A galvanometer, together with an unknown resistance in series, is connected to two identical batteries each of $1.5 V$. When the batteries are in series, the galvanometer registers a current of 1 ampere. When the batteries are in parallel, the current is $0.6 ampere$. What is the internal resistance of battery in $\Omega$?

Let  $R$ be the combined resistance of galvanometer and an unknown resistance in series and  $r$ be the internal resistance of each battery. When the batteries, each of emf $E$, are connected in series, the net emf $=2E$ and net internal resistance $=2r$.
Current,  ${\text{I}}_1=\frac{2E}{R+2r}\Rightarrow 1=\frac{3}{R+2r}$
When the batteries are connected in parallel, the emf remains $E$ and net internal resistance 
becomes $r/2$. Therefore, current is:

${\text{I}}_2=\frac{E}{R+\frac{r}{2}}\Rightarrow 0.6=\frac{1.5}{R+\frac{r}{2}}=\frac{3}{2R+r}$
 
Solving above equations, we get,  $R=\frac{7}{3}\Omega ,r=\frac{1}{3}\Omega$