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(D) The formula for the extension in a wire is given by $l = \frac{FL}{AY}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,a

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One-fourth for wire $A$ as for wire $B$

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(D) The formula for the extension in a wire is given by $l = \frac{FL}{AY}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since the wires are made of the same material,$Y$ is the same. Given that $F$ and $L$ are also identical,we have $l \propto \frac{1}{A}$.Since $A = \pi r^2$,where $r$ is the radius,we get $l \propto \frac{1}{r^2}$.Given that the diameter of wire $A$ is twice that of wire $B$,the radius $r_A = 2r_B$.Therefore,$\frac{l_A}{l_B} = \left( \frac{r_B}{r_A} \right)^2 = \left( \frac{r_B}{2r_B} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.This implies $l_A = \frac{1}{4} l_B$,which means the increase in length for wire $A$ is one-fourth that of wire $B$.

Two wires of equal lengths are made of the same material. Wire $A$ has a diameter that is twice as that of wire $B$. If identical weights are suspended from the ends of these wires,the increase in length is

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