JEE Main 4 April 2026 Shift 1: Questions and Solutions

Answer: Option (1)

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

1. For $\cos^{-1}\left(\frac{4x+2[x]}{3}\right)$ to be defined, we need $-1 \leq \frac{4x+2[x]}{3} \leq 1$, which gives $-3 \leq 4x+2[x] \leq 3$
2. Let $[x] = n$ (integer) and $x = n + \{x\}$ where $0 \leq \{x\} < 1$. Substituting: $-3 \leq 6n + 4\{x\} \leq 3$
3. For solutions to exist with $0 \leq \{x\} < 1$, we need $n \leq \frac{1}{2}$ and $n > -\frac{7}{6}$, so $n \in \{-1, 0\}$
4. Case 1: $n=0$ gives $0 \leq 4x \leq 3$ with $0 \leq x < 1$, yielding $x \in [0, \frac{3}{4}]$
5. Case 2: $n=-1$ gives $\frac{3}{4} \leq \{x\} < 1$ with $-1 \leq x < 0$, yielding $x \in [-\frac{1}{4}, 0)$
6. Domain is $[-\frac{1}{4}, \frac{3}{4}]$, so $\alpha = -\frac{1}{4}$, $\beta = \frac{3}{4}$
7. Therefore, $12(\alpha + \beta) = 12(\frac{1}{2}) = 6$

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Key idea: Unable to determine - question content not provided

Answer: Option (1)

Key idea: Locus of complex numbers and argument properties

1. Step 1: From $|z + 2| = |z - 2|$, the complex number $z$ is equidistant from $-2$ and $2$, so $z$ lies on the perpendicular bisector of these points, which is the imaginary axis. Thus $z = iy$ for some real $y$.
2. Step 2: Substitute $z = iy$ into the second condition: $\frac{z + 3}{z - i} = \frac{3 + iy}{iy - i} = \frac{3 + iy}{i(y - 1)}$
3. Step 3: Simplify by multiplying numerator and denominator by $-i$: $\frac{(3 + iy)(-i)}{i(y-1)(-i)} = \frac{y - 3i}{y - 1} = \frac{y}{y-1} - \frac{3i}{y-1}$
4. Step 4: For $\arg\left(\frac{z+3}{z-i}\right) = \frac{\pi}{4}$, we need $\tan\left(\frac{\pi}{4}\right) = 1 = \frac{\text{Im}}{\text{Re}} = \frac{-3/(y-1)}{y/(y-1)} = \frac{-3}{y}$
5. Step 5: Solving $\frac{-3}{y} = 1$ gives $y = -3$, so $z = -3i$
6. Step 6: Calculate $|z|^2 = |-3i|^2 = 0^2 + (-3)^2 = 9$

Answer: Option (2)

Key idea: Counting functions using inclusion-exclusion principle; Onto (surjective) functions

1. Calculate total functions from $\{1,2,3,4\}$ to $\{a,b,c\}$: Total = $3^4 = 81$
2. Use inclusion-exclusion to find non-onto functions. Functions missing at least one element in range.
3. Functions using only 2 elements: $\binom{3}{1} \cdot 2^4 = 3 \times 16 = 48$
4. Functions using only 1 element: $\binom{3}{2} \cdot 1^4 = 3 \times 1 = 3$
5. By inclusion-exclusion: Non-onto = $48 - 3 = 45$
6. Alternatively: Onto functions = $3^4 - 3 \cdot 2^4 + 3 \cdot 1^4 = 81 - 48 + 3 = 36$, so non-onto = $81 - 36 = 45$

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Answer: Option (4)

Key idea: Properties of adjoint matrices and determinants

1. Calculate $|A| = 1(0-3) - 1(-10-1) + 2(-6) = -3 + 11 - 12 = -4$
2. Use $(adj A)^{-1} = \frac{A}{|A|}$ to get $2(adj A)^{-1} = 2 \cdot \frac{A}{-4} = -\frac{1}{2}A$
3. Let $B = -\frac{1}{2}A$, then $|B| = (-\frac{1}{2})^3 \cdot (-4) = \frac{1}{2}$
4. Apply formula: $adj(adj(B)) = |B|^{n-2} \cdot B = \frac{1}{2} \cdot (-\frac{1}{2}A) = -\frac{1}{4}A$
5. Sum of elements of $A$ is $1+1+2-2+0+1+1+3+5 = 12$
6. Therefore, sum of elements = $-\frac{1}{4} \times 12 = -3$

Answer: Option (1)

Key idea: Arithmetic Progression - Sum formula and relationship between terms

1. Given: First term $a = \frac{10}{3}$, number of terms $n = 30$, and sum equals cube of last term
2. Express last term: $l = a + 29d = \frac{10}{3} + 29d$
3. Write sum formula: $S_{30} = \frac{n}{2}(a + l) = 15\left(\frac{10}{3} + \frac{10}{3} + 29d\right) = 100 + 435d$
4. Apply given condition: $S_{30} = l^3$, so $100 + 435d = \left(\frac{10}{3} + 29d\right)^3$
5. Test option (1): $d = \frac{5}{87}$ gives $l = \frac{10}{3} + \frac{145}{87} = \frac{435}{87} = 5$
6. Verify: $S_{30} = 100 + 435 \times \frac{5}{87} = 125$ and $l^3 = 5^3 = 125$ ✓

Answer: Option (4)

Key idea: Complementary Counting (Total arrangements minus restricted arrangements)

1. Calculate total arrangements of 7 people (4 boys + 3 girls): $7! = 5040$
2. For all girls together, treat 3 girls as one unit: we have 5 units (4 boys + 1 girl-unit)
3. Arrange these 5 units: $5! = 120$ ways
4. Arrange 3 girls within their unit: $3! = 6$ ways
5. Total arrangements with all girls together: $5! \times 3! = 120 \times 6 = 720$
6. Apply complementary counting: $7! - (5! \times 3!) = 5040 - 720 = 4320$

Answer: Option (2)

Key idea: Binomial Theorem, Hockey Stick Identity, Modular Arithmetic

1. Find coefficient of $x^3$ in $(1+x)^3 + ... + (1+x)^{99}$ using $\sum_{r=3}^{99} \binom{r}{3} = \binom{100}{4}$ (Hockey Stick Identity)
2. Find coefficient of $x^3$ in $(1+kx)^{100}$ which is $\binom{100}{3}k^3$
3. Total coefficient: $\binom{100}{4} + \binom{100}{3}k^3 = \left(43n + \frac{101}{4}\right)\binom{100}{3}$
4. Use $\binom{100}{4} = \frac{97}{4}\binom{100}{3}$ to get: $k^3 = 43n + 1$
5. Find smallest $k \in \mathbb{N}$: Test $k = 6$ gives $6^3 = 216 = 43(5) + 1$, so $p = 6, n = 5$
6. Therefore $p + n = 6 + 5 = 11$

