The total length of potentiometer wire AB is 50 text cm$50 \text{ cm}$ in the arrangement as shown in figure. If P is the point where the galvanometer shows zero reading then the length AP is ________ cm.

### Related Formula For a balanced Wheatstone bridge configuration: fracR_1R_2 = fracl_1l_2$$\frac{R_1}{R_2} = \frac{l_1}{l_2}$$ ### Core Logic When the galvanometer shows zero reading, it implies that the bridge is balanced. The resistance of the wire segments is proportional to their lengths. Let the length of segment AP be l_AP$l_{AP}$ and PB be l_PB$l_{PB}$. frac6\,OmegaR_AP = frac4\,OmegaR_PB$$\frac{6\,\Omega}{R_{AP}} = \frac{4\,\Omega}{R_{PB}}$$ Since R propto l$R \propto l$: fracR_APR_PB = fracl_APl_PB = frac64 = frac32$$\frac{R_{AP}}{R_{PB}} = \frac{l_{AP}}{l_{PB}} = \frac{6}{4} = \frac{3}{2}$$ ### Step 1: Applying Total Length Constraint We are given the total length: l_AP + l_PB = 50 text cm$$l_{AP} + l_{PB} = 50 \text{ cm}$$ Using the ratio l_AP = frac32 l_PB$l_{AP} = \frac{3}{2} l_{PB}$: frac32l_PB + l_PB = 50$$\frac{3}{2}l_{PB} + l_{PB} = 50$$ frac52l_PB = 50 implies l_PB = 20 text cm$$\frac{5}{2}l_{PB} = 50 \implies l_{PB} = 20 \text{ cm}$$ ### Step 2: Final Conclusion Solving for l_AP$l_{AP}$: l_AP = 50 - 20 = 30 text cm$$l_{AP} = 50 - 20 = 30 \text{ cm}$$ Alternatively, l_AP = frac33+2 times 50 = frac35 times 50 = 30 text cm$l_{AP} = \frac{3}{3+2} \times 50 = \frac{3}{5} \times 50 = 30 \text{ cm}$. ### Pattern Recognition Null deflection over a uniform wire directly translates resistance ratios into length ratios. Standard Wheatstone bridge application wrapped in potentiometer geometry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity