Correct Answer - Option 2 :

\(1 - {{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{T}}}}}\)
Let the standard first-order open-loop transfer function is:

\({\rm{G}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{sT}}}}\)

then,

The closed-loop transfer function is:

\({\rm{H}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT}}}}\) (assume unity negative feedback)

\({\rm{Y}}\left( {\rm{s}} \right){\rm{\;}} = {\rm{\;X}}\left( {\rm{s}} \right).{\rm{\;H}}\left( {\rm{s}} \right)\)

If input = unit step

\( {\rm{\;X}}\left( {\rm{s}} \right) = \frac{1}{{\rm{s}}}\)

\(\begin{array}{l} {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT\;}}}}.\frac{1}{{\rm{s}}} \\= \frac{1}{{\rm{s}}} + \frac{{ - {\rm{T}}}}{{\left( {1 + {\rm{sT}}} \right)}} \\= \frac{1}{{\rm{s}}} - \frac{1}{{\frac{1}{{\rm{T}}} + {\rm{s}}}}\\ {\rm{y}}\left( {\rm{t}} \right) = 1 - {{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{T}}}}} \end{array}\)

Hence, option 2 is correct.