A battery of e.m.f 6V and internal resistance  $1\Omega$  gives a p.d of 5.8V when connected to a resistance. The external resistance is  $x\Omega$  then the value of 29+x is

e.m.f of battery E = 6V; Internal resistance of battery  $r=1\Omega ;$  p.d across the resistance V = 5.8 V 
Let the external resistance be ‘R’
$E=V+Ir\Rightarrow E=V+\frac{V}{R}r\Rightarrow \frac{r}{R}=\frac{E-V}{V}\Rightarrow R=\frac{Vr}{E-V}=\frac{58\times 1}{6-5.8}=\frac{5.8}{0.2}=20+x\Omega =9$