• # question_answer A spherical steel ball released at the top of a long column of glycerine of length L, falls through a distance L/2 with accelerated motion and the remaining distance L/2 with a uniform velocity. If ${{t}_{1}}$ and ${{t}_{2}}$ denote the times taken to cover the first and second half and ${{W}_{1}}$ and ${{W}_{2}}$ the work done against gravity in the two halves, then A) ${{t}_{1}}<{{t}_{2}};{{W}_{1}}>{{W}_{2}}$  B) ${{t}_{1}}>{{t}_{2}};{{W}_{1}}<{{W}_{2}}$ C) ${{t}_{1}}={{t}_{2}};{{W}_{1}}={{W}_{2}}$ D) ${{t}_{1}}>{{t}_{2}};{{W}_{1}}={{W}_{2}}$

The average velocity in the first half of the distance $=\frac{0+v}{2}=\frac{v}{2};$ while in the second half, the average velocity is v. therefore,${{t}_{1}}>{{t}_{2}}$. The work done against gravity in both halves =$mgh=mgL/2$. Hence the correct choice is [D].