JEE Physics 2023 Question with Solution

Given: The wire has lengths $l_1$ and $l_2$ under tensions $100 \, \text{N}$ and $120 \, \text{N}$ respectively, and $l_1 = 11 \, l_0$ where $l_0$ is the natural length.

Find: The value of $x$ if the natural length is $\frac{1}{x} \, l_1$.

Using Hooke's law, the extension is proportional to the applied force:

$$F = k\Delta x$$

For tension $100 \, \text{N}$,

$$100 = k(l_1 - l_0)$$

Given $l_1 = 11l_0$,

$$100 = k(11l_0 - l_0) = k(10l_0)$$

So,

$$k = \frac{10}{l_0}$$

Now the natural length is written as $\frac{1}{x} \, l_1$. Since $l_1 = 11l_0$,

$$l_0 = \frac{l_1}{11} = \frac{1}{x} \, l_1$$

Hence,

$$x = 11$$

the solution concludes with "Therefore, $x = 2$", but that conclusion does not follow from the shown working. The defensible result from the given question statement is $x = 11$, and none of the listed options matches this value.

From the second tension, the extracted solution writes:

$$120 = k(l_2 - l_0)$$

Using $k = \frac{10}{l_0}$,

$$120 = \frac{10}{l_0}(l_2 - l_0)$$

so

$$l_2 - l_0 = 12l_0$$

and therefore

$$l_2 = 13l_0$$

This only gives the relation between $l_2$ and $l_0$. It does not determine $x$ from $l_0 = \frac{1}{x}l_1$ because that relation is already fixed by $l_1 = 11l_0$.

Thus,

$$l_0 = \frac{l_1}{11}$$

which means the required form is $\frac{1}{x}l_1$ with $x = 11$.

Therefore, the extracted option list is inconsistent with the algebra shown, and no option exactly matches the correct value.