A very small ball of radius r and mass m is released in a large tank having a liquid of density ρ and moving with an acceleration a as shown in figure. If coefficient of viscosity the liquid is η, then terminal speed of the ball will be (neglect buoyant force)

Engineering

Physics

Relative Motion

Viscosity and Terminal Velocity New

Forces and Types of Forces

Question

A very small ball of radius r and mass m is released in a large tank having a liquid of density ρ and moving with an acceleration a as shown in figure. If coefficient of viscosity the liquid is η, then terminal speed of the ball will be (neglect buoyant force)

$\frac{m}{6 πηr} \sqrt{g + a}$

$\frac{ma}{6 πηr}$

$\frac{m}{6 πηr} \sqrt{g^{2} + a^{2}}$

$\frac{mg}{6 πηr}$

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The terminal speed occurs when the net force on the ball is zero. The ball experiences three forces: its weight (mg) downward, the pseudo-force (ma) due to the accelerating frame (opposite to acceleration), and the viscous drag (6πηrv) opposing motion. The buoyant force is neglected. The effective gravity is the vector sum of g and a, so the net downward force is m√(g² + a²). At terminal velocity, this equals the Stokes' drag force.

Therefore, the equation is: $6 π η r v_{t} = m \sqrt{g^{2} + a^{2}}$ . Solving for terminal speed v_t gives the correct option.

Final Answer: $\frac{m}{6 π η r} \sqrt{g^{2} + a^{2}}$