Q.The terminal velocity of a copper ball of radius $2.0\ \text{mm}$ falling through a tank of oil at $20\,^{\circ}\text{C}$ is $6.5\ \text{cm s}^{-1}$. Compute the viscosity of the oil at $20\,^{\circ}\text{C}$. Density of oil is $1.5 \times 10^{3}\ \text{kg m}^{-3}$, density of copper is $8.9 \times 10^{3}\ \text{kg m}^{-3}$.

The key idea is Viscous Force Balance.

At terminal velocity, the net force on the copper ball is zero. The downward gravitational force is balanced by the upward buoyant force and the upward viscous drag force.

1. Force Balance: $F_{\text{gravity}} = F_{\text{buoyancy}} + F_{\text{viscous}}$
2. Substitute formulas: $\rho_{\text{copper}} \left(\frac{4}{3}\pi r^3\right) g = \rho_{\text{oil}} \left(\frac{4}{3}\pi r^3\right) g + 6\pi \eta r v_t$
3. Rearrange for viscosity ($\eta$): $6\pi \eta r v_t = \left(\frac{4}{3}\pi r^3\right) g (\rho_{\text{copper}} - \rho_{\text{oil}})$ $\eta = \frac{2 r^2 g (\rho_{\text{copper}} - \rho_{\text{oil}})}{9 v_t}$
4. Substitute values: $r = 2.0\ \text{mm} = 2.0 \times 10^{-3}\ \text{m}$ $v_t = 6.5\ \text{cm s}^{-1} = 6.5 \times 10^{-2}\ \text{m s}^{-1}$ $\rho_{\text{copper}} = 8.9 \times 10^{3}\ \text{kg m}^{-3}$ $\rho_{\text{oil}} = 1.5 \times 10^{3}\ \text{kg m}^{-3}$ $g = 9.8\ \text{m s}^{-2}$ …

When an object falls through a fluid, it experiences three main forces: its weight pulling it down, an upward buoyant force from the displaced fluid, and an upward viscous drag force opposing its motion. Initially, the object accelerates because its weight is greater than the sum of the buoyant and drag forces. As its speed increases, the viscous drag force also increases (since it depends on velocity). Eventually, the drag force becomes large enough that the total upward force (buoyancy + drag) exactly balances the downward weight. At this point, the net force on the object becomes zero, and it stops accelerating, continuing to fall at a constant maximum velocity called the terminal velocity.

This problem asks us to find the viscosity of the oil, given the terminal velocity of a copper ball. The key concept here is the force balance at terminal velocity. By setting the sum of the upward forces equal to the downward force, we can derive an expression for viscosity.

Let's break down the calculation:

1. Identify the forces acting on the copper ball.

   • Gravitational Force ($F_g$): This acts downwards and is the weight of the copper ball. $F_g = m_c g = \rho_c V g$ where $\rho_c$ is the density of copper, $V$ is the volume of the ball, and $g$ is the acceleration due to gravity.
   • Buoyant Force ($F_b$): This acts upwards and is equal to the weight of the fluid displaced by the ball (Archimedes' Principle). $F_b = m_o g = \rho_o V g$ where $\rho_o$ is the density of the oil.
   • Viscous Drag Force ($F_v$): This acts upwards, opposing the motion of the ball through the oil. For a small spherical object moving slowly through a viscous fluid, this force is given by Stokes' Law. $F_v = 6 \pi \eta r v_t$ where $\eta$ is the viscosity of the oil, $r$ is the radius of the ball, and $v_t$ is its terminal velocity.

2. Apply the condition for terminal velocity.

   At terminal velocity, the net force on the ball is zero. This means the downward force equals the sum of the upward forces:

   $F_g = F_b + F_v$

3. Substitute the force expressions into the balance equation.

   The volume of a sphere is $V = \frac{4}{3} \pi r^3$. Substituting this and the force formulas:

   $\rho_c \left(\frac{4}{3} \pi r^3\right) g = \rho_o \left(\frac{4}{3} \pi r^3\right) g + 6 \pi \eta r v_t$

4. Rearrange the equation to solve for viscosity ($\eta$).

   First, group the terms involving densities:

   $\left(\rho_c - \rho_o\right) \frac{4}{3} \pi r^3 g = 6 \pi \eta r v_t$

   Now, isolate $\eta$:

   $\eta = \frac{\left(\rho_c - \rho_o\right) \frac{4}{3} \pi r^3 g}{6 \pi r v_t}$

   We can simplify this expression by cancelling $\pi$ and one $r$:

   $\eta = \frac{\left(\rho_c - \rho_o\right) \frac{4}{3} r^2 g}{6 v_t}$

   $\eta = \frac{2 \left(\rho_c - \rho_o\right) r^2 g}{9 v_t}$

   The viscosity of the fluid can be calculated using the terminal velocity formula:

   $\eta = \frac{2 (\rho_c - \rho_o) r^2 g}{9 v_t}$

5. Convert all given values to SI units. …

Terminal velocity: gravity=buoyancy+Stokes drag. eta=2r^2g*(rho_Cu-rho_oil)/(9*v_t). Substituting r=2.0e- …