A uniform metallic wire is elongated by 0.04, | vedclass

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Question

$F = Y.A \frac{\Delta \ell}{\ell}$

$\Delta \ell=\frac{ F }{ Y.A } \cdot \ell$

$\Delta \ell=\frac{ F \ell}{ Y \cdot \pi r ^{2}}$

$\Delta \ell

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$F = Y.A \frac{\Delta \ell}{\ell}$

$\Delta \ell=\frac{ F }{ Y.A } \cdot \ell$

$\Delta \ell=\frac{ F \ell}{ Y \cdot \pi r ^{2}}$

$\Delta \ell \propto \frac{\ell}{ r ^{2}}$

$\frac{\Delta \ell_{2}}{\Delta \ell_{1}}=\left(\frac{\ell_{2}}{\ell_{1}}\right)\left(\frac{ r _{1}}{ r _{2}}\right)^{2}$

$=(2)(\frac{1}{2})^2$

$\frac{\Delta \ell_{2}}{\Delta \ell_{1}}=\frac{1}{2}$

$\Delta \ell_{2}=\frac{\Delta \ell_{1}}{2}$

$=\frac{0.04}{2}$

$=0.02\, m$

$\Delta \ell_{2}=2\, cm$

A uniform metallic wire is elongated by $0.04\, m$ when subjected to a linear force $F$. The elongation, if its length and diameter is doubled and subjected to the same force will be ..... $cm .$

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