Question

An aluminium rod, Young's modulus $7.0 \times 10^{9} Nm ^{-2}$, 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 $10^{4} N$ is

• A. $1 \times 10^{-2}$
• B. $1.4 \times 10^{-3}$
• C. $1.0 \times 10^{-3}$
• D. $7.1 \times 10^{-4}$

Answer 1

Answer: (D)

$A s, Y=\frac{F / A}{\text { Breaking strain }}$
or $ A =\frac{F}{Y \times \text { Breaking strain }}=\frac{10^{4} \times 100}{7 \times 10 \times 0.2} $
$=0.71 \times 10^{-3}=7.1 \times 10^{-4} $