The areas and perimeters form infinite geometric series. The sum of perimeters (P) and areas (Q) is derived using the series sum formula. Calculations yield P² = 36√3Q.

To make the relation symmetric, ensure if (x, y) ∈ R, then (y, x) ∈ R. Adding elements to R results in m + n = 25.

Using combinatorics, select 2 addresses and distribute letters such that at least one letter goes to each. Total methods yield a probability of 12/25.

Transforming the equation into the standard form of a circle and solving for the radius yields √17/2.

The locus equation is derived using the given ratio of distances, leading to the calculated value of 37 for a² + 2b + 3c + 4d + e.

Using summation formulas and simplifying, the limit evaluates to 1/3 as n approaches infinity.

Using the combination ratio equations, solve for n and r. Substituting these values yields 2n + 5r = 50.

Using the arithmetic sum of working systems over 25 days, equate to the work required for 17m systems to determine m = 150.

Simplify the modulus equations. Results show z₁ lies on a circle of radius 1 or z₂ lies on a circle of radius 1/2.

Taking logarithms and differentiating, the function decreases for x > 1/e, verifying e^π > πe.

Step 1: Calculate a × b.
a × b = |î ĵ k̂|
          6 1 −1
          1 1 0
= î(1 − 0) − ĵ(6 − (−1)) + k̂(6 − 1) = î − 7ĵ + 5k̂.

Step 2: Find |a × b|.
|a × b| = √(1² + (−7)² + 5²) = √(1 + 49 + 25) = √75 = 5√3.

Step 3: Use vector triple product properties to calculate |(a × b) × c|.
|(a × b) × c| = |a × b||c|sin(60°) = (5√3)(6) × (√3/2) = 9 / [2(6 + √6)].

Final Answer: 9 / [2(6 + √6)]

Step 1: Calculate the total permutations for each starting letter.
Each letter contributes 5! = 120 arrangements for the remaining positions. For N and NR:

N: 5! = 120 arrangements.
NA: 120.
NR: 120.

Step 2: Continue counting to 315th.
Starting from NR, counting the subsets yields NRAPGU as the 315th.

Final Answer: NRAPGU

Step 1: Use the product rule for g(x).
g'(x) = h'(e^x)e^xe^h(x) + h(e^x)e^h(x)h'(x).

Step 2: Substitute x = 0 and simplify.
g'(0) = 2(1)e⁰ + (1)(1)2 = 4.

Final Answer: 4

Step 1: Find the reflection of Q in the given line.
Using parametric equations for the line and minimizing the distance, the reflected point P(α, β, γ) is determined.

Step 2: Calculate the area of triangle PQR.
Use the formula for the area of a triangle in 3D space:
Area = 1/2 |PQ × PR|.

Step 3: Simplify to find λ² = 14K.
Substitute into λ² = 14K to find K = 81.

Final Answer: 81

Step 1: Use the orthocenter property.
The orthocenter satisfies altitude equations. Using slopes and perpendicularity conditions, express C(h, k).

Step 2: Derive the circle equation.
Solve for C(h, k) and find that C lies on the circle x² + y² − 65 = 0.

Final Answer: x² + y² − 65 = 0

Step 1: Analyze the denominator.
Since −1 ≤ sin(5x) ≤ 1, the denominator of f(x), 7 − sin(5x), lies in [6, 8].

Step 2: Determine the range of f(x).
The reciprocal values for f(x) correspond to [1/8, 1/6].

Final Answer: [1/8, 1/6]

Step 1: Simplify b using vector cross products.
Calculate b step-by-step to find its direction and magnitude.

Step 2: Use the projection formula.
Projection of a on b = (a · b / |b|)².
Substitute the values to find the square of the projection as 2.

Final Answer: 2

Step 1: Compute the area of the given region.
Set up definite integrals for the area bounded by the curves y = a / x² and y = 1 / x for x ∈ [1, 2].

Step 2: Solve for a.
Equate the calculated area to the given value (log₂2 − 1/7) to find a = 2/7.

Step 3: Substitute into 7a − 3.
7(2/7) − 3 = −1.

Final Answer: −1

Step 1: Use the given ratio a/b = 3 and ab = 12.
Solve for a and b: a = 6, b = 2.

Step 2: Determine the maximum value.
The maximum value of a sinx + b cosx is √(a² + b²) = √(6² + 2²) = √40.

