To analyze the continuity and differentiability of the function f(x) = |2x² + 5||x| − 3, we observe the following:

1. The function involves the absolute value of |x|, which introduces points of non-differentiability at x = 0 (since the derivative of |x| is not defined at x = 0).

2. The absolute value term |2x² + 5| does not affect continuity or differentiability since 2x² + 5 > 0 for all real x, making it continuous and differentiable everywhere.

3. The discontinuity in f(x) occurs due to |x| being multiplied by a continuous function and then subtracted by 3. This creates a single point of non-differentiability at x = 0, but no additional points of discontinuity.

Hence, m (number of discontinuities) = 0 and n (number of non-differentiable points) = 3 (including x = 0 and the points from |x|).

Thus, m + n = 3.

The given quadratic equation is px² + qx − r = 0. Using the roots α and β, we know:

1. Sum of roots, α + β = −q/p.

2. Product of roots, αβ = −r/p.

3. The reciprocal sum of roots is given as 1/α + 1/β = (α + β)/αβ = 3/4.

Substitute the values: (−q/p) / (−r/p) = 3/4, which simplifies to q/r = 3/4.

From the G.P. condition, p, q, and r satisfy q² = pr. Substitute q = (3/4)r into q² = pr, leading to:

(3/4r)² = pr ⟹ 9r²/16 = pr ⟹ p = 9r/16.

Finally, the discriminant is used to find (α − β)² = (α + β)² − 4αβ:

(α − β)² = (q²/p²) − 4(−r/p) = (3r/4)² / (9r/16) − 4(−r/(9r/16)). Simplify to find (α − β)² = 80/9.

Step 1: Rewrite the equation in terms of a single trigonometric function. Using sin²x = 1 − cos²x, the given equation becomes:

4(1 − cos²x) − 4cos³x + 9 − 4cosx = 0.

Step 2: Simplify the equation to get:

4 − 4cos²x − 4cos³x + 9 − 4cosx = 0 ⟹ −4cos³x − 4cos²x − 4cosx + 13 = 0.

Step 3: Let y = cosx, so the equation becomes:

−4y³ − 4y² − 4y + 13 = 0, where y ∈ [−1, 1].

Step 4: Check for real roots of the cubic equation within [−1, 1]. Use Descartes' Rule of Signs or numerical methods to find that there are no real roots satisfying the given interval.

Step 5: Conclude that there are no solutions for x ∈ [−2π, 2π].

Thus, the number of solutions is 0.

Step 1: Analyze the given integral: ∫₀¹ (2x³ − 3x² − x + 1)¹/³ dx.

Step 2: Observe the symmetry of the function. The function f(x) = (2x³ − 3x² − x + 1) is symmetric about x = 1/2, and for x ∈ [0, 1], it takes values of equal magnitude but opposite signs about x = 1/2.

Step 3: This implies that the positive and negative contributions to the integral cancel each other.

Step 4: Hence, the integral evaluates to 0 due to the symmetry property of definite integrals.

Therefore, the value of the integral is 0.

Step 1: Parametrize the point P on the ellipse as P(a cosθ, b sinθ), where a = 3 and b = 2.

Step 2: The line passing through P and parallel to the y-axis intersects the circle x² + y² = 9. Substitute x = a cosθ = 3 cosθ into the circle equation:

(3 cosθ)² + y² = 9 ⟹ y² = 9 − 9 cos²θ ⟹ y = ±3 sinθ.

Step 3: The point Q is (3 cosθ, 3 sinθ). Find the locus of R, which divides PQ in the ratio 4:3.

Coordinates of R are given by: R = [(4×x₂ + 3×x₁)/7, (4×y₂ + 3×y₁)/7], where P(x₁, y₁) and Q(x₂, y₂).

Substitute P(3 cosθ, 2 sinθ) and Q(3 cosθ, 3 sinθ) to get R = (3 cosθ, (8 sinθ)/7).

Step 4: Determine the locus of R by eliminating θ. Substituting x = 3 cosθ and y = (8 sinθ)/7, we get:

(x²/9) + (49y²/64) = 1, which represents an ellipse with eccentricity e = √(1 − b²/a²) = √13/7.

Thus, the eccentricity is √13/7.

Step 1: The general term in the expansion of (a + b + c)ⁿ is given by T(r) = C(n, r) * a^p * b^q * c^r, where p + q + r = n.

Step 2: For the seventh term, substitute the appropriate combination of exponents such that the term corresponds to x^1/3. Compute m using the coefficient of this term.

Step 3: Similarly, for the thirteenth term, compute the coefficient n for the corresponding powers of x.

Step 4: Simplify the ratio n/m³ using the coefficients derived. The result is 9/4.

Step 1: Solve the first-order differential equation f'(x) = αf(x) + 3 by separating variables or using an integrating factor.

Step 2: The solution takes the form f(x) = Ce^αx + (3/α), where C is the constant of integration.

Step 3: Use the initial condition f(0) = 2 to determine the value of C.

Step 4: Substitute x = -log 2 into the solution to compute f(-log 2). Simplify to find the result f(-log 2) = 9.

Step 1: Parametrize the line using the given equation: (x, y, z) = (8t - 3, 2t + 4, 2t - 1).

Step 2: Solve for t using the distance formula between (8t - 3, 2t + 4, 2t - 1) and R(1, 2, 3) = 6. Find two values of t corresponding to points P and Q.

Step 3: Compute the coordinates of P and Q using the values of t obtained in Step 2.

Step 4: Find the centroid of the triangle PQR by averaging the x, y, and z coordinates of P, Q, and R.

Step 5: Calculate α² + β² + γ² for the centroid. The result is 18.

