Correct Answer - Option 2 : 0.01

__Concept:__

The relation between new per-unit value & old per unit value of reactance

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

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

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

Where,

(Xpu)new = New per unit value of reactance

(Xpu)old = Old per unit value of reactance

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:__

Given that,

Xd(old) = 0.2 pu

MVA(new) = 100

MVA(old) = 500

KV(old) = 11

KV(new) = 22

∴ \({X_{d\left( {new} \right)}} = 0.2 \times \frac{{100}}{{500}} \times {\left( {\frac{{11}}{{22}}} \right)^2} = 0.01\;pu\)