A force is applied to a steel wire 'A',rigidl | vedclass

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(D) The Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta \ell / \ell}$,where $F$ is the force,$A$ is the cross-sectional area,$\ell$ is the orig

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$6.9$

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(D) The Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta \ell / \ell}$,where $F$ is the force,$A$ is the cross-sectional area,$\ell$ is the original length,and $\Delta \ell$ is the elongation.Since the material is the same (steel),$Y$ is constant. Rearranging the formula,we get $\Delta \ell = \frac{F \ell}{Y A}$.For wire '$A$': $\Delta \ell_A = \frac{F \ell_A}{Y A_A} = 0.2\,mm$.For wire '$B$': $\ell_B = 2 \ell_A$ and $d_B = 2.4 d_A$. Since $A = \pi (d/2)^2$,the area $A_B = (2.4)^2 A_A = 5.76 A_A$.Now,$\Delta \ell_B = \frac{F \ell_B}{Y A_B} = \frac{F (2 \ell_A)}{Y (5.76 A_A)} = \frac{2}{5.76} 	imes \left( \frac{F \ell_A}{Y A_A} ight)$.Substituting $\Delta \ell_A = 0.2\,mm$: $\Delta \ell_B = \frac{2}{5.76} 	imes 0.2 = \frac{0.4}{5.76} \approx 0.06944\,mm$.Converting to the required format: $0.06944\,mm = 6.944 	imes 10^{-2}\,mm$. Rounding to the nearest option,we get $6.9 	imes 10^{-2}\,mm$.

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