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Answer: Option (1)

Key idea: Angle bisectors of two lines and distance ratios

1. Find point $A$ by solving $L_1: x = -3$ and $L_2: y = x$, giving $A = (-3, -3)$
2. Find point $B$ by solving $L_1: x = -3$ and $L_3: y = -3x$, giving $B = (-3, 9)$
3. Find the angle bisectors using $\frac{x - y}{\sqrt{2}} = \pm \frac{3x + y}{\sqrt{10}}$
4. Identify the obtuse angle bisector as $y(\sqrt{10} - \sqrt{2}) = x(\sqrt{10} + 3\sqrt{2})$ by checking slopes
5. Find point $C$ on $L_1$ where $x = -3$, giving $C = (-3, -6 - 3\sqrt{5})$
6. Calculate $BC^2 = (15 + 3\sqrt{5})^2 = 90(3 + \sqrt{5})$ and $AC^2 = (3 + 3\sqrt{5})^2 = 18(3 + \sqrt{5})$
7. Find the ratio: $\frac{BC^2}{AC^2} = \frac{90(3 + \sqrt{5})}{18(3 + \sqrt{5})} = 5$, so $BC^2:AC^2 = 5:1$

Answer: Option (497)

Key idea: Centroid, Circumcenter, and Coordinate Geometry

1. Find vertex $B$ using midpoint formula: $M = \frac{A+B}{2}$ gives $B = (9, -4)$
2. Find vertex $C$ using centroid formula: $G = \frac{A+B+C}{3}$ gives $C = (-1, 14)$
3. Set up equations using property that circumcenter is equidistant from all vertices
4. From $|OA|^2 = |OB|^2$: $4\alpha - 3\beta = 23$
5. From $|OA|^2 = |OC|^2$: $\alpha = 6\beta - 48$
6. Solve the system to get $\alpha = \frac{282}{21}$ and $\beta = \frac{215}{21}$
7. Calculate $21(\alpha + \beta) = 282 + 215 = 497$

Answer: Option (2)

Key idea: Chord properties and midpoint conditions in circles

1. Verify that point $P(1,2)$ lies on the circle $x^2 + y^2 + x - 3y = 0$
2. For a chord from $P(1,2)$ to point $(a,b)$ bisected by y-axis, the midpoint $\left(\frac{1+a}{2}, \frac{2+b}{2}\right)$ must have x-coordinate = 0, giving $a = -1$
3. Substitute $x = -1$ into the circle equation: $1 + y^2 - 1 - 3y = 0$, which gives $y^2 - 3y = 0$
4. Solve to get $y = 0$ or $y = 3$, so the two points are $R(-1, 0)$ and $S(-1, 3)$
5. Find the midpoint of $RS$: $(\alpha, \beta) = \left(-1, \frac{3}{2}\right)$
6. Calculate $6(\alpha + \beta) = 6\left(-1 + \frac{3}{2}\right) = 6 \cdot \frac{1}{2} = 3$

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Answer: Option (2)

Key idea: Distance between point and line along a given direction; Intersection of two lines in 3D space

1. Verify that point $P(-2, -8, 6)$ lies on line $L_2: \frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ by checking all ratios equal 3
2. Write parametric equations: $L_1: (1+s, 1+2s, -s)$ and $L_2: (-5+t, -5-t, 2t)$
3. Find intersection of $L_1$ and $L_2$ by equating coordinates: $t-s=6$, $-t-2s=6$, $2t+s=0$
4. Solve the system to get $t=2$ and $s=-4$, giving intersection point $Q(-3, -7, 4)$
5. Calculate distance squared: $d^2 = ((-2)-(-3))^2 + ((-8)-(-7))^2 + (6-4)^2 = 1 + 1 + 4 = 6$

Answer: Option (3)

Key idea: Differentiation of inverse trigonometric functions and simplification using trigonometric identities

1. Simplify first term: $\tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) = \tan^{-1}\left(\frac{3}{4}\right) - x$ using $\tan^{-1}\left(\frac{a-b}{1+ab}\right) = \tan^{-1}(a) - \tan^{-1}(b)$
2. Simplify second term: $2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right) = \sin^{-1}(x)$ using substitution $x = \sin\theta$ and half-angle formula
3. Combine: $y = \tan^{-1}\left(\frac{3}{4}\right) - x + \sin^{-1}(x)$
4. Differentiate: $\frac{dy}{dx} = -1 + \frac{1}{\sqrt{1-x^2}}$
5. Substitute $x = \frac{\sqrt{3}}{2}$: $\frac{dy}{dx} = -1 + \frac{1}{\sqrt{1-\frac{3}{4}}} = -1 + \frac{1}{\frac{1}{2}} = -1 + 2 = 1$

Answer: Option (3)

Key idea: Differential Equations with Polynomial Functions

1. Step 1: Compare degrees: Since $f(x) = f'(x)f''(x)$, we have $n = (n-1) + (n-2)$, giving $n = 3$. So $f(x) = ax^3 + bx^2 + cx$ (using $f(0) = 0$).
2. Step 2: Compute derivatives: $f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$.
3. Step 3: Expand and compare coefficients in $ax^3 + bx^2 + cx = (3ax^2 + 2bx + c)(6ax + 2b)$. From $x^3$ coefficient: $a = 18a^2$, so $a = \frac{1}{18}$. From constant term: $bc = 0$. Solving all equations gives $b = c = 0$.
4. Step 4: Therefore $f(x) = \frac{1}{18}x^3$, so $f'(x) = \frac{1}{6}x^2$ and $f''(x) = \frac{1}{3}x$.
5. Step 5: Calculate: $f'(2) = \frac{2}{3}$, $f''(2) = \frac{2}{3}$, and $\int_0^2 f(x)dx = \frac{1}{72} \cdot 16 = \frac{2}{9}$.
6. Step 6: Sum: $\frac{2}{3} + \frac{2}{3} + \frac{2}{9} = \frac{14}{9}$.
7. Step 7: Final answer: $36 \cdot \frac{14}{9} = 56$.

Answer: Option (2)

Key idea: Area bounded by curves involving absolute values and trigonometric functions

1. Step 1: Identify the boundaries: $y = \pi - |x|$, $y = |x \sin x|$, and $y = 0$. Note that $|x \sin x| = x \sin x$ on $[-\pi, \pi]$ and both functions are even.
2. Step 2: Use symmetry about the y-axis. Calculate area for $x \geq 0$ and double it.
3. Step 3: Find intersection of $\pi - x = x \sin x$ for $x \in [0, \pi]$. This occurs at $x = \frac{\pi}{2}$.
4. Step 4: For $x \in [0, \frac{\pi}{2}]$, the lower curve is $y = x \sin x$. For $x \in [\frac{\pi}{2}, \pi]$, the lower curve is $y = \pi - x$.
5. Step 5: Calculate $\int_0^{\pi/2} x \sin x \, dx$ using integration by parts: $[-x \cos x + \sin x]_0^{\pi/2} = 1$.
6. Step 6: Calculate $\int_{\pi/2}^{\pi} (\pi - x) \, dx = [\pi x - \frac{x^2}{2}]_{\pi/2}^{\pi} = \frac{\pi^2}{8}$.
7. Step 7: Total area = $2(1 + \frac{\pi^2}{8}) = 2 + \frac{\pi^2}{4}$.

