If the terminal speed of a sphere of gold (density $=19.5\times10^{3}\ \mathrm{kg/m^{3}}$) is $0.2\ \mathrm{m/s}$ in a viscous liquid (density $=1.5\times10^{3}\ \mathrm{kg/m^{3}}$), then the terminal speed of a sphere of silver (density $=10.5\times10^{3}\ \mathrm{kg/m^{3}}$) of the same size in the same liquid is

(D) $0.1 \mathrm{~m} / \mathrm{s}$

(A) $0.2 \mathrm{~m} / \mathrm{s}$

(B) $0.4 \mathrm{~m} / \mathrm{s}$

(C) $0.133 \mathrm{~m} / \mathrm{s}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Fluid Mechanics

If the terminal speed of a sphere of gold (density

= 19.5 \times 1 0^{3} kg/ m^{3}

) is

0.2 m/s

in a viscous liquid (density

= 1.5 \times 1 0^{3} kg/ m^{3}

), then the terminal speed of a sphere of silver (density

= 10.5 \times 1 0^{3} kg/ m^{3}

) of the same size in the same liquid is

(A) $0.2 m / s$
(B) $0.4 m / s$
(C) $0.133 m / s$
(D) $0.1 m / s$ ✓ Correct

Answer: (D)

Solution

At terminal speed the net of weight, buoyancy and Stokes drag is zero, giving

v_{t} \propto (ρ_{sphere} - ρ_{liquid})

for spheres of equal radius in the same medium. Hence

\frac{v _{silver}}{v _{gold}} = \frac{ρ _{silver} - ρ _{ℓ}}{ρ _{gold} - ρ _{ℓ}} = \frac{10.5 - 1.5}{19.5 - 1.5} = \frac{9}{18} = \frac{1}{2}

. With

v_{gold} = 0.2 m/s

this gives

v_{silver} = 0.1 m/s

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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