A potentiometer $PQ$ is set up to compare two resistances as shown in the figure. The ameter $A$ in the circuit reads $1.0\, A$ when two way key $K_3$ is open. The balance point is at a length $l_1\, cm$ from $P$ when two way key $K_3$ is plugged in between $2$ and $1$ , while the balance point is at a length $l_2\, cm$ from $P$ when key $K_3$ is plugged in between $3$ and $1$ . The ratio of two resistances $\frac{{{R_1}}}{{{R_2}}}$ is found to be

Q: A potentiometer $PQ$ is set up to compare two resistances as shown in the figure. The ameter $A$ in the circuit reads $1.0\, A$ when two way key $K_3$ is open. The balance point is at a length $l_1\, cm$ from $P$ when two way key $K_3$ is plugged in between $2$ and $1$ , while the balance point is at a length $l_2\, cm$ from $P$ when key $K_3$ is plugged in between $3$ and $1$ . The ratio of two resistances $\frac{{{R_1}}}{{{R_2}}}$ is found to be

d When key is at point $( 1)$ $\mathrm{V}_{1}=\mathrm{i} \mathrm{R}_{1}=\mathrm{x} l_{1}$ When key is at $( 3)$ ${{\rm{v}}_2} = {\rm{i}}\left( {{{\rm{R}}_1} + {{\rm{R}}_2}} \right) = {\rm{x}}{l_2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}-l_{1}}$

Question

A$\frac{{{l_1}}}{{{l_1} + {l_2}}}$
B$\frac{{{l_2}}}{{{l_2} - {l_1}}}$
C$\frac{{{l_1}}}{{{l_1} - {l_2}}}$
D$\frac{{{l_1}}}{{{l_2} - {l_1}}}$

JEE MAIN 2017, Medium

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