The line L intersects the axes at A(5,0) and B(0,9). The lines l₁ and l₂ trisect AB. The points trisecting AB are P(10/3, 3) and Q(5/3, 6). m₁ = 9/10 and m₂ = 18/10 = 9/5. m₁ + m₂ = 9/2 The equation of the line passing through the origin with slope 9/2 is y = (9/2)x or 9x - 2y = 0. Solving the system of equations 9x + 5y = 45 and 9x - 2y = 0, we get y=45/7 and x=10/7. This point satisfies y - x = 5.

For A, B, C, and D to be coplanar, the scalar triple product of vectors AB, AC, and AD must be zero. AB = -4i - 3j + (3-2λ)k, AC = -7i + (x-5)j + (4-2λ)k, and AD = -6i + (6-2λ)k. The scalar triple product gives the equation -4λ² + 20λ - 24 = 0, which simplifies to λ² - 5λ + 6 = 0. Solving for λ gives λ=2 or λ=3. Therefore S={2,3}. Σ_λ∈S(λ+2)² = (2+2)² + (3+2)² = 16 + 25 = 41.

Integrate by parts with u = x² and dv = ^{(xsec2x + tanx)}⁄_{(xtanx + 1)²}. Then v = -1/(xtanx + 1). I(x) = ^-x2⁄_(xtanx+1) + ∫ ^2x⁄_{(xtanx + 1)} dx = ^-x2⁄_(xtanx+1) + 2ln|xsinx+cosx|. Since I(0) = 0, the constant of integration is 0. Substitute x = π/4 and evaluate.

The differences between consecutive terms form an arithmetic progression: 6, 8, 10, 12, ... The nth term of this progression is 2n+4. The nth term of the original series is given by T_n = n² + 3n + 1. The sum of the first 20 terms is Σ²⁰_n=1(n²+3n+1) = Σn² + 3Σn + Σ1 = 2870 + 630 + 20 = 3520.

The probability of getting a sum of 5 is 4/36 = 1/9. Let p = 1/9 and q = 8/9. The probability of at least 4 successes in 5 trials is given by P(X=4) + P(X=5) = ⁵C₄p⁴q¹ + ⁵C₅p⁵ = 5(1/9)⁴(8/9) + (1/9)⁵ = ⁴¹⁄_9⁵ = ¹²³⁄₃₁₁. Thus, k=123.

Let A = [[α, β], [γ, δ]]. Since A² = I, |A²|=|I| => |A|² = 1 => |A| = b = ±1. Also, A² = [[α² + βγ, αβ + βδ],[αγ + γδ, βγ + δ²]] = I. This leads to α + δ = 0, or a = 0. Therefore, 3a² + 4b² = 3(0)² + 4(±1)² = 4.

Rationalize each term in the summation: Σ^n-1_k=1 ^{(√a_k+1 - √a_k)}⁄_d = ^{(√a_n - √a₁)}⁄_d. Substituting a_n = a₁ + (n-1)d and simplifying the limit yields 1.

Using the combination formula, we get ^4(2n-1)⁄_(n-2) = 10. Solving gives n=8. Substitute n=8 into the ratio (n² + 3n) : (n² - 3n + 4) to get 88:44, which simplifies to 2:1.

For set A, consider the cases x ≥ -3, -4 ≤ x < -3, and x < -4. Solving the inequality in each case gives -5 ≤ x ≤ -2, so A = [-5, -2]. For set B, the summation is a geometric series. Simplifying the inequality gives x+[x] < 3. Let x=n+ε, where n is an integer and 0 < ε < 1. This gives -5 ≤ x ≤ -2, so B=[-5, -2]. Therefore A=B.

Let the edge parallel to the z-axis be through (3,0,0). The direction vector of OP is <3,4,5>. The direction vector of the edge is <0,0,1>. The shortest distance is given by the formula |(a2-a1)·(b1 x b2)|/|b1 x b2|. Here, a1=(0,0,0), b1=<3,4,5>, a2=(3,0,0) and b2=<0,0,1>. b1 x b2 = <4,-3,0>. |b1 x b2| = 5. (a2-a1)·(b1 x b2) = 12. So, the shortest distance is 12/5. If we consider the edge parallel to z-axis through (0,4,0), a2 becomes (0,4,0), and the shortest distance becomes 12/√5.

