Correct Answer - C
`(1)/(2)I_(1)omega_(1)^(2)=(1)/(2)I_(2)omega_(2)^(2) rArr (omega_(1)^(2))/(omega_(2)^(2))=(I_(2))/(I_(1))`
`rArr (omega_(1))/(omega_(2)) = sqrt((I_(2))/(I_(1)))=sqrt((9)/(1))= 3:1`
`therefore (L_(1))/(L_(2))=(omega_(2))/(omega_(1))=(1)/(3) rArr L_(1) : L_(2)`
`= 1:3`.