JEE Main 2025 Jan 22 Shift 2, Physics Q2: Hydrodynamic

JEE Main2025Jan 22, Shift 2Physics

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)

A
$\frac{3}{2} M g$
B
$\frac{M g}{2}$
C
$M g$
D
$2 M g$

Solution

At terminal velocity the net force on the ball is zero. Three forces act: weight $M g$ downward, buoyancy $F_{B}$ upward, and viscous drag $f$ upward.

Buoyancy on the ball equals the weight of displaced glycerine. Since $ρ_{glyc} = \frac{1}{2} ρ_{ball}$ and the volume is the same:

$F_{B} = \frac{1}{2} M g$ .

Force balance: $M g - F_{B} - f = 0 \Rightarrow f = M g - \frac{M g}{2} = \frac{M g}{2}$ .

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JEE Main 2025 Jan 22 Shift 2, Physics Q2: Hydrodynamic

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