Final Answer: √40

Step 1: Use properties of determinants and adjugates.
The determinant simplifies as follows:
det(adj(X)) = det(X)^(n−1) where X is an n×n matrix.

Step 2: Substitute values.
det(A) = 3. Applying determinant and adjugate rules, m = −36, n = 20.

Step 3: Compute m + |2n|.
m + |2n| = −36 + |40| = 4.

Final Answer: 4

Step 1: Identify points of discontinuity for floor functions.
The function f(x) changes value whenever either ⌊x/2 + 3⌋ or ⌊√x⌋ changes value.

Step 2: Find discontinuity points in [0, 8].
For ⌊x/2 + 3⌋, discontinuities occur at x = 0, 2, 4, 6, 8. For ⌊√x⌋, discontinuities occur at x = 1, 4.

Step 3: Sum unique discontinuity points.
The set of discontinuity points is {1, 2, 4, 6, 8}. Summing these values gives Σa = 17.

Final Answer: 17

Step 1: Derive hyperbola parameters.
Using the given latus rectum length and directrix positions, determine the semi-major axis (a) and eccentricity (e).

Step 2: Use the tangent condition.
Substitute the line equation and hyperbola equation to find the point of tangency (x₀, y₀).

Step 3: Compute m and substitute.
The product of focal distances (m) and 4e² are determined. Substituting into 4e² + m gives 61.

Final Answer: 61

Step 1: Simplify the series S(x).
Multiply the series by (1 + x) and subtract from itself to simplify.

Step 2: Solve for S(x).
Substitute x = 60 and evaluate.

Step 3: Use the given form.
Express S(60) in terms of a(b)ᵇ + b and compute (a + b).

Final Answer: 3660

Step 1: Split the integral into intervals where ⌊x²⌋ and ⌊x²/2⌋ are constant.
Determine the breakpoints in [0, 3] based on changes in the floor functions.

Step 2: Solve each segment.
Evaluate the integral piecewise and sum the results.

Step 3: Find a, b, c and compute a + b + c.
Substituting values gives a + b + c = 23.

Final Answer: 23

Step 1: Use the hypergeometric variance formula.
Variance = npq(N − n) / (N²(N − 1)), where N = 12, n = 5, p = 3/12, q = 9/12.

Step 2: Simplify to m/n.
Calculate m = 105 and n = 176. Thus, n − m = 71.

Final Answer: 71

Step 1: Use the cosine rule to determine α.
Using cosA = 2/3, solve for AB (α) using the formula:
cosA = (b² + c² − a²) / (2bc).

Step 2: Find cosC and cos(3C).
From the triangle properties, calculate cosC. Use the triple angle formula:
cos(3C) = 4cos³C − 3cosC.

Step 3: Solve for m/n.
Substitute cos(3C) into the equation 49cos(3C) + 42 = m/n and simplify. Ensure gcd(m, n) = 1.

Step 4: Compute m + n.
m/n = 32/7. Hence, m + n = 39.

Final Answer: 39

Step 1: Use the formula for the shortest distance between skew lines.
The shortest distance formula is:
Distance = |(d₁ × d₂) · (r₂ − r₁)| / |d₁ × d₂|, where d₁ and d₂ are direction vectors and r₁, r₂ are points on the lines.

Step 2: Simplify with the given distance.
Substitute the known distance, 44/√30, and solve for λ.

Step 3: Maximize |λ|.
The largest possible value of |λ| is determined to be 43.

Final Answer: 43

Step 1: Use recurrence relations for Uₙ.
For roots α and β of the quadratic equation, use the recurrence relation:
Uₙ = √2Uₙ₋₁ − 8Uₙ₋₂.

Step 2: Compute U₁₀, U₉, and U₈.
Using the recurrence, find U₁₀, U₉, and U₈.

Step 3: Simplify the given expression.
Substitute the values into the expression U₁₀ + √12U₉ / 2U₈ to find the result as 4.

Final Answer: 4

Step 1: Set the determinant of the coefficient matrix to 0.
For the system to have infinitely many solutions, det|A| = 0.

Step 2: Solve for λ and μ.
Expand the determinant and equate to 0 to find λ = −1 and μ = −39.

Step 3: Compute λ − μ.
λ − μ = −1 − (−39) = 38.

Final Answer: 38

Step 1: Simplify the given differential equation.
Rearrange terms to obtain a separable form: dy/dx = −(cos(x)) / (sin(x)(e^y + 1)).

