A man sitting firmly over a rotating stool has his arms stretched. If he folds his arms, the work done by the man is:
$
  {\text{A}}{\text{. zero}} \\
  {\text{B}}{\text{. positive}} \\
  {\text{C}}{\text{. negative}} \\
  {\text{D}}{\text{. may be positive or negative}} \\
$

Hint: Total angular momentum shall remain conserved in this case. Angular momentum depends on the moment of inertia. These will affect the work done.

Complete answer:
When the man spreads out his hands his inertial mass changes. As the man is rotating, he possesses an angular velocity $\omega $ .
The inertial mass varies as$I = m{r^2}$.
On spreading out arms r increases and so does the value of Inertial mass.
The work done will be the difference in the initial and final energies. So, the initial and final kinetic energies will be-
$
  K = \dfrac{1}{2}I{\omega ^2} \\
  \omega = \dfrac{L}{I} \\
  {K_i} = \dfrac{1}{2}{I_i}{\left( {\dfrac{L}{{{I_i}}}} \right)^2} = \dfrac{1}{2}.\dfrac{{{L^2}}}{{{I_i}}} \\
  {K_f} = \dfrac{1}{2}{I_f}{\left( {\dfrac{L}{{{I_f}}}} \right)^2} = \dfrac{1}{2}.\dfrac{{{L^2}}}{{{I_f}}} \\
  {I_i} < {I_f} \\
  {K_i} < {K_f} \\
  W = \Delta E = {K_f} - {K_i} > 0 \\
 $
Hence, the work done will be positive.

So, the correct answer is “Option B”.

Note:
On increment of I there will be an effect on the angular velocity since the total angular momentum has to remain conserved.
Therefore,
$
  L = I\omega \\
  \omega = \dfrac{L}{I} \\
  {I_i} < {I_f} \\
  \dfrac{L}{{{I_i}}} > \dfrac{L}{{{I_f}}} \\
  {\omega _i} > {\omega _f} \\
 $