A small ball bearing is released at the top of a long vertical column of glycerine of height `2h`. The bal bearing falls thorugh a height `h` in a time `t_(1)` and then the reamining height with the terminal velocity in time `t_(2)`. Let `W_(1)` and `W_(2)` be the done against viscous drag over these height. Therefore,

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