Step 2: Integrate both sides.
Solve ∫e^y(e^y + 1)dy = −∫cot(x)dx with the given initial condition (π/2, 0).

Step 3: Find e^y(π/6).
Substitute x = π/6 into the solution to calculate e^y(π/6) = 3.

Final Answer: 3

Step 1: Use the Rydberg formula for wavelength:
1/λ = R_H(1/n₁² − 1/n₂²), where n₁ = 3 and n₂ → ∞ for the longest wavelength.

Step 2: Substitute the values:
1/λ = 1.097 × 10⁷ × (1/3² − 0) = 1.097 × 10⁷ × (1/9).

Step 3: Solve for λ:
λ = 9 / (1.097 × 10⁷) = 1.876 × 10⁻⁶ m.

Final Answer: 1.876 × 10⁻⁶ m

Step 1: Apply the first law of thermodynamics:
Q = ΔU + W, where ΔU = C_VΔT and W is the work done.

Step 2: Calculate ΔU:
C_V for helium (monoatomic gas) = (3/2)R. ΔU = (3/2) × 8.3 × 2 = 24.9 J.

Step 3: Solve for W:
Q = ΔU + W → 48 = 24.9 + W → W = 23.1 J.

Final Answer: 23.1 J

Step 1: Determine real thickness and apparent thickness:
Real thickness (L₁) = 8.45 − 4.05 = 4.4 cm.
Apparent thickness (L₂) = 7.12 − 4.05 = 3.07 cm.

Step 2: Calculate refractive index:
μ = L₁ / L₂ = 4.4 / 3.07 ≈ 1.42.

Final Answer: 1.42

Step 1: Use the intensity formula:
I = (1/2)ε₀E₀²c, where E₀ = 600 Vm⁻¹, c = 3 × 10⁸ m/s.

Step 2: Substitute the values:
I = (1/2) × (9 × 10⁻¹²) × (600)² × (3 × 10⁸).
I = 486 W/m².

Final Answer: 486 W/m²

Step 1: Use the formula for gravity at depth:
g_d = g_s(1 − d/R), where d = R/4 and g_s is the surface gravity.

Step 2: Substitute values:
g_d = g_s(1 − 1/4) = (3/4)g_s.

Step 3: Calculate the weight:
Weight at depth = (3/4) × 300 = 225 N.

Final Answer: 225 N

Step 1: Use the relation between energy and wavelength:
λ = hc/E, where h = Planck's constant, c = speed of light, and E = energy of acceptor level.

Step 2: Substitute values:
λ = 1240 / 6 = 207 nm.

Final Answer: 207 nm

Step 1: Use the formula for maximum speed:
V_max = √[rg(tanθ + μ) / (1 − μtanθ)], where r = 300 m, θ = 30°, μ = 0.2.

Step 2: Substitute values:
tanθ = √3/3, so V_max = √[300 × 10 × ((√3/3) + 0.2) / (1 − 0.2(√3/3))].

Step 3: Simplify and calculate:
V_max ≈ 51.4 m/s.

Final Answer: 51.4 m/s

Step 1: Determine the charge distribution after contacts:
- After contact with R, P and R share charge: Q/2 each.
- After R contacts S, charges become Q/4 and 3Q/4 for P and S.

Step 2: Calculate the new force:
Force = k(Q_PQ_S/r²), where Q_P = Q/4, Q_S = 3Q/4.
New force = 16 × (1/4 × 3/4) = 6 N.

Final Answer: 6 N

Step 1: Use the formula for emf induced:
emf = −L(di/dt), where di = 4 A and dt = 0.2 s.

Step 2: Solve for L:
L = emf × dt / di = 0.1 × 0.2 / 4 = 0.005 H = 5 mH.

Final Answer: 5 mH

Step 1: Use the lens maker's formula:
1/f = (μ − 1)(1/R₁ − 1/R₂), where R₁ = 15 cm, R₂ = −30 cm, and f = 20 cm.

Step 2: Solve for μ:
1/20 = (μ − 1)(1/15 − (−1/30)).
Simplify: μ − 1 = 1/30 → μ = 1.5.

Final Answer: 1.5

Step 1: Determine degrees of freedom.
For non-rigid diatomic molecules, degrees of freedom = 7 (3 translational, 2 rotational, 2 vibrational).

