In the experiment to find internal resistance of Daniel cell using potentiometer with E as Emt of the cell 'L' the balancing length of the potentiometer wire when potentiometer circuit is closed. (when no current flows through cell. When $R= 5 \Omega $ is introduced the new balancing length is l.) Then the internal resistance of the cell is

Question

$(a)\;r= \large\frac{l}{(L-l)} \times R \\(b)\;r= \large\frac{L-l}{e \times R} \\(c)\;r= \large\frac{l}{(L)} \times R \\(d)\;r= \large\frac{L}{l} \times R$