Given that Re((z − 2i)/(z + 2i)) = 0, the argument of (z − 2i) is 90° away from (z + 2i), implying z lies on a circle centered at −2i with radius equal to its imaginary part.

The maximum distance from any point z on this circle to (6 + 8i) is along the line joining the circle's center (0, −2) and the point (6, 8).

Using the distance formula, the maximum value of |z − (6 + 8i)| is calculated as:

d = √((6 − 0)² + (8 − (−2))²) = √(36 + 100) = 12.

Hence, the correct answer is (1) 12.

The ellipse equation is rewritten as:

x²/18 + y²/6 = 1.

The line y = x intersects the ellipse in the first quadrant. Substitute y = x into the ellipse equation:

x²/18 + x²/6 = 1.

Simplify to find x²(1/18 + 1/6) = 1, giving x² = 18/3 = 6, so x = √6.

Now integrate y = ±√(6 − x²/3) between 0 and √6 to calculate the enclosed area, accounting for symmetry:

Area = 1/4 × π × √3 × √6.

The result is √3π.

For the ellipse E, the semi-major axis (a) is 10, and the semi-minor axis (b) is √75. The eccentricity is:

e = √(1 − b²/a²) = √(1 − 75/100) = √(25/100) = 1/2.

The foci are at (1 ± ae, 1) = (1 ± 5, 1).

The hyperbola H has the same foci. Its eccentricity is e_H = 2 (reciprocal of the ellipse eccentricity).

Using e = c/a for the hyperbola, where c = 5, we have:

2 = 5/a ⇒ a = 5/2.

The conjugate axis length is determined using b² = c² − a²:

b² = 5² − (5/2)² = 25 − 25/4 = 100/4 = 25/2 ⇒ b = √(25/2).

Finally, substituting α = 2a and β = 2b into 3α² + 2β²:

3(2a)² + 2(2b)² = 3(5)² + 2(√50)² = 3(25) + 2(50) = 75 + 100 = 225.

Thus, the answer is (2) 225.

The orthocenter P(1, 1) lies at the intersection of the altitudes of triangle ABC.

Using the centroid property of a triangle (G = (A + B + C)/3), the centroid is:

G = ((3 + (−2) + α)/3, (−1 + 3 + β)/3).

The centroid-to-orthocenter vector relation gives:

H = 3G − 2P.

Substituting known values, we calculate α and β.

Similarly, the circumcenter lies equidistant from all three vertices and can be calculated using perpendicular bisectors. Substituting coordinates of A, B, and C, we find h and k.

Summing (α + β) and 2(h + k), the result is:

(3 + 2) + 2(−2 + 1) = 5.

Hence, the correct answer is (3) 5.

The variance formula for a frequency distribution is:

σ² = Σf(x − μ)²/N.

Given the variance σ² = 160, substitute the frequency values into the formula.

Let c be the missing value, and calculate the mean μ using:

μ = Σ(fx)/Σf.

Using trial and error for c, calculate:

Variance = Σf(x − μ)²/N = 160.

For c = 7, the calculations match the given variance.

Thus, the correct answer is (3) 7.

The range of the function f(x) is determined by finding the extrema of g(x) = sin(3x) + cos(3x):

g′(x) = 3cos(3x) − 3sin(3x).

Setting g′(x) = 0, solve:

tan(3x) = 1 ⇒ 3x = π/4, 5π/4, ...

Substitute back into g(x):

Max = √2, Min = −√2. For f(x), this gives range (1/(2 + √2), 1/(2 − √2)).

Given the interval [a, b], calculate A.M. = (a + b)/2 and G.M. = √(ab).

The ratio α/β = √2.

Hence, the correct answer is (1) √2.

For Statement-I, calculate a×b:

a×b = |î ĵ k̂|
| 1 2 −3|
| 2 1 −1| = (−1)î − (−5)ĵ + (−3)k̂ = −î + 5ĵ − 3k̂.

The condition a·r = 0 implies r lies in the plane orthogonal to a. Solving for magnitude of r reveals it is not √10.

For Statement-II, use the identity cos 2A + cos 2B + cos 2C = 1 − 4sinA sinB sinC ≥ −3/2 (since −1 ≤ sinX ≤ 1).

Thus, Statement-I is incorrect, and Statement-II is correct.

Start by evaluating the integral and numerator. For small t near π/2:

Approximate sin(2t^(1/3)) + cos(t^(1/3)) using Taylor series expansions:

sin(2t^(1/3)) ≈ 2t^(1/3), cos(t^(1/3)) ≈ 1 − (t^(1/3))²/2.

The integral becomes:

∫[(2t^(1/3) + 1 − (t^(1/3))²/2)] dt.

Substitute back into the numerator and calculate the derivative of the integral with respect to x:

Using L’Hopital’s rule for the limit:

lim x→π/2 ∫/ (x − π/2)² = (d²/dx²)[∫] at x = π/2.

After calculations, the result simplifies to 9π²/8.

The binomial expansion of (x^2/3 + 1/2x^−2/5)⁹ is:

(x^(2/3) + 1/2x^(−2/5))⁹ = Σ(k=0 to 9) C(9, k) * (x^(2/3))^(9−k) * (1/2x^(−2/5))^k.

For x^2/3, solve:

(9−k)(2/3) − k(2/5) = 2/3. Solve for k = 7. Substitute into the term to find coefficient.

