JEE Physics 2025 Question with Solution

Given: Two strings are made of the same material and are stretched with the same tension. The first string has radius $R$ and wave speed $v_1$. The second string has radius $R/2$ and wave speed $v_2$.

Find: The value of $\frac{v_2}{v_1}$.

For a transverse wave on a stretched string,

$$v = \sqrt{\frac{T}{\mu}}$$

where $T$ is the tension and $\mu$ is the linear mass density.

Since both strings have the same tension and are made of the same material, the velocity depends on the linear mass density.

For a string of circular cross section,

$$A = \pi R^2$$

and

$$\mu = \rho A = \rho \pi R^2$$

So, for the first string,

$$\mu_1 = \rho \pi R^2$$

For the second string of radius $R/2$,

$$\mu_2 = \rho \pi \left(\frac{R}{2}\right)^2 = \rho \pi \frac{R^2}{4}$$

Now,

$$\frac{v_2}{v_1} = \sqrt{\frac{\mu_1}{\mu_2}}$$

Substituting the values,

$$\frac{v_2}{v_1} = \sqrt{\frac{\rho \pi R^2}{\rho \pi R^2/4}} = \sqrt{4} = 2$$

Therefore, $\frac{v_2}{v_1} = 2$ and the correct option is D.

Given: Wave speed in a string is determined by tension and linear mass density.

Find: The ratio $\frac{v_2}{v_1}$ when the radii are $R$ and $R/2$.

Use

$$v = \sqrt{\frac{T}{\mu}}$$

Since $T$ is the same for both strings,

$$v \propto \frac{1}{\sqrt{\mu}}$$

Also, for the same material,

$$\mu \propto A$$

and for a circular cross section,

$$A \propto r^2$$

Therefore,

$$\mu_1 : \mu_2 = R^2 : \left(\frac{R}{2}\right)^2 = R^2 : \frac{R^2}{4} = 4 : 1$$

Hence,

$$\frac{v_2}{v_1} = \sqrt{\frac{\mu_1}{\mu_2}} = \sqrt{4} = 2$$

Therefore, the correct option is D.

The solution also states Option: 2, which conflicts with the listed options. Since the worked value is clearly $2$ and option D equals $2$, the correct mapped answer is D.