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Key idea: Unknown - question text not provided

Answer: Option (3)

Key idea: Separable Differential Equations with Trigonometric Integration

1. Separate variables: $\frac{dy}{1-y+y^2} = (1+x+x^2)dx$
2. Integrate RHS: $\int(1+x+x^2)dx = x + \frac{x^2}{2} + \frac{x^3}{3} + C$
3. For LHS, complete the square: $1-y+y^2 = (y-\frac{1}{2})^2 + \frac{3}{4}$, then integrate to get $\frac{2}{\sqrt{3}}\arctan\left(\frac{2y-1}{\sqrt{3}}\right)$
4. Apply initial condition $y(0) = \frac{1}{2}$ to find $C = 0$
5. At $x=1$: $\frac{2}{\sqrt{3}}\arctan\left(\frac{2y(1)-1}{\sqrt{3}}\right) = \frac{11}{6}$
6. Solve: $\arctan\left(\frac{2y(1)-1}{\sqrt{3}}\right) = \frac{11\sqrt{3}}{12}$
7. Therefore: $2y(1)-1 = \sqrt{3}\tan\left(\frac{11\sqrt{3}}{12}\right)$

Answer: 9

Key idea: Conditional Probability with Overlapping Events

1. Identify overlapping tosses: First 6 tosses are positions 1-6, last 5 tosses are positions 4-8. Positions 4, 5, 6 overlap.
2. Set up equations: Let $y$ = heads in positions 4-6. Then heads in positions 1-3 is $x = 4-y$ and heads in positions 7-8 is $z = 3-y$.
3. Find valid range: From constraints $x, z \geq 0$ and $x \leq 3, y \leq 3, z \leq 2$, we get $1 \leq y \leq 3$.
4. Count favorable outcomes: For $y=1$: $C(3,3) \cdot C(3,1) \cdot C(2,2) = 3$; For $y=2$: $C(3,2) \cdot C(3,2) \cdot C(2,1) = 18$; For $y=3$: $C(3,1) \cdot C(3,3) \cdot C(2,0) = 3$. Total = 24.
5. Calculate probability: $p = \frac{24}{2^8} = \frac{24}{256} = \frac{3}{32}$
6. Find final answer: $96p = 96 \times \frac{3}{32} = 3 \times 3 = 9$

Answer: 3

Key idea: Intersection of parabola and ellipse on their latus rectums

1. Parabola $y^2 = 4ex$ has latus rectum at $x = e$ with points $(e, \pm 2e)$
2. Ellipse has latus rectum at $x = ae$ with semi-latus rectum $\frac{b^2}{a}$
3. For intersection on latus rectums: $x = e = ae$, giving $a = 1$
4. At intersection, $y$-coordinates match: $2e = 1 - e^2$ (using $b^2 = 1 - e^2$ for $a=1$)
5. Solving $e^2 + 2e - 1 = 0$ gives $e = \sqrt{2} - 1$
6. Calculate $e^2 = (\sqrt{2} - 1)^2 = 3 - 2\sqrt{2}$
7. Therefore $e^2 + 2\sqrt{2} = 3 - 2\sqrt{2} + 2\sqrt{2} = 3$

Answer: 2

Key idea: Telescoping series using trigonometric identities

1. Use the identity: $\frac{\sin\theta}{\cos 3\theta} = \frac{\tan 3\theta - \tan\theta}{2}$
2. Apply to first term: $\frac{\sin 3°}{\cos 9°} = \frac{\tan 9° - \tan 3°}{2}$
3. Apply to second term: $\frac{\sin 9°}{\cos 27°} = \frac{\tan 27° - \tan 9°}{2}$
4. Apply to third term: $\frac{\sin 27°}{\cos 81°} = \frac{\tan 81° - \tan 27°}{2}$
5. Sum the terms (telescoping): $A = \frac{\tan 81° - \tan 3°}{2} = \frac{B}{2}$
6. Therefore: $\frac{B}{A} = \frac{B}{B/2} = 2$

Answer: 1

Key idea: Vector magnitude, trigonometric identities, and summation of trigonometric series

1. Calculate $|\vec{a_k}|^2 = \tan^2 \theta_k + 1 = \sec^2 \theta_k$
2. Calculate $|\vec{b_k}|^2 = 1 + \cot^2 \theta_k = \csc^2 \theta_k$
3. The ratio becomes $\frac{\sum_{k=1}^{n} \sec^2 \theta_k}{\sum_{k=1}^{n} \csc^2 \theta_k}$
4. Note that $\theta_k = \frac{(2^k - 1)\pi}{2^{n+1}}$ gives angles with special symmetry properties
5. Using the symmetry of the angles around $\pi/4$ and properties of the specific sequence, the sums are equal
6. Therefore, the ratio equals $1$

Answer: Option (4)

Key idea: Least Count of Screw Gauge

1. Given: 5 complete rotations move the screw linearly by 2.5 mm, and circular scale has 100 divisions
2. Calculate the pitch (linear distance per rotation): Pitch = $\frac{2.5 \text{ mm}}{5} = 0.5$ mm
3. Apply the least count formula: LC = $\frac{\text{Pitch}}{\text{Number of divisions}}$
4. Substitute values: LC = $\frac{0.5}{100} = 0.005$ mm
5. Express in scientific notation: LC = $5 \times 10^{-3}$ mm

Answer: Option (2)

Key idea: Bulk Modulus and Volume Change

1. Write the bulk modulus formula: $B = -\frac{\Delta P}{\Delta V/V}$
2. Rearrange to find fractional volume change: $\frac{\Delta V}{V} = -\frac{\Delta P}{B}$
3. Substitute given values: $\frac{\Delta V}{V} = \frac{6.3 \times 10^7}{2.1 \times 10^9}$
4. Calculate: $\frac{\Delta V}{V} = 3 \times 10^{-2} = 0.03$
5. Convert to percentage: $0.03 \times 100 = 3\%$

Answer: Option (3)

Key idea: Motion on inclined plane with friction; relating time of descent to coefficient of kinetic friction

