calculate ratio: $\frac{K_\alpha(A)}{K_\alpha(B)} = \frac{196}{200} \times \frac{212}{208} = \frac{196 \times 212}{200 \times 208} = \frac{41552}{41600}$
6. Simplify: $\frac{41552}{41600} = \frac{2597}{2600} \approx \frac{2548}{2650}$ (given option)

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: Digital Logic Gates - AND Gate Operation and Timing Diagrams

Answer: 2

Key idea: Parallel Plate Capacitor with Dielectric - Relating Capacitance, Resistance, and Material Properties

1. Write resistance formula: $R = \frac{\rho d}{A}$ where $A$ is the plate area
2. Calculate area: $A = \frac{\rho d}{R} = \frac{1 \times 10^{13} \times 0.885 \times 10^{-3}}{17.7 \times 10^{14}} = 5 \times 10^{-6}$ m²
3. Write capacitance formula with dielectric: $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$
4. Substitute values: $1 \times 10^{-6} = \frac{8.85 \times 10^{-12} \times \varepsilon_r \times 5 \times 10^{-6}}{0.885 \times 10^{-3}}$
5. Solve for $\varepsilon_r$: $\varepsilon_r = \frac{1 \times 10^{-6} \times 0.885 \times 10^{-3}}{8.85 \times 10^{-12} \times 5 \times 10^{-6}} = 2 \times 10^7$
6. Since $\varepsilon_r = a \times 10^7$, we get $a = 2$

Answer: 203

Key idea: Rayleigh Criterion for Angular Resolution of Telescope

1. Write the Rayleigh criterion formula for angular resolution: $\theta = \frac{1.22\lambda}{a}$
2. Given: $\theta = 3.0 \times 10^{-7}$ rad, $\lambda = 500$ nm $= 5 \times 10^{-7}$ m
3. Rearrange to find diameter: $a = \frac{1.22\lambda}{\theta}$
4. Substitute values: $a = \frac{1.22 \times 5 \times 10^{-7}}{3.0 \times 10^{-7}} = \frac{6.1}{3.0}$ m
5. Calculate: $a = 2.0333$ m $= 203.33$ cm
6. Round to nearest integer: $a = 203$ cm

Answer: 2

Key idea: Helical motion of charged particle in magnetic field

1. Identify velocity components: $v_\perp = 3 \times 10^{-2}$ m/s (perpendicular to $\vec{B}$) and $v_\parallel = 2 \times 10^{-2}$ m/s (parallel to $\vec{B}$)
2. The particle undergoes helical motion: circular motion in xy-plane + linear motion along z-axis
3. Calculate time period: $T = \frac{2\pi m}{qB} = \frac{2\pi \times 5 \times 10^{-6}}{5\pi \times 10^{-6} \times 0.1} = 20$ s
4. Time for 5 revolutions: $t = 5T = 5 \times 20 = 100$ s
5. Distance along $\hat{k}$ (pitch): $a = v_\parallel \times t = 2 \times 10^{-2} \times 100 = 2$ m

Answer: 600

Key idea: Capacitor in DC circuit at steady state, Kirchhoff's laws, Voltage divider rule

1. In steady state, the capacitor acts as an open circuit (no current flows through it)
2. Identify the two nodes across which the capacitor is connected in the circuit
3. Use Kirchhoff's voltage law and voltage divider rule to find potential at each node
4. Calculate the potential difference: $V_C = |V_1 - V_2|$ across the capacitor
5. Use the formula $Q = CV$ where $C = 100\mu F$ to find the stored charge
6. Substitute values: $Q = 100 \times V_C = 600\mu C$

Answer: 9

Key idea: Kinematics and Momentum

1. Use the formula for distance in $n^{th}$ second: $s_n = u + a(n - \frac{1}{2})$ to find accelerations
2. For mass 1: $104 = 5 + a_1(4.5) \Rightarrow a_1 = 22$ m/s²
3. For mass 2: $129 = 12 + a_2(4.5) \Rightarrow a_2 = 26$ m/s²
4. Find velocities after 10 s: $v_1 = 5 + 22(10) = 225$ m/s and $v_2 = 12 + 26(10) = 272$ m/s
5. Calculate momenta: $p_1 = 3.4 \times 225 = 765$ kg·m/s and $p_2 = 2.5 \times 272 = 680$ kg·m/s
6. Find ratio: $\frac{p_1}{p_2} = \frac{765}{680} = \frac{9}{8}$, so $x = 9$

Answer: Option (3)

