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Two strings A and 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 v_A and on string B with speed v_B . The ratio $\frac {v_{A}} {v_{B}}$ is 

Answer:

Explanation:

 According to question, the velocity of wave travelling on string is given by

$v=n\lambda=\frac{\lambda}{2L}\sqrt{\frac{T}{\mu}}$                    $\left(\because n=\frac{1}{2L}\sqrt{\frac{T}{\mu}}\right)$

 $\therefore$  $v=\sqrt {\frac{T}{m/l}}\Rightarrow v=\sqrt{\frac{Tl}{m}}$

 and Young's modulus , $Y= \frac{T\times l}{A \triangle l}$

$\therefore$   $T \times l$=$Y A \triangle L$

$\therefore$      $V\propto \sqrt{A}$       (A= Area)

 For string A, radius is 2r and for string B radius  is r

 So,     $\frac{V_{A}}{V_{B}}=\sqrt{\frac{4 r^{2}}{r^{2}}}=\sqrt{4}=2$

 [ $\because$    Y for both the strings is same]