At what depth below Earth’s surface is g reduced to 7.35m/s2? (g0\=9.8m/s2,RE\=6.4×106m)
g(d) = g0(1−d/RE). 7.35 = 9.8(1−d/RE). 1−d/RE = 0.75. d/RE = 0.25. d = 0.25×6.4×106 = 1.6×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.6 × 10⁶ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.