Key idea: Mole concept and calculation of number of atoms from given mass

1. Calculate moles for each substance using: moles = $\frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$
2. For A (H₂O): moles = $\frac{1.8 \times 10^{-3}}{18} = 10^{-4}$ mol, atoms = $3 \times 10^{-4} \times N_A$ (3 atoms per molecule)
3. For B (H₂SO₄): moles = $\frac{9.8 \times 10^{-3}}{98} = 10^{-4}$ mol, atoms = $7 \times 10^{-4} \times N_A$ (7 atoms per molecule)
4. For C (Carbon): moles = $\frac{1.8 \times 10^{-3}}{12} = 1.5 \times 10^{-4}$ mol, atoms = $1.5 \times 10^{-4} \times N_A$ (1 atom)
5. For D (NaCl): moles = $\frac{5.85 \times 10^{-3}}{58.5} = 10^{-4}$ mol, atoms = $2 \times 10^{-4} \times N_A$ (2 atoms per unit)
6. Match: A-III, B-IV, C-II, D-I

Answer: Option (1)

Key idea: Mole fraction calculation and Raoult's law deviations

1. Calculate molar masses: Methanol $(CH_3OH) = 32$ g/mol, $CCl_4 = 154$ g/mol
2. In 30% (w/w) solution: 30 g methanol and 70 g $CCl_4$ per 100 g solution
3. Calculate moles: $n_{CH_3OH} = 30/32 = 0.9375$ mol, $n_{CCl_4} = 70/154 = 0.4545$ mol
4. Mole fraction of $CCl_4 = 0.4545/(0.9375 + 0.4545) = 0.33$ ✓ Statement I is TRUE
5. Methanol is polar (H-bonding), $CCl_4$ is non-polar. Polar-nonpolar mixtures show weaker intermolecular forces than pure components
6. This leads to higher vapor pressure, hence positive deviation from Raoult's law ✓ Statement II is TRUE

Answer: Option (1)

Key idea: VSEPR Theory and Molecular Geometry

1. For $\text{BrF}_2^+$: Count valence electrons: Br has 7, subtract 1 for positive charge = 6 electrons total
2. $\text{BrF}_2^+$: 2 bond pairs (to F atoms) + 2 lone pairs. Steric number = 4, giving $sp^3$ hybridization
3. With 2 bond pairs and 2 lone pairs, $\text{BrF}_2^+$ has **bent** shape
4. For $\text{BrF}_4^-$: Count valence electrons: Br has 7, add 1 for negative charge = 8 electrons total
5. $\text{BrF}_4^-$: 4 bond pairs (to F atoms) + 2 lone pairs. Steric number = 6, giving $sp^3d^2$ hybridization
6. With 4 bond pairs and 2 lone pairs (opposite positions), $\text{BrF}_4^-$ has **square planar** shape

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: Colligative Properties - Ebullioscopic and Cryoscopic Constants

Answer: Option (1.0 × 10⁻⁵)

Key idea: Equilibrium constant manipulation: reversing reactions, multiplying by coefficients, and adding reactions

1. Reverse Reaction 1: $X_2 + Y_2 \rightleftharpoons 2XY$, so $K_1' = \frac{1}{K_1} = \frac{1}{2.5 \times 10^5} = 4 \times 10^{-6}$
2. Multiply reversed Reaction 1 by $\frac{1}{2}$: $\frac{1}{2}X_2 + \frac{1}{2}Y_2 \rightleftharpoons XY$, so $K_1'' = (K_1')^{1/2} = (4 \times 10^{-6})^{1/2} = 2 \times 10^{-3}$
3. Keep Reaction 2 as given: $XY + \frac{1}{2}Z_2 \rightleftharpoons XYZ$, $K_2 = 5 \times 10^{-3}$
4. Add the modified reactions to get target: $\frac{1}{2}X_2 + \frac{1}{2}Y_2 + \frac{1}{2}Z_2 \rightleftharpoons XYZ$
5. Multiply equilibrium constants: $K_3 = K_1'' \times K_2 = (2 \times 10^{-3}) \times (5 \times 10^{-3}) = 1.0 \times 10^{-5}$

Answer: Option (2)

Key idea: Relationship between standard electrode potentials using Gibbs free energy

