A spherical steel ball released at the top of a long column of gylycerine 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 time taken to cover the first and second half and `W_(1)` and `W_(2)` the work done against gravity in the two halves, then

Options

A. `t_(1) lt t_(2)`
B. `t_(1) gt t_(2)`
C. `W_(1) = W_(2)`
D. `W_(1) lt W_(2)`

Answer

B, D

Explanation

Time `t_(t)` will be more then `t_(2)` because speed increase from zero to terminal speed in `t_(t)` duration and ball covers a distance `h`.
Work done against viscous force depends on magnitude of viscous force and displacement ball.
viscous force increases from zero to maximum value and then remains constant