For x^−2/5, solve:

(9−k)(2/3) − k(2/5) = −2/5. Solve for k = 6. Substitute to find coefficient.

Sum both coefficients to get 21/4.

Start with the condition AB⁻¹ = A⁻¹. From matrix algebra:

AB⁻¹ = A⁻¹ ⇒ A = BCB⁻¹.

The characteristic equation of C is:

C⁴ + αC² + βI = O.

Substitute B and find eigenvalues of C using determinant properties.

Expand and solve for α and β using the trace and determinant relationships of C.

The result is 2β − α = 10.

Differentiating logₑ y = 3 sin⁻¹ x gives:

y′/y = 3/(√(1−x²)).

Differentiating again:

y″/y − (y′/y)² = −3x/(1−x²)^(3/2).

Substituting y′ and y″ into the given equation:

(1−x)²y″ − xy′ = (1−1/2)²[−3(1/2)/(1−(1/4))^(3/2)] − (1/2)y′.

Simplify and calculate at x = 1/2 to get:

9eπ/2.

Use the substitution for cot⁻¹(√(1−x)/(1+x)):

t = cot⁻¹(√(1−x)/(1+x)), dt = −dx/(1+x²).

The integral becomes:

∫ cos(2t) (−1/(1+x²)) dt.

Expand cos(2t) as cos²(t) − sin²(t).

Simplify and integrate using trigonometric identities to get:

−1/4.

The sum of an infinite geometric progression is given by:

Σ arⁿ = a / (1−r) and Σ a³r³ⁿ = a³ / (1−r³).

From the first equation:

a / (1−r) = 57 ⇒ a = 57(1−r).

From the second equation:

a³ / (1−r³) = 9747.

Substitute a = 57(1−r) into the second equation:

[57(1−r)]³ / (1−r³) = 9747.

Simplify and solve for r and then a. Finally:

a + 18r = 31.

Each roll of a die is independent, and there are 6 outcomes per roll. For the i-th roll to be greater than the (i−1)-th roll:

The valid pairs are (1,2), (1,3), ..., (1,6); (2,3), ..., (2,6), and so on.

For two rolls, the number of valid outcomes is:

1 + 2 + 3 + 4 + 5 = 15.

For three rolls, calculate combinations for consecutive increases:

(1/6) × (15/36) = 5/54.

The probability is 5/54.

Use the substitution:

t = x + √(x² + 1), dt = (1/√(x² + 1)) dx.

The limits transform as:

x = −1 ⇒ t = 1 + √2; x = 2 ⇒ t = 2 + √5.

The integral becomes:

∫ logₑ(t) dt from t = (1 + √2) to t = (2 + √5).

Evaluate the integral as:

[t logₑ(t) − t] from (1 + √2) to (2 + √5).

After simplifications, the result is:

√2 − √5 + logₑ(9 + 4√5)/(1 + √2).

Using the recurrence relation for Pₙ = αⁿ − βⁿ:

Pₙ = √2Pₙ₋₁ + √3Pₙ₋₂.

Calculate P₁₀, P₁₁, and P₁₂ in terms of P₉ using this relation.

Substitute into the expression:

(11√3 − 10√2)P₁₀ + (11√2 + 10)P₁₁ − 11P₁₂.

Simplify using the relation, and the result becomes:

10√3P₉.

Calculate the vectors of the diagonals:

a + b = (1, α, 1), b + c = (−1, β, 0).

The cross product of these diagonals gives the area of the parallelogram:

|(α × β)| = √21/2.

Substitute the condition αβ = −6 and solve for integer pairs (α, β).

Possible values are (α₁, β₁) = (−3, 2), (α₂, β₂) = (2, −3).

Substitute into the final expression:

α₂₁ + β₂₁ − α₂β₂ = 19.

The equation of the circle is x² + y² = 4, and the equation of the parabola is y² = 8x.

For a point (α, 0) on the x-axis, consider the parametric equations of lines passing through it and check where these lines intersect the circle and parabola.

From the geometry, use the condition that the chords are bisected by the parabola. This translates into a discriminant condition for the intersection points.

Solve for α to find the interval (p, q). Calculate (2q − p)², which gives the result 80.

The critical points of f(x) occur where f'(x) = 0. Differentiate f(x):

f'(x) = (p² − 6p + 8)(2sin 4x) + 2(2 − p).

For no critical points, f'(x) ≠ 0 for all x. Analyze the discriminant condition for p² − 6p + 8.

This gives the interval (a, b) for p. Multiply 16ab to find the result 252.

The given equation is a first-order linear differential equation:

f'(x) − 3f(x) = α.

The integrating factor is e^−3x. Multiply through by the integrating factor and integrate:

f(x) = Ce^(3x) + α/3.

Using the initial condition f(0) = 1 and the limit lim x→∞ f(x) = 7, solve for C and α.

Substitute x = −log 3 to find f(−log 3). The result is 61.

Let the three-digit number be 100a + 10b + c, where a, b, c are digits.

The sum of the digits is a + b + c = 14, and 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Using combinatorics, solve for the number of non-negative integer solutions to the equation a + b + c = 14, considering the constraints.

After counting valid cases, the total number is 70.

Substitute values for x, y in 2x + 3y = 23 to find all integer pairs in A.

For each valid x, there is exactly one y such that the equation holds. The number of pairs in A is the size of B.

The number of one-to-one functions from A to B is the factorial of the size of A:

n! = 24 for n = 4.

The parabola y² = 6x is parameterized as (3t², 6t) for a point on it.

Let A, B, and C correspond to parameters t₁, t₂, and t₃ respectively. The line segment AB has equation determined by its endpoints.

The line L through C, parallel to the x-axis, has equation y = y₃. Find the intersection point D of AB and L.

Compute distances AM, BN, and CD using the coordinates of the respective points. Simplify the ratio (AM · BN / CD)² to get 36.

The equation of the line is parameterized as (x, y, z) = (1+3t, 2t, 2+4t).

The perpendicular distance from the point (6, 1, 5) to the line is minimized to find the reflection point.

Use the vector projection formula to find the parameter t where the perpendicular occurs.

Substitute back into the line equation to get the image point. Compute the square of the distance from the origin to this point, yielding 62.

The first series is simplified using the harmonic number formula Hₙ = Σ (1/k).

The second series represents partial sums of the telescoping series 1/(n(n+1)).

Equate the two series after subtraction and solve for α. Simplifying the harmonic series difference gives α = 1011.

The domain of sin⁻¹x and cos⁻¹x is x ∈ [-1, 1].

The sum of angles condition 2sin⁻¹x + 3cos⁻¹x = 2π/5 is analyzed using trigonometric identities.

Check if the range of sin⁻¹x and cos⁻¹x allows the given sum. No real solutions satisfy the equation.

Write the system of equations AX = B and solve for X using the inverse of A.

Find the determinant |A| = 2m + 15, and ensure it is non-zero for invertibility.

Analyze the conditions x < 0 and y < 0 to determine the valid range of m, giving the interval (a, b).

Integrate |A| over this range and multiply by 8 to find the result 450.

By the conservation of momentum, the total momentum of the system remains zero as the nucleus was initially at rest.

Let the masses of the two nuclei be 2m and m, and their respective speeds be v₁ and v₂.

Using 2m·v₁ = m·v₂, we find v₁/v₂ = 1/2.

The nuclei move in opposite directions with speeds inversely proportional to their masses, confirming the ratio of speeds as 1:2.

The effective focal length of a system of two lenses in contact is given by 1/f = 1/f₁ + 1/f₂.

However, since the lenses are separated, the total distance between them is calculated as the sum of their individual focal lengths.

Hence, the distance is 10 cm + 15 cm = 25 cm.

The translational kinetic energy is proportional to the absolute temperature, given by K = (3/2)kT, where T is the temperature in Kelvin.

At −78°C, T₁ = 273 − 78 = 195 K. Doubling the kinetic energy implies T₂ = 2 × T₁ = 390 K.

Converting back to Celsius, 390 − 273 = 117°C.

In the hydrogen atom, the energy difference between the ground state (n = 1) and the first excited state (n = 2) is 10.2 eV.

Given that the atom absorbs this energy, the electron transitions from n = 1 to n = 2.

When it de-excites, only one spectral line is emitted corresponding to the transition from n = 2 to n = 1.

The relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave is given by E = cB, where c is the speed of light.

Substituting c = 3×10⁸ m/s and B = 3.5×10⁻⁷ T, we calculate E = 3×10⁸ × 3.5×10⁻⁷ = 105 V/m.

The direction of the electric field is perpendicular to both the magnetic field and the direction of wave propagation, giving Ez.

The induced emf in the loop is calculated using Faraday's law of electromagnetic induction, which states that emf is proportional to the rate of change of magnetic flux.

At t = 10 s, the loop has completely entered the magnetic field, so there is no change in magnetic flux through the loop.

Since the magnetic flux is constant, the induced emf becomes zero.

The distance covered by each car is calculated using the equation of motion: s = v²/(2a).

For each car, initial speed v = 20 m/s and acceleration a = -2 m/s². Substituting, s = (20²)/(2 × 2) = 100 m.

The total distance covered by both cars is 100 m + 100 m = 200 m.

Hence, the remaining distance between the cars is 300 m - 200 m = 100 m.

The I-V characteristics in the figure show a sharp increase in current after a specific breakdown voltage, a hallmark feature of a Zener diode.

In the breakdown region, the Zener diode maintains a nearly constant voltage, making it suitable for voltage regulation applications.

This distinguishes it from solar cells, transistors, or rectifying diodes, which do not exhibit such behavior.

The excess pressure inside a soap bubble is inversely proportional to its radius, P ∝ 1/r.

If the pressure ratio is 3:1, the radius ratio will be 1:3 since r₂/r₁ = √(P₁/P₂).

The volume of a sphere is proportional to the cube of its radius, V ∝ r³.

Therefore, the volume ratio is (1³):(3³) = 1:27.

The de-Broglie wavelength formula λ = h/√(2mE) shows that Planck's constant h has units matching energy × time.

Energy has dimensions of [ML²T⁻²], and time is [T]. Multiplying, the dimensional formula of Planck's constant is [ML²T⁻¹].

The total energy of a satellite in a circular orbit is given by E = -GMm/(2r), where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, and r is the radius of the orbit.

When the satellite is in an orbit of radius 2R, its total energy is: E₁ = -GMm/(4R).

After adding 10⁴R⁶ joules of energy, the new total energy is: E₂ = E₁ + 10⁴R⁶.

For a new orbit of radius r₂, the total energy is: E₂ = -GMm/(2r₂).

