For the same voltage drop, increasing the voltage of a distributor n-times:

Correct option is A

​$$\begin{aligned}R &= \dfrac{\rho l}{A} \\[10pt]\text{Where:} \\[6pt]&\bullet \; R = \text{Resistance}, \\ &\bullet \; l = \text{Length of the conductor}, \\ &\bullet \; A = \text{Area of cross-section}, \\ &\bullet \; \rho = \text{Resistivity of the material}. \\[12pt]\text{Voltage drop:} \\[6pt]&\text{Voltage drop } = I \cdot R \\[12pt]\text{Active power:} \\[6pt]&P = V \cdot I \cdot \cos \phi \\[14pt]1 \;\;\; \text{Given: Voltage is increased } n\text{-times:} \\[6pt]&V_2 = n \cdot V_1 \;\; \Rightarrow \;\; \dfrac{V_2}{V_1} = n \\[14pt]2 \;\;\; \text{To maintain constant power transfer:}\end{aligned}$$

$$\begin{aligned}&\dfrac{P_2}{V_2 \cdot I_2 \cdot \cos \phi_2} = \dfrac{P_1}{V_1 \cdot I_1 \cdot \cos \phi_1} \\[12pt]3 \;\;\; &\text{Since the connected load and power factor remain constant:} \\[6pt]&I_2 = \dfrac{I_1}{n} \\[12pt]4 \;\;\; &\text{Given that the voltage drop remains the same:} \\[6pt]&I_2 \cdot R_2 = I_1 \cdot R_1 \;\; \Rightarrow \;\; R_2 = n \cdot R_1 \\[12pt]5 \;\;\; &\text{Substituting } R = \dfrac{\rho l}{A}: \\[6pt]&\dfrac{\rho l}{A_2} = n \cdot \dfrac{\rho l}{A_1} \;\; \Rightarrow \;\; A_2 = \dfrac{A_1}{n}\end{aligned}$$​​​