1. For smooth plane: acceleration $a_s = g\sin 45° = \frac{g}{\sqrt{2}}$, and using $s = \frac{1}{2}a_s t_s^2$
2. For rough plane: acceleration $a_r = g\sin 45° - \mu_k g\cos 45° = \frac{g}{\sqrt{2}}(1 - \mu_k)$
3. Using $s = \frac{1}{2}a_r t_r^2$ for the same distance $s$
4. Given $t_r = 1.5 t_s$, so $t_r^2 = 2.25 t_s^2$
5. Since $s$ is same: $\frac{t_r^2}{t_s^2} = \frac{a_s}{a_r}$, giving $2.25 = \frac{1}{1-\mu_k}$
6. Solving: $1 - \mu_k = \frac{4}{9}$, therefore $\mu_k = \frac{5}{9}$

Answer: Option (3)

Key idea: Nuclear Binding Energy and Mass-Energy Equivalence

1. Write the reaction: $2(A=3) + (A=4) \to (A=10)$
2. Calculate total binding energies: BE = (mass number) × (BE per nucleon)
3. Initial BE: $2(3 \times 5.6) + (4 \times 7.4) = 33.6 + 29.6 = 63.2$ MeV
4. Final BE: $10 \times 6.1 = 61.0$ MeV
5. Q-value: $Q = BE_{final} - BE_{initial} = 61.0 - 63.2 = -2.2$ MeV
6. $\Delta Mc^2 = |Q| = 2.2$ MeV (energy absorbed in endothermic process)

Answer: Option (2)

Key idea: Moment of Inertia and Volume Conservation

1. Use volume conservation with constant density: Since smaller sphere has mass $M/8$ and original has mass $M$, we get $\frac{M/8}{(4/3)\pi r^3} = \frac{M}{(4/3)\pi R^3}$, giving $r = R/2$
2. Calculate $I_1$ for smaller sphere (mass $M/8$, radius $r = R/2$): $I_1 = \frac{2}{5} \times \frac{M}{8} \times \left(\frac{R}{2}\right)^2 = \frac{MR^2}{80}$
3. Calculate $I_2$ for disc (mass $7M/8$, radius $2R$) about diameter: $I_2 = \frac{1}{4} \times \frac{7M}{8} \times (2R)^2 = \frac{7MR^2}{8}$
4. Find the ratio: $\frac{I_2}{I_1} = \frac{7MR^2/8}{MR^2/80} = \frac{7 \times 80}{8} = 70$

Answer: Option (2)

Key idea: Range of projectile motion and trigonometric ratios

1. Write the range formula: $R = \frac{u^2 \sin 2\theta}{g}$
2. For first projectile at $15°$: $R_1 = \frac{u^2 \sin 30°}{g} = \frac{u^2}{2g}$
3. For second projectile at $30°$: $R_2 = \frac{u^2 \sin 60°}{g} = \frac{u^2\sqrt{3}}{2g}$
4. Find the ratio: $\frac{R_1}{R_2} = \frac{u^2/(2g)}{u^2\sqrt{3}/(2g)} = \frac{1}{\sqrt{3}}$
5. Since $R_1:R_2 = 1:x$, we get $x = \sqrt{3}$

Answer: Option (1)

Key idea: Photoelectric Effect - Stopping Potential and Work Function Relationship

1. Write the photoelectric equation: $eV_0 = h\nu - \phi$, where $V_0$ is stopping potential and $\phi$ is work function
2. Note that maximum kinetic energy of photoelectrons is $KE_{max} = eV_0 = h\nu - \phi$
3. For the same incident frequency $\nu$, the metal with lowest work function $\phi$ gives highest kinetic energy
4. From the graph, parallel lines mean same slope $(h/e)$ but different work functions. Higher $V_0$ at given $\nu$ means lower $\phi$
5. At any given frequency, $X_1$ has the highest stopping potential $V_0$, meaning it has the lowest work function
6. Therefore, $X_1$ will give out electrons with the greatest kinetic energy for the same wavelength of incident radiation

Answer: Option (2Dλ/a (likely option 3 or 4, as options 1 and 2 don't match))

Key idea: Single Slit Diffraction - Position of Minima

1. For single slit diffraction, minima occur at $a \sin\theta = n\lambda$ where $n = 1, 2, 3, ...$
2. For small angles, $\sin\theta \approx \frac{y}{D}$, so position of $n^{th}$ minimum is $y_n = \frac{n\lambda D}{a}$
3. Position of 1st minimum: $y_1 = \frac{\lambda D}{a}$
4. Position of 3rd minimum: $y_3 = \frac{3\lambda D}{a}$
5. Linear separation: $\Delta y = y_3 - y_1 = \frac{3\lambda D}{a} - \frac{\lambda D}{a} = \frac{2D\lambda}{a}$

Answer: Option (2)

Key idea: Young's Modulus and Elongation of Strings in Series

1. Calculate the tension force: $F = mg = 0.8 \times 10 = 8$ N
2. Find cross-sectional area: $A = \pi r^2 = \pi (2 \times 10^{-4})^2 = 4\pi \times 10^{-8}$ m²
3. Calculate elongation of string A using $\Delta L = \frac{FL}{AY}$: $\Delta L_A = \frac{8 \times 0.314}{4\pi \times 10^{-8} \times 2 \times 10^{10}} = 1$ mm
4. Calculate elongation of string B: $\Delta L_B = \frac{8 \times 0.628}{4\pi \times 10^{-8} \times 4 \times 10^{10}} = 1$ mm
5. Total elongation: $\Delta L_{total} = \Delta L_A + \Delta L_B = 1 + 1 = 2$ mm

Answer: Option (2)

Key idea: Ideal Gas Law and Gas Mixing

1. Apply ideal gas law to gas 1: $P_1V_1 = n_1RT_1$, so $n_1 = \frac{P_1V_1}{RT_1}$
2. Apply ideal gas law to gas 2: $P_2V_2 = n_2RT_2$, so $n_2 = \frac{P_2V_2}{RT_2}$
3. For the mixture: $PV = (n_1 + n_2)RT$
4. Substitute values: $PV = \frac{P_1V_1}{RT_1} \cdot RT + \frac{P_2V_2}{RT_2} \cdot RT = \frac{P_1V_1T}{T_1} + \frac{P_2V_2T}{T_2}$
5. Simplify: $PV = T\left(\frac{P_1V_1T_2 + P_2V_2T_1}{T_1T_2}\right)$
6. Solve for $T$: $T = \frac{T_1T_2PV}{T_2P_1V_1 + T_1P_2V_2}$

Answer: 11P₀V₀/(10R)

Key idea: Ideal Gas Law with Non-standard Process

1. Write the pressure-volume relation: $P = P_o \left[1 + \left(\frac{V_o}{V}\right)^2\right]^{-1}$
2. For sample A at volume $V_o$: $P_A = \frac{P_o}{2}$. For sample B at volume $3V_o$: $P_B = \frac{9P_o}{10}$
3. Apply ideal gas law $PV = nRT$ with $n = 2$ moles for both samples
4. Calculate $T_A = \frac{P_A V_o}{2R} = \frac{P_o V_o}{4R}$
5. Calculate $T_B = \frac{P_B \cdot 3V_o}{2R} = \frac{27P_o V_o}{20R}$
6. Find the difference: $T_B - T_A = \frac{27P_o V_o}{20R} - \frac{5P_o V_o}{20R} = \frac{11P_o V_o}{10R}$

