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Sol.  Angular width of central maxima $\theta_{cm} = \frac{2\lambda}{d}$

= $\frac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}}$ rad.

= $(628 \times 10^{-5}) \times \frac{180}{\pi}$ degrees

$\approx 36 \times 10^{-2}$ degrees

$\therefore$ Answer : $\alpha = 36$

Sol.  $\gamma = \frac{C_p}{C_v} = \frac{3}{2}$

$\because P_1 V_1^\gamma = P_2 V_2^\gamma$

$\Rightarrow V_2 = \left( \frac{P_1}{P_2} \right)^{1/\gamma} V_1$

$\Rightarrow V_2 = \left( \frac{8}{1} \right)^{2/3} \times 0.15$

$\Rightarrow V_2 = 0.6 \text{ m}^3$

$\because W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$

= $\left( \frac{8 \times 0.15 - 1 \times 0.6}{1.5 - 1} \right) \times 10^5 = 1.2 \times 10^5$

$\Rightarrow W = 120 \text{ kJ}$

Sol.  $\vec{F} = q(\vec{v} \times \vec{B})$

= $1 \times 10^{-6} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 3 & -5 \end{vmatrix}$

= $10^{-6} [\hat{i} + 11\hat{j} + 7\hat{k}]$

$\Rightarrow \vec{F} = \sqrt{171} \times 10^{-6} \text{ N}$

$\therefore \alpha = 171$

Sol.  

$\sigma = \frac{T}{A} = \frac{W + \frac{w}{3}}{A}$

$\Rightarrow \sigma = \frac{W}{A} + \frac{2}{6} \frac{w}{A}$

$\therefore \gamma = 6$

Sol.  

Weight of coin = Additional upthrust

$\Rightarrow m_c g = \rho_l V_{add} g$

$\Rightarrow m_c = 1 \times 3.87 \times (10)^2$

$\Rightarrow m_c = 387 \text{ gram}$

CONCEPT:

Aromaticity and Hückel's Rule

• Aromatic compounds are cyclic, planar molecules with a conjugated π-electron system.
• They must follow Hückel’s rule: a compound is aromatic if it contains (4n + 2) π-electrons, where n is a non-negative integer (n = 0, 1, 2, ...).
• The compound must be fully conjugated (i.e., every atom in the ring has a p-orbital) and must be planar to allow continuous overlap of p-orbitals.

EXPLANATION:

• (a) Cyclopentadienyl cation: Has 6 π-electrons → follows Hückel’s rule → Aromatic
• (b) Cyclopentadienyl anion: Has 4 π-electrons → does not satisfy (4n + 2) rule → Not Aromatic
• (c) 4-membered ring with 6 π-electrons → satisfies Hückel’s rule → Aromatic
• (d) 4-membered ring with 6 π-electrons (similar to c) → Aromatic
• (e) Cycloheptatrienyl cation (tropylium ion) → has 6 π-electrons → Aromatic
• (f) Cycloheptatrienyl anion: Has 8 π-electrons → does not follow Hückel’s rule → Not Aromatic
• (g) 4n π-electron system (non-planar rectangle) → Not Aromatic
• (h) Cyclopropyl cation: Has 2 π-electrons → satisfies Hückel’s rule for n = 0 → Aromatic

Therefore, the correct answer is: a, c, d, e, h are aromatic and b, f, g are not aromatic.

CONCEPT:

Reaction Pathway Involving Alkyl Halide → Alkyne → Ketone

• Alkyl halides can undergo elimination reactions with alcoholic KOH to form alkenes.
• Alkenes react with bromine to give vicinal dibromides (bromination across the double bond).
• Vicinal dibromides treated with strong bases like NaNH₂ in alcohol can undergo double elimination to yield alkynes.
• Alkynes undergo hydration in the presence of H₂SO₄ and HgSO₄ to yield ketones (Markovnikov addition of water).

EXPLANATION:

• Given compound [A] is an optically active C₄H₉Br, most likely 2-bromobutane.
• On treatment with hot alcoholic KOH, [A] undergoes dehydrohalogenation to give but-2-ene [B].
• [B] reacts with Br₂ to form vicinal dibromide [C], i.e., 2,3-dibromobutane.
• [C] reacts with alcoholic NaNH₂ to undergo two eliminations, forming but-2-yne [D].
• [D] undergoes hydration with dilute H₂SO₄ and HgSO₄ at 333 K to form butan-2-one [E].

Therefore, the IUPAC name of compound [E] is Butan-2-one.

CONCEPT:

Trends in Group-17 Elements (Halogens)

• The halogens include Fluorine (F), Chlorine (Cl), Bromine (Br), and Iodine (I).
• Covalent Radius: Increases down the group due to the addition of new shells (F < Cl < Br < I).
• Ionic Radius: Also increases as you go down the group (F⁻ < Cl⁻ < Br⁻ < I⁻).
• First Ionization Energy: Decreases down the group due to increasing atomic size (F > Cl > Br > I).
• Electron Affinity: Typically increases down the group, but fluorine shows an irregularity—it has lower electron affinity than chlorine despite being smaller.

EXPLANATION:

• Electron Affinity: The irregularity arises because fluorine is very small, and adding an extra electron leads to high electron–electron repulsion, reducing the energy released.
• All other properties—covalent radius, ionic radius, and first ionization energy—follow regular trends down the group without deviation in the first four elements.

Therefore, the correct answer is: B only

CONCEPT:

Henry’s Law and Temperature Dependence

• Henry’s Law Constant (K_H): It relates the solubility of a gas in a liquid to the pressure of the gas above the liquid.
• According to Henry's Law: C = K_H × P, where C is the concentration (solubility), P is the partial pressure, and K_H is Henry’s constant.
• Temperature Dependence: Solubility of gases in liquids generally decreases with increase in temperature, hence K_H values typically increase with temperature.
• The increase in K_H with temperature is not linear and varies for different gases. Light gases like He show a steeper increase.

EXPLANATION:

• Among the options, only graph 4 shows the correct qualitative trend: Henry’s constant (K_H) increases with temperature for all gases.
• Also, gases like He (non-polar and light) exhibit lower solubility and hence higher K_H, followed by N₂ and CH₄.
• This trend is consistent with the observation that lighter gases have higher Henry’s constants and are less soluble in water.

Therefore, the correct answer is: Option 4

CONCEPT:

Bohr’s Model and Radius of Electron Orbits

r ∝ n²

rₙ = r₁ × n²

• According to Bohr’s model of the hydrogen atom, the radius of the nth orbit is proportional to the square of the principal quantum number n:
• This means the radius of orbit n can be expressed as:
• where r₁ is the radius of the first orbit (ground state).

EXPLANATION:

• Radius of 3^rd orbit relative to 1^st orbit:

r₃ / r₁ = (3)² / (1)² = 9 / 1 = 9

Statement 1 is correct.

• Radius of 8^th orbit relative to 4^th orbit:

r₈ / r₄ = (8)² / (4)² = 64 / 16 = 4

Statement 2 is correct.

• Radius of 6^th orbit relative to 4^th orbit:

r₆ / r₄ = (6)² / (4)² = 36 / 16 = 2.25

Statement 3 says “three times larger”, but the actual ratio is 2.25.

Therefore, Statement 3 is incorrect.

• Radius of 4^th orbit relative to 2^nd orbit:

r₄ / r₂ = (4)² / (2)² = 16 / 4 = 4

Statement 4 is correct.

Therefore, the incorrect statement is: Radius of 6^th orbit is three times larger than that of 4^th orbit (Option 3).