Q. If the interatomic spacing in a steel wire is $3.0\mathring{\mathrm{A}}$ and $Y_{steel} = 20 \times 10^{10}N/m^{2}$ then force constant is
A) $6 \times 10^{- 2}N/\mathring{\mathrm{A}}$
B) $6 \times 10^{- 9}N/\mathring{\mathrm{A}}$
C) $4 \times 10^{- 5}N/\mathring{\mathrm{A}}$
D) $6 \times 10^{- 5}N/\mathring{\mathrm{A}}$
Explanation
(b) $K = Yr_{0} = 20 \times 10^{10} \times 3 \times 10^{- 10} = 60\ N/m$
$= 6 \times 10^{- 9}N/\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.