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Step-by-Step Solution
Step 1: Identify the Known Variables
• Density of gold, $ \rho_g = 19.5\ \mathrm{kg/m^3} $
• Density of silver, $ \rho_s = 10.5\ \mathrm{kg/m^3} $
• Density of the liquid, $ \sigma = 1.5\ \mathrm{kg/m^3} $
• Terminal velocity of the gold sphere, $ v_g = 0.2\ \mathrm{m/s} $
• We need to find the terminal velocity of the silver sphere, $ v_s $.
Step 2: Write Down the Formula for Terminal Velocity
According to Stokes' law, for a sphere of radius $ r $ moving through a viscous fluid:
$$
6\pi \eta\, r\, v_t \;=\;\left(\frac{4}{3}\pi r^3\right)\,(\rho - \sigma)\,g
$$
Here, $ \eta $ is the coefficient of viscosity of the fluid, $ \rho $ is the density of the sphere, $ \sigma $ is the density of the liquid, and $ g $ is the acceleration due to gravity.
Solving for the terminal velocity $ v_t $, we get:
$$
v_t = \frac{2\,r^2\,(\rho - \sigma)\,g}{9\,\eta}
$$
Step 3: Apply the Formula to Both Gold and Silver Spheres
Let $ v_g $ be the terminal velocity for the gold sphere and $ v_s $ be that for the silver sphere. From the formula, notice that all parameters except for the density of the material remain the same (liquid density $ \sigma $, radius $ r $, viscosity $ \eta $, and $ g $).
Therefore, the ratio of their terminal velocities will depend only on the difference in densities:
$$
\frac{v_g}{v_s}
= \frac{\rho_g - \sigma}{\rho_s - \sigma}
$$
Step 4: Substitute the Numerical Values and Solve
Substitute $ \rho_g = 19.5 \ \mathrm{kg/m^3}, \rho_s = 10.5 \ \mathrm{kg/m^3}, \sigma = 1.5 \ \mathrm{kg/m^3} $:
$$
\frac{v_g}{v_s}
= \frac{(19.5 - 1.5)}{(10.5 - 1.5)}
= \frac{18}{9}
= 2
$$
Hence,
$$
v_s = \frac{v_g}{2}.
$$
Given $ v_g = 0.2\ \mathrm{m/s} $, we find:
$$
v_s = \frac{0.2}{2} = 0.1\ \mathrm{m/s}.
$$
Step 5: State the Final Answer
The terminal velocity of the silver sphere in the same liquid is $ 0.1 \ \mathrm{m/s} $.