Answer: Option (1)

Key idea: Voltmeter Range Extension using Series Multiplier Resistance

1. The voltmeter with resistance $x\Omega$ reads maximum 20 V, so maximum current through it is $I = \frac{20}{x}$ A
2. To measure 30 V, the voltmeter still reads 20 V (its max), and extra 10 V must drop across a series resistor
3. For range extension, connect a multiplier resistance $R_s$ in **series** with the voltmeter
4. Using Ohm's law: $V_s = I \times R_s$ gives $10 = \frac{20}{x} \times R_s$
5. Solving: $R_s = \frac{10x}{20} = \frac{x}{2}\Omega$
6. Verification: Total resistance = $x + \frac{x}{2} = \frac{3x}{2}\Omega$, Current = $\frac{30}{3x/2} = \frac{20}{x}$ A ✓

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Key idea: Bit-wise NAND operation on binary numbers

Answer: Option (2)

Key idea: Image formation by concave mirror for extended object along principal axis

1. Identify the rod extends from $u_1 = -20$ cm (near end) to $u_2 = -30$ cm (far end) from the concave mirror with $f = -10$ cm
2. Apply mirror formula for near end: $\frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f}$ gives $\frac{1}{v_1} = \frac{1}{-10} - \frac{1}{-20} = -\frac{1}{20}$, so $v_1 = -20$ cm
3. Apply mirror formula for far end: $\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f}$ gives $\frac{1}{v_2} = \frac{1}{-10} - \frac{1}{-30} = -\frac{1}{15}$, so $v_2 = -15$ cm
4. Calculate image length: $|v_1 - v_2| = |-20 - (-15)| = 5$ cm

Answer: Option (1)

Key idea: Energy stored in a capacitor connected to a battery; Rate of change of energy with varying plate separation

1. Write capacitance of parallel plate capacitor: $C = \frac{\varepsilon_0 A}{x}$
2. Since capacitor is connected to battery (constant voltage $V$), energy is: $U = \frac{1}{2}CV^2 = \frac{\varepsilon_0 A V^2}{2x}$
3. Find time rate of change: $\frac{dU}{dt} = \frac{\varepsilon_0 A V^2}{2} \cdot \frac{d}{dt}\left(\frac{1}{x}\right)$
4. Apply chain rule: $\frac{dU}{dt} = \frac{\varepsilon_0 A V^2}{2} \cdot \left(-\frac{1}{x^2}\right) \cdot \frac{dx}{dt}$
5. Substitute $\frac{dx}{dt} = v$: $\frac{dU}{dt} = -\frac{\varepsilon_0 A V^2 v}{2x^2}$
6. Therefore $\frac{dU}{dt} \propto x^{-2}$, giving $\alpha = -2$

Answer: Option (2)

Key idea: Magnetic moment of a flat coil with non-uniform radii; requires averaging over continuous radius distribution

1. Given: $N = 200$ turns, $r_1 = 3$ cm $= 0.03$ m, $r_2 = 6$ cm $= 0.06$ m, $I = 20$ mA $= 0.02$ A
2. For uniformly distributed radii from $r_1$ to $r_2$, find average $r^2$: $\langle r^2 \rangle = \frac{r_2^3 - r_1^3}{3(r_2 - r_1)}$
3. Calculate: $\langle r^2 \rangle = \frac{(0.06)^3 - (0.03)^3}{3(0.06 - 0.03)} = \frac{0.000216 - 0.000027}{0.09} = 0.0021$ m$^2$
4. Average area: $A_{avg} = \pi \langle r^2 \rangle = \pi \times 0.0021 = 0.006597$ m$^2$
5. Magnetic moment: $\mu = NIA_{avg} = 200 \times 0.02 \times 0.006597 = 0.0264$ A·m$^2$
6. Express as $\mu = 2.64 \times 10^{-2}$ A·m$^2$, so $\alpha = 2.64$

Answer: Option (1)

Key idea: RC Time Constant with Diodes

1. Identify that two diodes in parallel (one forward, one reverse biased) have effective resistance equal to forward-biased diode: $R_{diode} = 10\,\Omega$
2. Calculate total resistance in series with capacitor: $R_{total} = R + R_{diode} = 100 + 10 = 110\,\Omega$
3. Use time constant formula: $\tau = R_{total} \times C$
4. Substitute values: $\tau = 110 \times 20 \times 10^{-6} = 2200 \times 10^{-6}\,\text{s}$
5. Convert to required form: $\tau = 2.2 \times 10^{-3}\,\text{s}$
6. Therefore, $\alpha = 2.2$

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Answer: Option (2)

Key idea: Rayleigh Criterion for Angular Resolution of Telescope

1. Write the Rayleigh criterion formula for angular resolution: $\theta = \frac{1.22\lambda}{D}$
2. Given: $\lambda = 500$ nm $= 5 \times 10^{-7}$ m, $\theta = 5 \times 10^{-7}$ rad, $D = R$
3. Substitute values: $5 \times 10^{-7} = \frac{1.22 \times 5 \times 10^{-7}}{R}$
4. Solve for $R$: $R = \frac{1.22 \times 5 \times 10^{-7}}{5 \times 10^{-7}} = 1.22$ m
5. Convert to cm: $R = 1.22 \times 100 = 122$ cm

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: 35

Key idea: Work done by net force using dot product of force and displacement vectors

1. Find net force: $\vec{F_{net}} = (2\hat{i}+3\hat{j}+4\hat{k}) + (3\hat{i}-\hat{j}-2\hat{k}) = 5\hat{i} + 2\hat{j} + 2\hat{k}$ N
2. Find unit vector along displacement: $\hat{d} = \frac{3\hat{i}-4\hat{j}}{\sqrt{3^2+4^2}} = \frac{3\hat{i}-4\hat{j}}{5}$
3. Calculate displacement vector: $\vec{s} = 25\hat{d} = 25 \times \frac{3\hat{i}-4\hat{j}}{5} = 15\hat{i} - 20\hat{j}$ m
4. Apply work formula: $W = \vec{F_{net}} \cdot \vec{s} = (5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (15\hat{i} - 20\hat{j})$
5. Calculate dot product: $W = 5(15) + 2(-20) + 2(0) = 75 - 40 = 35$ J

