If the interatomic spacing in a steel wire is $3\,{A^{\rm{o}}}\,\,\,{\rm{ and}}\,\,\,{\rm{ }}{y_{{\rm{steel}}}} = 20 \times {10^{10}}N/{m^2}$ then force constant is

$6 \times {10^{ - 9}}\frac{N}{{{A^{\rm{o}}}}}$

$6 \times {10^{ - 2}}\frac{N}{{{A^{\rm{o}}}}}$

$4 \times {10^{ - 5}}\frac{N}{{{A^{\rm{o}}}}}$

$6 \times {10^{ - 5}}\frac{N}{{{A^{\rm{o}}}}}$

JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

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Application +4 | -1

If the interatomic spacing in a steel wire is

3\,{A^\mathrm{o}}\,\,\,\mathrm{ and}\,\,\,\mathrm{ }{y_{\mathrm{steel}}} = 20 \times {10^{10}}N/{m^2}

then force constant is

View answer and explanation

Correct option: B

6 \times {10^{ - 9}}\frac{N}{{{A^\mathrm{o}}}}

Option B (

6 \times {10^{ - 9}}\frac{N}{{{A^\mathrm{o}}}}}

) is correct because it correctly calculates the force constant using the Young's modulus and interatomic spacing.

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l

L

l

\mathrm{A}

1 \mathrm{~cm}

Y=\frac{F l}{A \Delta l},