Correct Answer - Option 2 :
\(\frac{{102}}{{{s^2} + 10s + 100}}\)
Standard 2nd order system is \({\rm{T}}\left( {\rm{s}} \right) = \frac{{{\rm{K\omega }}_{\rm{n}}^2}}{{{{\rm{s}}^2} + 2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}}{\rm{s}} + {\rm{\omega }}_{\rm{n}}^2}}\)
Given \({\rm{\xi \;}} = {\rm{\;}}0.5{\rm{\;and\;}}{{\rm{\omega }}_{\rm{n}}}{\rm{\;}} = {\rm{\;}}10\)
Steady state value \({\left. {{\rm{T}}\left( {\rm{s}} \right)} \right|_{{\rm{s}} = 0}} = {\rm{K}} = 1.02\)
\(\therefore {\rm{T}}\left( {\rm{s}} \right) = \frac{{\left( {1.02} \right)\left( {100} \right)}}{{{{\rm{s}}^2} + 10{\rm{s}} + 100}} = \frac{{102}}{{{{\rm{s}}^2} + 10{\rm{s}} + 100}}\)