The equation of the plane passing through the intersection of the two given planes is (2x - y + z - 3) + λ(4x - 3y + 5z + 9) = 0. This plane is parallel to the line with direction vector <-2, 4, 5>. The normal vector of the plane is <2+4λ, -1-3λ, 1+5λ>. Since the plane is parallel to the line, the normal vector is perpendicular to the direction vector of the line. Therefore, -2(2+4λ) + 4(-1-3λ) + 5(1+5λ) = 0, which gives λ=3/5. Substituting this value of λ into the equation of the plane and comparing it to ax + by + cz + 6 = 0 gives a=11, b=-7, and c=10. Thus, a+b+c = 14.

The fifth term from the beginning is ⁿC₄(√2)^n-4(1/√3)⁴. The fifth term from the end is ⁿC_n-4(√2)⁴(1/√3)^n-4. The ratio of these terms is given as √6. Simplifying and solving for n, we get n=10. The third term from the beginning is ¹⁰C₂(√2)⁸(1/√3)² = 60√3.

We consider two cases: x² - 8x + 15 ≥ 0 and x² - 8x + 15 < 0. In the first case, x² - 10x + 22 = 0, which gives x = 5±√3. Only 5+√3 satisfies the condition. In the second case, -x² + 6x - 8 = 0, which gives x=2 or x=4. Only x=4 satisfies the condition. The sum of the roots is 4 + (5+√3) = 9 + √3.

Let BQ = y. tan60° = AB/BQ => y = 10√3. Let AP = x. tan15° = x/y => x = 15(2-√3). PQ = 30-x = 10(3√3-3). The area of the trapezium BCPQ is (1/2)(BC+PQ) * BQ = (1/2)(15 + 10(3√3-3)) * 10√3 = 600(√3-1).

Since d is perpendicular to both b and c, d is parallel to b x c. b x c = 2i - j + 2k. Let d = λ(2i - j + 2k). Since a·d = 18, we have 2(2λ) + 3(-λ) + 4(2λ) = 18 => λ=2. So, d = 4i - 2j + 4k. |a|² = 29, |d|² = 36. |a x d|² = |a|²|d|² - (a·d)² = 29*36 - 18² = 720.

Differentiate both sides with respect to x using logarithmic differentiation: 2^xy(ylnx + x(1/y)^dy⁄_dx) + 3^yx(lny*y^x+y^x*xlny(1/y)^dy⁄_dx) = 0. Substituting x=2 and y=2 and simplifying gives ^dy⁄_dx = ^{-(8 + 12ln2)}⁄_{(12 + 8ln2)} = ^-(2+3ln2)⁄_(3+2ln2) = ^-(2+ln8)⁄_(3+ln4).

For infinitely many solutions, the determinant of the coefficient matrix and the determinants of the matrices obtained by replacing each column with the constants must be zero. The coefficient matrix determinant is 7-a. Setting this to zero gives a=7. The determinant Δ_x (replacing the x column with the constants) is 11b-33. Setting this to zero gives b=3. Therefore, 2a+3b = 2(7)+3(3) = 23.

(P⇒Q)∧(R⇒Q) is equivalent to (¬P∨Q)∧(¬R∨Q), which can be written as (¬P∧¬R)∨Q. Using De Morgan's Law, this becomes ¬(P∨R)∨Q, which is equivalent to (P∨R)⇒Q.

Substitute 1/x for x to get another equation: 5f(¹⁄_x) + 4f(x) = x + 3. Solve this system of equations to get f(x) = ^{(5 - 4x)}⁄_(9x) + ¹⁄₃. Integrate 18f(x) from 1 to 2 to get 10loge2 - 6.

The combined variance is given by σ²_c = ^{(n₁σ2₁ + n₂σ2₂)}⁄_(n1+n2) + ^{(n₁n₂(m₁-m₂)2)}⁄_(n1+n2)². Substituting n₁=n₂=15, m₁=12, m₂=14, σ²₁=14, and σ²_c = 13, and solving for σ²₂ = σ², we get σ²=10.

The given curve can be rewritten as (x+1)² = 4(y-2). The equation of the tangent at P(1,3) is x - y + 2 = 0, so A is (0,2). The line through P parallel to x - 3y = 6 is x - 3y = -8. Substituting x = (3y-8) into y²=4x, we get y² - 12y + 32 = 0, so y=4 or y=8. Thus the intersection points are (4,4) and (16,8). Since B lies on 2x-3y=8, B must be (16,8). Then (AB)² = (16-0)² + (8-2)² = 292.

