In an experiment to find emf of a cell using potentiometer, the...

We know, the condition where the galvanometer shows a null point is,

$\dfrac{{E}_{1}}{{E}_{2}} = \dfrac{{l}_{1}}{{l}_{2}}$

So, $\dfrac{1 . 5}{{E}_{2}} = \dfrac{60}{60 + 40} = \dfrac{6}{10} = \dfrac{3}{5}$

$⇒ {E}_{2} = \dfrac{5}{2}V$

Hence, the value of $x = 25$.