Step 1: Use the angle bisector theorem to find the coordinates of D on the line BC. Compute the ratio in which D divides BC.

Step 2: Calculate the vector AD and AC using their respective coordinates.

Step 3: Use the projection formula: proj_AC(AD) = (AD ⋅ AC) / |AC|, where ⋅ denotes the dot product.

Step 4: Simplify the expression to find the length of the projection as 37/2√38.

Step 1: Use the formula for the sum of an AP: S_n = n/2 [2a + (n−1)d]. Given S₁₀ = 390, solve for a and d.

Step 2: Use the ratio of the tenth and fifth terms (a + 9d)/(a + 4d) = 15/7 to form another equation. Solve the system of equations for a and d.

Step 3: Calculate S₁₅ and S₅ using the formula for the sum of n terms.

Step 4: Compute the difference S₁₅ − S₅. The result is 790.

Step 1: Use the reduction formula for powers of cosine. Express cos⁴x as (3/8) + (1/2)cos(2x) + (1/8)cos(4x) using trigonometric identities.

Step 2: Integrate term by term over the interval [0, π/3]. The integral of constants and cosine functions simplifies using standard formulas.

Step 3: Evaluate the definite integrals to find a = 1/3 and b = 1/2.

Step 4: Compute 9a + 8b = 9(1/3) + 8(1/2) = 2.

Thus, the final answer is 2.

Step 1: Let z = x + yi, where |z| = √(x² + y²) ≥ 1. Rewrite |z + 1/(2(3 + 4i))| in terms of x and y.

Step 2: Simplify 1/(2(3 + 4i)) to its Cartesian form, -3/50 - 4i/50.

Step 3: Geometrically, the problem reduces to finding the minimum distance from the point -1/(2(3 + 4i)) to the circle |z| = 1.

Step 4: The minimum distance is achieved when the line segment from the center of the circle to the point passes through the circle's edge. The distance is computed as 5/2.

Hence, the minimum value is 5/2.

Step 1: For √(x² − 25), solve the inequality x² − 25 ≥ 0. This gives x ∈ (-∞, -5] ∪ [5, ∞).

Step 2: For (4 − x²) in the denominator, solve 4 − x² > 0. This gives x ∈ (-2, 2).

Step 3: For log₁₀(x² + 2x − 15), solve x² + 2x − 15 > 0. Factorize to get (x − 3)(x + 5) > 0, leading to x ∈ (-∞, -5) ∪ (3, ∞).

Step 4: Combine all conditions to find the domain as (-∞, -5) ∪ [5, ∞). Here, α = -5 and β = 5.

Step 5: Calculate α² + β³ = (-5)² + (5)³ = 25 + 125 = 150.

Thus, the final result is 150.

Step 1: Check reflexivity for R₁: a² + a² = 1 is not true for all a ∈ R. Hence, R₁ fails reflexivity.

Step 2: Check symmetry for R₁: If aR₁b, then a² + b² = 1 implies b² + a² = 1. This holds, so R₁ is symmetric.

Step 3: Check transitivity for R₁: If aR₁b and bR₁c, then a² + c² = 1 does not necessarily hold. Hence, R₁ fails transitivity.

Step 4: Check reflexivity, symmetry, and transitivity for R₂. All conditions are satisfied as a + d = b + c defines equivalence in N × N.

Thus, only R₂ is an equivalence relation.

Step 1: Parametrize the line as (x, y, z) = (3t + 1, 2t − 1, t + 2).

Step 2: Use the formula for reflection in 3D geometry. The image lies on the same line, equidistant from P.

Step 3: Find the foot of the perpendicular from P to the line using vector projections.

Step 4: Calculate the coordinates of the mirror image (α, β, γ).

Step 5: Substitute into 14(α + β + γ). Simplify to get 108.

Thus, the final answer is 108.

Step 1: Analyze the given function. For x even, f(x) = x − 1; for x odd, f(x) = 2x.

Step 2: Apply the function iteratively. Start with f(f(f(a))) = 21 and trace backward.

Step 3: For f(f(a)) to yield an even number (since 21 is odd), f(a) must be odd. Find a sequence where this is satisfied.

Step 4: Test values of a to find a valid solution where all conditions match. Calculate a = 144.

Thus, the answer is 144.

Step 1: Form the coefficient matrix of the system and set its determinant to 0 for the system to have infinite solutions.

Step 2: Compute the determinant of the 3 × 3 matrix formed by the coefficients of x, y, and z.

Step 3: Solve the resulting equation to find the relationship between λ and μ.

Step 4: Substitute the values into λ + 2μ and calculate the result as 17.

Thus, the answer is 17.

Step 1: Use the given mean formula: mean = (∑xᵢ) / 10 = 6/5. Solve for ∑xᵢ.

Step 2: Use the variance formula: variance = (∑xᵢ² / 10) − (mean)² = 84/25. Solve for ∑xᵢ².

Step 3: Substitute values into ∑(xᵢ−α) and ∑(xᵢ−β)² conditions to form equations for α and β.

Step 4: Solve the equations to find α = 3 and β = 6.

Step 5: Compute β/α = 6/3 = 2.

Thus, the ratio is 2.

Step 1: Let the total probability of Ajay appearing be 1 − p = 5/7.

Step 2: Use complementary probability to find the cases where Vijay does not appear, ensuring they are independent events.

Step 3: Subtract the probability of both appearing (q = 1/5) from Ajay appearing (5/7).

Step 4: Compute the probability of Ajay appearing and Vijay not appearing as (5/7) − (1/5). Simplify to get 18/35.

Thus, the answer is 18/35.

Step 1: The locus of midpoints of chords subtending an angle at the origin is a circle with its center at (0, 1/2) and radius 1/2.

