The length of metallic wire is $l$. The tension in the wire is $T_1$ for length $l_1$ and tension in the wire is $T_2$ for length $l_2$. Find the original length.
The length of metallic wire is $l$. The tension in the wire is $T_1$ for length $l_1$ and tension in the wire is $T_2$ for length $l_2$. Find the original length.
- [JEE MAIN 2021]
Increase in length under tension $\mathrm{T}_{1}=l_{1}-l$ Increase in length under tension $\mathrm{T}_{2}=l_{2}-l$ $\mathrm{Y}=\frac{\mathrm{T}_{1}}{\mathrm{~A}} \times \frac{l}{l_{1}-l}$ and $\mathrm{Y}=\frac{\mathrm{T}_{2}}{\mathrm{~A}} \times \frac{l}{l_{2}-l}$
Since material of wire is same hence $\mathrm{Y}$ is same.
$\therefore \frac{\mathrm{T}_{1}}{\mathrm{~A}} \times \frac{l}{l_{1}-l}=\frac{\mathrm{T}_{2}}{\mathrm{~A}} \times \frac{l}{l_{2}-l}$
$\therefore \mathrm{T}_{1}\left(l_{2}-l\right)=\mathrm{T}_{2}\left(l_{1}-l\right)$
$\therefore \mathrm{T}_{1} l_{2}-\mathrm{T}_{1} l=\mathrm{T}_{2} l_{1}-\mathrm{T}_{2} l$
$\therefore \mathrm{T}_{1} l_{2}-\mathrm{T}_{2} l_{1}=\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right) l$
$\therefore l=\frac{\mathrm{T}_{1} l_{2}-\mathrm{T}_{2} l_{1}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}$ or $\frac{\mathrm{T}_{2} l_{1}-\mathrm{T}_{l} l_{2}}{\mathrm{~T}_{2}-\mathrm{T}_{1}}$
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- [JEE MAIN 2024]