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(D) The formula for Young's modulus $Y$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the length,$A$ is the cross

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$\frac{1}{16} \ mm$

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(D) The formula for Young's modulus $Y$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,and $\Delta L$ is the elongation.Since $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we can write $\Delta L = \frac{F \cdot L}{Y \cdot A} = \frac{4 \cdot F \cdot L}{Y \cdot \pi d^2}$.Given that $F$,$L$,and $Y$ are constant for both wires,we have $\Delta L \propto \frac{1}{d^2}$.Let $\Delta L_1 = 1 \ mm$ and $d_1 = d$. For the second wire,$d_2 = 4d$.Therefore,$\frac{\Delta L_2}{\Delta L_1} = \left( \frac{d_1}{d_2} \right)^2 = \left( \frac{d}{4d} \right)^2 = \left( \frac{1}{4} \right)^2 = \frac{1}{16}$.Thus,$\Delta L_2 = \frac{1}{16} \cdot \Delta L_1 = \frac{1}{16} \cdot 1 \ mm = \frac{1}{16} \ mm$.

$A$ wire is stretched $1 \ mm$ by a force $F$. If a second wire of same material,same length and $4$ times the diameter of the first wire is stretched by the same force $F$,then the elongation of the second wire is

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