An aluminum rod has a breaking strain 0.4%. The minimum cross-sectional area

of the rod in $m^2$  in order to support a load of ${10}^{5}N$  is if Young’s modulus is

$7\times {10}^{9}N{m}^{-2}$ (Nearly)

• A

  $1.5\times {10}^{-4}$

• B

  $2.5\times {10}^{-3}$

• C

  $3.6\times {10}^{-3}$ 

• D

  $2.5\times {10}^{-4}$