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 ( in `Omega`) has the value.

Options

A. 3
B. 6
C. 9
D. 12

Answer

C

Explanation

(c ) Let `S` be larger and `R` be smaller resistance connected in two gaps of meter bridge.
`:. S = ((100 - 1)/(l)) R = (100 - 20)/(20) R = 4 R`
When `5 Omega` resistance is added to resistance `R` then
`S = ((100 - 40)/(40)) (R + 15) = (6)/(4) (R + 15)`
From equaitons (i) and (ii) `R = 9 Omega`