Two resistances are connected in the two gaps of a meter bridge. The balance point is `20 cm` from the zero end. When a resistance `15 Omega` is connected in series with the smaller of two resistance, the null point+ shifts to `40 cm`. The smaller of the two resistance has the value.

Options

A. 12
B. 24
C. 36
D. 48

Answer

B

Explanation

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`