Two solid spheres of same radius and materials of specific gravities 4 and 8 are released in water. The terminal velocity attained by the heavier sphere is:

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TestHub · Accepted Answer

At terminal speed, viscous drag balances weight minus buoyancy.

$6 π η R v_{t} = \frac{4 π R ^{3}}{3} (ρ_{s} - ρ_{w}) g$

$v_{t} = \frac{2 R ^{2} ( ρ _{s} - ρ _{w} ) g}{9 η}$

R is the common radius; rho_s and rho_w are sphere and water densities, and eta is dynamic viscosity (Pa s).

$\frac{v _{2}}{v _{1}} = \frac{ρ _{s, 2} - ρ _{w}}{ρ _{s, 1} - ρ _{w}}$

$\frac{v _{2}}{v _{1}} = \frac{8 - 1}{4 - 1} = \frac{7}{3}$

The denser sphere falls at 7/3 times the lighter sphere’s terminal speed (B).

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