Physics Q34 – JEE-MAIN 2025

Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section R is

v_{1}

, and that in the other string having radius of cross section R/2 is

v_{2}

. Then

\frac{v_{2}}{v_{1}} =

Clue 1 of 3: Wave Velocity on a String

Recall the formula for the velocity of a transverse wave on a stretched string.

🥷

Ninja Strategy•Proportionality Analysis

Recognize that velocity is inversely proportional to the square root of the linear mass density, and linear mass density is proportional to the square of the radius. This implies velocity is inversely proportional to the radius ( $v \propto 1 / r$ ). Therefore, $v_{2} / v_{1} = r_{1} / r_{2} = R / (R / 2) = 2$ .

🧠 Full Solution Path

Step 1: Identify the formula for wave velocity and linear mass density✦ Active

The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{μ}}$ , where $T$ is the tension and $μ$ is the linear mass density. The linear mass density $μ$ for a string of circular cross-section is $μ = ρ A = ρ (π r^{2})$ , where $ρ$ is the material density and $r$ is the radius.

Step 2: Express velocity in terms of given parameters○ Expand

Substituting $μ$ into the velocity formula, we get $v = \sqrt{\frac{T}{ρ π r^{2}}}$ . Since $T$ and $ρ$ are constant for both strings, $v \propto \frac{1}{r}$ .

v_{1} = \sqrt{\frac{T}{ρ π R^{2}}}

v_{2} = \sqrt{\frac{T}{ρ π (R / 2)^{2}}} = \sqrt{\frac{T}{ρ π (R^{2} / 4)}} = \sqrt{\frac{4 T}{ρ π R^{2}}}

Step 3: Calculate the ratio

\frac{v_{2}}{v_{1}}

○ Expand

\frac{v_{2}}{v_{1}} = \frac{\sqrt{\frac{4 T}{ρ π R^{2}}}}{\sqrt{\frac{T}{ρ π R^{2}}}} = \sqrt{\frac{4 T / (ρ π R^{2})}{T / (ρ π R^{2})}} = \sqrt{4} = 2

💡 Teacher's Secret Hint

Notice that the ratio simplifies directly due to common terms under the square root.

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Physics Question 34 – JEE-MAIN 2025

Clue 1 of 3: Wave Velocity on a String

🧠 Full Solution Path