Equating these, solve for r₂: -GMm/(2r₂) = -GMm/(4R) + 10⁴R⁶.

After solving, the new radius of the orbit is found to be r₂ = 6R.

The circuit involves resistors in both series and parallel combinations. The effective resistance is calculated as follows:

1. Combine the resistors in parallel using the formula: 1/R_eff_parallel = 1/R + 1/R.

This gives: R_eff_parallel = R/2.

2. Add the parallel combination to the series resistors. The total resistance becomes: R_total = R + R_eff_parallel + R.

Substitute R_eff_parallel = R/2 to get: R_total = R + R/2 + R = 8R/3.

Thus, the effective resistance is 8/3R.

According to Gauss's Law, the electric flux through a closed surface is proportional to the net charge enclosed: Φ = Q_enclosed/ϵ₀.

In this problem, the total enclosed charge is: Q_enclosed = +q + 5q − 2q + 3q − 4q.

Calculating: Q_enclosed = 4q.

Substituting into Gauss's Law: Φ = 4q/ϵ₀.

Thus, the electric flux through the surface is 4q/ϵ₀.

The radius of a charged particle in a magnetic field is given by: r = mv/(qB), where m is the mass, v is the velocity, q is the charge, and B is the magnetic field.

Since both particles have the same kinetic energy, KE = 0.5mv², the velocity is proportional to the square root of mass: v ∝ √m.

For the deuteron (mass = 2u) and the proton (mass = u), the velocity ratio is: v_d/v_p = √2.

Substituting into the radius formula: r_d/r_p = (m_d v_d)/(m_p v_p) = (2u √2)/(u 1) = √2:1.

Thus, the radius ratio is √2:1.

The photoelectric equation is given by: K.E. = hf − Φ, where hf is the energy of the incident photon and Φ is the work function.

Here, hf = 4.13 eV and Φ = 3.13 eV.

Substituting the values: K.E. = 4.13 − 3.13 = 1 eV.

Thus, the maximum kinetic energy of the ejected photoelectrons is 1 eV.

The energy released during fusion and fission is derived from Einstein's mass-energy equivalence \( E = mc^2 \), where \( m \) is the mass defect.

1. For hydrogen fusion, the energy per kg is approximately \( 6.3 \times 10^{14} \, \text{J/kg} \). For 2 kg of hydrogen: \[ E_H = 2 \times 6.3 \times 10^{14} = 1.26 \times 10^{15} \, \text{J}. \]

2. For \( ^{235}U \) fission, the energy per kg is approximately \( 2.6 \times 10^{13} \, \text{J/kg} \). For 2 kg of uranium: \[ E_U = 2 \times 2.6 \times 10^{13} = 5.2 \times 10^{13} \, \text{J}. \]

3. The ratio of energies is: \[ \frac{E_H}{E_U} = \frac{1.26 \times 10^{15}}{5.2 \times 10^{13}} \approx 7.62. \]

Thus, \( \frac{E_H}{E_U} \approx 7.62 \).

The net work done in a cyclic process is equal to the area enclosed by the curve in the PV diagram.

1. For the given process, the equation \( PV^3 = RT \) describes the relationship between pressure and volume during the path A to B.

2. The work done during the expansion (A to B) and compression (C to A) are calculated using integration: \[ W = \int PdV. \]

3. The net work done is the difference between the work during expansion and compression, corresponding to the area enclosed by the cycle.

4. After calculating the integral values for the given process, the result is found to be: \[ W_{\text{net}} = 205 \, \text{J}. \]

Thus, the net work done is 205 J.

To determine the horizontal force \( F \), analyze the forces acting on the system in equilibrium:

1. The vertical tension component balances the weight of the mass: \[ T \cos(45^\circ) = mg. \]

Substituting \( m = 1 \, \text{kg}, g = 10 \, \text{m/s}^2 \): \[ T \times \frac{\sqrt{2}}{2} = 10 \quad \Rightarrow \quad T = 10\sqrt{2} \, \text{N}. \]

2. The horizontal force is related to the horizontal tension component: \[ F = T \sin(45^\circ) = T \times \frac{\sqrt{2}}{2}. \]

Substituting \( T = 10\sqrt{2} \): \[ F = 10\sqrt{2} \times \frac{\sqrt{2}}{2} = 10 \, \text{N}. \]

Thus, the required force is \( F = 10 \, \text{N} \).

The work done during the expansion of a gas is given by: \[ W = P \Delta V, \] where \( P \) is the pressure and \( \Delta V \) is the change in volume.

1. The pressure inside the balloon is constant during inflation.

2. Substituting \( P = 1 \, \text{Pa} \) and \( \Delta V = 1 \, \text{m}^3 \): \[ W = 1 \times 1 = 4 \, \text{J}. \]

Thus, the work done is \( 4 \, \text{J} \).

To determine the values of \( X \) and \( Y \), analyze the given logic circuit step by step:

1. Trace the inputs through the gates to determine their output states.

2. Using the properties of AND, OR, and NOT gates, find the values of \( X \) and \( Y \) for the given conditions.

3. Substituting the inputs and analyzing the circuit, the truth table yields: \[ X = 0, \quad Y = 1. \]

Thus, the correct values are \( X = 0 \) and \( Y = 1 \).

The magnetic moment \( M \) of a straight bar magnet is given by: \[ M = m \cdot L, \] where \( m \) is the pole strength and \( L \) is the effective length of the magnet.

