The ratio of radii of two wires is $1: 2$ and the density of their materials are in the ratio $1: 4$. If…

The ratio of radii of two wires is $1: 2$ and the density of their materials are in the ratio $1: 4$. If same tension is applied to both the wires then the ratio of the speed of transverse waves produced in them is
  1. $1: 16$
  2. $16: 1$
  3. $1: 4$
  4. $4: 1$

Solution

$\begin{aligned} & \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{1}{2} ; \frac{\rho_1}{\rho_2}=\frac{1}{4} \\ \mathrm{v}= & \sqrt{\frac{\mathrm{T}}{\mu}}\left(\text { where } \mu=\frac{\text { mass }}{\text { Length }}=\frac{\rho \mathrm{V}}{\mathrm{L}}=\rho \mathrm{A}=\rho \pi \mathrm{R}^2\right) \\ \mathrm{v} & =\sqrt{\frac{\mathrm{T}}{\rho \pi \mathrm{R}^2}} \\ & \frac{\mathrm{v}_1}{\mathrm{v}_2}=\sqrt{\frac{\mathrm{T}_2}{\rho_1 \pi \mathrm{R}_1^2}} \times \sqrt{\frac{\rho_2 \pi \mathrm{R}_2^2}{\mathrm{~T}}} \\ = & \sqrt{\frac{\rho_2}{\rho_1} \times \frac{\mathrm{R}_2^2}{\mathrm{R}_1^2}}=\sqrt{4 \times(2)^2}=\frac{4}{1}\end{aligned}$

Asked in: MHT CET Full Test 13

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