b
\(\begin{array}{l}
Given\,situation\,is\,shown\,in\,the\,figure.\\
Velocity\,of\,ship\,A,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{v_A} = 10\,km\,{h^{ - 1}}\,towards\,west\\
Velocity\,of\,ship\,B,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{v_B} = 10\,km\,{h^{ - 1}}\,towards\,north\\
\,\,\,\,\,\,\,\,\,\,OS = 100\,km\\
\,\,\,\,\,\,\,\,\,OP = \,shortest\,{\rm{distance}}\\
{\rm{Relative}}\,{\rm{velocity}}\,{\rm{between}}\,{\rm{A}}\,{\rm{and}}\,{\rm{B}}\,{\rm{is}}\\
\,\,\,\,\,\,\,\,\,{{\rm{v}}_{AB}} = \sqrt {v_A^2 + v_B^2}  = 10\sqrt 2 \,km\,{h^{ - 1}}
\end{array}\)

\(\begin{array}{l}
\cos {45^ \circ } = \frac{{OP}}{{OS}};\frac{1}{{\sqrt 2 }} = \frac{{OP}}{{100}}\\
\,\,\,OP = \frac{{100}}{{\sqrt 2 }} = \frac{{100\sqrt 2 }}{2} = 50\sqrt 2 \,km\\
The\,time\,after\,which\,{\rm{distance}}\,{\rm{between}}\,{\rm{them}}\\
{\rm{equals}}\,{\rm{to}}\,{\rm{OP}}\,is\,given\,by\\
\,\,\,\,\,t = \frac{{OP}}{{{v_{AB}}}} = \frac{{50\sqrt 2 }}{{10\sqrt 2 }} \Rightarrow t = 5h
\end{array}\)