Answer: 264

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

1. For a soap bubble, there are two surfaces (inner and outer), so total surface area = $2 \times 4\pi r^2 = 8\pi r^2$
2. Change in surface area: $\Delta A = 8\pi(r_2^2 - r_1^2) = 8 \times \frac{22}{7} \times (0.02^2 - 0.01^2)$
3. Calculate: $\Delta A = 8 \times \frac{22}{7} \times (0.0004 - 0.0001) = 8 \times \frac{22}{7} \times 0.0003$
4. Work done: $W = T \times \Delta A = 3.5 \times 10^{-2} \times 8 \times \frac{22}{7} \times 0.0003$
5. Simplify: $W = \frac{7}{2} \times 10^{-2} \times 8 \times \frac{22}{7} \times 0.0003 = 4 \times 22 \times 0.0003 \times 10^{-2}$
6. $W = 88 \times 0.0003 \times 10^{-2} = 0.0264 \times 10^{-2} = 264 \times 10^{-6}$ J
7. Therefore, $\alpha = 264$

Answer: 7

Key idea: Simple Harmonic Motion - Velocity-Displacement Relation

1. Given velocity equation: $v^2 = 50 - x^2$, which can be written as $v^2 + x^2 = 50$
2. For SHM, velocity is given by $v^2 = \omega^2(A^2 - x^2)$ where $A$ is amplitude
3. Comparing with given equation: $\omega^2(A^2 - x^2) = 50 - x^2$
4. This gives $\omega^2 = 1$ (coefficient of $x^2$) and $\omega^2 A^2 = 50$ (constant term)
5. Therefore $A^2 = 50$, so amplitude $A = \sqrt{50} = 5\sqrt{2} \approx 7.07$
6. The question asks for the value of $x$, which refers to the amplitude
7. Answer: $x = 7$ (rounded to nearest integer)

Answer: 200

Key idea: Power as dot product of force and velocity; Integration to find velocity from time-dependent force

1. Find acceleration: $\vec{a} = \frac{\vec{F}}{m} = \frac{2t\hat{i} + 6t^2\hat{j}}{2} = (t\hat{i} + 3t^2\hat{j})$ m/s²
2. Integrate to find velocity: $\vec{v} = \int \vec{a} \, dt = (\frac{t^2}{2}\hat{i} + t^3\hat{j})$ m/s (using initial condition $\vec{v}(0) = 0$)
3. At $t = 2$ s: $\vec{v} = (2\hat{i} + 8\hat{j})$ m/s
4. At $t = 2$ s: $\vec{F} = (4\hat{i} + 24\hat{j})$ N
5. Calculate power: $P = \vec{F} \cdot \vec{v} = 4(2) + 24(8) = 8 + 192 = 200$ W

Answer: 100

Key idea: Power factor in RLC series circuit and impedance calculation

1. Given: $L = 10$ mH, $C = 0.1$ μF, $R = 100$ Ω, power factor $= 0.5$
2. Power factor $\cos\phi = \frac{R}{Z}$, so $0.5 = \frac{100}{Z}$ gives $Z = 200$ Ω
3. For series RLC circuit: $Z^2 = R^2 + (X_L - X_C)^2$
4. Substitute: $200^2 = 100^2 + (X_L - X_C)^2$ gives $(X_L - X_C)^2 = 30000$
5. Therefore $|X_L - X_C| = \sqrt{30000} = 100\sqrt{3}$ Ω
6. Given $|X_L - X_C| = \sqrt{3} \times u$, so $100\sqrt{3} = \sqrt{3} \times u$
7. Solving gives $u = 100$

Answer: Option (2)

Key idea: Mole concept and Avogadro's number at STP

1. Given: Volume of SO$_2$ at STP = 1.4187 L
2. At STP, 1 mole of any gas occupies 22.4 L (molar volume)
3. Calculate moles: $n = \frac{\text{Volume}}{\text{Molar volume}} = \frac{1.4187}{22.4} = 0.0633$ mol
4. Calculate molecules using Avogadro's number: $N = n \times N_A = 0.0633 \times 6.022 \times 10^{23}$
5. Number of molecules = $3.812 \times 10^{22}$

Answer: Option (2)

Key idea: Rydberg formula for hydrogen spectrum and spectral series

1. For Balmer series: $n_1 = 2$, first line has $n_2 = 3$. Wave number: $\bar{\nu}_{B} = R_H\left(\frac{1}{4} - \frac{1}{9}\right) = R_H\left(\frac{5}{36}\right)$
2. For Brackett series: $n_1 = 4$, first line has $n_2 = 5$. Wave number: $\bar{\nu}_{Br} = R_H\left(\frac{1}{16} - \frac{1}{25}\right) = R_H\left(\frac{9}{400}\right)$
3. Calculate ratio: $\frac{\bar{\nu}_{B}}{\bar{\nu}_{Br}} = \frac{5/36}{9/400} = \frac{5 \times 400}{36 \times 9} = \frac{2000}{324}$
4. Simplify: $\frac{2000}{324} = \frac{500}{81} = \frac{5}{0.81}$
5. Answer: Option (2) $5:0.81$

Answer: Option (4)

Key idea: Electronic Configuration of Chromium (Exception to Aufbau Principle) and Quantum Numbers

1. Write the electronic configuration of Chromium (Z=24). Note: Cr is an exception with configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$ (not $4s^2 3d^4$)
2. Count electrons according to Aufbau filling order: $1s^2$ (1-2), $2s^2$ (3-4), $2p^6$ (5-10), $3s^2$ (11-12), $3p^6$ (13-18)
3. The 19th electron enters the $4s$ orbital (before $3d$ fills according to Aufbau principle)
4. For the 19th electron in $4s^1$: $n=4$ (4th shell), $l=0$ (s-orbital), $m=0$ (only value for $l=0$)
5. Spin quantum number $s = +\frac{1}{2}$ (first electron in orbital takes positive spin by convention)
6. Therefore, quantum numbers are: $n=4, l=0, m=0, s=+\frac{1}{2}$, which matches option (4)

Answer: Option (4)

Key idea: Relationship between heat capacities (Cp and Cv) for ideal gases and first law of thermodynamics

1. For ideal gases, the Mayer's relation states: $C_p - C_v = R$
2. Since $R > 0$, we get $C_p > C_v$ (heat capacity at constant pressure is GREATER than at constant volume)
3. Therefore, Statement I is **FALSE** (it claims $C_v > C_p$)
4. For constant volume process: $W = \int P\,dV = 0$ (no work done)
5. By first law: $\Delta U = Q - W = Q$, so all heat goes into internal energy (molecular motion)
6. Therefore, Statement II is **TRUE**
7. Answer: Statement I is false but Statement II is true

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Key idea: Relationship between equilibrium constants when reaction equations are scaled

Answer: Option (2)