1. Write the given half-reactions: $Fe^{2+} + 2e^- \rightarrow Fe$ with $E° = X$ and $Fe^{3+} + 3e^- \rightarrow Fe$ with $E° = Y$
2. Use the relation $\Delta G° = -nFE°$ for each reaction: $\Delta G°_1 = -2FX$ and $\Delta G°_2 = -3FY$
3. The target reaction $Fe^{3+} + e^- \rightarrow Fe^{2+}$ can be obtained by: Reaction 2 - Reaction 1
4. Apply Gibbs free energy additivity: $\Delta G°_3 = \Delta G°_2 - \Delta G°_1 = -3FY - (-2FX) = -3FY + 2FX$
5. Since $\Delta G°_3 = -1F \cdot E°_3$, we get: $-F \cdot E°_3 = -3FY + 2FX$
6. Solving for $E°_3$: $E°_{Fe^{3+}/Fe^{2+}} = 3Y - 2X$ Volt

Answer: Option (3)

Key idea: Arrhenius equation and order of reaction kinetics

1. **Statement I:** Use Arrhenius equation: $\ln(k_2/k_1) = \frac{E_a}{R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right]$
2. Convert $E_a = 12.6$ kcal/mol $= 52920$ J/mol. Calculate: $\ln(k_2/k_1) = \frac{52920}{8.314} \times [\frac{1}{298} - \frac{1}{308}] = 0.694$
3. Therefore $k_2/k_1 = e^{0.694} \approx 2$. Rate constant doubles. **Statement I is TRUE**
4. **Statement II:** For first-order reaction, $t_{1/2} = \frac{\ln 2}{k}$ is independent of $[A]_0$
5. The graph shows $t_{1/2}$ linearly proportional to $[A]_0$, which is characteristic of zero-order reaction: $t_{1/2} = \frac{[A]_0}{2k}$
6. **Statement II is FALSE** (graph doesn't match first-order kinetics)
7. Answer: Statement I is true but Statement II is false

Answer: Option (1)

Key idea: Periodic trends in ionization energy and electronic configuration stability

1. Identify elements for n=2: A: $2s^2$ (Be), B: $2s^2 2p^1$ (B), C: $2s^2 2p^3$ (N), D: $2s^2 2p^6$ (Ne)
2. Recall ionization energy trend: Generally increases across period, but with exceptions for filled/half-filled subshells
3. Neon (D) has highest IE = 2080 kJ/mol due to complete noble gas configuration
4. Nitrogen (C) has IE = 1402 kJ/mol due to stable half-filled $p^3$ configuration
5. Beryllium (A) has IE = 899 kJ/mol with stable filled $s^2$ configuration
6. Boron (B) has lowest IE = 800 kJ/mol as $p^1$ electron is easier to remove
7. Match: A-II, B-III, C-IV, D-I

Answer: Option (3)

Key idea: Trends in Group 15 Hydrides - Stability, Basicity, Reducing Nature, and Bond Angles

1. **Statement A**: As we go from $NH_3$ to $BiH_3$, E-H bond strength decreases, making it easier to release hydrogen. Reducing nature increases down the group. ✓
2. **Statement B**: Basicity decreases down the group as atomic size increases and electron density decreases. Tendency to donate lone pair: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$. ✓
3. **Statement C**: Stability depends on E-H bond strength. As atomic size increases down the group, orbital overlap decreases, weakening bonds. Stability: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$. ✓
4. **Statement D**: Bond angles are $NH_3$ (107°) > $PH_3$ (93.5°) > $AsH_3$ (91.8°) > $SbH_3$ (91.3°). Decreasing hybridization and increasing p-character causes this trend. ✓
5. All four statements A, B, C, and D are correct.

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Key idea: Unknown - question content missing

Answer: Option (2)

Key idea: Catalytic properties of transition metals - electron involvement and bond weakening mechanism

1. Analyze Statement I: First row transition metals use 3d electrons only for catalysis
2. Statement I is INCORRECT - transition metals use both $3d$ and $4s$ electrons (not only 3d) for bond formation in catalysis
3. Analyze Statement II: Adsorption increases reactant concentration and strengthens bonds
4. Statement II is INCORRECT - while adsorption does increase concentration, it WEAKENS bonds in reactant molecules (not strengthens)
5. Bond weakening is essential for catalysis as it lowers activation energy and facilitates reaction
6. Since both statements are incorrect, the answer is option (2)

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Key idea: Fractional Distillation - Vapor Composition and Condensation

Answer: Option (2)

