Current Electricity Part 4

Concept:A meter bridge works on the balanced Wheatstone bridge principle.The ratio of resistances equals the ratio of the wire lengths from the null point.Explanation:Case 1: Left gap: $2\,\Omega$, right gap: $3\,\Omega$.Null point at $l$ cm from left, so right length = $(100 - l)$ cm.Balance condition: $\frac{2}{3} = \frac{l}{100 - l}$.Cross multiply: $2(100 - l) = 3l$ → $200 - 2l = 3l$ → $5l = 200$ → $l = 40$ cm.Case 2: Unknown $x\,\Omega$ is connected in parallel with $3\,\Omega$ in the right gap.New right gap resistance: $S' = \frac{3x}{3 + x}$.The null point shifts 10 cm to the right, so new length $l' = 40 + 10 = 50$ cm.Right length = $100 - 50 = 50$ cm.New balance condition: $\frac{2}{S'} = \frac{50}{50} = 1$.So $2 = S'$ → $2 = \frac{3x}{3 + x}$.Multiply: $2(3 + x) = 3x$ → $6 + 2x = 3x$ → $x = 6\,\Omega$.Answer:The value of unknown resistance $x$ is $6\,\Omega$.Hence, the correct answer is 6.