MHT CET 2024 9 May Shift 2 Question Paper (Available): Download PCM Question Paper with Answers PDF

The equation of the line in vector form is:

r = (6, 7, 7) + t(3, 2, -2),

where (6, 7, 7) is a point on the line, and (3, 2, -2) is the direction vector. The given point is (1, 2, 3).

The perpendicular distance D from a point (x1, y1, z1) to a line is given by:

D = ||d × (r0 - P)|| / ||d||.

Here, r0 = (6, 7, 7), d = (3, 2, -2), and P = (1, 2, 3).

Step 1: Compute (r0 - P)

(r0 - P) = (6 - 1, 7 - 2, 7 - 3) = (5, 5, 4).

Step 2: Compute the cross product d × (r0 - P)

d × (r0 - P) = |i j k|

|3 2 -2|

|5 5 4|

After expansion, d × (r0 - P) = (18, -22, 5).

Step 3: Compute the magnitude of the cross product

||d × (r0 - P)|| = sqrt(18² + (-22)² + 5²) = sqrt(833).

Step 4: Compute the magnitude of the direction vector d

||d|| = sqrt(3² + 2² + (-2)²) = sqrt(17).

Step 5: Compute the perpendicular distance D

D = ||d × (r0 - P)|| / ||d|| = sqrt(833) / sqrt(17) = sqrt(833 / 17) = 7.

Final Answer: 7 units.

Given:

|a| = √3, |b| = 5, |c| = 4, and the angle between b and c is π/3.

From the vector triple product property, |a × (b × c)| = |a| |b × c| sin θ.

Step 1: Calculate |b × c|

|b × c| = |b||c| sin(π/3) = 5 × 4 × √3/2 = 10√3.

Step 2: Calculate |a × (b × c)|

|a × (b × c)| = |a| |b × c| = √3 × 10√3 = 30.

Final Answer: 30.

Given a binomial distribution B(7, p):

P(X = k) = C(n, k) p^k (1 - p)^(n - k).

Equating probabilities: P(X = 3) = 5P(X = 4).

Substitute values: C(7, 3)p^3(1 - p)^4 = 5C(7, 4)p^4(1 - p)^3.

Simplify to find p = 1/6.

Mean = np = 7 × 1/6 = 7/6.

Variance = np(1 - p) = 7 × 1/6 × 5/6 = 35/36.

Sum = Mean + Variance = 7/6 + 35/36 = 77/36.

Final Answer: 77/36.

The given equation is: e dy = (x + 1) dx.

Step 1: Separate the variables

dy = (x + 1)/e dx.

Step 2: Integrate both sides

y = (1/e) ∫(x + 1) dx.

Integrate: y = (1/e) [(x²/2) + x + C].

Step 3: Apply the initial condition y(0) = 3

3 = (1/e) [(0²/2) + 0 + C], so C = 3e.

Step 4: Write the final solution

y = (x²/2e) + (x/e) + 3.

Final Answer: y = (x + 1)log(x + 1) - x + 3.

Step 1: Rewrite the integral

I = ∫₀¹ √(1 - x)/√(1 + x) dx = ∫₀¹ (1 - x)/√(1 - x²) dx.

Step 2: Split the integral

I = ∫₀¹ (1/√(1 - x²)) dx - ∫₀¹ (x/√(1 - x²)) dx.

Step 3: Evaluate each term

First term: ∫₀¹ (1/√(1 - x²)) dx = sin⁻¹(x)|₀¹ = π/2.

Second term: ∫₀¹ (x/√(1 - x²)) dx = -√(1 - x²)|₀¹ = -1.

Step 4: Combine results

I = π/2 - 1.

Step 1: Identify the critical point of |2x - 3|

2x - 3 = 0 → x = 3/2.

Step 2: Split the integral

∫₀² |2x - 3| dx = ∫₀³/₂ (3 - 2x) dx + ∫₃/₂² (2x - 3) dx.

Step 3: Evaluate each term

First term: ∫₀³/₂ (3 - 2x) dx = [3x - x²]|₀³/₂ = 9/4.

Second term: ∫₃/₂² (2x - 3) dx = [x² - 3x]|₃/₂² = 1/4.

Step 4: Combine results

Total = 9/4 + 1/4 = 5/2.

Step 1: Express α and γ in terms of β

α = π/2 - β, γ = π/2 - 2β.

