When a force F_1 is applied on a metallic wir | vedclass

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(C) According to Hooke's Law, the change in length of a wire is directly proportional to the applied force within the elastic limit.Let $L$ be the ori

Vedclass · Accepted Answer

$\frac{F_1 L_2-F_2 L_1}{F_1-F_2}$

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(C) According to Hooke's Law, the change in length of a wire is directly proportional to the applied force within the elastic limit.Let $L$ be the original length and $K$ be the force constant of the wire.For force $F_1$, the extension is $(L_1 - L)$, so $F_1 = K(L_1 - L)$  --- $(i)$For force $F_2$, the extension is $(L_2 - L)$, so $F_2 = K(L_2 - L)$  --- (ii)Dividing equation $(i)$ by equation (ii):$\frac{F_1}{F_2} = \frac{K(L_1 - L)}{K(L_2 - L)}$$\frac{F_1}{F_2} = \frac{L_1 - L}{L_2 - L}$Cross-multiplying gives:$F_1(L_2 - L) = F_2(L_1 - L)$$F_1 L_2 - F_1 L = F_2 L_1 - F_2 L$Rearranging to solve for $L$:$F_2 L - F_1 L = F_2 L_1 - F_1 L_2$$L(F_2 - F_1) = F_2 L_1 - F_1 L_2$$L = \frac{F_2 L_1 - F_1 L_2}{F_2 - F_1} = \frac{F_1 L_2 - F_2 L_1}{F_1 - F_2}$

When a force $F_1$ is applied on a metallic wire, the length of the wire is $L_1$. If a force $F_2$ is applied on the same wire, the length of the wire is $L_2$. The original length of the wire $L$ is

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