Physics Elasticity Question 151
Question: If the interatomic spacing in a steel wire is 3.0Å and $ Y _{steel} $ = $ 20\times 10^{10}N/m^{2} $ then force constant is
[RPET 2001]
Options:
A) $ 6\times {{10}^{-2}},N/{\AA} $
B)$ 6\times {{10}^{-9}}N/{\AA} $
C) $ 4\times {{10}^{-5}},N/{\AA} $
D)$ 6\times {{10}^{-5}}N/{\AA} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ K=Yr _{0}=20\times 10^{10}\times 3\times {{10}^{-10}}=60\ N/m $
$ =6\times {{10}^{-9}}N/{\AA} $
BETA