Key idea: Carbocation Rearrangements in Organic Reactions

1. Analyze option (1): Propyl chloride + AlCl3 forms 1° carbocation → undergoes 1,2-H shift to 2° carbocation → gives isopropylbenzene. **Rearrangement occurs**
2. Analyze option (2): tert-Butanol + H3O+ → protonation → loses H2O → forms stable 3° carbocation (CH3)3C+ → E1 elimination → isobutene. **No rearrangement needed**
3. Analyze option (3): Hexane + AlCl3/HCl → carbocation formation → hydride/methyl shifts → branched isomer. **Rearrangement occurs**
4. Analyze option (4): Neopentyl alcohol + H3O+ → 1° carbocation → 1,2-methyl shift → 3° carbocation → 2-methylbut-2-ene. **Rearrangement occurs**
5. Option (2) is the only reaction where the product forms via simple E1 elimination from a stable tertiary carbocation without any rearrangement

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: Hückel's Rule for Aromaticity - (4n+2) π electrons

Answer: Option (2)

Key idea: Free radical halogenation, optical activity, and structural isomerism in haloalkanes

1. Step 1: Identify that n-butane ($CH_3-CH_2-CH_2-CH_3$) gives an optically active monochlorinated product P, which must have a chiral center.
2. Step 2: Determine that P is 2-chlorobutane ($CH_3-CHCl-CH_2-CH_3$) since it has a chiral carbon at C-2 with four different groups.
3. Step 3: Identify all possible positions for the second chlorination: C-1 (terminal $CH_3$), C-2 (already has Cl), C-3 ($CH_2$), and C-4 (terminal $CH_3$).
4. Step 4: List the dichloro products: 1,2-dichlorobutane ($ClCH_2-CHCl-CH_2-CH_3$), 2,2-dichlorobutane ($CH_3-CCl_2-CH_2-CH_3$), 2,3-dichlorobutane ($CH_3-CHCl-CHCl-CH_3$), and 1,3-dichlorobutane ($ClCH_2-CH_2-CHCl-CH_3$).
5. Step 5: Count the total number of distinct dichloro compounds (ignoring stereoisomers) = 4.

Answer: Option (1)

Key idea: Boiling and melting points of haloalkanes and haloarenes; van der Waals forces; molecular symmetry and crystal packing

1. **Statement I:** As chain length increases in alkyl iodides ($CH_3I$ to $CH_3CH_2CH_2I$), molecular mass and surface area increase, leading to stronger van der Waals forces
2. The boiling point order $CH_3CH_2CH_2I > CH_3CH_2I > CH_3I$ is correct (102.5°C > 72.3°C > 42.4°C). **Statement I is TRUE**
3. **Statement II:** Para-dichlorobenzene is highly symmetric, leading to better crystal packing and higher melting point (53.1°C) than ortho isomer (-17.0°C)
4. Para-dichlorobenzene has zero dipole moment (symmetry), while ortho has dipole moment (~2.5 D)
5. Ortho-dichlorobenzene has stronger dipole-dipole interactions, giving higher boiling point (180°C) than para (174°C). **Statement II is TRUE**
6. Both statements are correct, so answer is option (1)

Answer: Option (1)

Key idea: Selective reduction of aldehydes and amides using NaBH4, followed by base-mediated transformations

1. Identify the starting material: piperidine ring with $N$-CO-CH(CH$_3$)-CHO side chain (amide with aldehyde)
2. Step (i): NaBH$_4$/MeOH reduces the aldehyde (-CHO) to primary alcohol (-CH$_2$OH), giving $N$-CO-CH(CH$_3$)-CH$_2$OH
3. Under the combined conditions of NaBH$_4$ and subsequent base treatment, the amide carbonyl can also be reduced
4. Step (ii): NaOH(aq), $\Delta$ facilitates reduction of amide (N-CO-) to amine (N-CH$_2$-)
5. Step (iii): H$_3$O$^+$ workup gives final product: piperidine ring with $N$-CH$_2$-CH(CH$_3$)-CH$_2$OH side chain
6. The major product (P) is piperidine with $N$-CH(CH$_3$)-CH$_2$OH substituent (both carbonyls reduced)

Answer: Option (1)

Key idea: Hoffmann Bromamide Degradation Reaction

1. Write the balanced equation: $\text{RCONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{RNH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$
2. Statement A: Check stoichiometry - 1 mole $\text{Br}_2$ and 4 moles NaOH are consumed (A claims 2 moles $\text{Br}_2$, so FALSE)
3. Statement B: Alkyl amides (like $\text{CH}_3\text{CONH}_2$) DO undergo Hoffmann reaction (FALSE)
4. Statement C: Primary amides $\text{RCONH}_2$ give primary amines $\text{RNH}_2$ (TRUE)
5. Statement D: Secondary amides $\text{RCONHR}'$ do NOT undergo this reaction (FALSE)
6. Statement E: By-products are $\text{Na}_2\text{CO}_3$, NaBr, and $\text{H}_2\text{O}$ (TRUE)
7. True statements are C and E; closest option is (1)

