QJEEM23Ph9a8e: JEE Main 2023 Physics - Class 11 JEEM23Ph9a8e Solution: The length of a wire becomes and when and tensions are applied…

Given:

1. When tension $T_1 = 100 \mathrm{~N}$ , extension $= l_1 - l_0$ .
2. When tension $T_2 = 120 \mathrm{~N}$ , extension $= l_2 - l_0$ .

Now, let's write the equations using Hooke's Law:

$100 = k(l_1 - l_0)$

$120 = k(l_2 - l_0)$

Divide the first equation by the second equation:

$\frac{5}{6} = \frac{l_1 - l_0}{l_2 - l_0}$

Given the relationship between $l_1$ and $l_2$ :

$10l_2 = 11l_1$

Now, let's solve for $l_0$ :

$5l_2 - 5l_0 = 6l_1 - 6l_0$

$l_0 = 6l_1 - 5l_2$

Substitute the relationship between $l_1$ and $l_2$ :

$l_0 = 6l_1 - 5\left(\frac{11l_1}{10}\right)$

$l_0 = 6l_1 - \frac{11l_1}{2}$

$l_0 = \frac{l_1}{2}$

Therefore, the natural length of the wire is $\frac{1}{x}l_1 = \frac{2}{1}l_1 = 2l_1$ .

The value of $x$ is $2$ .