A spherical steel ball released at the top of along column of glycerin of length `l` falls through a distance `l//2` with accelerated motion and the remaining distance `l//2` with uniform velocity let `t_(1)` and `t_(2)` denote the times taken to cover the first and second half and `w_(1)` and `w_(2)` are the work done against gravity in the two halves, then compare times and work done.

Explanation

Average velocity in first half of the distance `lt v`, while in the second half the average velocity is v.
Therefore, `t_(1) gt t_(2)`. The work done against gravity in both halves is mg `l//2 therefore t_(1) gt t_(2) therefore w_(1)=w_(2)`