1. Forces acting on the steel ball in glycerine

   When the ball is dropped in glycerine, three forces act on it:

   • Weight: $mg$ downward
   • Buoyant force: $U$ upward
   • Viscous drag: $kv$ upward, which increases with speed

2. Initial motion

   At the moment of release, the velocity is zero, so viscous drag is also zero.

   Hence net downward force initially is $F = mg - U$ so the ball starts accelerating downward.

3. As speed increases

   As the ball gains speed, viscous force $kv$ increases upward. Therefore net force becomes $F = mg - U - kv$ and the acceleration gradually decreases.

4. Terminal velocity

   After some time, the viscous force becomes large enough that $mg = U + kv_t$ where $v_t$ is the terminal velocity.

   Then net force becomes zero, so acceleration becomes zero and the ball moves with constant velocity.

5. Shape of the $v$-$t$ graph

   Therefore the velocity-time graph must:

   • start from zero,
   • rise with decreasing slope,
   • finally become horizontal at terminal velocity.

   So the correct graph is the one that approaches a constant value asymptotically.

6. Option check

   • Any graph showing velocity increasing linearly forever is wrong.
   • Any graph showing velocity first increasing and then decreasing is wrong.
   • The correct one is the graph where velocity increases and then levels off.

Thus, the correct option is B.