Concept:

Induced EMF in a Moving Conductor:

• When a conducting rod moves in a uniform magnetic field, an emf is induced across its ends.
• The induced emf (e) is given by the formula:

         e = vBℓ

Where:

• e = Induced emf (V)
• v = Velocity of the rod (m/s)
• B = Magnetic field strength (T)
• ℓ = Effective length of the rod (m)

Calculation:

Given:

Induced EMF (e) = 5 V

Magnetic field strength (B) = 10 T

Time (t) = 0.4 sec

The equation of the circle is:

         (x - 2)² + y² = 4

At t = 0.4 s, using the parametric motion:

         y = v × t

         y = v × 0.4 = 0.4v

From the circle equation:

         (x - 2)² + (0.4v)² = 4

         (x - 2)² = 4 - 0.16v²

         (x - 2) = √(4 - 0.16v²)

         x = 2 + √(4 - 0.16v²)

Effective length of the rod (ℓ) = 2x

         ℓ = 2 × (2 + √(4 - 0.16v²))

Now, using the emf formula:

         e = vBℓ

         5 = v × 10 × (2 + √(4 - 0.16v²))

         5 / 10 = v × (2 + √(4 - 0.16v²))

         0.5 = v × (2 + √(4 - 0.16v²))

Solving for v:

         v = 0.5 / (2 + √(4 - 0.16v²))

Squaring both sides:

         v² = (0.5)² / (2 + √(4 - 0.16v²))²

Approximating for small v:

         v ≈ 0.125 m/s

Final Answer:

The required velocity of the rod is 0.125 m/s

Concept:

Breaking Stress in Rotational Motion:

• When a rod rotates about its center, the stress is zero at the free ends and maximum at the middle.
• According to Newton’s Second Law, the force equation is:

F - (F + dF) = dm × ω² x

⇒ -∫ dF = ∫ ρ A ω² x dx

Where:

• σ = Breaking strength (Pa)
• ρ = Density of material (kg/m³)
• ω = Angular velocity (rad/s)
• ℓ = Length of the rod (m)

Calculation:

Given:

Breaking Strength (σ) = 8 × 10¹¹ Pa

Mass (m) = 8.7 kg

Radius of the rod (r) = 0.02 m

Angular frequency (ω) = 10⁶ rad/s

Density (ρ) = m/V = 8.7 / (π × 0.02² × ℓ)

⇒ Using the formula:

(ρ ω² ℓ²) / 8 = σ

⇒ ℓ = (8 σ × π × r²) / (m × ω²)

⇒ ℓ = ((8× 8 × 10¹¹ × π × (0.02)²) / (8.7 × (10⁶)²))

⇒ ℓ ≈ 0.092 cm

∴ The maximum allowable length of the rod is 0.092 cm.

Concept:

Internal Energy and Molar Specific Heat:

• The internal energy (U) of an ideal gas is related to its pressure (P) and volume (V) as per the given equation:

         U = PV/2

We use the first law of thermodynamics and the specific heat relation:

         U = PV/2

Calculation:

U=RT/2

⇒ dU=R/2dT

⇒  dU/dT=R/2 =CV

Using the formula for molar specific heat at constant pressure:

         C_p = R+C_V

Substituting values:

         C_p = R+R/2

         C_p = 3/2 R

         C_p = 1.5R

The molar specific heat of the gas in an isobaric process is 1.5R.

CONCEPT:

Reaction Mechanism and Degree of Unsaturation

• This reaction involves a series of steps with different reagents leading to the formation of products from the given starting material.
• The starting compound contains two bromine atoms attached to a bicyclic structure. The reaction proceeds through various steps: dehydrobromination, hydrolysis, and decarboxylation.
• Understanding the degree of unsaturation involves determining the number of double bonds and rings in the structure. The degree of unsaturation for a compound can be calculated using the formula:

  Degree of Unsaturation (DU) = (2C + 2 - H) / 2

• The reactions in question involve common organic transformations such as elimination (base-induced dehydrobromination), hydrolysis, and decarboxylation, each altering the degree of unsaturation of the product.

• Monobromination of C: When C undergoes monobromination, it forms two products due to the possibility of two distinct sites for substitution in the bicyclic structure. This leads to the formation of two different products depending on where the substitution occurs.
• Degree of Unsaturation of C: The compound C contains a bicyclic structure with a carboxyl group (-COOH), indicating a high degree of unsaturation. The degree of unsaturation of C is calculated to be five based on its structure, which accounts for the presence of double bonds and rings.
• Degree of Unsaturation of B: In compound B, after the base treatment, a decarboxylation step occurs, resulting in the loss of a CO₂ group. This leads to a decrease in the degree of unsaturation, reducing it compared to C.

Therefore, the correct answer is: "Degree of unsaturation of C is five" (Option 2).

Concept:

Solubility of Alkaline Earth Metal Sulphates

• The solubility of salts in water depends on the balance between two competing energies: lattice energy and hydration energy.
• Lattice energy is the energy required to break the ionic lattice of a salt, and it generally decreases down the group as the ionic radius increases.
• Hydration energy is the energy released when water molecules surround and solvate an ion. This energy generally decreases down the group due to increasing ionic size, which reduces the electrostatic attraction between the ion and water molecules.
• For alkaline earth metal sulphates (BeSO₄, MgSO₄, CaSO₄, SrSO₄, BaSO₄), the solubility trend is affected by the interplay between these two energies.

Explanation:

• A: Hydration energy is inversely proportional to size. Down the group, size increases, and hydration energy decreases. This is incorrect.
• B: While both lattice energy and hydration energy decrease down a group, the rate at which they decrease is different.  Hydration energy is generally more sensitive to the size of the ion than lattice energy.  This means that as ions get larger, the decrease in hydration energy is more pronounced than the decrease in lattice energy. This is correct.
• C: Lattice energy is inversely proportional to size. Down the group, size increases, and lattice energy decreases. This is incorrect.
• D: The strong attraction results in a large amount of energy being released when these ions are hydrated.  Water molecules are tightly bound to the Be²⁺ and Mg²⁺ ions, forming a relatively stable hydrated ion. Beryllium (Be²⁺) and Magnesium (Mg²⁺) have exceptionally high hydration energies compared to other alkaline earth metals and even many other metal ions. This is correct.
• E: Covalency is inversely proportional to size. Down the group, size increases, and covalency decreases. This is incorrect.

Therefore, the correct option is 3.

Concept:

Semi Pinacol-Pinacolone Rearrangement:

The semi Pinacol-Pinacolone rearrangement is a chemical reaction that involves the rearrangement of an alcohol into a carbonyl compound (usually a ketone) via a carbocation intermediate. This transformation is similar to the classic Pinacol-Pinacolone rearrangement but doesn't start from a diol hence the term "semi."

Here's the general mechanism:

• Formation of the Carbocation: The reaction typically starts with an alcohol that undergoes protonation (if necessary) to form a good leaving group such as water in the case of a secondary or tertiary alcohol or through activation by a Lewis acid or some other method. The loss of the leaving group generates a carbocation at the carbon holding the hydroxyl group.
• Carbocation Rearrangement: The intermediate carbocation then undergoes a 12-alkyl shift (or hydride shift) to create a more stable carbocation. This shift moves an adjacent alkyl group (or hydrogen) to the positively charged carbon transferring the positive charge to the position where the alkyl group (or hydrogen) was previously located.
• Formation of the Carbonyl Compound: Finally the rearranged carbocation loses a proton (in the case of a hydride shift) or gains an electron pair from a nucleophile (in the case of alkyl shift) to form a ketone (or aldehyde if the shift involves hydrogen).