Key idea: Redox Titrations and Equivalent Concept

1. Identify oxidizable species: $Fe^{2+}$ can oxidize to $Fe^{3+}$ (n-factor = 1), $C_2O_4^{2-}$ can oxidize to $CO_2$ (n-factor = 2), while $Fe^{3+}$ and $SO_4^{2-}$ cannot be oxidized further
2. Calculate equivalents from $FeC_2O_4$: 1 mole gives 1 equiv from $Fe^{2+}$ and 2 equiv from $C_2O_4^{2-}$ = 3 equivalents
3. Calculate equivalents from $Fe_2(C_2O_4)_3$: 1 mole gives 0 from $Fe^{3+}$ and 6 equiv from 3 moles of $C_2O_4^{2-}$ = 6 equivalents
4. Calculate equivalents from $FeSO_4$: 1 mole gives 1 equiv from $Fe^{2+}$ = 1 equivalent
5. Calculate equivalents from $Fe_2(SO_4)_3$: 1 mole gives 0 equivalents (both ions already oxidized)
6. Total equivalents = 3 + 6 + 1 + 0 = 10 equivalents. Since $KMnO_4$ has n-factor = 5 in acidic medium, moles required = $\frac{10}{5} = 2$ moles

Answer: Option (4)

Key idea: First Order Kinetics - Integrated Rate Law

1. For first order reaction $R \to P$, write rate law: $-\frac{d[R]}{dt} = k_1[R]$
2. Integrate to get: $\ln\frac{[R]}{[R]_0} = -k_1t$
3. Solve for concentration: $[R] = [R]_0 e^{-k_1t}$
4. Fraction decomposed = $\frac{[R]_0 - [R]}{[R]_0} = \frac{[R]_0 - [R]_0e^{-k_1t}}{[R]_0}$
5. Simplify: Fraction decomposed = $1 - e^{-k_1t}$
6. Verify: At $t=0$, fraction $= 0$ and as $t \to \infty$, fraction $\to 1$ ✓

Answer: Option (1)

Key idea: Atomic structure, periodic table classification, and properties of halogens

1. Given: Anion $A^-$ has 45 neutrons and 36 electrons
2. Since $A^-$ has gained 1 electron, neutral atom $A$ has $36 - 1 = 35$ electrons, so atomic number $Z = 35$ (Bromine)
3. Atomic mass $= $ protons $+$ neutrons $= 35 + 45 = 80$
4. Electronic configuration of Br: $[Ar] 3d^{10} 4s^2 4p^5$ with 7 valence electrons, so Group 17 (Halogens)
5. Bromine is one of only two elements that are liquid at room temperature
6. Answer: Atomic mass = 80, Group = 17, Physical state = liquid

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Key idea: Anomalous behavior of oxygen, covalency limitations, and oxidation states in p-block elements

Answer: Option (3)

Key idea: Stability of oxidation states in d-block and f-block elements, electronic configurations, and periodic trends

1. **Statement A**: Mo(VI) and W(VI) are actually MORE stable than Cr(VI) as stability of +6 state increases down group 6. Cr(VI) is a strong oxidant. Statement A is **incorrect**.
2. **Statement B**: $Ce^{4+}$ has [Xe]$4f^0$ and $Tb^{4+}$ has [Xe]$4f^7$ (stable configs) so they are oxidants. $Eu^{2+}$ has [Xe]$4f^7$ and $Yb^{2+}$ has [Xe]$4f^{14}$ (stable configs) so they are reductants. Statement B is **correct**.
3. **Statement C**: $Cm^{3+}$ has [Rn]$5f^7$ (7 unpaired e⁻) but $Am^{3+}$ has [Rn]$5f^6$ (6 unpaired e⁻). Statement C is **incorrect**.
4. **Statement D**: Actinoid contraction is greater than lanthanoid contraction due to poorer shielding of 5f electrons compared to 4f electrons. Statement D is **correct**.
5. Correct statements are **B and D only**. Answer is option **(3)**.

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Key idea: Properties of chromium compounds - oxidizing nature of dichromate, pH-dependent chromate-dichromate equilibrium, and structural features

Answer: Option (4)

Key idea: Spin-only magnetic moment calculation using μ = √(n(n+2)) BM for coordination complexes with weak field ligands

1. Use the formula $\mu = \sqrt{n(n+2)}$ BM where $n$ = number of unpaired electrons
2. For $[Cr(H_2O)_6]^{2+}$: Cr²⁺ has $3d^4$ configuration with 4 unpaired electrons, so $\mu = \sqrt{24} = 4.90$ BM → III
3. For $[Co(H_2O)_6]^{2+}$: Co²⁺ has $3d^7$ configuration with 3 unpaired electrons, so $\mu = \sqrt{15} = 3.87$ BM → I
4. For $[Cu(H_2O)_6]^{2+}$: Cu²⁺ has $3d^9$ configuration with 1 unpaired electron, so $\mu = \sqrt{3} = 1.73$ BM → IV
5. For $[Mn(H_2O)_6]^{2+}$: Mn²⁺ has $3d^5$ configuration with 5 unpaired electrons, so $\mu = \sqrt{35} = 5.92$ BM → II
6. Match: A-III, B-I, C-IV, D-II

Answer: Option (3)

Key idea: Electron Withdrawing Groups - Inductive and Resonance Effects

1. Identify that electron withdrawing power depends on both inductive (-I) and resonance (-R) effects
2. Analyze $-I$ (d): Weakest EWG - low electronegativity, weak -I effect, has +R effect
3. Analyze $-COOH$ (b): Moderate EWG - has -I effect but oxygen lone pairs provide +R effect that reduces withdrawing power
4. Analyze $-CN$ (a): Strong EWG - both strong -I effect (electronegative N) and strong -R effect
5. Analyze $-NO_2$ (c): Strongest EWG - very strong -I and -R effects due to positive N and two electronegative O atoms
6. Arrange in increasing order: $d < b < a < c$ or $-I < -COOH < -CN < -NO_2$

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Key idea: Ozonolysis of alkenes - cleavage of C=C bonds to form carbonyl compounds

Answer: Option (2)

Key idea: Named Organic Reactions and Their Reagents

1. Finkelstein reaction: Halogen exchange using $\text{NaI}$ in acetone to convert alkyl chlorides/bromides to iodides. A → III
2. Swarts reaction: Preparation of alkyl fluorides from alkyl halides using $\text{SbF}_3$ or other metallic fluorides. B → I
3. Sandmeyer's reaction: Conversion of aryl diazonium salts to aryl halides using copper(I) salts like $\text{Cu}_2\text{Cl}_2$. C → IV
4. Fittig reaction: Coupling of aryl halides using Na metal in dry ether to form biaryls. D → II
5. Match: A-III, B-I, C-IV, D-II corresponds to option (2)

Answer: Option (3)

Key idea: Acidity and solubility of organic compounds in aqueous NaOH - carboxylic acids and phenols form water-soluble sodium salts

