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(B) The velocity of a transverse wave in a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass

Vedclass · Accepted Answer

$1 : 2\sqrt{2}$

Vedclass · Answer

(B) The velocity of a transverse wave in a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Since $\mu = \pi r^2 ho$ (where $r$ is the radius and $ho$ is the density of the material),the velocity is $v = \sqrt{\frac{T}{\pi r^2 ho}}$.Since both wires are made of steel,$ho$ is constant. Thus,$v \propto \frac{\sqrt{T}}{r}$.Given: $d_A = 2d_B \implies r_A = 2r_B$ and $T_A = \frac{1}{2}T_B$.The ratio of velocities is $\frac{v_A}{v_B} = \sqrt{\frac{T_A}{T_B}} 	imes \frac{r_B}{r_A}$.Substituting the values: $\frac{v_A}{v_B} = \sqrt{\frac{1}{2}} 	imes \frac{1}{2} = \frac{1}{\sqrt{2}} 	imes \frac{1}{2} = \frac{1}{2\sqrt{2}}$.Therefore,the ratio is $1 : 2\sqrt{2}$.

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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