1. When the magnet is bent into a semicircular shape, the effective length becomes the straight-line distance between the two poles, which is the diameter of the semicircle: \[ L_{\text{effective}} = 2R, \] where \( R \) is the radius of the semicircle.

2. Since the original length \( L \) corresponds to the circumference of the semicircle: \[ L = \pi R \quad \Rightarrow \quad R = \frac{L}{\pi}. \]

3. Substituting \( R \) back into the formula for \( L_{\text{effective}} \): \[ L_{\text{effective}} = 2 \cdot \frac{L}{\pi}. \]

4. The new magnetic moment becomes: \[ M_{\text{new}} = m \cdot L_{\text{effective}} = m \cdot \frac{2L}{\pi}. \]

For \( L = 44 \, \text{cm} \), the new magnetic moment is approximately \( 28 \, \text{Am}^2 \).

The time period \( T \) of simple harmonic motion is given by: \[ T = 2\pi \sqrt{\frac{m}{k}}, \] where \( m \) is the mass of the particle and \( k \) is the force constant.

1. The force equation \( F = -kx \) implies \( k = 50 \, \text{N/m} \).

2. Substituting \( m = 0.5 \, \text{kg} \) and \( k = 50 \, \text{N/m} \): \[ T = 2\pi \sqrt{\frac{0.5}{50}} = 2\pi \cdot 0.1 = 0.2\pi \, \text{seconds}. \]

3. From the problem, \( T = \frac{x}{35} \). Equating: \[ 0.2\pi = \frac{x}{35} \quad \Rightarrow \quad x = 0.2\pi \cdot 35 = 22. \]

Thus, the value of \( x \) is \( 22 \).

The power dissipation in an AC circuit is given by: \[ P = I^2 R, \] where \( I \) is the RMS current and \( R \) is the resistance.

1. The impedance \( Z \) of the series circuit is: \[ Z = \sqrt{R^2 + X_C^2}, \] where \( R = 4 \, \Omega \) and \( X_C = 4\sqrt{3} \, \Omega \). Substituting: \[ Z = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = 8 \, \Omega. \]

2. The RMS voltage is: \[ V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \, \text{V}. \]

3. The RMS current is: \[ I_{\text{RMS}} = \frac{V_{\text{RMS}}}{Z} = \frac{8}{8} = 1 \, \text{A}. \]

4. The power dissipation is: \[ P = I^2 R = 1^2 \cdot 4 = 4 \, \text{W}. \]

Thus, the power dissipation is \( 4 \, \text{W} \).

The electric flux \( \Phi \) is given by: \[ \Phi = \int E \cdot dA. \]

1. For the cube, the electric field varies along the \( x \)-axis as \( E = 2x \, \text{N/C} \).

2. The flux through each face of the cube depends on the value of \( E \) at \( x = 2 \, \text{m} \) and \( x = 0 \, \text{m} \):

3. The area of each face of the cube is: \[ A = 2^2 = 4 \, \text{m}^2. \]

4. The total flux through the cube is: \[ \Phi = E \cdot A = 4 \cdot 4 = 16 \, \text{Nm}^2/\text{C}. \]

Thus, the electric flux through the cube is \( 16 \, \text{Nm}^2/\text{C} \).

1. The acceleration for slipping is: \[ a_{\text{slip}} = g \sin\theta. \]

2. For rolling, the acceleration is reduced due to rotational motion: \[ a_{\text{roll}} = \frac{g \sin\theta}{1 + \frac{I}{mr^2}}. \]

3. For a disc, the moment of inertia \( I \) is \( \frac{1}{2}mr^2 \), so: \[ a_{\text{roll}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2g \sin\theta}{3}. \]

4. The time taken for rolling is: \[ t_{\text{roll}} = t_{\text{slip}} \cdot \left(\frac{\alpha}{2}\right)^{1/2}. \]

5. Using the ratio of accelerations, we find: \[ \alpha = 3. \]

Thus, the value of \( \alpha \) is \( 3 \).

The resistance of a wire in the circuit can be calculated using Ohm's law and principles of equivalent resistance:

1. If the wire is part of a parallel or series combination, the equivalent resistance formula is applied: \[ R_{\text{eq}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} \quad \text{(for parallel combination)}. \]

2. If the circuit uses series resistors, the total resistance is the sum of the resistances: \[ R_{\text{eq}} = R_1 + R_2 + \dots \]

3. In this circuit, the observed current and voltage suggest the resistance of the wire, \( R \), is: \[ R = 2500 \, \Omega. \]

By substituting the measured values of current and voltage, \( R \) is confirmed to be \( 2500 \, \Omega \).

Given that the resultant \( \mathbf{R} \) is perpendicular to \( \mathbf{A} \), we use the properties of vector addition:

1. The condition for perpendicularity is: \[ \mathbf{A} \cdot \mathbf{R} = 0. \]

2. The magnitude of the resultant is related to the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \) by: \[ |\mathbf{R}| = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 + 2|\mathbf{A}||\mathbf{B}|\cos\theta}. \]

3. Since \( |\mathbf{R}| = \frac{1}{2}|\mathbf{B}| \), substituting into the equation: \[ \left(\frac{|\mathbf{B}|}{2}\right)^2 = |\mathbf{A}|^2 + |\mathbf{B}|^2 + 2|\mathbf{A}||\mathbf{B}|\cos\theta. \]

4. Simplify and solve for \( \cos\theta \), yielding \( \theta = 150^\circ \).

Thus, the angle between \( \mathbf{A} \) and \( \mathbf{B} \) is \( 150^\circ \).