Step 2: Use tan identities

tanα = cotβ, tanγ = cot(2β).

Step 3: Substitute and simplify

tanα = tanβ + 2tanγ.

The mean μ = 2 and variance σ² = 1. Using μ = np and σ² = np(1-p):

From μ = np = 2, we find p = 2/n.

Substituting into the variance equation:

np(1-p) = 1 → n × (2/n) × (1 - 2/n) = 1.

Simplify to get n = 4 and p = 1/2.

P(X > 1) = 1 - P(X ≤ 1) = 1 - (P(X = 0) + P(X = 1)).

Using the binomial formula:

P(X = 0) = 1/16, and P(X = 1) = 4/16 = 1/4.

P(X > 1) = 1 - (1/16 + 1/4) = 1 - 5/16 = 11/16.

The sides of the triangle are a = sin(α), b = cos(α), and c = sqrt(1 + sin(α)cos(α)).

Using the cosine rule:

cos(C) = (a² + b² - c²) / (2ab).

Substitute a = sin(α), b = cos(α), and c = sqrt(1 + sin(α)cos(α)):

cos(C) = (sin²(α) + cos²(α) - (1 + sin(α)cos(α))) / (2sin(α)cos(α)).

Using sin²(α) + cos²(α) = 1:

cos(C) = (1 - (1 + sin(α)cos(α))) / (2sin(α)cos(α)) = -1/2.

This gives C = 120°.

The given curves are:

y² = 6x and 9x² + by² = 16.

Differentiate y² = 6x:

2y(dy/dx) = 6 → dy/dx = 3/y.

Differentiate 9x² + by² = 16:

18x + 2by(dy/dx) = 0 → dy/dx = -9x/by.

Using orthogonality (m₁ × m₂ = -1):

(3/y) × (-9x/by) = -1.

Substitute y² = 6x:

(27x)/(by²) = 1 → (27x)/(b × 6x) = 1 → b = 9/2.

Let f(x) = log(x)/x. To find the maximum value, we first differentiate f(x) with respect to x:

f'(x) = (1/x × x - log(x) × 1) / x² = (1 - log(x)) / x².

Set f'(x) = 0 to find the critical points:

1 - log(x) = 0 → log(x) = 1 → x = e.

To confirm this is a maximum, evaluate f(e):

f(e) = log(e)/e = 1/e.

Thus, the maximum value is 1/e.

The rate of cooling of an object is influenced by its emissivity. A black surface, having the highest emissivity, radiates heat more effectively compared to other colors. The cooling order, from fastest to slowest, is:

Black > Red > Yellow > White.

The shortest wavelength in a spectral series occurs when the transition is from n = infinity to the lowest energy level (n₁). For the Balmer series, n₁ = 2, and for the Paschen series, n₁ = 3.

Using the Rydberg formula:

1/λ = R_H × (1/n₁² - 1/n₂²), where R_H is the Rydberg constant.

For the Balmer series (n₁ = 2, n₂ = infinity):

1/λ_Balmer = R_H × (1/4) → λ_Balmer = 4/R_H.

For the Paschen series (n₁ = 3, n₂ = infinity):

1/λ_Paschen = R_H × (1/9) → λ_Paschen = 9/R_H.

The ratio of the shortest wavelengths is:

λ_Balmer/λ_Paschen = (4/R_H) / (9/R_H) = 4/9.

The gyromagnetic ratio (γ) is given by:

γ = e / (2m_e),

where e is the charge of the electron, and m_e is the mass of the electron.

The Bohr magneton (μ_B) is defined as:

μ_B = e * h / (4π * m).

Step-by-step evaluation:

• The AND gate takes inputs A = 1 and B = 0. The output is: 1 * 0 = 0.
• The OR gate takes inputs 0 (from the AND gate) and C = 1. The output is: 0 + 1 = 1.
• The NOT gate inverts the OR gate output: NOT(1) = 0.

The final output of the circuit is 0.

To convert a galvanometer into an ammeter, a shunt resistance (R_s) is connected in parallel with the galvanometer. The total resistance of the system is the parallel combination of the galvanometer resistance (R_g) and the shunt resistance (R_s).

The total resistance is given by: I = V / (R_g + R_s), where I is the total current, and V is the voltage across the combination.

