TG EAMCET 2 May 2025 Shift 2 Paper

First setup: Let the resistance in the left gap be $R^{\prime}$, and in the right gap there are two resistors with resistance $R$, connected in series. So, the right gap has a total resistance of $R+R=2R$.The balancing point is at 50 cm . In a metre bridge, the ratio of resistances equals the ratio of the lengths on both sides: $\frac{R^{\prime}}{2R}=\frac{50}{50}$ So, $\frac{R^{\prime}}{2R}=1$ which means $R^{\prime}=2R$Second setup:Now, one resistor from the right gap is taken out and connected in parallel with the resistor in the left gap.Now, the left gap contains $R^{\prime}$ and $R$ in parallel. The resistance of two resistors $A$ and $B$ in parallel is: $R^{\prime\prime}=\frac{A \times B}{A+B}$ The parallel resistance is: $R^{\prime\prime}=\frac{R^{\prime} \times R}{R^{\prime}+R}$From earlier, we know $R^{\prime}=2R$. Substitute this value: $R^{\prime\prime}=\frac{2R \times R}{2R+R}=\frac{2R^{2}}{3R}=\frac{2R}{3}$Now, the right gap just has $R$.Let the new balancing point be $l$. The bridge principle gives us: $\frac{R^{\prime\prime}}{R}=\frac{l}{100-l}$ Substitute $R^{\prime\prime}=\frac{2R}{3}$ :$\frac{\frac{2R}{3}}{R}=\frac{l}{100-l}\;\frac{2}{3}=\frac{l}{100-l}$Solve for $l: 3l=2(100-l) \quad 3l=200-2l \quad 3l+2l=200 \quad 5l=200 \quad l=40\,\text{cm}$