#### Two potentiometer wires W1 and W2 of equal length $l$ connected to a battery of emf EP and internal resistance ‘r’ through two switches S1 and S2. A battery of emf E is balanced on these potentiometer wires. If potentiometer wire W1 is of resistance 2r and balancing length on W1 is $l/2$ when only switch S1 is closed and S2 is open. On closing S2 and opening S1 the balancing length on W2 is found to be $\left ( \frac{2l}{3} \right )$ , the resistance of the potentiometer W2 will beOption: 1 $r$  Option: 2 $\frac{2r}{5}$Option: 3 $3r$  Option: 4 $\frac{2r}{3}$

When $\fn_cm S_{1}$ is closed and $\fn_cm S_{2}$ is opened

$\fn_cm I=\frac{E_{P}}{3r}$

$\fn_cm E=Ir_{1}$

$\fn_cm E=I\times\frac{1}{2}\times\frac{2r}{I}$

$\fn_cm E=\frac{E_{P}}{3}$

When $\fn_cm S_{2}$ is opened and $\fn_cm S_{1}$ is closed

$\fn_cm I=\frac{E_{P}}{r_{1}+r}$

$\fn_cm E= I \times r_{2}$

$\fn_cm =I \times \frac{2I}{3}\times \frac{r_{1}}{I}$

$\fn_cm =\frac{2E_{P}r_{1}}{3\left ( r_{1}+r \right )}$

$\fn_cm \frac{E_{P}}{3}=\frac{2E_{P}r_{1}}{3\left ( r_{1}+r \right )}$

$\fn_cm r_{1}+r=2r_{1}$

$\fn_cm r_{1}=r$