For the point (p,p+1) to lie inside E, we must have 3-p ≤ p+1 ≤ √(9-p²). The first inequality gives p ≥ 1. The second inequality gives p² + p - 4 ≤ 0, which means ^(-1-√17)⁄₂ ≤ p ≤ ^(-1+√17)⁄₂. Combining these gives 1 ≤ p < ^(-1+√17)⁄₂. Therefore, a=1 and b= ^(-1+√17)⁄₂. b² + b - a² = ^{(1-17+2√17+4)}⁄₄ + ^(-1+√17)⁄₂ - 1 = 3.

Rewrite the differential equation as ^dy⁄_dx + (^{(xsinx + cosx)}⁄_(xcosx))y = ¹⁄_(xcosx). The integrating factor is xsecx. Multiplying the equation by the integrating factor gives d(xysecx)/dx = sec²x, so xysecx = tanx + C. Using y(π/3) = √3, we get C = √3. Thus y = ^{(sinx + √3cosx)}⁄_x. Differentiating twice and substituting x=π/6 gives |y"(π/6) + 2y'(π/6)| = 2.

Since 0 < x < π, we have 0 < sinx ≤ 1, and thus 0 < 13sinx ≤ 13. The greatest integer function is discontinuous (and hence not differentiable) at integer values. Thus, f(x) is not differentiable when 13sinx is an integer from 1 to 13. The equation 13sinx = k has two solutions for k=1,2,...,12 and one solution for k=13 in (0,π). Thus there are 2*12 + 1 = 25 points.

The region S is defined by the inequalities 2y-y² ≤ x ≤ 2y and x ≥ y. This region lies between the parabola x = 2y-y² and the lines x=2y and x=y. The area of the required region can be calculated by integrating from y=0 to y=1 (between the parabola and x=2y) and subtracting the area from y=0 to y=1 between x=y and x=2y. This gives ∫¹₀ (2y - (2y-y²))dy = ¹⁄₃. The area of the triangle formed by x=y, x=2y, and y=1 is 1/2. Therefore the required area is 1/2 - 1/3 = 1/6. Equating this to ^(1-1⁄_n)π⁄₄ gives n=5.

The total number of ways to distribute 20 oranges among 3 children is 3²⁰. The number of ways where at least one child gets no oranges is ³C₁ * 2²⁰ - ³C₂ * 1²⁰. The number of ways where each child gets at least one orange is 3²⁰ - (3*2²⁰ - 3) = 3483638676.

Let Q be (α, β, γ). The midpoint of PQ, M(^(α+1)⁄₂, ^(β+2)⁄₂, ^(γ+3)⁄₂), lies on the plane. The direction ratios of PQ are proportional to the normal of the plane, so ^(α-1)⁄₂ = ^(β-2)⁄_-1 = ^(γ-3)⁄₁ = k. This gives α=2k+1, β=-k+2, γ=k+3. Substituting in the equation of the plane gives k=2, so Q is (5,0,5). Now, PQ = <4,-2,2> and PR = <5,8,4>. The area of triangle PQR is (1/2)|PQ x PR|. PQ x PR = <-24, -6, 42>. The magnitude of this cross product is 2√594. Thus, the square of the area is 594.

The equation of a circle touching both axes is (x-a)² + (y-a)² = a². Since P(α,β) lies on the circle, (α-a)² + (β-a)² = a², which simplifies to α² + β² - 2a(α+β) + a² = 0. The equation of line AB is x+y=a. The distance PQ = |α + β - a|/√2 = 11. Thus, (α + β - a)² = 242. Substituting this into the simplified circle equation gives 2αβ=242, so αβ=121.

