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$6 \times {10^{ - 5}}N/{\mathring A}$
Similar Questions
A $5\, m$ long aluminium wire ($Y = 7 \times {10^{10}}N/{m^2})$ of diameter $3\, mm$ supports a $40\, kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times {10^{10}}N/{m^2})$ of the same length under the same weight, the diameter should now be, in $mm.$
A $5\, m$ long aluminium wire ($Y = 7 \times {10^{10}}N/{m^2})$ of diameter $3\, mm$ supports a $40\, kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times {10^{10}}N/{m^2})$ of the same length under the same weight, the diameter should now be, in $mm.$
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A bar is subjected to axial forces as shown. If $E$ is the modulus of elasticity of the bar and $A$ is its crosssection area. Its elongation will be
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The maximum elongation of a steel wire of $1 \mathrm{~m}$ length if the elastic limit of steel and its Young's modulus, respectively, are $8 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}$ and $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$, is:
The maximum elongation of a steel wire of $1 \mathrm{~m}$ length if the elastic limit of steel and its Young's modulus, respectively, are $8 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}$ and $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$, is:
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