If the terminal speed of a sphere of gold (density = 19.5 kg/m3) is 0.2 m/s in a viscous liquid (density = 1.5 kg/m3), find the terminal speed of a sphere of silver (density = 10.5 kg/m3) of the same size in the same liquid.

Question

If the terminal speed of a sphere of gold (density = 19.5 kg/m3) is 0.2 m/s in a viscous liquid (density = 1.5 kg/m3), find the terminal speed of a sphere of silver (density = 10.5 kg/m3) of the same size in the same liquid. (A) 0.4 m/s (B) 0.133 m/s (C) 0.1 m/s (D) 0.2 m/s

Tardigrade Admin · Accepted Answer

Terminal speed of spherical body in a viscous liquid is given by

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

where

ρ =

density of substance of body

σ =

density of liquid
From given data

\frac{v _{T} ( A g )}{v _{T} ( G o l d )} = \frac{ρ _{A g} - σ _{l}}{ρ _{G o l d} - σ _{l}}

\Rightarrow v_{T} (A g) = \frac{10.5 - 1.5}{19.5 - 1.5} \times 0.2 = \frac{9}{18} \times 0.2

= 0.1 m / s

Q. If the terminal speed of a sphere of gold (density $= 19.5 k g / m^{3}$ ) is $0.2 m / s$ in a viscous liquid (density $= 1.5 k g / m^{3}$ ), find the terminal speed of a sphere of silver (density $= 10.5 k g / m^{3}$ ) of the same size in the same liquid.

$0.4 m / s$

$0.133 m / s$

$0.1 m / s$

$0.2 m / s$

Q. If the terminal speed of a sphere of gold (density =19.5kg/m3) is 0.2m/s in a viscous liquid (density =1.5kg/m3), find the terminal speed of a sphere of silver (density =10.5kg/m3) of the same size in the same liquid.

0.4m/s

0.133m/s

0.1m/s

0.2m/s

Terminal speed of spherical body in a viscous liquid is given by vr​=9η2r2(ρ−σ)g​ where ρ= density of substance of body σ=density of liquid From given data vT​(Gold)vT​(Ag)​=ρGold​−σl​ρAg​−σl​​ ⇒vT​(Ag)=19.5−1.510.5−1.5​×0.2=189​×0.2 =0.1m/s

Q. If the terminal speed of a sphere of gold (density $= 19.5 k g / m^{3}$ ) is $0.2 m / s$ in a viscous liquid (density $= 1.5 k g / m^{3}$ ), find the terminal speed of a sphere of silver (density $= 10.5 k g / m^{3}$ ) of the same size in the same liquid.

$0.4 m / s$

$0.133 m / s$

$0.1 m / s$

$0.2 m / s$