Both the strings, shown in figure, are made of same material and have same cross-section. The pulleys are light. The wave speed of a transverse wave in the string AB is $v_{1}$ and in CD it is $v_{2}$ the $v_{1} / v_{2}$ is

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\sqrt{2}

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\frac{1}{\sqrt{2}}

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Solution

The correct option is D $\frac{1}{\sqrt{2}}$
$\begin{matrix} T_{A B} = T T_{C D} = 2 T w h e r e, T_{A B} i s t h e t e n s i o n i n t h e s t r i n g A B T_{C D} i s t h e t e n s i o n i n t h e s t r i n g C D T h e a c c e l e r a t i o n b e t w e e e n t e n s i o n a n d t h e w a v e s p e e d i s g i v e n b y v = \sqrt{\frac{T}{μ}} v α \sqrt{T} w h e r e, v i s t h e w a v e s p e e d o f t h e t r e n s v e r s e w a v e μ i s t h e m a s s p e r u n i t l e n g t h o f t h e s t r i n g \frac{v_{1}}{v_{2}} = \sqrt{\frac{T}{2 T}} = \frac{1}{\sqrt{2}} \end{matrix}$
Hence,
option $(D)$ is correct answer.

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