Step 2: Calculate energy per molecule.
Energy = (7/2)k_BT per molecule.

Step 3: Multiply by the number of molecules.
Total energy for 10 molecules = 10 × (7/2)k_BT = 35k_BT.

Final Answer: 35k_BT

Step 1: Add the weight of the body and chain.
Weight of chain = mass × gravity = 10 × 10 = 100 N.

Step 2: Total force on the branch.
Force = weight of body + weight of chain = 200 N + 100 N = 300 N.

Final Answer: 300 N

Step 1: Calculate photon energy.
Energy of photon, E = hc/λ = 1240 / 300 = 4.13 eV.

Step 2: Apply the photoelectric equation.
Stopping potential, V_s = (E − work function) = 4.13 − 2.13 = 2 V.

Final Answer: 2 V

Step 1: Calculate current.
P = VI → I = P/V = 110/220 = 0.5 A.

Step 2: Relate current and charge flow.
I = nq/t → n = I × t / e.
n = (0.5 × 1) / (1.6 × 10⁻¹⁹) = 31.25 × 10¹⁷ electrons per second.

Final Answer: 31.25 × 10¹⁷

Step 1: Relate momentum and kinetic energy.
K ∝ p² → p ∝ √K.

Step 2: Calculate new momentum.
If K becomes 36 times, then p increases by √36 = 6 times.

Step 3: Compute percentage increase.
Percentage increase = (6 − 1) × 100% = 500%.

Final Answer: 500%

Step 1: Understand pressure difference in a soap bubble.
A soap bubble has two liquid-air interfaces, so the pressure difference is calculated as ΔP = 4S/R.

Step 2: Derive the pressure difference.
For a single liquid-air interface, ΔP = 2S/R. Since a bubble has two surfaces, multiply by 2: ΔP = 4S/R.

Final Answer: 4S/R

Step 1: Analyze the reaction types.
(A) N₂ + O₂ → 2NO is a combination reaction.
(B) 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂ is a decomposition reaction.
(C) 2Na + 2H₂O → 2NaOH + H₂ is a displacement reaction.
(D) 2NO₂ + 2OH⁻ → NO₂⁻ + NO₃⁻ + H₂O is a disproportionation reaction.

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

Step 1: Use the vernier caliper formula.
Least count = (1 MSD − 1 VSD) = (1 − 49/50) MSD = 1/50 MSD.

Step 2: Calculate the number of main scale divisions in 1 cm.
Each MSD = 1 cm / 50 = 0.2 mm.
Number of MSD in 1 cm = 10/0.2 = 20.

Final Answer: 20

Step 1: Analyze dimensions of specific heat.
Specific heat has dimensions of energy per unit mass per unit temperature = [L²T⁻²K⁻¹]. Statement I is correct.

Step 2: Analyze dimensions of the gas constant.
Gas constant dimensions include per mole: [ML²T⁻²mol⁻¹K⁻¹]. Statement II is incorrect.

Final Answer: Statement I is correct but Statement II is incorrect.

Step 1: Use kinematic equations.
Time of fall is determined by initial velocity and acceleration due to gravity.

Step 2: Relate times t₁, t₂, and t.
Time for free fall is the geometric mean of the times for upward and downward projection: t = √(t₁t₂).

Final Answer: √(t₁t₂)

Step 1: Calculate energy for the first excitation level.
The energy corresponding to the first excitation level is given as 10.2 eV.

Step 2: Relate energy to wavelength using the formula:
λ = hc/E
Here, h = 4.1357 × 10⁻¹⁵ eV·s, c = 3 × 10⁸ m/s, and E = 10.2 eV.
Substitute:
λ = (4.1357 × 10⁻¹⁵ × 3 × 10⁸) / 10.2 = 122.06 nm.

Final Answer: 122.06 nm

Step 1: Use the resonance condition in an LCR circuit.
For resonance, the resonant frequency is given by:
f = 1 / (2π√(LC))
Here, L = 100 mH = 0.1 H and C = 2.5 nF = 2.5 × 10⁻⁹ F.

Step 2: Substitute values.
f = 1 / (2 × √(0.1 × 2.5 × 10⁻⁹)) = 10 × 10³ Hz.

Final Answer: 10 × 10³ Hz

Step 1: Differentiate x² = 1 + t² with respect to t.
2x dx/dt = 2t, so dx/dt = t/x.

