MHT CET 2024 2 May Shift 2 Question Paper: Download PCM Question Paper with Answers PDF

We are given that B is the adjoint of a matrix A, and the determinant of A, |A| = 4. The adjoint B is related to the determinant of A as follows:

B = adj(A) = |A| × A⁻¹.

The elements of B are cofactors of the corresponding elements in A, so the element B₁₂ (second element in the first row) is the cofactor of A₁₂, which is α.

From the question, we are told that |A| = 4. For α, we compute the cofactor corresponding to the matrix element. The matrix B suggests that for A, the cofactor corresponding to A₁₂ is 1.

Conclusion: Therefore, α = 1.

The IUPAC name of an ether is formed by identifying the two alkyl groups attached to the oxygen atom. The name is derived by prefixing the alkyl groups with "methoxy," "ethoxy," etc., depending on the alkyl groups, and adding the suffix "ether."

In this case, the ether consists of a methoxy group (CH₃) and an ethane group (C₂H₅) attached to the oxygen atom.

Conclusion: The correct IUPAC name is "Methoxy ethane."

To find A⁻¹, we calculate det(A) and adj(A). The determinant of A is:

det(A) = 0 * ((2 * 1) - (3 * 1)) - 1 * ((1 * 1) - (3 * 3)) + 2 * ((1 * 1) - (2 * 3)) = -2

Since det(A) ≠ 0, A⁻¹ exists. Using the formula:

A⁻¹ = (1/det(A)) * adj(A)

The adjugate matrix is:

adj(A) = [[1, 2, 1], [-8, 6, 2], [5, 3, 1]]

Thus, A⁻¹ is:

A⁻¹ = ½ * [[1, 2, 1], [-8, 6, 2], [5, 3, 1]].

The Clemmensen reduction is a chemical process where aldehydes or ketones are reduced to alkanes. This transformation is carried out using zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl).

Reaction:

R-CO-R' + Zn(Hg) + HCl → R-CH₂-R'

This reduction is effective for aldehydes and ketones stable under acidic conditions.

Among the 3d transition elements, Zinc (Zn) exhibits a lower oxidation state compared to other elements in the series. This is because, unlike most other transition metals, Zinc has a completely filled d¹⁰-orbital configuration, which restricts its ability to exhibit higher oxidation states.

Electronic Configuration of Zn:

Zn: [Ar] 3d10 4s2

Due to this fully occupied d-orbital, Zinc predominantly shows a +2 oxidation state, derived from the loss of its two 4s-electrons. This is in contrast to other transition metals, which typically display a range of oxidation states due to the availability of partially filled d-orbitals.

Conclusion: Zinc is unique in the 3d series as it predominantly exhibits a +2 oxidation state due to its filled d¹⁰-orbitals.

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation, which is given by:

pH = pKa + log([salt]/[acid])

Where:

• pK_a is the negative logarithm of the acid dissociation constant (K_a),
• [salt] is the concentration of the conjugate base (salt), and
• [acid] is the concentration of the weak acid.

Conclusion: The pH of the buffer solution is calculated by applying the Henderson-Hasselbalch equation using the concentrations of salt and acid along with pK_a.

The pH of a solution is related to the concentration of hydrogen ions (H⁺) by the equation:

pH = -log[H+]

Given that the pH is 2.7, we rearrange the equation to solve for [H⁺]:

[H+] = 10-pH = 10-2.7

Calculating 10^-2.7:

10-2.7 ≈ 1.99 × 10-3 M

Thus, the concentration of hydrogen ions is approximately 1.99 × 10^-3 M.

According to Henry’s law, the solubility of a gas (S) in a liquid is directly proportional to the external pressure (P) at a constant temperature:

S ∝ P

Conclusion: The solubility of a gas increases with increasing pressure, following Henry's law.

In a Body-Centered Cubic (BCC) unit cell, the unit particles consist of atoms located at the corners and at the center of the unit cell.

• Atoms at the corners contribute 1/8 of an atom each, with 8 atoms at the corners. Thus, the total contribution from the corners is 8 × 1/8 = 1.
• The atom in the center contributes a whole atom (1).

Thus, the total number of atoms in the BCC unit cell is:

1 (from corners) + 1 (from the center) = 2

Pyridinium chlorochromate (PCC) is commonly used to oxidize primary alcohols (R-CH₂-OH) into aldehydes (R-CHO) without further oxidizing them to carboxylic acids.

Conclusion: The correct reagent for converting R-CH₂-OH to R-CHO is PCC.

For a BCC unit cell, the relationship between the edge length (a) and the atomic radius (r) is given by:

a = 4r/√3

This relationship is derived from the geometry of the BCC structure, where the body diagonal of the cube equals the sum of two atomic radii. This can be expressed in terms of the edge length.

Conclusion: The edge length of a BCC unit cell is 4r/√3.

Preliminary characterization of nanoparticles often involves techniques such as Transmission Electron Microscopy (TEM) and Dynamic Light Scattering (DLS). These methods are more commonly used than the ones listed in the options. X-ray diffraction and scanning electron microscopy (SEM) are useful for characterizing the structure, but they are not considered preliminary tests for nanoparticles.

Conclusion: The correct answer is None of these.

To derive the IUPAC name of a haloarene, start by adding the appropriate halogen prefix, such as chloro, bromo, or iodo, to the name of the parent hydrocarbon, such as benzene, toluene, etc. In this case, the prefix "halo-" combined with the parent hydrocarbon results in the name.

Conclusion: The correct name is based on the halogen present and the parent hydrocarbon.

To find the converse of a logical statement (A → B), we simply swap A and B. In this case, for ((~p) ∧ q) → r, the converse is r → ((~p) ∧ q).

Using logical equivalence, this becomes:

(p ∨ (~q)) → (~r)

Conclusion: The converse of ((~p) ∧ q) → r is (p ∨ (~q)) → (~r).

To compute the negation of (p ∧ (~q)) ∨ (~p), we apply De Morgan's laws:

~[(p ∧ (~q)) ∨ (~p)] = [~(p ∧ (~q)) ∧ ~(~p)]

Using De Morgan's laws:

(~p ∨ q) ∧ p

Simplify further:

p ∧ q

Conclusion: The negative of (p ∧ (~q)) ∨ (~p) is p ∧ q.

To find the variance of a probability distribution, we use the formula:

Variance = E(X^2) - (E(X))^2

Step 1: Calculate E(X), the expected value:

E(X) = Σ x * P(X)
E(X) = (0 * 9/16) + (1 * 3/8) + (2 * 1/16)
E(X) = 0 + 3/8 + 2/16
E(X) = 4/8 = 1/2
            

Step 2: Calculate E(X^2), the expected value of X^2:

E(X^2) = Σ x^2 * P(X)
E(X^2) = (0^2 * 9/16) + (1^2 * 3/8) + (2^2 * 1/16)
E(X^2) = 0 + 3/8 + 4/16
E(X^2) = 5/8
            

Step 3: Calculate the variance:

Variance = E(X^2) - (E(X))^2
Variance = 5/8 - (1/2)^2
Variance = 5/8 - 2/8 = 3/8
            

Conclusion: The variance of the given probability distribution is 3/8.