Correct Answer - Option 1 : Increase by 25%
At constant torque load slip is inversely proportional to square of the voltage.
\(\begin{array}{l}
s \propto \frac{1}{{{V^2}}}\\
\Rightarrow \frac{{{s_1}}}{{{s_2}}} = {\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^2}\\
\Rightarrow \frac{{{s_1}}}{{{s_2}}} = {\left( {\frac{{0.9\;{V_1}}}{{{V_1}}}} \right)^2}\\
\Rightarrow {s_2} = 1.23\;{s_1}
\end{array}\)
Hence the slip will increase by 25% approximately.