Question

An aluminium rod (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 order to support a load of ${10}^4$ Newton's is:

• A. $1\times {10}^{-2}{m}^2$
• B. $1.4\times {10}^{-3}{m}^2$
• C. $3.5\times {10}^{-3}{m}^2$
• D. $7.1\times {10}^{-4}{m}^2$

Answer 1

Answer: (C)

$3.5\times {10}^{-3}{m}^2$

No correct option.

Given,

Young's modulus $=7\times {10}^{9}N/m^2$ Breaking strain $=0.2\%f={10}^{4}N$

So from young's modulus formula.

$y=\frac{f{I}^{\prime }}{AI}$ Where $I^{\prime }=\frac{2}{1000}$

By putting the value then we get,

$7\times {10}^9={10}^4\times \frac{2}{1000}\times A\Rightarrow A=3.5\times {10}^8{m}^2$