Correct Answer - B
In first case, `I_(x) = 40` cm, `I_(R) = 60` cm
`x/R = (I_(x))/(I_(R)) = 40/60 = 2/3 "…"(i)`
In second case,
`:. (x+30)/(R ) = 60/40 = 3/2`
`rArr R = (2(x + 30))/(3) "…"(iii)`
From Eqs. (i) and (ii), we get
`(mu)/(2((x+30)/(3))) = 2/3`
`rArr (3x)/(2(x+30))= 2/3`
`9x = 4 x + 120`
`rArr 5x = 120 rArr x = 24 Omega`