The general term in the expansion of (1+x²+2x)¹⁵ is given by ¹⁵C_r1,r2,r3 (1)^r₁(x²)^r₂(2x)^r₃, where r₁ + r₂ + r₃ = 15. To find the coefficient of x¹⁸, we need 2r₂ + r₃ = 18. Since r₁, r₂, r₃ are non-negative integers, we can have r₂=9, r₃=0, r₁=6; r₂=8, r₃=2, r₁=5; ... ; r₂=0, r₃=18, r₁=-3 (not possible). Since r₁+r₂+r₃ = 15, when r₃=18, r₂ must be negative or zero since r₁ is nonnegative. If r₂=0, r₁ = -3 which is not allowed. Therefore, the maximum value of r₃ is 2r₂+r₃=18 along with r₁+r₂+r₃ = 15, from which r₁ = 15-18 = -3, so not allowed. Only one possible solution is r₁ = 6, r₂=6, and r₃=3. If r₃=0, then 2r₂=18 => r₂=9 and r₁=6. Term = ¹⁵C₆*⁹C₀ = 5005. This is the only case.

Let k = a-b. Then we need to find the number of pairs (a,b) such that 2k²+3k ∈ B. Since a,b ∈ A, -9 ≤ k ≤ 9. If k=0, 2k²+3k = 0 ∈ B. This gives a=b. Since there are 10 possible values for a (and hence b), there are 10 pairs. If k=-2, 2k²+3k = 2 ∈ B. This gives a=b-2. Since 1 ≤ a ≤ 10 and 1 ≤ b ≤ 10, we have 3 ≤ b ≤ 10, which gives 8 possible values for b, and hence 8 pairs. No other values of k give a result in B. Therefore, the total number of elements in R is 10 + 8 = 18.

Step 1: De Broglie Wavelength Formula
The de Broglie wavelength is given by: λ = h/p, where λ is the wavelength, h is Planck’s constant, and p is the momentum.

Step 2: Relate Momentum and Kinetic Energy
The momentum is related to kinetic energy (KE) by: p = √(2mKE), where m is the mass.

Step 3: Calculate Wavelengths for Electron, Proton, and Alpha Particle
For the electron: λ_e = h/√(2m_e * 4K)
For the proton: λ_p = h/√(2m_p * K)
For the alpha particle: λ_α = h/√(2 * 4m_p * 2K)
Comparing the expressions, we get λ_α < λ_p < λ_e.

Step 1: Escape Velocity Formula
Escape velocity (V_esc) is the minimum speed required to escape a planet’s gravitational pull: V_esc = √(2GM/r).

Step 2: Explanation
Earth has a higher escape velocity than the moon, allowing it to retain its atmosphere. The moon’s lower escape velocity means lighter gases can escape more easily.

Step 1: First Law of Thermodynamics
The change in internal energy (ΔU) is given by: ΔU = Q - W.

Step 2: Calculate Rate of Internal Energy Increase
Given heat rate Q = 1000 W and work rate W = 200 W:
ΔU = 1000 W - 200 W = 800 W.

Step 1: Forces Acting on the Ball
The forces are: Weight (Mg), Buoyant force (B), and Viscous force (f).

Step 2: Solve for Viscous Force
At terminal velocity: Mg = f + B, solving gives: F = Mg(1 − ρ_o/ρ).

Step 1: Centripetal Force
The tension in the spring provides the necessary centripetal force for circular motion.

Step 2: Calculate Tension
Using the given values: T = 0.75 N.

Step 1: Instantaneous Velocity
The instantaneous velocity is tangential to the circular path with a constant magnitude.

Step 2: Average Velocity
The average velocity is the displacement divided by the time taken.

Step 3: Ratio Calculation
The ratio of instantaneous velocity to average velocity is π:√2, giving x = 2.

Step 1: Equivalent Resistance Formula
For parallel resistances: 1/R_eq = 1/R₁ + 1/R₂.

Step 2: Error Propagation
The percentage error is calculated using error propagation rules. The final error is 4.33%.

Step 1: Electric Potential Inside and Outside the Shell
Inside the shell, the potential is constant. Outside the shell, it decreases as 1/r.

Step 2: Graphical Representation
The graph shows a constant potential inside and a decreasing potential outside.

Step 1: Ampere’s Law
Using Ampere’s law, the magnetic field inside and outside the wire is calculated.

Step 2: Final Expression
Inside the wire, B ∝ r. Outside the wire, B ∝ 1/r.

Step 1: Transmission Range Formula
The range is proportional to the square root of the tower’s height.

Step 2: Percentage Increase
A 21% increase in height results in a 10% increase in range.

Step 1: Collision Frequency Formula
Collision frequency is directly proportional to the number density of molecules.

