A cell of negligible resistance and emf $2$ volt is connected to a series combination of $2, 3$ and $5$ ohm. The potential difference in volts between the terminals of $3$-ohm resistance will be:

Given,

$V=2 V$

$\begin{aligned} \text { Total Desistance } & =(2+3+5) \Omega \\ \therefore \text { Req } & =10 \Omega \end{aligned}$

Using voltage divider rule,

$\begin{aligned} V_{3} & =\left(\frac{3}{10} \times 2\right) v \\ & =\frac{6}{10} \mathrm{~V} \\ & =0.6 \mathrm{~V} \end{aligned}$