"The length of a wire becomes $l_{1}$ and $l_{2}$ when $100 \mathrm{~N}$ and $120 \mathrm{~N}$ tensions are applied respectively. If $10 ~l_{2}=11~ l_{1}$, the natural length of wire will be $\frac{1}{x} ~l_{1}$. Here the value of $x$ is _____________."

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"

Given:

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

"

The length of a wire becomes $l_{1}$ and $l_{2}$ when $100 \mathrm{~N}$ and $120 \mathrm{~N}$ tensions are applied respectively. If $10 ~l_{2}=11~ l_{1}$, the natural length of wire will be $\frac{1}{x} ~l_{1}$. Here the value of $x$ is _____________.

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The length of a wire becomes $l_{1}$ and $l_{2}$ when $100 \mathrm{~N}$ and $120 \mathrm{~N}$ tensions are applied respectively. If $10 ~l_{2}=11~ l_{1}$, the natural length of wire will be $\frac{1}{x} ~l_{1}$. Here the value of $x$ is _____________.

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