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(D) Given: $T_A = T_B = T$,$r_A = 2r_B$.The frequency of the $n$-th overtone for a string is given by $f_n = (n+1) \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,

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$3: 1$

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(D) Given: $T_A = T_B = T$,$r_A = 2r_B$.The frequency of the $n$-th overtone for a string is given by $f_n = (n+1) \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 ho$.First overtone of $A$ $(n=1)$: $f_{A,1} = 2 	imes \frac{1}{2L_A} \sqrt{\frac{T}{\pi r_A^2 ho}} = \frac{1}{L_A r_A} \sqrt{\frac{T}{\pi ho}}$.Second overtone of $B$ $(n=2)$: $f_{B,2} = 3 	imes \frac{1}{2L_B} \sqrt{\frac{T}{\pi r_B^2 ho}} = \frac{3}{2L_B r_B} \sqrt{\frac{T}{\pi ho}}$.Equating the two frequencies: $\frac{1}{L_A r_A} = \frac{3}{2L_B r_B}$.Substituting $r_A = 2r_B$: $\frac{1}{L_A (2r_B)} = \frac{3}{2L_B r_B}$.Simplifying: $\frac{1}{2L_A} = \frac{3}{2L_B} \implies \frac{L_B}{L_A} = \frac{3}{1}$.

Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$,the ratio of the length of string $B$ to that of $A$ is

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