Step 2: Ratio Calculation
The ratio is 0.25 based on the change in number density.

Step 1: Energy and Wavelength Relationship
The transition with the largest energy difference produces the shortest wavelength.

Step 1: Energy Density Formula
The electric and magnetic energy densities are equal in an electromagnetic wave.

Step 1: Gravitational Force Calculation
Weight depends on the mass and radius of the planet. The radius increases as 2^1/3.

Step 1: Resistivity vs. Temperature
For semiconductors, resistivity decreases with increasing temperature.

Step 1: Use Snell's Law
Snell's law is given by: n₁ sin θ₁ = n₂ sin θ₂. Given that θ₁ = 45° and θ₂ = 30°, we solve for the refractive index ratio.

Step 2: Calculate Wavelength and Frequency
The frequency remains constant during refraction, and the wavelength changes according to the refractive index. The calculated values show that λ₂ = (1/√2) λ₁ and ν₂ = ν₁.

Step 1: Equivalent Spring Constant
Since the springs are in parallel, the equivalent spring constant is K = K₁ + K₂.

Step 2: Time Period Formula
The time period for oscillation is given by T = 2π √(m/K). Substituting the equivalent spring constant gives the final answer.

Step 1: Analyze the Circuit
The circuit shows a configuration that behaves as a NOR gate, with both inputs needing to be 0 to give an output of 1.

Step 2: Truth Table
Constructing the truth table for the circuit confirms that it is a NOR gate.

Step 1: Faraday's Law
An induced emf is produced when there is a change in magnetic flux through a coil. Rotating the coil or changing its area will change the flux.

Step 2: Analyze the Options
Only options C (rotating the coil) and D (changing the area of the coil) cause a change in flux.

Step 1: Horizontal Range Formula
The horizontal range is given by R = (u² sin 2θ) / g. The maximum range occurs when sin 2θ = 1, which is at θ = 45°.

Step 2: Explanation
Both the assertion and the reason are correct, and the reason accurately explains the assertion.

Step 1: Magnetic Field due to a Circular Loop
The magnetic field at the center of a circular loop of radius r carrying current i is given by:
B = μ₀i/2r, where μ₀ = 4π × 10^-7 T·m/A.
Step 2: Magnetic Field due to Each Loop
Given: r = 0.2 m, i = √2 A. Substituting these values:
B = (4π × 10^-7)(√2)/(2 × 0.2).
Step 3: Net Magnetic Field
The magnetic fields are perpendicular, so the net field is calculated using Pythagoras' theorem:
B_net = √2 × (μ₀i/2r).
Simplifying: B_net = 628 × 10^-8 T.

Step 1: Young's Modulus and Strain
Young’s modulus is defined as stress/strain.
Step 2: Calculate Strain
Stress = Force/Area. Using A = πr², strain = stress/Young’s modulus.
Substitute: strain = (62.8 × 10³)/(π × (20 × 10^-3)² × 2 × 10¹¹) = 25 × 10^-5.

Step 1: Resistance Formula
Resistance R = ρl/A.
Step 2: New Resistance
New length: l' = 1.2l. New area: A' = 0.96A. Calculate R': R'/R = (l'/l) / (A'/A) = 1.2/0.96 = 1.25.
Step 3: Percentage Change
Change = (1.25 - 1) × 100 = 25%.

Step 1: Bohr's Model Formula
r_n = (0.51 × n²)/Z Å. Substituting: n = 5, Z = 3.
Step 2: Calculate Radius
r₅ = (0.51 × 25)/3 = 425 × 10^-12 m.

Step 1: Work-Energy Theorem
Use acceleration a = -2x and integrate: v dv = -2x dx.
Step 2: Solve for Energy Loss
Energy loss = m x², giving (10/x)^-2.

Step 1: Transformer Voltage Ratio
Voltage ratio: V_s/V_p = N_s/N_p. Given V_p = 12,000 V, V_s = 120 V, N_s/N_p = 1/100.
Step 2: Power Conservation
Input power = output power. Given output power P = 60 kW.
Step 3: Secondary Current
I_s = P/V_s = 60,000/120 = 500 A.
Step 4: Resistive Load
R_s = V_s/I_s = 120/500 = 0.24 Ω = 240 mΩ.