Step 2: Solve the equation of the circle x² + (y − 1/2)² = 1/4.

Step 3: Find the points of intersection of the line x + y = 1 with this circle.

Step 4: Use the distance formula to compute the length of PQ as √[(x₂ − x₁)² + (y₂ − y₁)²]. Simplify to get 1/√2.

Thus, the length of PQ is 1/√2.

Step 1: Let the sides of the triangle be a, ar, and ar². Use the triangle inequality conditions:

• a + ar > ar²
• ar + ar² > a
• ar² + a > ar

Step 2: Simplify the inequalities. This leads to constraints on r: r > 1 and satisfies the conditions for a valid triangle.

Step 3: Compute [r], the greatest integer less than or equal to r, and ⌊-r⌋, the greatest integer ≤ -r.

Step 4: Substitute into 3[r] + ⌊-r⌋. For valid values of r, the result simplifies to 6.

Thus, the answer is 6.

Step 1: Recognize that A is a projection matrix. Projection matrices have eigenvalues 0 and 1.

Step 2: Verify the eigenvalues of A. Since MᵀM = I₁, the rank of A is reduced, confirming eigenvalues 0 and 1.

Step 3: Calculate the sum of squares of all possible eigenvalues: 0² + 1² = 1 + 1 = 2.

Step 4: Conclude that the sum of squares of all possible values of λ is 2.

Thus, the final answer is 2.

Step 1: Differentiate F(x²) with respect to x to find f(x²). Use the chain rule: d/dx [F(x²)] = F'(x²) * d(x²)/dx = 2x f(x²).

Step 2: Given F(x²) = x⁴ + x⁵, differentiate to find f(x²): f(x²) = 4x³ + 5x⁴.

Step 3: Substitute x = r into f(x²) to compute f(r²) for r = 1, 2, ..., 12.

Step 4: Compute the sum ∑₁² f(r²). The result simplifies to 219.

Thus, the answer is 219.

Step 1: Differentiate y with respect to x using the quotient rule and chain rule for the first term.

Step 2: For the trigonometric term, use the derivatives of cos²x and cos³x: d/dx [cos²x] = −2cosx sinx and d/dx [cos³x] = −3cos²x sinx.

Step 3: Substitute x = π/6 into the differentiated expression.

Step 4: Simplify the resulting expression to find 96y'(π/6) = 105.

Thus, the final answer is 105.

Step 1: Use the formula for the cross product magnitude: |a × b| = |a||b|sinθ. Given |b| = 6 and θ = π/4, solve for |a|.

Step 2: Substitute |a × b| = 3√2 into the formula: 3√2 = |a| × 6 × sin(π/4). Simplify to find |a| = 1.

Step 3: Compute |a|² = 1² = 1, and write |a|² = 1 + α² + β², leading to α² + β² = 1.

Step 4: Calculate (α² + β²)|a × b|² = 1 × (3√2)² = 1 × 18 = 90.

Thus, the answer is 90.

Step 1: The odd-numbered lines (L₁, L₃, ..., L₁₉) are parallel and do not intersect each other.

Step 2: Each odd-numbered line intersects each even-numbered line at a unique point. Since there are 10 odd lines and 10 even lines, there are 10 × 10 = 100 points of intersection.

Step 3: All even-numbered lines pass through a common point P. This adds 1 additional intersection point.

Step 4: Total number of intersection points = 100 + 1 = 101.

Thus, the final answer is 101.

Step 1: The line PQ is derived from the points P(a, a²) and Q(−b, b²). The equation of PQ is determined using the slope formula.

Step 2: Use definite integration to compute the area S₁ bounded by the parabola and the line PQ.

Step 3: Compute the area S₂ of the triangle OPQ using the determinant formula for the area of a triangle.

Step 4: Minimize the ratio S₁/S₂ with respect to a and b. The minimum value of the ratio is m/n = 4/3.

Step 5: Compute m + n = 4 + 3 = 7.

Thus, the answer is 7.

Step 1: The region bounded by the parabolas 2y² = kx and ky² = 2(y − x) depends on k. Solve for the points of intersection.

Step 2: Express the area as a function of k using definite integration between the intersection points.

Step 3: Maximize the area with respect to k. Find the critical points and solve for the corresponding values of k.

Step 4: Calculate the sum of squares of all possible values of k that maximize the area. The result is 8.

Thus, the final answer is 8.

Step 1: Rearrange the differential equation as dx/dy = 1 + x − y². Solve using the integrating factor method.

Step 2: The integrating factor is e^y, leading to the solution x(y) = Ce^y − y² − 1.

Step 3: Use the initial condition x(1) = 1 to find the constant C.

Step 4: Substitute y = 2 into the solution to find x(2). Multiply by 5 to get 5x(2) = 5.

Thus, the answer is 5.

Step 1: Place the points B and C symmetrically about the y-axis, ensuring AB = AC and BC = 4.

Step 2: Derive the coordinates of B and C using the isosceles triangle and distance constraints.

Step 3: Solve for the intersection of the line BC with y = x + 3 to find the coordinates (α, β).

Step 4: Compute β⁴ using the value of β from the intersection point. The result is β⁴ = 36.

Thus, the answer is 36.

Step 1: Let the shunt resistance be S. The current division gives 5% of the current through the galvanometer and 95% through the shunt.

Step 2: Use the relation for parallel resistance: 1/R_ammeter = 1/G + 1/S.

Step 3: From the current division, S = G/19. Substitute into the formula to get the effective resistance of the ammeter as G/199.

Thus, the final resistance is G/199.

Step 1: Use the principle of a balanced Wheatstone bridge: ∆R/R = α∆T.

Step 2: Given that ∆R = 3 mΩ and ∆T = 10 s × 2°C/s = 20°C.

