We know, volume = cross sectional area × length
e.g., V = A × l = πr²l
or, logV = log(πr² × l)
or, logV = logπ + 2logr + logl
differentiate both sides,
± dV/V =± 2dr/r ± dl/l
so, for getting maximum error , dl/l = 2dr/r + dV/V
hence, % change in l = 2 × % change in r + % change in V
=2 × 0.002 % + 0.2 % = 0.204%
so, dl/l = strain = 0.204/100
now, stress = Y × strain
= 2 × 10¹¹ × 0.204/100
= 0.408 × 10^9 N/m² or 4.08 × 10^8 N/m²
hence, tensile stress = 4.08 × 10^8 N/m²
