Two unknown resistances are connected in two gaps of a meter-bridge. The null point is obtained at 40 cm from left end. A 30 Ω resistance is connected in series with the smaller of the two resistances, the null point shifts by 20 cm to the right end. The value of smaller resistance in Ω is

Step-by-step solution

l x = 40   cm ,   l R = 60   cm x R = l x l R = 40 60 = 2 3 ....... (i) ∴           x + 30   R = 60 40 = 3 2 ∴         x + 30 R = 3 2 ∴       2x + 30 = 3 R R =   2 ( x + 30 ) 3 ........ (ii) From (i) and (ii), 3 x 2 ( x + 30 ) = 2 3 9 x = 4 x + 120 5 x = 120     ⇒     x = 24   Ω