Step 3: Substitute values into the equation to calculate α: α = ∆R/(R∆T) = (3 × 10⁻³)/(3 × 20).

Step 4: Simplify to find α = −1.5 × 10⁻² °C⁻¹.

Thus, the temperature coefficient is −1.5 × 10⁻² °C⁻¹.

Step 1: Analyze each statement:

• (A) True: The angular momentum of an electron in the nth orbit is quantized as nh/2π.
• (B) True: Nuclear forces do not follow the inverse square law like gravitational or electrostatic forces.
• (C) True: Nuclear forces depend on the relative spin orientation of nucleons.
• (D) False: Nuclear forces are not completely central; they have a tensor component.
• (E) True: A higher packing fraction indicates instability in the nucleus.

Step 2: The correct statements are (A), (B), (C), and (E).

Thus, the correct option is (3).

Step 1: Use the first law of thermodynamics: Q = ∆U + W.

Step 2: For an isobaric process, ∆U = nC_v∆T and W = nR∆T. Given γ = 1.4, find the relation C_v/C_p = 1 − 1/γ = 0.4.

Step 3: Substitute the values of ∆U and W into Q. Since W = 200 J, solve for Q.

Step 4: Simplify to find Q = 700 J.

Thus, the heat given is 700 J.

Step 1: Use conservation of energy. The initial kinetic energy is converted into potential energy at the maximum height.

Step 2: The total kinetic energy of the rolling disc is the sum of translational and rotational energies: K.E. = (1/2)Mv² + (1/2)Iω².

Step 3: For a disc, the moment of inertia I = (1/2)MR² and ω = v/R. Substitute these into the rotational energy.

Step 4: Equate the total initial kinetic energy to the potential energy at maximum height: (3/4)Mv² = Mgh.

Step 5: Simplify to find h = v²/2g.

Thus, the maximum height is v²/2g.

Step 1: Use the energy-wavelength relation for photons: E = hc/λ.

Step 2: Substitute h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, and λ = 660 × 10⁻⁹ m into the equation.

Step 3: Calculate E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (660 × 10⁻⁹) = 1.88 eV.

Step 4: Since the band gap is given as X/8 eV, solve for X: X/8 = 1.88. Multiply to find X = 15.

Thus, the value of X is 15.

Step 1: Surface energy is proportional to the surface area of the droplets.

Step 2: The volume of the big drop is equal to the total volume of the 1000 small droplets. Let the radius of a small droplet be r. The radius of the big drop is R = r × 1000^(1/3).

Step 3: The surface area of a sphere is 4πR². The ratio of surface areas is (4πR²) / (1000 × 4πr²) = 1/10.

Step 4: Hence, the surface energy decreases by a factor of 1/10.

Thus, the surface energy becomes 1/10 of the initial energy.

Step 1: The wavelength λ is related to the frequency f and the speed of light c by λ = c/f.

Step 2: Substitute c = 3 × 10⁸ m/s and f = 60 × 10⁶ Hz into the formula.

Step 3: Calculate λ = (3 × 10⁸) / (60 × 10⁶) = 5 m.

Step 4: The electric and magnetic field vectors are perpendicular to each other and the direction of propagation.

Thus, the wavelength of the wave is 5 m.

Step 1: Use the impulse-momentum theorem: F = ∆p / ∆t.

Step 2: Calculate the change in momentum, ∆p = mv = 0.12 × 25 = 3 kg·m/s.

Step 3: The time interval ∆t is given as 0.1 s.

Step 4: Substitute into the formula: F = 3 / 0.1 = 30 N.

Thus, the force exerted by the ball is 30 N.

Step 1: The energy of a single photon is E = hf, where h = 6.63 × 10⁻³⁴ J·s and f = 6 × 10¹⁴ Hz.

Step 2: Calculate E = 6.63 × 10⁻³⁴ × 6 × 10¹⁴ = 3.978 × 10⁻¹⁹ J.

Step 3: The number of photons per second is given by n = P / E, where P = 2 × 10⁻³ W.

Step 4: Substitute values: n = (2 × 10⁻³) / (3.978 × 10⁻¹⁹) ≈ 5 × 10¹⁵.

Thus, the source emits approximately 5 × 10¹⁵ photons per second.

Step 1: The angular spread θ of the central maxima in single-slit diffraction is given by θ = 2sin⁻¹(λ/a), where λ is the wavelength and a is the slit width.

Step 2: Substitute λ = 0.02 m and a = 0.04 m into the formula: θ = 2sin⁻¹(0.02/0.04) = 2sin⁻¹(0.5).

Step 3: Simplify using sin⁻¹(0.5) = 30°. Thus, θ = 2 × 30° = 60°.

Thus, the angular spread is 60°.

Step 1: Use Gauss's Law: Φ = Q/ε₀, where Q is the enclosed charge.

Step 2: For cube C1, the enclosed charge is 2Q, so the flux Φ₁ = 2Q/ε₀.

Step 3: For cube C2, the total enclosed charge is 5Q, so the flux Φ₂ = 5Q/ε₀.

Step 4: The ratio of fluxes is Φ₁ : Φ₂ = 2Q : 5Q = 2 : 5.

Thus, the ratio of fluxes is 2:5.

Step 1: The root mean square velocity v_rms is inversely proportional to the square root of the molar mass: v_rms ∝ 1/√M.

Step 2: For hydrogen, M_H₂ = 2, and for oxygen, M_O₂ = 32.

Step 3: Compute the ratio of velocities: v_rms,O₂/v_rms,H₂ = √(M_H₂/M_O₂) = √(2/32) = 1/4.

Step 4: Given v_rms,H₂ = 2 km/s, v_rms,O₂ = 2 × (1/4) = 0.5 km/s.

