At what height above Earth’s surface is g reduced to 7.84m/s2? (g0\=9.8m/s2,RE\=6.4×106m)
g(h) = g0(1+h/RE)2. 7.84 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.25. 1+h/RE = 1.25≈1.118. h/RE = 0.118. h = 0.118×6.4×106≈7.55×105m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.6 × 10⁵ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.