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(B) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{l_1}{l_2} - 1 \right)$,where $l_1$ is

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$10$

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(B) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{l_1}{l_2} - 1 ight)$,where $l_1$ is the balancing length without shunt and $l_2$ is the balancing length with shunt $R$.Given:Case $1$: $R_1 = 5\,\Omega$,$l_1 = 2\,m$Case $2$: $R_2 = 10\,\Omega$,$l_2 = 3\,m$Let $L$ be the balancing length of the cell without any shunt.For the first case: $r = 5 \left( \frac{L}{2} - 1 ight) = 2.5L - 5$ ... $(i)$For the second case: $r = 10 \left( \frac{L}{3} - 1 ight) = \frac{10}{3}L - 10$ ... $(ii)$Equating $(i)$ and $(ii)$:$2.5L - 5 = \frac{10}{3}L - 10$$5 = \frac{10}{3}L - 2.5L$$5 = \frac{10L - 7.5L}{3} = \frac{2.5L}{3}$$15 = 2.5L \implies L = 6\,m$Substituting $L = 6$ into equation $(i)$:$r = 5 \left( \frac{6}{2} - 1 ight) = 5(3 - 1) = 5 	imes 2 = 10\,\Omega$.

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