Q 11-09-042JEE MainJEE Main 2025 (22 Jan, Shift 2)Easy
A small rigid spherical ball of mass $M$ is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider $g$ as acceleration due to gravity)
Answer: (D) $\dfrac{Mg}{2}$
At terminal velocity the net force is zero:
$$Mg = F_B + F_v$$
The buoyant force is the weight of displaced glycerine. Its density is half that of the ball and the volume is the same, so $F_B = \dfrac{Mg}{2}$.
$$F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}$$
Solution by Sreeraj P, M.Sc Physics