Question

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

Answer 1

Correct Answer - B
`l_(x)=40cm, I_(R)=60cm`
`x/R=(l_(x))/(l_(R))=(40)/(60)=2/3 .....(i)`
`therefore (x+30)/(R)=(60)/(40)=3/2`
`therefore (x+30)/(R)=3/2`
`therefore 2(X+30)rArr 3R`
`R=(2(x+30))/(3).........(ii)`
From (i) and (iii)
`(X)/(2((x+30))/(3)))=2/3`
`(3x)/(2(x+30))=2/3`
`9x=4x+120`
`5x=120 rArr x=24Omega`