Q74 / 75 · JEE Main 2025 (8 Apr Shift 2) · Physics

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 $\frac{v_2}{v_1}=$

  • A
    $\sqrt{2}$
  • Correct answer
    B
    2
  • C
    8
  • D
    4

Solution

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

Hence, the correct answer is option: 'B'.

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