A spherical ball of mass 4m, density σ and radius r is attached to a pulley-mass system as shown in figure. The ball is released in a liquid of coefficient of viscosity η and density ρ (σ > 2ρ). If the length of the liquid column is sufficiently long, the terminal velocity attained by the ball is given by (assume all pulleys and string to be massless) :

Engineering

Physics

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Viscosity and Terminal Velocity New

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Question

A spherical ball of mass 4m, density σ and radius r is attached to a pulley-mass system as shown in figure. The ball is released in a liquid of coefficient of viscosity η and density ρ (σ > 2ρ). If the length of the liquid column is sufficiently long, the terminal velocity attained by the ball is given by (assume all pulleys and string to be massless) :

$\frac{2}{9} \frac{r^{2} (σ - 4 ρ) g}{η}$

$\frac{2}{9} \frac{r^{2} (σ - ρ) g}{η}$

$\frac{2}{9} \frac{r^{2} (2 σ - ρ) g}{η}$

$\frac{2}{9} \frac{r^{2} (σ - 2 ρ) g}{η}$

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At terminal velocity, net force on the system is zero. The ball experiences three vertical forces: downward weight (4mg), upward buoyant force (F_b = (4m/σ)ρg = (4ρ/σ)mg), and upward viscous drag (F_d = 6πηrv). The pulley system gives a mechanical advantage: the mass m moves with velocity v/2. For the hanging mass m to be at rest, tension T = mg.

Balancing forces on the ball: 4mg + T (downward) = F_b + F_d (upward). Substituting T=mg and F_b:

$4 m g + m g = \frac{4 ρ}{σ} m g + 6 π η r v$

Using mass m = (4/3)πr³σ, solve for v. The correct terminal velocity is:

$\frac{2}{9} \frac{r^{2} (σ - 2 ρ) g}{η}$