Two springs of force constant 2000 N/m and 4000 N/m are stretched by same force. The ratio of their respective potential energies  is

PE of mass attached to a spring is PE=12kx2, where k= force constant, x= displacement restoring force of a spring is F=-kx,                                                ⇒ x=-Fk,substitute in PE equation                                                   PE=12kFk2                                               ⇒PE=12kF2k2                                               ⇒PE =12F2k                                               ⇒PEα1k since stretched by same force                                               ⇒PE1PE2=k2k1=40002000=21  obtained after substituting the given values of force constant k1,k2