• B (Period 2): smallest
• Al (Period 3): larger than B
• Ga (Period 4): slightly smaller than Al due to d-block contraction (3d¹⁰ electrons offer poor shielding)
• In (Period 5): larger than Ga
• Tl (Period 6): largest

The three slabs are placed side-by-side (parallel combination), each occupying one-third of the area $A/3$ and the full separation $d$.

Total capacitance:

$C = \frac{\varepsilon_0 (A/3)}{d}(K_1 + K_2 + K_3) = \frac{\varepsilon_0 A}{3d}(10+12+14) = \frac{36\varepsilon_0 A}{3d} = \frac{12\varepsilon_0 A}{d}$

For a single dielectric $K$:

$C = \frac{K\varepsilon_0 A}{d}$

$\Rightarrow K = 12$

The equivalent dielectric constant is simply the arithmetic mean: $(10+12+14)/3 = 12$.

Option A ($10,000) is correct.

Variable cost per unit = ($130,000 + $15,000) ÷ 5,000 = $29

Variable cost of sales (4,000 units) = $116,000

Contribution = $180,000 − $116,000 = $64,000

Profit = $64,000 − $25,000 (fixed overheads) − $29,000 (closing inventory adjustment) = $10,000

Secondary alcohols oxidize to ketones (e.g., isopropanol → acetone). Primary alcohols → aldehydes → carboxylic acids; tertiary alcohols resist oxidation.

Initially, beam A's plane is parallel to the polaroid ($\theta_A = 0^\circ$) and B's is perpendicular ($\theta_B = 90^\circ$).

After $30^\circ$ rotation, the angles become $30^\circ$ and $60^\circ$ respectively.

Equal brightness means: $I_A \cos^2(30^\circ) = I_B \cos^2(60^\circ)$

$I_A \left( \frac{\sqrt{3}}{2} \right)^2 = I_B \left( \frac{1}{2} \right)^2 \implies I_A \cdot \frac{3}{4} = I_B \cdot \frac{1}{4} \implies \frac{I_A}{I_B} = \frac{1}{3}$.