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Transverse waves of the same frequency are generated in two steel wires $A$ and $B$. The diameter of $A$ is twice that of $B$ and the tension in $A$ is half that in $B$. The ratio of the velocities of the waves in $A$ and $B$ is
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$1: 2 \sqrt{2}$
$1: 2 \sqrt{2}$
The velocity of transverse waves is given by $v=\sqrt{\frac{T}{m}}$ where, $T=$ Tension and $m=$ mass per unit length of the wire. If $r$ is the radius of the wire and $\rho$ its density then,
$m=\pi r^2 \rho$
$\therefore \quad v=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{\pi r^2 \rho}}$
$\therefore \quad v_A=\frac{\sqrt{T_A}}{r_A \sqrt{\pi \rho}}$
and $\quad v_B=\frac{\sqrt{T_B}}{r_B \sqrt{\pi \rho}}$
Now, $\quad \frac{v_A}{v_B}=\sqrt{\frac{T_A}{T_B}} \cdot \frac{r_B}{r_A}$
$\because \quad r_A=2 r_B$ and $T_A=\frac{1}{2} T_B$
Hence, $\quad \frac{v_A}{v_B}=\frac{1}{2 \sqrt{2}}$
$m=\pi r^2 \rho$
$\therefore \quad v=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{\pi r^2 \rho}}$
$\therefore \quad v_A=\frac{\sqrt{T_A}}{r_A \sqrt{\pi \rho}}$
and $\quad v_B=\frac{\sqrt{T_B}}{r_B \sqrt{\pi \rho}}$
Now, $\quad \frac{v_A}{v_B}=\sqrt{\frac{T_A}{T_B}} \cdot \frac{r_B}{r_A}$
$\because \quad r_A=2 r_B$ and $T_A=\frac{1}{2} T_B$
Hence, $\quad \frac{v_A}{v_B}=\frac{1}{2 \sqrt{2}}$
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