8.Mechanical Properties of Solids
medium

If the interatomic spacing in a steel wire is $3.0\mathring A$ and ${Y_{steel}}$= $20 \times {10^{10}}N/{m^2}$ then force constant is

A$6 \times {10^{ - 2}}\,N/{\mathring A}$
B$6 \times {10^{ - 9}}N/{\mathring A}$
C$4 \times {10^{ - 5}}\,N/{\mathring A}$
D$6 \times {10^{ - 5}}N/{\mathring A}$

Solution

(b)$K = Y{r_0} = 20 \times {10^{10}} \times 3 \times {10^{ – 10}} = 60\;N/m$
$ = 6 \times {10^{ – 9}}N/{\mathring A}$
Standard 11
Physics

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