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 $m^2$ in order to support a load of
${10}^4$ N is

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

Answer 1

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

$Y=\frac{\frac{F}{A}}{Breakingstrain}$

or $a=\frac{F}{Y\times Breakingstrain}=\frac{{10}^4\times 100}{7\times 10\times 0.2}$
$=0.71\times {10}^{-3}=7.1\times {10}^{-4}$