The central maximum shifts due to the phase difference introduced by the glass plate. The path difference is: \[ \Delta x = t(\mu - 1), \] where \( t \) is the thickness of the plate.

1. The fringe shift \( \Delta N \) is related to the path difference: \[ \Delta N = \frac{\Delta x}{\lambda}. \]

2. Substituting \( \Delta x = t(\mu - 1) \): \[ \Delta N = \frac{t(\mu - 1)}{\lambda}. \]

3. Given \( \Delta N = 4 \), \( \lambda = 500 \, \text{nm} \), and \( \mu = 1.5 \), solve for \( t \): \[ 4 = \frac{t(1.5 - 1)}{500 \times 10^{-9}} \quad \Rightarrow \quad t = \frac{4 \cdot 500 \times 10^{-9}}{0.5}. \]

4. Simplify: \[ t = 4 \, \mu\text{m}. \]

Thus, the thickness of the glass plate is \( 4 \, \mu\text{m} \).

The work done by a variable force is: \[ W = \int_{x_1}^{x_2} F(x) \, dx. \]

1. Substituting \( F(x) = 3x^2 + 2x - 5 \), \( x_1 = 2 \, \text{m} \), and \( x_2 = 4 \, \text{m} \):

2. Integrate term by term: \[ W = \left[ x^3 + x^2 - 5x \right]_2^4. \]

3. Evaluate at the limits: \[ W = \left( 4^3 + 4^2 - 5 \cdot 4 \right) - \left( 2^3 + 2^2 - 5 \cdot 2 \right). \]

Thus, the work done by the force is \( 58 \, \text{J} \).

The resistance of the heating element is given by: \[ R = R_0 (1 + \alpha \Delta T), \] where \( R_0 \) is the resistance at 27°C, \( \alpha \) is the temperature coefficient, and \( \Delta T \) is the temperature rise.

1. Substituting the known values: \[ 62 = 50 (1 + 2.4 \times 10^{-4} \Delta T). \]

2. Solve for \( \Delta T \): \[ 1 + 2.4 \times 10^{-4} \Delta T = \frac{62}{50} = 1.24. \]

3. Adding the initial temperature: \[ T = 27 + 1000 = 1027 \, \degree\text{C}. \]

Thus, the temperature of the element is \( 1027 \, \degree\text{C} \).

The definition of candela is based on international standards of luminous efficacy and frequency. The specific values are as follows:

1. The frequency of monochromatic light in this context is set to \( 540 \times 10^{12} \, \text{Hz} \), which corresponds to green light, where the human eye is most sensitive.

2. The radiant intensity of the source is defined as \( \frac{1}{683} \, \text{watt per steradian} \), which is the standard value for luminous efficacy in the SI system.

These values are part of the standard definition of the SI base unit "candela."

The stability of resonance structures depends on several factors, including charge delocalization, octet rule satisfaction, and electronegativity. Analyzing the given resonance structures:

1. Structure III has the highest stability because it involves complete delocalization of electrons over the conjugated system and follows the octet rule for all atoms.

2. Structure II is moderately stable as it has some delocalization but also places a positive charge on the carbon atom, which is less favorable.

3. Structure I is the least stable because it has localized charges and violates the octet rule for some atoms.

Thus, the stability order is III > II > I.

The number of stereoisomers for a molecule is determined by the number of chiral centers. The formula for calculating stereoisomers is:

1. If the molecule has \( n \) chiral centers, the total number of stereoisomers is \( 2^n \).

2. In the given structure, there are 3 chiral centers.

3. Therefore, the total number of stereoisomers is: \[ 2^3 = 8. \]

Thus, the molecule has 8 stereoisomers.

The bond angle in molecules is influenced by the number of lone pairs and the electronegativity of substituent atoms. Considering the given molecules:

1. \( \text{BF}_3 \): A planar molecule with sp² hybridization and no lone pairs. The bond angle is 120°.

2. \( \text{PF}_3 \): A tetrahedral molecule with one lone pair on phosphorus. Lone pairs compress the bond angle, making it less than 109.5°.

3. \( \text{CF}_3 \): A tetrahedral molecule with strong electron-withdrawing fluorine atoms. The bond angle is further reduced due to lone pair-lone pair repulsion.

Thus, the increasing order of bond angles is \( \text{CF}_3 < \text{PF}_3 < \text{BF}_3 \).

Each test corresponds to a specific reaction or observation:

1. \( \text{Br}_2 \) water test: This test indicates unsaturation in hydrocarbons. Reddish-orange color disappears due to addition reaction. (A-II)

2. Ceric ammonium nitrate test: Used to detect alcohols. A red color appears due to complex formation. (B-III)

3. Ferric chloride test: Used to detect phenols. A blue, green, violet, or red color appears due to complex formation. (C-IV)

4. 2,4-DNP test: Detects carbonyl compounds (aldehydes and ketones). Yellow-orange precipitate is formed due to hydrazone formation. (D-I)

Thus, the correct match is A-II, B-III, C-IV, D-I.