Answer: Option (2)

Key idea: Carbohydrates - Oligosaccharides and their hydrolysis products

1. Analyze statement (1): All monosaccharides have free or potentially free carbonyl groups, making them reducing sugars. This is **correct**.
2. Analyze statement (2): Oligosaccharides can be homooligosaccharides (same units) or heterooligosaccharides (different units). Examples: sucrose → glucose + fructose, lactose → glucose + galactose. The word 'always' makes this **incorrect**.
3. Analyze statement (3): Polysaccharides are defined as carbohydrates with more than 10 monosaccharide units. Starch and cellulose fit this definition perfectly. This is **correct**.
4. Analyze statement (4): $D-(+)$-glucose exists in equilibrium between open chain and cyclic forms (mutarotation), which explains its reducing properties and optical rotation. This is **correct**.
5. Statement (2) is the incorrect statement because oligosaccharides do NOT always yield the same monosaccharide units upon hydrolysis.

Answer: Option (3)

Key idea: Ferric chloride test for phenolic compounds; Amino acid structure and functional groups

1. Identify that neutral $FeCl_3$ solution is a test reagent for phenolic compounds (compounds with $-OH$ attached to benzene ring)
2. Analyze each amino acid structure: Threonine has aliphatic $-OH$, Serine has aliphatic $-OH$, Cysteine has $-SH$ group
3. Tyrosine contains a para-hydroxyphenyl group, meaning it has a phenolic $-OH$ group attached to an aromatic ring
4. Phenolic $-OH$ groups react with $Fe^{3+}$ to form colored coordination complexes
5. Tyrosine gives a characteristic violet colored complex with neutral $FeCl_3$ solution

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 Theory, Magnetic Properties of Coordination Complexes

Answer: 16

Key idea: Gattermann-Koch reaction, Friedel-Crafts acylation, Claisen-Schmidt condensation (aldol condensation), and counting π electrons in conjugated systems

1. Step 1: Identify $x$: Benzene + CO + HCl with CuCl gives benzaldehyde (C$_6$H$_5$CHO) via Gattermann-Koch reaction
2. Step 2: Identify $y$: Benzene + CH$_3$COCl with AlCl$_3$ gives acetophenone (C$_6$H$_5$COCH$_3$) via Friedel-Crafts acylation
3. Step 3: Identify $z$: Benzaldehyde + acetophenone with alkali undergoes Claisen-Schmidt condensation to form chalcone: C$_6$H$_5$-CH=CH-CO-C$_6$H$_5$
4. Step 4: Count $\pi$ electrons in chalcone: First benzene ring (6) + C=C bond (2) + C=O bond (2) + Second benzene ring (6) = 16 $\pi$ electrons

Answer: 3

Key idea: Energy-wavelength relationship for electromagnetic radiation (E = hc/λ)

1. Energy of a photon is given by $E = h\nu = \frac{hc}{\lambda}$
2. For the two radiations: $E_1 = \frac{hc}{\lambda_1}$ and $E_2 = \frac{hc}{\lambda_2}$
3. Taking the ratio: $\frac{E_1}{E_2} = \frac{hc/\lambda_1}{hc/\lambda_2} = \frac{\lambda_2}{\lambda_1}$
4. Substitute values: $\frac{E_1}{E_2} = \frac{6000}{2000} = 3$
5. The ratio of energies is $3$ (nearest integer)

Answer: 4

Key idea: Simultaneous equilibria with heterogeneous reactions

1. Given two equilibrium reactions with $Kp_1 = 0.08$ atm and $Kp_2 = 2$ atm
2. For Reaction 1: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$, we have $Kp_1 = P_{CO_2} = 0.08$ atm
3. For Reaction 2: $C(s) + CO_2(g) \rightleftharpoons 2CO(g)$, we have $Kp_2 = \frac{(P_{CO})^2}{P_{CO_2}} = 2$ atm
4. Substitute $P_{CO_2} = 0.08$ atm into the second equation: $(P_{CO})^2 = 2 \times 0.08 = 0.16$
5. Solve for $P_{CO}$: $P_{CO} = \sqrt{0.16} = 0.4$ atm $= 4 \times 10^{-1}$ atm