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(B) Let $E$ be the $EMF$ and $r$ be the internal resistance of the cell. Let $K$ be the potential gradient of the potentiometer wire.When the cell is

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$4$

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(B) Let $E$ be the $EMF$ and $r$ be the internal resistance of the cell. Let $K$ be the potential gradient of the potentiometer wire.When the cell is shunted with resistance $R_1 = 4 \ \Omega$, the terminal voltage is $V_1 = \frac{E \cdot R_1}{r + R_1} = \frac{E \cdot 4}{r + 4}$.The balance length is $l_1 = 120 \ cm$, so $V_1 = K \cdot l_1 \Rightarrow \frac{E \cdot 4}{r + 4} = 120K$  --- $(1)$When the cell is shunted with resistance $R_2 = 12 \ \Omega$, the terminal voltage is $V_2 = \frac{E \cdot R_2}{r + R_2} = \frac{E \cdot 12}{r + 12}$.The balance length is $l_2 = 180 \ cm$, so $V_2 = K \cdot l_2 \Rightarrow \frac{E \cdot 12}{r + 12} = 180K$  --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{4}{r + 4} \cdot \frac{r + 12}{12} = \frac{120K}{180K}$$\frac{1}{3} \cdot \frac{r + 12}{r + 4} = \frac{2}{3}$$\frac{r + 12}{r + 4} = 2$$r + 12 = 2(r + 4)$$r + 12 = 2r + 8$$r = 4 \ \Omega$.

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