If the terminal speed of a sphere of gold (de | vedclass

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(D) The terminal velocity $v_T$ of a sphere of radius $r$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is give

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(D) The terminal velocity $v_T$ of a sphere of radius $r$ and density $ho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by the formula: $v_T = \frac{2r^2(ho - \sigma)g}{9\eta}$.Since the radius $r$, the liquid density $\sigma$, the viscosity $\eta$, and the acceleration due to gravity $g$ are constant for both spheres, we have $v_T \propto (ho - \sigma)$.Therefore, $\frac{v_{T, 	ext{silver}}}{v_{T, 	ext{gold}}} = \frac{ho_{	ext{silver}} - \sigma}{ho_{	ext{gold}} - \sigma}$.Given $ho_{	ext{gold}} = 19.5 	imes 10^3 \ kg/m^3$, $ho_{	ext{silver}} = 10.5 	imes 10^3 \ kg/m^3$, $\sigma = 1.5 	imes 10^3 \ kg/m^3$, and $v_{T, 	ext{gold}} = 0.2 \ m/s$.Substituting the values: $\frac{v_{T, 	ext{silver}}}{0.2} = \frac{10.5 - 1.5}{19.5 - 1.5} = \frac{9}{18} = 0.5$.Thus, $v_{T, 	ext{silver}} = 0.2 	imes 0.5 = 0.1 \ m/s$.

If the terminal speed of a sphere of gold (density $= 19.5 \times 10^3 \ kg/m^3$) is $0.2 \ m/s$ in a viscous liquid (density $= 1.5 \times 10^3 \ kg/m^3$), find the terminal speed (in $m/s$) of a sphere of silver (density $= 10.5 \times 10^3 \ kg/m^3$) of the same size in the same liquid.

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