Step 1: Capacitance with Dielectric
For a partially filled capacitor: C = ϵ₀A/[d - x + x/K].
Step 2: Capacitance C₁
For x = d/3: C₁ = ϵ₀A/[2d/3 + d/12] = 2 µF.
Step 3: Capacitance C₂
For x = 2d/3: C₂ = ϵ₀A/[d/3 + d/6]. C₂ = 3 µF.

Step 1: Moment of Inertia of a Solid Sphere
For a solid sphere about its diameter: I_sphere = (2/5)mR².
Step 2: Parallel Axis Theorem
I = I_cm + md². Using d = 20 cm = 0.2 m, R = 0.1 m, and m = 2 kg.
Step 3: Total Moment of Inertia
I_total = 2 × [(2/5)(2)(0.1)² + (2)(0.2)²] = 176 × 10^-3 kg·m².

Step 1: Doppler Effect for Reflection
The observed frequency: f′ = [(v + v_s)/(v - v_s)]f₀.
Step 2: Frequency Change
Given v_s = 15 m/s, v = 330 m/s, and Δf = 40 Hz.
Solving: f₀ = 420 Hz.

Step 1: Snell’s Law
Using n₁sini = n₂sinr to calculate the angle of refraction.
Step 2: Calculate Tan r
tanr = (sinr/cosr) = 0.85.
Step 3: Height of the Pole
tan30° = h/0.875. Solving: h = 50 cm.

The one-letter codes for the given amino acids are:

• Arginine: R
• Aspartic acid: D
• Asparagine: N
• Alanine: A

Matching these with the options:

(A) Arginine - (IV) R
(B) Aspartic acid - (I) D
(C) Asparagine - (II) N
(D) Alanine - (III) A

Lead nitrate, Pb(NO₃)₂, is a soluble colorless compound. Confirmatory tests for ions usually involve the formation of a precipitate or a distinctly colored complex. Lead sulphate (white precipitate), lead chromate (yellow precipitate), and lead iodide (yellow precipitate) are all used as confirmatory tests for lead(II) ions. Since lead nitrate is colorless and soluble, it’s not used as a confirmatory test.

For neutralization, the milliequivalents of acid must equal the milliequivalents of base.

Milliequivalents = Molarity × n-factor × Volume

For HBr, n-factor = 1. For Ba(OH)₂, n-factor = 2.

Let V be the volume of HBr required.

0.02 × 1 × V = 0.01 × 2 × 10

0.02V = 0.2
V = 0.2 ÷ 0.02 = 10 mL.

As we move down Group 13, the metallic character increases, and hence the electropositive character increases. As electropositivity increases, the basicity of the oxides increases.

The order of basicity for Group 13 oxides is:

B₂O₃ < Al₂O₃ < Ga₂O₃ < In₂O₃ < Tl₂O₃.

Therefore, Tl₂O₃ is the most basic oxide.

The oxalate anion (C₂O₄²⁻) is a bidentate ligand, and the prefix “tris” indicates there are three oxalate ligands.

The complex anion is [Co(C₂O₄)₃]³⁻, and since the potassium ion has a +1 charge, there must be three potassium ions to balance the charge. The oxidation state of cobalt is +3, indicated by the Roman numeral (III).

Step 1: Recall the de Broglie principle
The circumference of an electron's orbit in an atom is an integer multiple of the electron's wavelength:
2πr = nλ, where r is the radius of the orbit, n is the principal quantum number, and λ is the de Broglie wavelength.

Step 2: Use the formula for the radius of the nth orbit
r_n = n²a₀/Z, where Z is the atomic number. For hydrogen (Z = 1) and n = 3, r₃ = 9a₀.

Step 3: Substitute into the de Broglie equation
2π(9a₀) = 3λ
λ = 18πa₀/3 = 6πa₀.

DDT (Dichlorodiphenyltrichloroethane) and Aldrin are well-known pesticides.
PAN (Peroxyacetyl nitrate) is a component of photochemical smog and not a pesticide.
Sodium chlorate is a herbicide, while Sodium arsinite has been used as an insecticide and rodenticide. Tetrachloroethene is used in dry cleaning.
Thus, the correct pair is DDT and Aldrin.

Key Observations:

• In column chromatography, the polarity of the compound determines the retention time.
• More polar compounds elute later, spending more time in the column.

Order of polarity: a > b > c.
Order of elution: c > b > a.
Incorrect statements: A, B, and C.

