Question

An aluminum 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'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

The correct option is D $7.1\times {10}^{-4}{m}^2$

We have,

breaking strain$\left(\frac{\Delta l}{l}\right)=$

we have,

breaking strain of aluminium rod$=0.2\%$

so, $\frac{\Delta l}{l}=\frac{0.2}{100}$

Let us take, the minimum cross-sectional area be 'A' to support ${10}^4$N

So,

the stress$\left(P\right)=\frac{{10}^4}{A}$

We know that, young's nodulus(y)

$y=\frac{p}{\Delta l/l}$

$\Rightarrow 7\times {10}^9=\frac{{10}^4}{A\left(\frac{0.2}{100}\right)}$

$\Rightarrow A=\frac{{10}^4\times 100}{0.2\times 7\times {10}^9}$

$A=7.1\times {10}^{-4}$ $m^2$.