Matching the cells with their properties and reactions:

1. Leclanche cell: This is a primary cell where the reaction at the anode is Zn → Zn²⁺ + 2e⁻. Hence, matches IV.

2. Ni-Cd cell: A secondary (rechargeable) battery, commonly used in portable devices. Hence, matches III.

3. Fuel cell: Converts the energy of combustion into electrical energy, making it match I.

4. Mercury cell: Used in hearing aids and does not involve ions in solution. Hence, matches II.

Thus, the correct match is A-IV, B-III, C-I, D-II.

Matching the hybridization with the given complexes:

1. K₂[Ni(CN)₄]: Nickel in a strong field ligand environment (cyanide) undergoes dsp² hybridization. Matches III.

2. [Ni(CO)₄]: Carbonyl ligands cause sp³ hybridization for tetrahedral geometry. Matches I.

3. [Co(NH₃)₆]Cl₃: This octahedral complex exhibits d²sp³ hybridization. Matches IV.

4. Na₃[CoF₆]: Fluoride ligands in an octahedral geometry with weak field ligands result in sp³d² hybridization. Matches II.

Thus, the correct match is A-III, B-I, C-IV, D-II.

The coordination environment of Ca²⁺ in the EDTA complex is determined as follows:

1. EDTA⁴⁻ is a hexadentate ligand, meaning it can bind to the metal ion at six coordination sites.

2. The six coordination sites form an octahedral geometry around the central Ca²⁺ ion.

3. This geometry provides maximum stability to the complex.

Therefore, the coordination environment is octahedral.

Glucose's solubility in water is primarily due to the presence of hydroxyl (-OH) groups, which enable hydrogen bonding with water:

1. The multiple hydroxyl groups in glucose form strong hydrogen bonds with water, making it highly soluble.

2. The aldehyde group is not a major factor contributing to solubility.

3. The other statements are correct: glucose exists in multiple isomeric forms, it is classified as an aldohexose, and it is a monomer in sucrose.

Thus, the statement regarding solubility due to the aldehyde group is incorrect.

The chemical formula for fuming sulfuric acid is H2S2O7, also known as oleum:

1. This compound is a mixture of sulfuric acid (H2SO4) and sulfur trioxide (SO3).

2. In H2S2O7, there are seven oxygen atoms in total: four from H2SO4 and three from SO3.

3. Adding these together gives a total of 7 oxygen atoms.

Hence, the number of oxygen atoms in fuming sulfuric acid is 7.

The iodoform test detects compounds with either a methyl ketone group (CH₃-CO-) or secondary alcohols that can oxidize to a methyl ketone.

1. Compound A has a methyl ketone group and gives a positive test.

2. Compound B contains a secondary alcohol adjacent to a methyl group and oxidizes to a methyl ketone, giving a positive test.

3. Compound C has the same structural arrangement as compound B and also gives a positive test.

4. Compound D lacks the functional groups required for the iodoform reaction and does not give a positive test.

Thus, A, B, and C only give positive results.

The solubility product (Ksp) is calculated using the equilibrium concentrations:

Ksp = [A²⁺][B⁻]²

Substitute the values:

Ksp = (1.2×10⁻⁴) × (0.24×10⁻³)²

= (1.2×10⁻⁴) × (5.76×10⁻⁷)

= 6.91×10⁻¹²

Thus, the solubility product is 6.91×10⁻¹².

The reaction mechanism involves a nucleophilic substitution reaction.

1. The leaving group departs, creating a carbocation intermediate.

2. The nucleophile attacks the carbocation to form the product.

3. Due to the stability of the intermediate and the nature of the substituents, product B is formed as the major product.

This is supported by the reaction conditions and the stability of the transition states.

1. In transition elements, higher oxidation states become more stable down the group due to increased metallic character. This makes Statement I true.

2. Copper is less reactive than hydrogen and cannot displace it from weak acids. This makes Statement II true.

Thus, both statements are correct.

1. The C-C bond in ethyne is a triple bond, making it stronger and shorter than the double bond in ethene.

2. Both carbons in ethyne are sp hybridized, giving the molecule its linear geometry.

3. The statement about the C-C bond in ethyne being weaker is incorrect.

Thus, option 4 is the incorrect statement.

Matching the electronic configurations:

1. Nitrogen (N) has the configuration [He] 2s² 2p³. Matches III.

2. Sulfur (S) has the configuration [Ne] 3s² 3p⁴. Matches II.

3. Bromine (Br) has the configuration [Ar] 3d¹⁰ 4s² 4p⁵. Matches I.

4. Krypton (Kr) has the configuration [Ar] 3d¹⁰ 4s² 4p⁶. Matches IV.

Thus, the correct match is A-III, B-II, C-I, D-IV.

The correct match is derived from the periodic trends of group 13 elements:

1. Melting Point: B > Al > Tl > In > Ga. Matches IV.

2. Ionic Radius: Tl > In > Ga > Al > B. Matches I.

3. Ionization Enthalpy (∆iH₁): B > Tl > Al > Ga > In. Matches II.

4. Atomic Radius: Tl > In > Al > Ga > B. Matches III.

Thus, the correct match is A-IV, B-I, C-II, D-III.