Step 1: Understand hydration enthalpy
Hydration enthalpy depends on the charge density (charge/size) of the ion. Smaller ions with higher charge have greater hydration enthalpy.

Step 2: Analyze the given ions
Be²⁺ is the smallest ion among the given alkaline earth metals, so it has the highest charge density and hence the highest hydration enthalpy.

Step 1: Stability of the conjugate base
The strength of an acid is determined by the stability of its conjugate base. Electron-withdrawing groups stabilize the conjugate base, increasing acidity.

Step 2: Analyze substituents
- NO₂ is a strong electron-withdrawing group (-I and -M effects), stabilizing the conjugate base the most.
- CH₃ is electron-donating, reducing acidity.
Thus, nitrophenol is the strongest acid.

Step 1: Protonation of the alcohol
The hydroxyl group is protonated by H₃O⁺, making it a better leaving group.

Step 2: Carbocation formation
Water leaves, generating a secondary carbocation.

Step 3: Cyclization
The double bond attacks the carbocation, forming a six-membered ring. This results in a tertiary carbocation adjacent to the oxygen in the newly formed ring.

Step 4: Deprotonation
A proton is lost to give the final cyclic ether product.

Thus, the correct product is Option 4.

Solid state: BeCl₂ exists as a polymeric structure with each beryllium atom bonded to four chlorine atoms, forming a chain-like structure.

Vapor phase: Below 1200 K, BeCl₂ exists as a dimer (Be₂Cl₄) with chloro-bridging.

High temperature: Above 1200 K, BeCl₂ breaks into monomeric units with a linear structure.

Step 1: Analyze intermediates and activated complexes
Intermediates are represented by valleys in the energy diagram (2 valleys). Activated complexes correspond to peaks (3 peaks).

Step 2: Identify the rate-determining step
The slowest step corresponds to the highest energy barrier, which is Step II.

Electrolysis reactions:

At cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻
At anode: 2Cl⁻ → Cl₂ + 2e⁻

Products formed: H₂, Cl₂, and NaOH. HCl is not formed during electrolysis.

Step 1: Analyze melting and boiling points
Boiling water temperature = 100°C.
- Li and Cs react with water.
- Br₂: Boiling point is 58.8°C (gas at 100°C).
- Ga: Melting point = 29.8°C, Boiling point = 2400°C (liquid at 100°C).

Step 1: Analyze Assertion A
In Ni(CO)₄ and Fe(CO)₅, the carbonyl ligand (CO) is neutral. Thus, the oxidation state of Ni and Fe is zero. Assertion A is correct.

Step 2: Analyze Reason R
Low oxidation states are stabilized by synergistic bonding, involving σ-donation from the ligand and π-backdonation from the metal. CO is a π-acceptor ligand, not a π-donor. Therefore, Reason R is incorrect.

Step 1: Analyze Statement I
Morphine is a narcotic analgesic that relieves pain but induces sleep and drowsiness. Hence, Statement I is false.

Step 2: Analyze Statement II
Morphine and its derivatives are indeed obtained from the opium poppy, making Statement II true.

Step 1: Perform 1,4-addition
The organocuprate reagent (MeMgBr/CuI) adds to the α, β-unsaturated ketone in a 1,4-conjugate addition manner.

Step 2: Alkylate the product
The resulting ketone reacts with n-propyl iodide (nPrI) via an alkylation reaction at the α-carbon.

The final product is Option 3.

Step 1: Reaction in a neutral medium
In a neutral medium, permanganate is reduced to manganese(IV) oxide (MnO₂), with a change in oxidation state of 3.

Step 2: Reaction in an acidic medium
In an acidic medium, permanganate is reduced to Mn²⁺, with a change in oxidation state of 5. The condition specified in the question aligns with a neutral medium.

Step 1: Composition of Nessler’s reagent
Nessler’s reagent is an alkaline solution of potassium tetraiodomercurate(II), K₂[HgI₄].

Step 2: Identify absent element
The reagent contains potassium (K), mercury (Hg), and iodine (I) but not nitrogen (N).

Step 1: Understand oxidation and reduction potentials
For a metal to be oxidized, the reduction potential of NO^-₃ (0.97 V) must be greater than the metal's oxidation potential.

