Two strings with circular cross section and made of same material, are stretched to have same amount of…

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 $\mathrm{v}_1$, and that in the other string having radius of cross section $R / 2$ is $v_2$. Then $\frac{v_2}{v_1}=$
  1. $\sqrt{2}$
  2. 2
  3. 8
  4. 4

Solution

$\begin{aligned} & v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{T}{f \pi R^2}} \\ & \frac{v_2}{v_1}=\frac{R_1}{R_2}=2\end{aligned}$

Asked in: JEE Main 2025 (08 Apr Shift 2)

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