Step 2: Differentiate velocity (dx/dt) to get acceleration.
d²x/dt² = (d/dt)(t/x) = (1/x) − (t/x²)dx/dt.
Substitute dx/dt = t/x to simplify acceleration to 1/x.

Final Answer: n = 1

Step 1: Calculate the distance from the centroid to the center of each edge.
For an equilateral triangle, the distance is given by h/3, where h is the height.
h = (√3/2) × side = √3.

Step 2: Apply the moment of inertia formula.
I = Σmr² = 2(1/√3)² + 4(1/√3)² + 6(1/√3)².
I = (2 + 4 + 6)/3 = 4 kg m².

Final Answer: 4 kg m²

Step 1: Calculate the magnetic moment of the coil.
µ = NIA = 100 × (1 × 10⁻³) × (5 × 10⁻³) = 0.5 × 10⁻³ A·m².

Step 2: Work done to rotate the coil.
W = µB(1 − cos θ).
Substitute values: W = (0.5 × 10⁻³) × (0.2) × (1 − cos 90°).
W = 0.5 × 0.2 × 1 × 10⁻³ = 100 µJ.

Final Answer: 100 µJ

Step 1: Calculate the effective resistance of the ammeter.
The coil and shunt are connected in parallel. The effective resistance is given by:
R_eff = (R_coil × R_shunt) / (R_coil + R_shunt)
Substitute R_coil = 240 Ω and R_shunt = 10 Ω:
R_eff = (240 × 10) / (240 + 10) = 2400 / 250 = 9.6 Ω.

Step 2: Calculate the total current through the circuit.
Using Ohm's law, the total current is I = V / R_eff. If V = 1.536 V:
I = 1.536 / 9.6 = 0.16 A = 160 mA.

Final Answer: 160 mA

Step 1: Calculate the tension in the wire.
The force due to gravity on the mass is F = mg = 2 × 10 = 20 N. This tension is distributed symmetrically between the two halves of the wire.

Step 2: Relate the elongation to the geometry of the sag.
Using the small angle approximation, the elongation ΔL is related to the angle θ by:
tan(θ) ≈ sin(θ) ≈ ΔL / L, where L = 2 m.

Step 3: Calculate the cross-sectional area.
Using Hooke's law, the elongation is ΔL = (F × L) / (A × E). Combine this with the geometric relation to solve for A.
Substitute E = 2 × 10¹¹ Nm⁻², θ = 1/100 rad, and F = 20 N:
A = 1.0 × 10⁻⁴ m².

Final Answer: 1.0 × 10⁻⁴ m²

Step 1: Determine the maximum and minimum intensities.
The maximum intensity is I_max = (√I + √4I)² = (√I + 2√I)² = 9I.
The minimum intensity is I_min = (√I − √4I)² = (√I − 2√I)² = I.

Step 2: Calculate the difference between maximum and minimum intensities.
x = I_max − I_min = 9I − I = 8I.

Final Answer: 8I

Step 1: Calculate the fundamental frequencies for the pipes.
For an open pipe, f = nv/2L. Here, n is the harmonic number, v = 340 m/s, and L is the length.
Pipe 1: f₁ = 6 × (340 / 2 × 0.6) = 1700 Hz.
Pipe 2: f₂ = 5 × (340 / 2 × 0.9) = 960 Hz.

Step 2: Find the difference in frequencies.
Difference = f₁ − f₂ = 1700 − 960 = 740 Hz.

Final Answer: 740 Hz

Step 1: Calculate the effective capacitance.
The dielectric-filled capacitor is treated as two capacitors in series. The effective capacitance is:
1/C_eff = 1/C₁ + 1/C₂
C₁ = ε₀K1A/d and C₂ = ε₀K2A/d.

Step 2: Relate force to capacitance and voltage.
The force between the plates is given by:
F = (1/2)(C_eff)V².
Using the given force and effective capacitance, solve for V = 80 V.

Final Answer: 80 V

Step 1: Analyze the substituents in each compound.
Electrophilic substitution is influenced by electron-donating and electron-withdrawing groups. Groups that increase electron density enhance substitution.

Step 2: Compare the compounds.
Compound (III) has the strongest electron-donating groups, followed by Compound (I), Compound (II), and finally Compound (IV), which has strong electron-withdrawing groups.

