JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Fluid Mechanics

At terminal velocity the ball moves with constant speed, so the net force on it is zero. Three forces act on it: weight $Mg$ downward, buoyant force $F_B$ upward, and viscous drag $f$ upward:
$Mg=F_B+f$

The buoyant force equals the weight of glycerine displaced:
$F_B={\rho }_{gly}Vg$

Since the ball has mass $M={\rho }_{ball}V$ and the glycerine's density is half the ball's density, ${\rho }_{gly}={\rho }_{ball}/2$:
$F_B=\frac{{\rho }_{ball}}{2}Vg=\frac{1}{2}({\rho }_{ball}V)g=\frac{Mg}{2}$

So the viscous force is
$f=Mg-F_B=Mg-\frac{Mg}{2}=\frac{Mg}{2}$