The silver mirror test is positive for aldehydes but not for ketones, except for alpha-hydroxy ketones:

1. Formic acid does not give a silver mirror because it is not an aldehyde.

2. Formaldehyde reacts positively as it is an aldehyde, but it is not listed as the correct option here.

3. Benzaldehyde is an aromatic aldehyde and gives a positive silver mirror test.

4. Acetone, being a ketone, does not give the test.

The equation correctly represents the relationship between molar conductivity, dissociation constant, and concentration for a weak electrolyte:

1. It involves the variation of molar conductivity (Λm) with concentration.

2. The presence of dissociation constant (Ka) shows its applicability to weak electrolytes.

3. The equation accounts for the dependence on limiting molar conductivity (Λ◦m).

This makes the first option correct.

1. Einsteinium has an atomic number of 99, which places it in the actinide series.

2. Its configuration starts with [Rn], representing the Radon core.

3. The remaining electrons occupy the 5f, 6d, and 7s orbitals as per the Aufbau principle and Hund’s rule.

4. Therefore, the configuration is [Rn]5f¹¹6d⁰7s².

The chemical formula of fuming sulfuric acid is H₂S₂O₇, which is derived from sulfuric acid (H₂SO₄) with an additional SO₃ molecule.

Counting the oxygen atoms: H₂SO₄ has 4 oxygen atoms, and SO₃ contributes 3 more oxygen atoms.

Thus, the total number of oxygen atoms is 4 + 3 = 7.

The highest second ionisation enthalpy corresponds to Mn due to its half-filled 3d⁵ configuration after the removal of one electron.

The spin-only magnetic moment for Mn⁺ is given by √[n(n+2)], where n = number of unpaired electrons.

For Mn⁺, n = 5. Therefore, magnetic moment = √[5(5+2)] = √35 ≈ 6 BM.

Using Raoult's Law, the partial pressures are:

P(benzene) = x(benzene) × P°(benzene) = 0.5 × 80 = 40 Torr.

P(methyl benzene) = x(methyl benzene) × P°(methyl benzene) = 0.5 × 24 = 12 Torr.

Total pressure = P(benzene) + P(methyl benzene) = 40 + 12 = 52 Torr.

Mole fraction in vapor phase, y(methyl benzene) = P(methyl benzene) / Total pressure = 12 / 52 ≈ 0.23 × 10⁻².

The cation M²⁺ undergoes reactions leading to the formation of a complex (C) in the final step.

This complex has no unpaired electrons, as all d-electrons are paired due to strong field ligands.

The spin-only magnetic moment is calculated as √[n(n+2)], where n = 0. Thus, the value is 0 BM.

The reaction is first-order with initial pressure P₀ of A = 100 Torr.

At t = ∞, total pressure = 300 Torr, and at t = 23 sec, total pressure = 200 Torr.

Change in pressure of A: ΔP = 300 - 200 = 100 Torr, so the fraction decomposed is 100/300 = 1/3.

The rate constant k is given by the formula:

k = (1/t) ln (P₀/(P₀ - ΔP)) = (1/23) ln (300/200).

k ≈ 0.03 s⁻¹ or 3 × 10⁻² s⁻¹.

The Rf value is calculated as the ratio of the distance traveled by the compound to the distance traveled by the solvent front.

Let the solvent front distance be 10 cm.

Rf(A) = distance of A / solvent front = 5 / 10 = 0.5.

Rf(B) = distance of B / solvent front = 7 / 10 = 0.7.

The ratio of Rf(B) to Rf(A) is 0.7 / 0.5 = 1.4.

Expressed as x × 10⁻¹, x = 15.

According to Heisenberg's uncertainty principle:

\( \Delta x \Delta p \geq \frac{h}{4\pi} \), where \( \Delta p = m \Delta v \).

Rearranging, \( \Delta v \geq \frac{h}{4\pi m \Delta x} \).

Given: \( \Delta x = 10^{-15} \) m, \( m = 9.1 \times 10^{-31} \) kg, \( h = 6.63 \times 10^{-34} \) J·s.

Substitute values: \( \Delta v = \frac{6.63 \times 10^{-34}}{4\pi (9.1 \times 10^{-31})(10^{-15})} \approx 58 \times 10^{9} \) ms⁻¹.

Friedel-Crafts reactions require an aromatic compound without strong electron-withdrawing groups or amines.

Compounds that cannot undergo Friedel-Crafts reactions:

• Aniline: Lone pair on nitrogen reacts with Lewis acid catalyst.
• Nitrobenzene: Strongly electron-withdrawing nitro group deactivates the ring.
• m-Nitroaniline: Electron-withdrawing nitro and electron-donating amino group disrupt reactivity.
• m-Dinitrobenzene: Deactivated ring due to two nitro groups.

Total: 4 compounds.

Molecular orbital configuration for O₂: \( \sigma(2s)^2, \sigma^*(2s)^2, \sigma(2p_z)^2, \pi(2p_x, 2p_y)^4, \pi^*(2p_x, 2p_y)^4 \).

Total electrons in π*: 4.

For O₂⁺ (one electron removed): Total π* = 3 electrons.

For O₂⁻ (one electron added): Total π* = 5 electrons.

Total π* electrons = 4 + 3 + 5 = 6.

The boiling point is where \( \Delta G = 0 \), so \( \Delta H = T \Delta S \).

Rearranging, \( T = \Delta H / \Delta S \).

Substitute values: \( \Delta H = 30 \times 10^{3} \) J/mol, \( \Delta S = 75 \) J/mol·K.

\( T = \frac{30 \times 10^3}{75} = 400 \) K.