The person is rotating at the rate of\({\omega _1} = 0.90\;{\rm{rev/s}}\).

The speed of rotation decreases to \({\omega _2} = 0.60\;{\rm{rev/s}}\).

In this problem, consider the person and the platform as a system to analyze angular momentum. Also, in this condition, the system’s angular momentum is totally conserved.

In this case, the force applied and the torque to raise the hand of the person is inherent to the system; so the raising as well as dropping of the hands will not affect the net angular momentum. However, the moment of inertia of the arms will increase, and due to the conservation of angular momentum, the rise in the moment of inertia will be followed by a reduction in the angular velocity of the person.

\(\begin{aligned}{c}L &= L'\\I{\omega _1} &= I'{\omega _2}\\I' &= \left( {\frac{{{\omega _1}}}{{{\omega _2}}}} \right)I\end{aligned}\)

Here, \(L\) and \(L'\) are the initial and final angular momentum, \(I\) and \(I'\) are the initial and final moment of inertia.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}I' &= \left( {\frac{{0.90\;{\rm{rev/s}}}}{{0.60\;{\rm{rev/s}}}}} \right)I\\I' &= 1.5I\end{aligned}\)

Thus, \(I' = 1.5I\) is the factor by which moment of inertia changes.