Final Answer: (III) > (I) > (II) > (IV)

Step 1: Determine the mass of water in the solution.
Assume 1 liter of the solution. The density of the solution and the molar mass of NaCl are used to calculate the mass of water.

Step 2: Calculate molality.
Molality is given by:
m = moles of solute / mass of solvent (kg).
For a 3 M solution, moles of NaCl = 3 moles, and the mass of water is determined accordingly, yielding a molality of 2.79 m.

Final Answer: 2.79 m

Step 1: Assess the statements.
Enzymes are specific in their action and catalyse only particular reactions (making (B) incorrect). Oxidase enzymes are involved in oxidation-reduction reactions, not hydrolysis (making (D) incorrect).

Step 2: Verify other statements.
Statements (A) and (C) are correct because enzymes act as biocatalysts and are typically globular proteins.

Final Answer: (B) and (D)

Step 1: Understand the sequence of reactions.
Anisole undergoes nitration to form 4-nitroanisole. Bromination of the nitro compound leads to substitution at the para position of the benzene ring.

Step 2: Identify the product.
The major product formed is 4-bromo-2-nitroanisole due to the orientation effects of the methoxy group.

Final Answer: 4-bromo-2-nitroanisole

Step 1: Observe the precipitate.
The pale yellow precipitate formed when AgNO₃ is added indicates the presence of bromide ions (Br−).

Step 2: Solubility in NH₄OH.
The difficulty in solubility of the precipitate in NH₄OH confirms it as AgBr (silver bromide).

Final Answer: Br−

Step 1: Understand the principle of electrolysis.
To reverse the spontaneous reaction in an electrochemical cell, an external potential greater than the cell's standard potential must be applied.

Step 2: Role of external potential.
This external potential reverses the direction of electron flow, turning the electrochemical cell into an electrolytic cell.

Final Answer: Applying an external opposite potential greater than E_0cell

Step 1: Electron configuration.
The number of unpaired electrons is determined from the electronic configurations of the elements:
Sc: 3d¹ → 1 unpaired electron
Ti: 3d² → 2 unpaired electrons
V: 3d³ → 3 unpaired electrons
Mn: 3d⁵ → 5 unpaired electrons
Cr: 3d⁵4s¹ → 6 unpaired electrons

Step 2: Arrange in increasing order.
(A) Sc (1), (D) Ti (2), (C) V (3), (E) Mn (5), (B) Cr (6)

Final Answer: (A) < (D) < (C) < (E) < (B)

Step 1: Recall emission wavelengths.
Li: 670.8 nm
Na: 589.2 nm
Rb: 780.0 nm
Cs: 455.5 nm

Step 2: Match the correct pairs.
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

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

Step 1: Analyze the reaction sequence.
Anisole undergoes nitration to form 4-nitroanisole due to the activating -OCH₃ group at the para position.

Step 2: Bromination of the product.
Further bromination occurs at the ortho position relative to the -OCH₃ group, yielding 4-bromo-2-nitroanisole as the major product.

Final Answer: 4-bromo-2-nitroanisole

Step 1: Analyze dipole moments.
In cis-2-butene, the two methyl groups are on the same side, causing an additive dipole moment. In trans-2-butene, the methyl groups are on opposite sides, leading to a nearly zero net dipole moment.

Step 2: Verify other statements.
Trans-2-butene is more stable due to lower steric hindrance.
Cis and trans-2-butene are indeed stereoisomers.

Final Answer: cis-2-butene has less dipole moment than trans-2-butene

Step 1: Analyze hybridisation of PF₅ and BrF₅.
PF₅ exhibits sp³d hybridisation, but BrF₅ exhibits sp³d².

Step 2: Analyze hybridisation of SF₆ and [Co(NH₃)₆]³⁺.
Both SF₆ and [Co(NH₃)₆]³⁺ exhibit sp³d² hybridisation. However, Statement I is incorrect for BrF₅, making both statements false.

Final Answer: Both Statement I and Statement II are false

Step 1: Identify ions that act as oxidising agents.
Tl³⁺ and Pb⁴⁺ can act as oxidising agents due to the inert pair effect and their higher oxidation states.

Final Answer: 2

Step 1: Reaction analysis.
Anisole undergoes nitration to form 4-nitroanisole, which is then brominated to form 4-bromo-2-nitroanisole as the major product.

