Correct Answer - A
Fundamental frequency of the first wire is given as,
`f = 1/(2L_(1)) sqrt(T/m)= (1)/(2L_(1)) sqrt((T)/(pir_(1)^(2)rho)) = 1/(2L_(1)r_(1)) sqrt((T)/(pirho))`
The first overtone,
`f_(1) = 2f = (2)/(2L_(1)r_(1)) sqrt((T)/(pirho)) = 1/(L_(1)r_(1)) sqrt((T)/(pirho)) "..."(i)`
The second overtone of the second wire is given as
`f_(2) = 3/(2L_(2)r_(2)) sqrt((T)/(pirho))`
Given , `f_(1) = f_(2)`
`rArr 1/(L_(1)r_(1)) sqrt((T)/(pirho)) = 3/(2L_(2)r_(2)) sqrt((T)/(pirho))`
`:. 3L_(1)r_(1) = 2L_(2)r_(2)`
`rArr (L_(1))/(L_(2)) = 2/(3).(r_(2))/(r_(1)) = 2/(3).(r_(2))/(2r_(2))` [given, `r_(1) = 2r_(2)`]
`= 1/3`