Current Electricity Part 4

Concept:The meter bridge works on the Wheatstone bridge principle.The balancing condition gives the ratio of resistances equal to the ratio of the corresponding lengths of the wire.Explanation:Let $l_1$ be the initial balancing length from end X.Given $R_1 = 2\ \Omega$ and $R_2 = 3\ \Omega$.Using the bridge formula: $\displaystyle \frac{R_1}{R_2} = \frac{l_1}{100 - l_1}$.Substitute: $\displaystyle \frac{2}{3} = \frac{l_1}{100 - l_1}$.Cross-multiply: $2(100 - l_1) = 3l_1$.Simplify: $200 - 2l_1 = 3l_1 \Rightarrow 200 = 5l_1$, so $l_1 = 40\ \text{cm}$.Initial null point is at 40 cm from X.An unknown resistor $X$ is connected in parallel with the $3\ \Omega$ resistor in the right gap.New effective right gap resistance: $R_2' = \frac{3X}{3+X}$.The null point shifts 22.5 cm toward Y, so new length $l_2 = l_1 + 22.5 = 62.5\ \text{cm}$.Apply bridge condition again: $\displaystyle \frac{R_1}{R_2'} = \frac{l_2}{100 - l_2}$.Substitute: $\displaystyle \frac{2}{R_2'} = \frac{62.5}{37.5} = \frac{5}{3}$.Hence $R_2' = 2 \times \frac{3}{5} = \frac{6}{5} = 1.2\ \Omega$.Now $1.2 = \frac{3X}{3+X}$. Multiply: $1.2(3+X) = 3X \Rightarrow 3.6 + 1.2X = 3X$.So $3.6 = 1.8X$, thus $X = \frac{3.6}{1.8} = 2\ \Omega$.Answer:The unknown resistance is $2\ \Omega$, which corresponds to option D.