Step 2: Reverse reduction potentials for oxidation potentials
- V: E⁰ (oxidation) = 1.19 V
- Fe: E⁰ (oxidation) = 0.04 V
- Ag: E⁰ (oxidation) = -0.80 V
- Au: E⁰ (oxidation) = -1.40 V

Step 3: Compare with NO^-₃
Metals oxidized: Fe, Ag, Au (E⁰ < 0.97 V).

Step 1: Define body-centered cubic (BCC)
A BCC structure has atoms at each corner of a cube and one atom at the center.

Step 2: Identify systems
BCC unit cells are found in cubic, tetragonal, and orthorhombic systems.

Step 3: Exclude others
Hexagonal, rhombohedral, monoclinic, and triclinic do not have BCC arrangements.

Step 1: Understand the iodoform test
The test is positive for methyl ketones (R-CO-CH₃) and secondary alcohols with CH₃CH(OH)- groups.

Step 2: Analyze compounds
Positive test: 1, 3, 4, 6 (methyl ketones or secondary alcohols with required groups).

Step 1: Analyze the molecular formula
C₈H₁₁N implies a benzene ring (degree of unsaturation = 4) with an amine group.

Step 2: Consider positions of substituents
Possible isomers: Aniline, ortho-, meta-, para-toluidine, and their derivatives.

Step 1: Define isotonic solutions
Solutions with the same osmotic pressure have the same number of solute particles.

Step 2: Analyze dissociation
Each pair has equal particle concentrations after dissociation.

Step 1: Identify colloidal systems
A colloid consists of a dispersed phase and a dispersion medium. For this question, focus on systems where the dispersion medium is liquid.

Step 2: Analyze each system
- Paints: Liquid medium
- Milk: Liquid medium
- Hair cream: Liquid medium
- Froth: Liquid medium
- Soap lather: Liquid medium
Others (gemstones, smoke, cheese, insecticide sprays) do not have a liquid dispersion medium.

Step 3: Count the systems
There are 5 systems with liquid as the dispersion medium: paints, milk, hair cream, froth, and soap lather.

Step 1: Understand hydrogen bonding in ice
Each water molecule in ice has two hydrogen atoms and two lone pairs of electrons. These allow it to form hydrogen bonds with four neighboring water molecules.

Step 2: Analyze the structure
- Each hydrogen atom forms one hydrogen bond as a donor.
- Each lone pair forms one hydrogen bond as an acceptor.
Thus, each water molecule is hydrogen bonded to four others, forming a tetrahedral arrangement.

Step 1: Write the target equation
The target reaction is: 2C(s) + 3H₂(g) + 0.5O₂(g) → C₂H₅OH(l).

Step 2: Use Hess's Law
Combine the given reactions:
- Combustion of carbon: 2C(s) + 2O₂(g) → 2CO₂(g), ΔH = -787 kJ
- Combustion of hydrogen: 3H₂(g) + 1.5O₂(g) → 3H₂O(l), ΔH = -725.4 kJ
- Combustion of ethanol: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH = -1234.7 kJ
Reverse the ethanol combustion reaction to form ethanol.

Step 3: Calculate heat of formation
ΔH_formation = (-787) + (-725.4) + (+1234.7) ≈ -278 kJ mol^-1.

Step 1: Write the equilibrium expression
K_c = [PCl₅]/([PCl₃][Cl₂]) = 20.

Step 2: Update concentrations
After adding 0.2 mol of Cl₂:
[PCl₃] = 0.2 - x, [Cl₂] = 0.3 - x, [PCl₅] = 0.4 + x.

Step 3: Solve for x
20 = (0.4 + x)/[(0.2 - x)(0.3 - x)]. Solve for x to find x ≈ 0.086 mol.

Step 4: Calculate [PCl₅]
[PCl₅] = 0.4 + 0.086 ≈ 0.486 mol L^-1 = 49 × 10^-2 mol L^-1.

Step 1: Identify species with square planar geometry
- XeF₄: Square planar due to lone pairs in axial positions.
- BrF^-₄: Square planar for similar reasons.
- [Cu(NH₃)₄]²⁺: Square planar due to Jahn-Teller distortion.
- [PtCl₄]^2-: Square planar due to dsp² hybridization.
Others (SF₄, SiF₄, [FeCl₄]^2-) do not have square planar geometry.

Step 2: Count the species
There are 4 species with square planar geometry.