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(D) The force $F$ required to produce an extension $\Delta l$ in a wire of length $l$,area of cross-section $A$,and Young's modulus $Y$ is given by $F

Vedclass · Accepted Answer

$8: 1$

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(D) The force $F$ required to produce an extension $\Delta l$ in a wire of length $l$,area of cross-section $A$,and Young's modulus $Y$ is given by $F = \frac{Y A \Delta l}{l}$.Since the area $A = \pi \left(\frac{d}{2}ight)^2 = \frac{\pi d^2}{4}$,we can write the force as $F = \frac{Y \pi d^2 \Delta l}{4 l}$.For wires of the same material,$Y$ is constant. Thus,$F \propto \frac{d^2 \Delta l}{l}$.Given $\Delta l_A = \Delta l_B$,the ratio of forces is $\frac{F_A}{F_B} = \frac{d_A^2}{d_B^2} 	imes \frac{l_B}{l_A}$.Given $\frac{l_A}{l_B} = \frac{1}{2} \implies \frac{l_B}{l_A} = 2$ and $\frac{d_A}{d_B} = \frac{2}{1}$.Substituting these values: $\frac{F_A}{F_B} = \left(\frac{2}{1}ight)^2 	imes 2 = 4 	imes 2 = 8$.Therefore,the ratio $F_A : F_B = 8: 1$.

Two wires $A$ and $B$ are of the same material. Their lengths are in the ratio $1: 2$ and diameters are in the ratio $2: 1$. When stretched by forces $F_{A}$ and $F_{B}$ respectively,they get equal increase in their lengths. Then the ratio $F_{A} / F_{B}$ is

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