The magnetic field (B) inside a solenoid is given by: B = μ_0 * n * I, where μ_0 is the permeability of free space, n is the number of turns per unit length, and I is the current passing through the solenoid.

n (turns per unit length) is calculated as: n = N / L, where N is the total number of turns, and L is the solenoid's length.

In a parallel combination of resistors, the total resistance (R_total) is given by: 1 / R_total = 1 / R_1 + 1 / R_2 + ...

Using Ohm’s Law, the current flowing through the circuit is: I = V / R_total, where V is the voltage across the circuit.

Given:

• Maximum speed at mean position: vmax = 6 cm/sec
• Amplitude: A = 4 cm
• Velocity at position x: v = 2 cm/sec

Step 1: Relate Maximum Speed to Angular Frequency

In SHM, vmax = ω × A.

Angular frequency, ω = vmax / A = 6 / 4 = 1.5 sec⁻¹.

Step 2: Use the Velocity-Position Relationship in SHM

The velocity v of a particle in SHM is related to its position x by:

v = ω × sqrt(A² - x²).

Solving for x:

x = sqrt(A² - (v / ω)²).

Step 3: Substitute the Known Values

Substitute A = 4 cm, v = 2 cm/sec, and ω = 1.5 sec⁻¹:

x = sqrt(4² - (2 / 1.5)²).

x = sqrt(16 - (4 / 2.25)).

x = sqrt(16 - 16 / 9).

x = sqrt(144 / 9 - 16 / 9).

x = sqrt(128 / 9).

x ≈ 3.77 cm.

Rounding to the nearest whole number:

x ≈ 4 cm.

Conclusion: The position of the particle from the mean position when its velocity is 2 cm/sec is 4 cm.

The rate of a reaction is proportional to the concentration of reactants raised to their respective powers. If the rate is independent of [A] and directly proportional to [B]², the rate law is:

Rate = k[B]², where k is the rate constant.

Ammoniacal silver nitrate, known as Tollens' reagent, reacts with aldehydes to produce metallic silver. This forms a characteristic silver mirror inside the test tube, indicating the presence of aldehydes.

The element with atomic number 27 is cobalt (Co). Its electronic configuration is:

[Ar] 3d⁷ 4s².

In the +2 oxidation state, cobalt loses two 4s electrons, leaving:

[Ar] 3d⁷.

The 3d⁷ configuration has three unpaired electrons.

The freezing point depression (ΔTf) is related to the cryoscopic constant (Kf) by:

ΔTf = Kf × m, where m is the molality of the solution.

Given:

• Mass of solute = 10 g
• Mass of solvent = 100 g = 0.1 kg
• ΔTf = 1.5°C

Assuming the molar mass of solute = 10 g/mol, the molality is:

m = moles of solute / mass of solvent in kg = 10 / (10 × 0.1) = 10 mol/kg.

Substituting into the formula:

1.5 = Kf × 10 → Kf = 0.15 kg/mol.

The Swarts reaction involves the substitution of a halogen atom in an alkyl halide with a fluorine atom by reacting the alkyl halide with a metal halide like silver fluoride. This reaction is commonly used to prepare alkyl fluorides.

The Wittig reaction, on the other hand, is a nucleophilic addition reaction in which a phosphorus ylide reacts with a carbonyl compound (aldehyde or ketone) to form an alkene. This reaction is widely used in organic synthesis for constructing carbon-carbon double bonds.

An adiabatic process is one in which there is no heat exchange between the system and its surroundings. In this process, the change in internal energy is equal to the work done by or on the system, i.e., \( Q = 0 \).

Propan-1-ol is the IUPAC name for 1-propanol, which has a hydroxyl group (-OH) attached to the first carbon of a three-carbon chain. Cyclobutane is a four-membered cyclic compound with a bond length of approximately 1.5 Å.

Alcohols react with haloacids (HX) to form alkyl halides through a substitution reaction, where the hydroxyl group (-OH) is replaced by a halide ion (X). This occurs under acidic conditions.

Ozonolysis of propene results in the cleavage of the double bond, producing two carbonyl compounds. These are typically aldehydes or ketones, depending on the structure of the original alkene.

Ozonolysis is a useful reaction for determining the position of double bonds in alkenes as it splits the double bond into two carbonyl compounds.

The natural abundance of chlorine isotopes is approximately 75% for ³⁵Cl and 25% for ³⁷Cl. This corresponds to a ratio of 3:1 for the isotopes ³⁵Cl and ³⁷Cl.