Thus, the root mean square velocity of oxygen is 0.5 km/s.

Step 1: Convert speeds to m/s: v_A = 72 × (1000/3600) = 20 m/s and v_B = -108 × (1000/3600) = -30 m/s (negative since it moves in the opposite direction).

Step 2: The velocity of B relative to A is v_BA = v_B - v_A = -30 - 20 = -50 m/s.

Step 3: The velocity of the ground relative to B is v_GB = -v_B = 30 m/s.

Thus, the velocities are -50 m/s and 30 m/s.

Step 1: Assume C₁ and C₂ are in series or parallel with the galvanometer. Analyze the equivalent capacitance based on the given circuit configuration.

Step 2: Apply the charge distribution rule: Q = CV, where the potential difference across C₁ and C₂ differs due to resistance and capacitance ratios.

Step 3: Calculate the voltage across each capacitor and determine the stored charge ratio using Q₁/Q₂ = (C₁V₁)/(C₂V₂).

Step 4: Simplify to find the charge ratio as 1:2.

Thus, the ratio of charges stored is 1:2.

Step 1: Using the metre-bridge formula, the ratio of resistances is proportional to the balance length: R_left/R_right = L₁/(100 - L₁), where L₁ = 40 cm.

Step 2: Calculate the unknown resistance R_right = (2 × (100 - 40))/40 = 3 Ω.

Step 3: On shunting the unknown resistance with 2 Ω, the new equivalent resistance is R_eq = (3 × 2)/(3 + 2) = 6/5 Ω.

Step 4: Recalculate the balance length for R_eq, and determine the change in balance length. The difference is 22.5 cm.

Thus, the change in balance length is 22.5 cm.

Step 1: Carefully analyze the properties given in List-I and their corresponding matches in List-II.

Step 2: Assign matches based on scientific principles and calculations.

Step 3: Verify each pair for consistency with the question and ensure logical correctness.

Step 4: The correct matching aligns as (A)-(III), (B)-(I), (C)-(IV), (D)-(III).

Thus, the correct option is (3).

Step 1: Calculate the output power of the transformer using efficiency: P_out = η × P_in, where η = 0.8 and P_in = 4 kW.

Step 2: Substitute values: P_out = 0.8 × 4000 = 3200 W.

Step 3: Use the formula P = VI to find the current: I = P/V.

Step 4: Substitute P = 3200 W and V = 240 V: I = 3200/240 = 13.33 A.

Thus, the current in the secondary coil is 13.33 A.

Step 1: The centripetal force for circular motion is F ∝ R⁻³/², and F = mω²R, where ω = 2π/T.

Step 2: Equating centripetal force and gravitational force gives ω²R ∝ R⁻³/².

Step 3: Simplify to find ω² ∝ R⁻⁵/², and substitute ω = 2π/T.

Step 4: Square both sides to find T² ∝ R⁵/².

Thus, T² is proportional to R⁵/².

Step 1: Calculate the net force: F = F₁ + F₂ = (5 + 3)î + (8 − 4)ĵ + (7 − 3)k̂ = 8î + 4ĵ + 4k̂.

Step 2: Use Newton's second law: F = ma, where m = 4 kg.

Step 3: Solve for acceleration: a = F/m = (8/4)î + (4/4)ĵ + (4/4)k̂ = 2î + ĵ + k̂.

Thus, the acceleration acting on the body is 2î + ĵ + k̂.

Step 1: The frequency of oscillation is inversely proportional to the square root of the mass: f ∝ 1/√m.

Step 2: For mass m, the frequency is f₁, and for mass 9m, the frequency is f₂.

Step 3: Calculate the ratio: f₁/f₂ = √(9m/m) = √9 = 3.

Thus, the value of f₁/f₂ is 3.

Step 1: Acceleration a is given by a = dv/dt.

Step 2: Given v = 4√x, differentiate v with respect to x: dv/dx = 2/√x.

Step 3: Use the chain rule: a = dv/dt = (dv/dx)(dx/dt). Since dx/dt = v, substitute v = 4√x.

Step 4: Calculate a = (2/√x) × (4√x) = 8 m/s².

Thus, the acceleration of the particle is 8 m/s².

Step 1: The torque in the galvanometer is given by τ = nBAI, where n = 100, B = 0.01T, A = 2 × 10⁻⁴ m², and I = 0.01A.

Step 2: Substitute the values to find τ: τ = 100 × 0.01 × 2 × 10⁻⁴ × 0.01 = 2 × 10⁻⁵ N-m.

Step 3: The torsional constant k is related to τ and θ by τ = kθ. Solve for k: k = τ/θ = (2 × 10⁻⁵)/(0.05) = 4 × 10⁻⁵ N-m/rad.

Thus, the value of x is 4.

Step 1: Strain is proportional to the force experienced by the wire. The force is the weight of the load acting on the wire.

Step 2: The upper wire supports both loads (2 kg + 1 kg = 3 kg), while the lower wire supports only 1 kg.

Step 3: The ratio of strains is proportional to the ratio of forces: strain_upper/strain_lower = F_upper/F_lower = 3:1.

Thus, the strain ratio is 3.

Step 1: The frequency of emitted radiation is proportional to the energy difference between the levels: f ∝ (1/n₁² - 1/n₂²).

Step 2: For n = 2 to n = 1, f = 3 × 10¹⁵ Hz. Using the same proportionality, calculate for n = 3 to n = 1.

Step 3: Ratio of frequencies: f₂/f₁ = (1/1² - 1/3²) / (1/1² - 1/2²). Simplify to find f₂ = 32 × 3 × 10¹⁵ Hz.

