A uniform metallic wire is elongated by 0.04, | vedclass

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(A) The formula for Young's modulus $Y$ is given by $Y = \frac{F \cdot \ell}{A \cdot \Delta \ell}$,where $A$ is the cross-sectional area.Rearranging f

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$2$

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(A) The formula for Young's modulus $Y$ is given by $Y = \frac{F \cdot \ell}{A \cdot \Delta \ell}$,where $A$ is the cross-sectional area.Rearranging for elongation $\Delta \ell$,we get $\Delta \ell = \frac{F \cdot \ell}{Y \cdot A}$.Since $A = \pi r^2$ (where $r$ is the radius,which is half the diameter $D$),we have $\Delta \ell = \frac{F \cdot \ell}{Y \cdot \pi r^2} = \frac{4 F \cdot \ell}{Y \cdot \pi D^2}$.From this,we see that $\Delta \ell \propto \frac{\ell}{D^2}$.Let the initial length be $\ell_1$ and initial diameter be $D_1$. Let the final length be $\ell_2 = 2\ell_1$ and final diameter be $D_2 = 2D_1$.Then,$\frac{\Delta \ell_2}{\Delta \ell_1} = \left( \frac{\ell_2}{\ell_1} \right) \left( \frac{D_1}{D_2} \right)^2$.Substituting the values: $\frac{\Delta \ell_2}{\Delta \ell_1} = (2) \left( \frac{1}{2} \right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2}$.Given $\Delta \ell_1 = 0.04\, m$,we have $\Delta \ell_2 = \frac{0.04}{2} = 0.02\, m$.Converting to centimeters,$\Delta \ell_2 = 2\, cm$.

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

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