Question

In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5 \,V$ is found to be $60 \,cm$. If this cell is replaced by another cell of emf $E$, the length-of null point increases by $40 \,cm$. The value of $E$ is $\frac{x}{10} V$. The value of $x$ is _______

Answer 1

$\frac{E_1}{E_2}=\frac{l_1}{l_2}$
$\frac{1.5}{E_2}=\frac{60}{60+40}=\frac{6}{10}=\frac{3}{5}$
$E_2=\frac{5}{2}=\frac{x}{10}$
$x=25$