Question

Overall changes in volume and radius of a uniform cylindrical

Answer 1

Answer:

Tensile Stress = 4.08 × 10 ^8 N per m²

Step-by-step explanation:

volume =  area × length  

As area = πr²

V = A × l = πr²l  

log V = log(πr² × l)  

log V = logπ + 2 log r + log l  

differentiating both sides,  

± dV/V =± 2dr/r ± dl/l  

dl/l = 2dr/r + dV/V  

Thus,

% change in l = 2 × % change in r + % change in V

=2 × 0.002 % + 0.2 % = 0.204%  

dl/l = strain = 0.204/100  

stress = Y × strain  

= 2 × 10¹¹ × 0.204/100  

= 0.408 × 10^9 N/m² or 4.08 × 10^8 N per m²  

Tensile Stress = 4.08 × 10 ^8 N per m².