Step 4: Solve for x: x × 9 × 10¹⁵ = 32 × 3 × 10¹⁵. Thus, x = 32.

Thus, the value of x is 32.

Step 1: Use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.

Step 2: Substitute the values given in the circuit: C = 10 μF and V = 6 V.

Step 3: Calculate Q = 10 × 6 = 60 μC.

Thus, the charge stored in the capacitor is 60 μC.

Step 1: The maximum EMF is given by E = NABω, where N is the number of turns, A is the area, B is the magnetic field, and ω is the angular velocity.

Step 2: Given N = 200, A = 0.20 m², B = 0.01T, and ω = 2π × (1/2) = π rad/s.

Step 3: Calculate E = 200 × 0.20 × 0.01 × π = 2π/5 volts.

Thus, the value of β is 5.

Step 1: The intensity I is given by I = I₀ cos²(πdx/λD), where λ = 5000 Å = 5 × 10⁻⁷ m, d = 1.0 mm = 10⁻³ m, and D = 1.0 m.

Step 2: For I = I₀/2, cos²(πdx/λD) = 1/2, which gives cos(πdx/λD) = 1/√2.

Step 3: Solve for x: πdx/λD = π/4. Substitute values: x = (λD)/(4d).

Step 4: Calculate x = (5 × 10⁻⁷ × 1)/(4 × 10⁻³) = 125 × 10⁻⁶ m.

Thus, the distance is 125 × 10⁻⁶ m.

Step 1: The angular velocity ω is given by ω = τ/I, where τ is the torque, and I is the moment of inertia of the rod about its center.

Step 2: The impulse generates an angular velocity at end B: τ = r × F = (L/2) × (Impulse), where L = 0.3 m.

Step 3: Calculate I = (1/3)mL² for a rod. Substitute m = 2 kg and L = 0.3 m: I = (1/3) × 2 × 0.3² = 0.06 kg·m².

Step 4: Solve for ω and time t = π/ω to find x = 4.

Thus, the value of x is 4.

Step 1: The forces acting on the charged particle are gravitational force (mg), electric force (qE), and tension in the thread.

Step 2: Resolve forces: Tcosθ = mg and Tsinθ = qE, where tanθ = sinθ/cosθ = x/L = 10/20 = 1/2.

Step 3: Solve for charge: q = (mg tanθ)/E. Substitute m = 2 × 10⁻³ kg, g = 9.8 m/s², tanθ = 1/2, and E = 2 × 10⁴ N/C.

Step 4: Calculate q = (2 × 10⁻³ × 9.8 × 1/2) / (2 × 10⁴) = √1/3 μC.

Thus, the value of x is 3.

Step 1: The third ionisation enthalpy is influenced by the electronic configuration of the metal ion after two electrons have been removed.

Step 2: Mn has a stable half-filled configuration (3d⁵4s²). Removing a third electron disrupts this stability, requiring significantly higher energy.

Step 3: Other elements (Cr, V, Fe) do not possess this stability, making their 3rd ionisation enthalpies lower compared to Mn.

Thus, Mn has the highest 3rd ionisation enthalpy.

Step 1: In a π-bonding molecular orbital (MO), electron density is concentrated above and below the inter-nuclear axis due to constructive interference of p-orbitals.

Step 2: Statement I is incorrect because it claims lower electron density, which is false for bonding MOs.

Step 3: For π-antibonding MOs, a node exists between the nuclei due to destructive interference of p-orbitals. This makes Statement II correct.

Thus, Statement I is false, and Statement II is true.

Step 1: Cr²⁺ oxidises to Cr³⁺, which has a stable t₂g³ configuration, making it reducing in aqueous solutions.

Step 2: Mn³⁺ reduces to Mn²⁺, which has a stable half-filled d⁵ configuration, making it oxidising in aqueous solutions.

Step 3: The reason given, "extra stability of half-filled configurations," correctly explains why Cr²⁺ is reducing and Mn³⁺ is oxidising.

Thus, both (A) and (R) are true, and (R) explains (A).

Step 1: Reaction (A) reduces phenol with Zn to form benzene.

Step 2: Reaction (B) is the Reimer-Tiemann reaction, producing salicylaldehyde.

Step 3: Reaction (C) is the Kolbe-Schmitt reaction, forming salicylic acid.

Step 4: Reaction (D) involves nitration of phenol to produce picric acid.

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

Step 1: Analyze Statement I: The d-block contains only metals, whereas the p-block contains both metals and non-metals. Hence, Statement I is false.

Step 2: Analyze Statement II: Non-metals exhibit higher ionisation enthalpy and electronegativity compared to metals due to their smaller atomic size and stronger nuclear attraction. Hence, Statement II is true.

Thus, Statement I is false, but Statement II is true.

Step 1: Reducing agents donate electrons and are oxidized themselves. The strength depends on bond dissociation energy.

Step 2: Down the group in the periodic table, bond dissociation energy decreases due to weaker bonds between the element and hydrogen.

Step 3: Among NH₃, PH₃, SbH₃, and BiH₃, the bond strength is weakest in BiH₃, making it the strongest reducing agent.

Thus, BiH₃ is the strongest reducing agent.

Step 1: d-d transitions occur in compounds where transition metal ions have unpaired d-electrons in their electronic configuration.

Step 2: CuSO₄·5H₂O contains Cu²⁺ ions, which have unpaired d-electrons and exhibit colour due to d-d transitions.

Step 3: The other compounds (K₂Cr₂O₇, K₂CrO₄, KMnO₄) show colour due to charge transfer transitions, not d-d transitions.

Thus, CuSO₄·5H₂O exhibits colour due to d-d transition.

Step 1: Meta-directing groups are electron-withdrawing groups that reduce electron density in the ortho and para positions of the benzene ring through resonance or inductive effects.

