Two wires A and B are of same materials. Thei | vedclass

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Question

(d) $F = Y 	imes A 	imes \frac{l}{L}$$&rArr;$ $F \propto \frac{{{r^2}}}{L}$      $(Y$ and $l$ are constant$)$

$\frac{{{F_A}}}{{{F_

Vedclass · Accepted Answer

$8:1$

Vedclass · Answer

(d) $F = Y 	imes A 	imes \frac{l}{L}$$&rArr;$ $F \propto \frac{{{r^2}}}{L}$      $(Y$ and $l$ are constant$)$

$\frac{{{F_A}}}{{{F_B}}} = {\left( {\frac{{{r_A}}}{{{r_B}}}} ight)^2} 	imes \left( {\frac{{{L_B}}}{{{L_A}}}} ight) = {\left( {\frac{2}{1}} ight)^2} 	imes \left( {\frac{2}{1}} ight) = \frac{8}{1}$

Two wires $A$ and $B$ are of same materials. Their lengths are in the ratio $1 : 2$ and diameters are in the ratio $2 : 1$ when stretched by force ${F_A}$ and ${F_B}$ respectively they get equal increase in their lengths. Then the ratio ${F_A}/{F_B}$ should be

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