Question

The length of wire is increased by $0.06\%$ by a load of $40N$ whose tensile modulus is $20\times {10}^{10}N/M^2$.The subjected stress is

• A. $12\times {10}^{10}N/m^2$
• B. $1.2\times {10}^{8}N/m^2$
• C. $120N/m^2$
• D. $1.25\times {10}^{6}N/m^2$

Answer 1

The correct option is B $1.2\times {10}^{8}N/m^2$

$\frac{△l}{l}\times 100=0.06$

$⟹\frac{△l}{l}=\frac{0.06}{100}$

Now stress $=\frac{20\times {10}^{10}\times 0.06}{1000}=1.2\times {10}^{8}N/m^2$