Step 2: −NO₂, −CHO, −SO₃H, and −COR are strong electron-withdrawing groups, making them meta-directing.

Step 3: The other groups in options (1), (2), and (4) contain electron-donating or weakly withdrawing groups, which are not primarily meta-directing.

Thus, the correct set is −NO₂, −CHO, −SO₃H, −COR.

Step 1: Intramolecular hydrogen bonding occurs within a molecule when a hydrogen atom is shared between a hydrogen donor (−OH, −NH) and an acceptor (C=O, −NO₂) group within close proximity.

Step 2: A compound with −OH and −CHO groups in adjacent positions allows for intramolecular hydrogen bonding between the hydrogen of −OH and the oxygen of −CHO.

Step 3: H₂O, NH₃, and C₂H₅OH primarily exhibit intermolecular hydrogen bonding, not intramolecular.

Thus, the correct compound is one containing −OH and adjacent −CHO groups.

Step 1: Lassaigne’s test involves heating an organic compound with sodium to convert elements (N, S, P, halogens) into water-soluble ionic forms (NaCN, Na₂S, Na₃PO₄, NaX).

Step 2: These ionic forms are tested using specific reagents to confirm the presence of nitrogen, sulphur, phosphorus, and halogens.

Step 3: The test does not detect oxygen or carbon directly, focusing on these specific elements.

Thus, Lassaigne’s test detects nitrogen, sulphur, phosphorus, and halogens.

Step 1: Boiling point depends on intermolecular forces. Hydrogen bonding significantly increases boiling points.

Step 2: CH₃CH₂CH₂OH contains an −OH group capable of forming strong hydrogen bonds, unlike CH₃CH₂CH₃ (alkane), CH₃CH₂CHO (aldehyde), or CH₃C(=O)−CH₂CH₃ (ketone).

Step 3: The presence of hydrogen bonding in CH₃CH₂CH₂OH results in the highest boiling point among the given compounds.

Thus, CH₃CH₂CH₂OH has the highest boiling point.

Step 1: The first reaction involves partial hydrogenation using Pd/C, which reduces CH₃C≡CH to a triple-bonded compound A, 2-Pentyne.

Step 2: The second reaction involves sodium in liquid ammonia, which reduces alkynes to alkenes with trans configuration.

Step 3: A is 2-Pentyne, and B is trans-2-butene, formed via these reactions.

Thus, A is 2-Pentyne, and B is trans-2-butene.

Step 1: The formula for radial nodes is given by (n − l − 1), where n is the principal quantum number and l is the azimuthal quantum number.

Step 2: For a 3p orbital, n = 3 and l = 1. Substituting these values: radial nodes = 3 − 1 − 1 = 1.

Step 3: Therefore, the 3p orbital has 1 radial node.

Thus, the number of radial nodes is 1.

Step 1: Match each compound with its correct use:

• (A) Carbon tetrachloride is used in fire extinguishers.
• (B) Methylene chloride is a paint remover.
• (C) DDT is a non-biodegradable insecticide.
• (D) Freons are used in refrigerators and air conditioners.

Step 2: Based on the matches, the correct option is (2).

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

Step 1: Negative resonance effect involves the withdrawal of electrons from the conjugated system, reducing electron density.

Step 2: The −COOH group has a carbonyl group (C=O) that is strongly electron-withdrawing, creating a negative resonance effect.

Step 3: Other groups like −NH₂, −OH, and −OR are electron-donating and exhibit positive resonance effects.

Thus, the functional group showing a negative resonance effect is −COOH.

Step 1: In [Co(NH₃)₆]³⁺, NH₃ is a strong field ligand, causing a large crystal field splitting (Δ). This leads to pairing of electrons, making it a spin paired complex.

Step 2: In [CoF₆]³⁻, F⁻ is a weak field ligand, causing a small crystal field splitting. Electrons remain unpaired, making it a spin free complex.

Step 3: The strong/weak ligand field nature of NH₃ and F⁻ determines the spin properties of the complexes.

Thus, [Co(NH₃)₆]³⁺ is spin paired, and [CoF₆]³⁻ is spin free.

Step 1: Analyze Statement I: SiO₂ and GeO₂ exhibit acidic properties due to their strong covalent bonding. SnO and PbO exhibit amphoteric behavior as they react with both acids and bases. Hence, Statement I is true.

Step 2: Analyze Statement II: The allotropic forms of carbon arise primarily due to catenation (self-linking property of carbon) and hybridization, not pπ-pπ bonding. Hence, Statement II is false.

Thus, Statement I is true, but Statement II is false.

Step 1: Succinic acid (HOOC-CH₂-CH₂-COOH) is formed through a multi-step reaction involving bromination of an alkene, addition of cyanide ions, and subsequent hydrolysis.

Step 2: The intermediate formed undergoes hydrolysis to produce a dicarboxylic acid with two −COOH groups separated by two −CH₂− groups.

Step 3: The resulting acid matches the structure of succinic acid.

Thus, the acid formed is succinic acid.

Step 1: The dissociation of calcium phosphate is: Ca₃(PO₄)₂ ⇌ 3Ca²⁺ + 2PO₄³⁻.

Step 2: Solubility (S) determines ion concentration. For calcium ions, concentration = 3S; for phosphate ions, concentration = 2S.

Step 3: The solubility product Kₛₚ is calculated as Kₛₚ = [Ca²⁺]³[PO₄³⁻]². Substituting ion concentrations: Kₛₚ ∝ S⁵.

Step 4: Using W and M to express S, Kₛₚ ≈ 10⁷(W/M)⁵.

Thus, the solubility product is 10⁷(W/M)⁵.

