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Home » Two strings and of same material are stretched by same tension. The radius of the string A is double the radius of string B. Transverse wave travels on string A with speed ‘ ‘ and on string with speed ‘ ‘. The ratio isA) B) C) 2D) 4
Two strings \mathrm{A} and \mathrm{B} of same material are stretched by same tension. The radius of the string A is double the radius of string B. Transverse wave travels on string A with speed ‘ \mathrm{V}_{\mathrm{A}} ‘ and on string B with speed ‘ V_{B} ‘. The ratio \frac{V_{A}}{V_{B}} is
A) \frac{1}{4}
B) \frac{1}{2}
C) 2
D) 4
Physics
Two strings \mathrm{A} and \mathrm{B} of same material are stretched by same tension. The radius of the string A is double the radius of string B. Transverse wave travels on string A with speed ‘ \mathrm{V}_{\mathrm{A}} ‘ and on string B with speed ‘ V_{B} ‘. The ratio \frac{V_{A}}{V_{B}} is
A) \frac{1}{4}
B) \frac{1}{2}
C) 2
D) 4

Answer is (C)
The velocity of wave travelling on string is
\begin{array}{l} v=n \lambda=\frac{\lambda}{2 L} \sqrt{\frac{T}{\mu}}=\sqrt{\frac{T}{\mu}} \\ \because \quad n=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \\ \therefore \quad \quad \vee=\sqrt{\frac{T}{m \ell^{2}}} \Rightarrow v=\sqrt{\frac{T \times \ell}{m}} \\ Y=\frac{T \times \ell}{A \Delta L} \\ \therefore \quad T \times \ell=Y A \Delta L \\ \therefore \quad V=\sqrt{\frac{Y \cdot A \cdot \Delta L}{m}} \\ \therefore \quad V \propto \sqrt{A} \quad \text { (A is area) } \end{array}
For string A, radius is 2 \mathrm{r}, and for string B, radius is \mathrm{r}
\therefore \frac{V_{\mathrm{A}}}{\mathrm{A}_{\mathrm{B}}}=\sqrt{\frac{4 r^{2}}{\mathrm{r}^{2}}}=\sqrt{4}=2
\because Y is same for both and in is same for both.

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