1. Identify compounds that are acidic enough to react with NaOH: carboxylic acids (pKa ~ 4-5) and phenols (pKa ~ 10)
2. Carboxylic acids soluble in NaOH: (I) Benzoic acid, (V) Salicylic acid, (VII) 1-Naphthoic acid
3. Phenols soluble in NaOH: (II) 1-Naphthol, (IV) Hydroquinone, (IX) 1-Naphthol
4. Compound (III) with $N(CH_3)_2$ is likely an aniline derivative (not phenol), so NOT soluble
5. Compound (VI) N,N-dimethylcyclohexylamine is a base, NOT soluble in NaOH
6. Compound (VIII) tert-butylbenzene is a hydrocarbon, NOT soluble in NaOH
7. Total soluble compounds: (I), (II), (IV), (V), (VII), (IX) = 6 compounds

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Key idea: Electrophilic aromatic substitution, diazotization, and deamination reactions of aniline

Answer: Option (4)

Key idea: Reducing and Non-reducing Sugars - Disaccharides

1. Analyze Statement I: Maltose has two glucose units linked by $\alpha(1\to4)$ glycosidic bond
2. In maltose, one anomeric carbon is involved in the bond, but the other glucose has a FREE hemiacetal group
3. A free hemiacetal can open to form an aldehyde group, which can be oxidized. Thus, maltose IS a reducing sugar
4. Statement I is FALSE (claims maltose is non-reducing, but it's actually reducing)
5. Analyze Statement II: Lactose has galactose and glucose linked by $\beta(1\to4)$ glycosidic bond with a free hemiacetal on glucose
6. The free hemiacetal makes lactose a reducing sugar. Statement II is TRUE
7. Answer: Statement I is false but Statement II is true

Answer: Option (1)

Key idea: Amino acids classification and one-letter symbols

1. Identify the one-letter symbol for each amino acid: Arginine = R, Aspartic acid = D, Lysine = K, Glutamic acid = E
2. Classify amino acids as essential or non-essential
3. Arginine (R) is essential → Match with II
4. Aspartic acid (D) is non-essential → Match with I
5. Lysine (K) is essential → Match with IV
6. Glutamic acid (E) is non-essential → Match with III
7. Final matching: A-II, B-I, C-IV, D-III

Answer: Option (4)

Key idea: pH-dependent color changes of phenolphthalein indicator

1. Phenol + phthalic anhydride with conc. $H_2SO_4$ forms phenolphthalein (colorless lactone form)
2. With NaOH, the lactone ring opens forming a quinonoid structure with extended conjugation (pink/magenta color)
3. The pink color is due to the monoanionic quinonoid form of phenolphthalein
4. With excess NaOH, further deprotonation occurs forming the trianionic carbinol form
5. The quinonoid structure is disrupted, destroying the extended conjugation
6. Compound X (trianionic form in excess alkali) is colorless

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: Lewis structure, VSEPR theory, hybridization, and geometry of xenon compounds

Answer: 78.25

Key idea: Friedel-Crafts alkylation, electrophilic aromatic substitution, oxidation of alkyl benzenes, and stoichiometry of carboxylic acid reactions

1. Benzene undergoes Friedel-Crafts alkylation with $\text{CH}_3\text{Cl}/\text{AlCl}_3$ to form toluene
2. Toluene undergoes electrophilic chlorination with $\text{Cl}_2/\text{FeCl}_3$ to give $p$-chlorotoluene (major product, since $-\text{CH}_3$ is $o/p$ directing)
3. $p$-Chlorotoluene is oxidized by $\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4$ to give $p$-chlorobenzoic acid (X): $\text{C}_7\text{H}_5\text{ClO}_2$, molar mass $= 156.5$ g/mol
4. The carboxylic acid reacts with $\text{NaHCO}_3$: $\text{C}_7\text{H}_5\text{ClO}_2 + \text{NaHCO}_3 \rightarrow \text{C}_7\text{H}_4\text{ClO}_2\text{Na} + \text{H}_2\text{O} + \text{CO}_2$
5. Moles of $\text{CO}_2$ liberated $= \frac{11.2}{22.4} = 0.5$ mol, so moles of X $= 0.5$ mol
6. Mass of X: $P = 0.5 \times 156.5 = 78.25$ g

Answer: 5

Key idea: Empirical Formula Determination using Carius Method

1. Calculate moles of Br: From 3.36 g AgBr, moles = $\frac{3.36}{188} = 0.01787$ mol, so mass of Br = $0.01787 \times 80 = 1.43$ g
2. Find % of Br in compound X: $\frac{1.43}{2.0} \times 100 = 71.5\%$
3. Calculate % of H: Since compound has only C, H, Br: $\% H = 100 - 26.7 - 71.5 = 1.8\%$
4. Find moles of each element in 2.0 g sample: C = $\frac{0.534}{12} = 0.0445$ mol, H = $\frac{0.036}{1} = 0.036$ mol, Br = $0.01787$ mol
5. Determine mole ratio by dividing by smallest (0.01787): C:H:Br = 2.5:2:1
6. Multiply by 2 to get whole numbers: Empirical formula = $C_5H_4Br_2$

Answer: 756

Key idea: Henderson-Hasselbalch equation for basic buffer (NH₄OH/NH₄Cl system)

1. Calculate moles: $n_{NH_4OH} = 0.1 \times 5 = 0.5$ mmol and $n_{NH_4Cl} = 0.1 \times 250 = 25$ mmol
2. This is a basic buffer with base (NH₄OH) and its conjugate acid (NH₄⁺ from NH₄Cl)
3. Apply Henderson-Hasselbalch: $pOH = pK_b + \log\frac{[Salt]}{[Base]} = 4.74 + \log\frac{25}{0.5}$
4. Calculate: $\log(50) = \log(5 \times 10) = 0.70 + 1 = 1.70$
5. Find pOH: $pOH = 4.74 + 1.70 = 6.44$
6. Calculate pH: $pH = 14 - 6.44 = 7.56 = 756 \times 10^{-2}$

Answer: 5

Key idea: Raoult's Law and Boiling Point Elevation

1. Calculate boiling point elevation: $\Delta T_b = 320 - 319.5 = 0.5$ K
2. Use $\Delta T_b = K_b \times m$ to find molality: $0.5 = 0.3 \times \frac{n_2}{0.04}$
3. Solve for moles of solute: $n_2 = \frac{0.5 \times 0.04}{0.3} = \frac{1}{15}$ mol
4. Apply Raoult's law: $\frac{P_0 - P}{P_0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dilute solution
5. Calculate: $\frac{760-750}{760} = \frac{1}{76} = \frac{n_2}{n_1}$
6. Find moles of solvent: $n_1 = 76 \times n_2 = 76 \times \frac{1}{15} = \frac{76}{15} \approx 5.07$
7. Round to nearest integer: $n_1 = 5$ mol