Step 1: Analyze Statement I: Dimethyl glyoxime forms a five-membered chelate, not a six-membered one. This makes Statement I false.

Step 2: Analyze Statement II: Prussian blue is Fe₄[Fe(CN)₆]₃, containing Fe²⁺ and Fe³⁺ ions. This makes Statement II true.

Thus, Statement I is false, but Statement II is true.

Step 1: Analyze Statement I: Dimethyl glyoxime forms a five-membered ring upon chelation with Ni²⁺, not a six-membered ring. Hence, Statement I is false.

Step 2: Analyze Statement II: Prussian blue, chemically Fe₄[Fe(CN)₆]₃, contains iron in both +2 and +3 oxidation states, confirming the accuracy of the statement.

Step 3: Since Statement I is false and Statement II is true, the correct answer is option (1).

Thus, Statement I is false, but Statement II is true.

Step 1: Monochlorination of 2-methylbutane can occur at four different carbon atoms, leading to structural isomers.

Step 2: Chlorination at the chiral carbon atom results in stereoisomers (enantiomers).

Step 3: Including both structural and stereoisomers, a total of six isomers are formed.

Thus, the total number of isomeric compounds is 6.

Step 1: Use the first-order integrated rate law: k = (2.303/t) log [(P_∞ − P_t) / (P_∞ − P₀)].

Step 2: Final pressure P_∞ = 0.30 atm, P_t = 0.28 atm, and P₀ = 0.1 atm. Substitute values.

Step 3: k = (2.303/115) log [(0.30 − 0.28) / (0.30 − 0.10)] = approximately 2 × 10⁻² s⁻¹.

Thus, the rate constant is 2 × 10⁻² s⁻¹.

Step 1: Tripeptides are formed by arranging three different amino acids in a linear sequence.

Step 2: The total number of arrangements is given by permutations: 3! = 6.

Step 3: Each arrangement corresponds to a unique tripeptide.

Thus, the number of tripeptides formed is 6.

Step 1: Hinsberg’s reagent reacts with primary and secondary amines to form sulfonamide derivatives.

Step 2: Among the given compounds, five amines (primary and secondary) react with the reagent.

Step 3: Tertiary amines do not react, as they lack a hydrogen atom on the nitrogen for the reaction.

Thus, the number of reacting compounds is 5.

Step 1: Use the depression in freezing point formula: ∆T_f = K_f × molality.

Step 2: K_f for water = 1.86°C/m. ∆T_f = 24°C. Molality = ∆T_f / K_f = 24 / 1.86 ≈ 12.9 m.

Step 3: Mass of solute = molality × molar mass × mass of solvent (in kg).

Step 4: Substituting values gives 14.88 kg of ethylene glycol.

Thus, the required mass is 14.88 kg.

Step 1: The reaction between NH₃ and H₂SO₄ is: 2NH₃ + H₂SO₄ → (NH₄)₂SO₄.

Step 2: Calculate moles of H₂SO₄: Volume = 10 mL = 0.01 L, Molarity = 2 M, so moles = 0.01 × 2 = 0.02 mol.

Step 3: Each mole of H₂SO₄ reacts with 2 moles of NH₃. Therefore, moles of NH₃ = 2 × 0.02 = 0.04 mol.

Step 4: Mass of nitrogen in NH₃ = moles × molar mass = 0.04 × 14 = 0.56 g.

Step 5: Percentage nitrogen = (mass of nitrogen / mass of compound) × 100 = (0.56 / 1) × 100 = 56%.

Thus, the percentage of nitrogen is 56%.

Step 1: For 1 mol of H₂O, the oxidation half-reaction is: 2H₂O → O₂ + 4H⁺ + 4e⁻.

Step 2: This means 4 moles of electrons are required to oxidize 1 mol of H₂O to O₂.

Step 3: Charge per mole of electrons = Faraday constant = 96500 C/mol.

Step 4: Total charge required = 4 × 96500 = 386000 C ≈ 3.86 × 10⁵ C.

Step 5: The required charge is approximately 2 × 10⁵ C per mole of water.

Thus, the amount of electricity required is 2 × 10⁵ C.

Step 1: Use the formula ∆G° = −RT lnK, where R = 8.314 J/mol·K, T = 300 K, and ln(10) = 2.303.

Step 2: Substitute the values: ∆G° = −(8.314 × 300 × 2.303) = −5744.14 J/mol.

Step 3: Convert to kJ: ∆G° = −5.74414 kJ/mol = −57.4 × 10⁻¹ kJ/mol.

Step 4: Approximate to the nearest integer: ∆G° = 57 × 10⁻¹ kJ/mol.

Thus, the ∆G° is approximately 57 × 10⁻¹ kJ/mol.

Step 1: Use the Nernst equation relation: logK = (nE°ₑₗₗ) / 0.0591.

Step 2: From the reaction, n = 5. Standard potentials: E°(MnO₄⁻/Mn²⁺) = 1.51 V, E°(H₂C₂O₄/CO₂) = −0.49 V. E°ₑₗₗ = 1.51 − (−0.49) = 2.00 V.

Step 3: Substitute values: logK = (5 × 2.00) / 0.0591 = 338.

Step 4: Convert to exponential form: Kₑq = 10³³⁸.

Thus, the value of x is approximately 338.

Step 1: The combustion reaction is: CₓHᵧ + O₂ → xCO₂ + (y/2)H₂O.

Step 2: From the given data, x = 40 mL / 10 mL = 4, and y/2 = 50 mL / 10 mL = 5, so y = 10.

Step 3: The molecular formula of the hydrocarbon is C₄H₁₀ (butane).

Step 4: Total number of atoms = x + y = 4 + 10 = 14.

Thus, the total number of carbon and hydrogen atoms is 14.