Faraday's first law states that the mass of a substance deposited at an electrode is proportional to the quantity of charge passed, expressed as m = (Q × M)/(n × F), where Q is charge, M is molar mass, n is the number of electrons per ion, and F is the Faraday constant (96500 C). The charge passed is Q = I × t = 9.65 A × (100 × 60) s = 9.65 × 6000 = 57900 C. For aluminium, Al^3+ + 3e^- → Al, so n = 3. Therefore m = (57900 × 27)/(3 × 96500) = 1563300/289500 = 5.4 g. Option 16.2 g wrongly uses n = 1. Option 2.7 g halves the correct value. Option 8.1 g uses an incorrect time conversion. This calculation reflects the industrial Hall-Héroult deposition pattern tested in JEE electrolysis problems. This concept also bridges to Some Basic Concepts in Chemistry and d- and f-Block Elements, so mastering it strengthens performance on linked questions from those topics as well. Plausibility check: 57900 C is about 0.6 faraday; depositing aluminium needs 3 faradays per mole, so roughly 0.2 mol, which is 0.2 × 27 = 5.4 g, consistent.