Question

An aluminum rod of Young's modulus $7\times {10}^{9}N{m}^{-}2$ has a breaking strain of 0.2%. The minimum cross-sectional area of the rod in $m^2$ in order to support a load of $7\times {10}^{3}N$ is:

• A. $5\times {10}^{-5}{m}^2$
• B. $5\times {10}^{-4}{m}^2$
• C. $5\times {10}^{-3}{m}^2$
• D. $5\times {10}^{-2}{m}^2$

Answer 1

The correct option is B $5\times {10}^{-4}{m}^2$

Given : $Y=7\times {10}^9$ $N/m^2$ $F=7\times {10}^{3}N$ $\frac{\Delta l}{l}=0.002$ $F=7\times {10}^{3}N$

Let the cross-section area of the wire be $A$.

From Hooke's law $Y=\frac{F}{A}\times \frac{l}{\Delta l}$

$\therefore$ $7\times {10}^9=\frac{7\times {10}^3}{A}\times \frac{1}{0.002}$ $⟹A=5\times {10}^{-4}{m}^2$