Question

A metallic rod breaks when strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^{9} N / m ^{2}$. The area of cross section of support a load of $10^{4} N$ is

• A. $7.1 \times 10^{-4} m ^{2}$
• B. $7.1 \times 10^{-2} m ^{2}$
• C. $7.1 \times 10^{-8} m ^{2}$
• D. $7.1 \times 10^{-6} m ^{2}$

Answer 1

Answer: (A)

Stress $=$ Strain $\times Y$
Thus, maximum stress $=\frac{0.2}{100} \times 7 \times 10^{9}=1.4 \times 10^{7}$
Now, $F$ orce $=$ Stress $\times$ Area
Thus, $10^{4}=1.4 \times 10^{7} \times A$, or, $A =7.14 \times 10^{-4} m ^{2}$