Explanation:
Potential drop across \(3\,m\)
long resistance wire is \(6\;V.\)
\(.i.e.\,\,\,\,\,\,\,\,\,\,\,{E_1} = 6\;V\) and \({l_1} = 3\;m\),
Also, \(\,\,\,\,\,\,\,{l_2} = 50\;cm = 0.5\;m\)
As we know that, \(E \propto l\)
\( \Rightarrow \,\,\,\,\,\,\,\,\,\,\frac{{{E_1}}}{{{E_2}}} = \frac{{{l_1}}}{{{l_2}}}\,\) or \(\,{E_2} = {E_1} \times \frac{{{l_2}}}{{{l_1}}}\)
Substituting the given values, we get
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E_2} = \frac{{6 \times 0.5}}{3} = 1\;V\)