Question

The Young's modulus for steel is 2 × 10^11 N/m^2. If the interatomic spacing for the metal is 2.8 angstrom, find the increase in the interatomic spacing for a stress of 10^9N/m^2​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• Thus, interatomic space increase by $0.014A$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

☆ɢɪᴠᴇɴ☆

• Young's Modulus,

$\:\:\:\:\:\:\:\:Y = 2\times {10}^{11} N{m}^{-2}$

• Intermolecular Space,

$\:\:\:\:\:\:\:\:l_{0} = 2.8A$

• Tensile Stress,

$\:\:\:\:\:\:\:\:\sigma = {10}^{9} N{m}^{-2}$

☆ᴛᴏ ғɪɴᴅ☆

• Increase in intermolecular space.

☆ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ☆

By Hooke's Law, stress and strain within propotional limit are related as

$\to \sigma = Y\epsilon$

$\to \epsilon = \frac{\sigma}{Y}$

$\to \frac{Δl}{l_{0}} = \frac{{10}^{9}}{2\times {10}^{11}}$

$\to \frac{Δl}{2.8\times {10}^{-10}} = \frac{{10}^{9}}{2\times {10}^{11}}$

$\to Δl = \frac{{10}^{9}\times 2.8\times {10}^{-10}}{2\times {10}^{11}}$

$\to Δl = 1.4\times {10}^{-12}\:m$

$\to Δl = 0.014A$

Thus, interatomic space increase by $0.014A$