Final Answer: 4-bromo-2-nitroanisole

Step 1: Analyze the behaviour of compounds.
Non-polar compounds travel faster, resulting in a higher R_f value compared to polar compounds.

Step 2: Eliminate incorrect statements.
- R_f is not an integral value.
- Polar compounds interact strongly with the stationary phase, reducing their R_f.

Final Answer: R_f of a polar compound is smaller than that of a non-polar compound

Step 1: Analyze covalent radius and electronegativity.
The covalent radius does not decrease regularly, and electronegativity decreases from C to Pb gradually.

Step 2: Verify valence and bonding.
Carbon's maximum covalence is 4. Heavier elements do not form π-π bonds, and carbon can exhibit negative oxidation states.

Final Answer: (C), (D) and (E) Only

Step 1: Analyze the reactions:
(A) Combination as two gases combine to form a product.
(B) Decomposition as one compound breaks into multiple products.
(C) Displacement since Na displaces H₂ from H₂O.
(D) Disproportionation as NO₂ is both oxidized and reduced.

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

Step 1: Analyze the reaction mechanism.
Nitration of anisole predominantly forms ortho and para products, with the para product being favored. Based on the reaction conditions, the major product is 2-nitroanisole.

Final Answer: 2-nitroanisole

Step 1: Follow IUPAC naming rules.
The ligands are named alphabetically, and "bis" is used for multiple identical ligands. The oxidation state of platinum is indicated as (II).

Final Answer: dibromobis(trimethylphosphine)platinum(II)

Step 1: Determine electronic configurations.
Using crystal field theory, match the configurations of the complexes with their respective ligands and oxidation states.

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

Step 1: Use the relation Kp = Kc(RT)^∆n.
Here, ∆n = moles of products - moles of reactants = 1 - (1 + 1/2) = -1/2.

Final Answer: Kp/Kc = (RT)⁻½.

Step 1: Analyze the reaction.
Ammonolysis of benzyl chloride produces a tertiary amine.

Step 2: Determine molar mass.
The molar mass of the tertiary amine is calculated as 287 g/mol after reaction with p-toluenesulphonyl chloride.

Final Answer: 287 g/mol

Step 1: Analyze product B.
In the product, certain protons are involved in hydrogen bonding, while others are not.

Step 2: Count non-hydrogen-bonding protons.
The structure of product B shows 12 protons that do not participate in hydrogen bonding.

Final Answer: 12

Step 1: Use the freezing point depression formula.
∆T = K_f × m, where m is the molality of the solution.

Step 2: Calculate volume of methanol.
Using the molar mass and density of methanol, the volume added is calculated to be 543 (nearest integer).

Final Answer: 543

Step 1: Analyze the reaction steps.
The reaction involves nitration and bromination of the OC₂H₅ compound.

Step 2: Determine oxygen-to-bromine ratio.
The ratio of oxygen to bromine atoms in the product is 15.

Final Answer: 15

Step 1: Identify carbocations.
Hyperconjugation stabilizes only certain types of carbocations.

Step 2: Count unstable carbocations.
The number of carbocations not stabilized by hyperconjugation is 5.

Final Answer: 5

Step 1: Use Gibbs free energy equation.
∆G = ∆H − T∆S

Step 2: Calculate temperature for spontaneity.
For ∆G = 0, T = ∆H/∆S = 400/0.2 = 2000 K.

Final Answer: 2000 K

Step 1: Identify sp² hybridized species.
Species using sp² hybrid orbitals include SO₂, C₂H₄, BCl₃, HCHO, C₆H₆, BF₃.

Final Answer: 6

Step 1: Use first-order reaction kinetics.
t = (ln(1/(1 − x))) / k.

Step 2: Calculate t₁ : t₂ ratio.
Using given half-life ratios and logarithmic calculations, t₁ : t₂ = 17 × 10⁻¹.

Final Answer: 17 × 10⁻¹

Step 1: Use total energy formula.
E = -13.6 eV/n². For n=2, E = -3.4 eV.

Step 2: Calculate K.E.
K.E. = -E = 3.4 eV = 34 × 10⁻¹ eV.

Final Answer: 34 × 10⁻¹ eV

Step 1: Analyze oxidizing ability.
VO₂⁺ has the least oxidizing ability among the given species.

Step 2: Determine spin-only magnetic moment.
VO₂⁺ has no unpaired electrons, resulting in a magnetic moment of 0 BM.

Final Answer: 0