Correct Answer - Option 4 : 0.200

__Concept:__

The relation between new per-unit value & old per unit value impedance

\({({Z_{pu}})_{new}}\; = {({Z_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)

Also \({Z_{pu}} = \frac{{{Z_{Actual}}}}{{{Z_{base}}}}\)

\( {Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

Where,

(Zpu)new = New per unit value of impedance

(Zpu)old = Old per unit value of impedance

kVbase = Old base value of voltage

kVnew = New base value of voltage

MVAnew = New base value of power

MVAold = Old base value of power

__Calculation__:

When the voltage bases is increased by 1.1 times (i.e. kVnew = 1.1kVbase), the new per unit impedance is

\((Z_{pu})_{new}=0.242\times \frac{1}{1.1^2}=0.2\ \ pu\)