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Step-by-Step Solution
Step 1: List the Forces Acting on the Sphere
When a sphere of radius $R$ falls through a liquid of density $\sigma$, the following forces act on it:
Gravitational Force (Weight),
$W = \frac{4}{3} \pi R^3 \,\rho g$,
where $\rho$ is the density of the sphere and $g$ is the acceleration due to gravity.
Buoyant Force,
$F_b = \frac{4}{3} \pi R^3 \,\sigma g$,
opposing the weight because of the displaced liquid.
Viscous Drag Force,
$F_\text{viscous} = 6 \pi \eta R \, v$,
where $\eta$ is the coefficient of viscosity of the liquid and $v$ is the velocity of the sphere.
Step 2: Terminal Velocity for the First Sphere
At terminal velocity, the net force on the sphere is zero. For a sphere of density $\rho_1$ moving with terminal velocity $v_1$:
$6 \pi \eta R \, v_1 = \frac{4}{3} \pi R^3 (\rho_1 - \sigma) g.$
Step 3: Terminal Velocity for the Second Sphere
For the second sphere of the same radius $R$ but density $\rho_2$, let its terminal velocity be $v_2$. Similarly,
$6 \pi \eta R \, v_2 = \frac{4}{3} \pi R^3 (\rho_2 - \sigma) g.$
Step 4: Relate the Two Terminal Velocities
Divide the second equation by the first to eliminate the common factors:
$\frac{v_2}{v_1}
= \frac{\frac{4}{3} \pi R^3 (\rho_2 - \sigma) g}{\frac{4}{3} \pi R^3 (\rho_1 - \sigma) g}
= \frac{\rho_2 - \sigma}{\rho_1 - \sigma}.$
Therefore,
$v_2 = \biggl(\frac{\rho_2 - \sigma}{\rho_1 - \sigma}\biggr) \, v_1.$
Step 5: Final Answer
The terminal velocity of the second sphere is
$v_2 = \left|\frac{\rho_2 - \sigma}{\rho_1 - \sigma}\right| v_1,$
which can be written without the absolute value (assuming $\rho_2 > \sigma$) as:
$v_2 = \biggl(\frac{\rho_2 - \sigma}{\rho_1 - \sigma}\biggr) v_1.$
