Sure! Here’s my step-by-step breakdown based on the image: --- ## Step 1: Identify the Question The question asks: *"The drag force \( F \) on a small sphere of radius \( r \) moving with speed \( v \) through a fluid with viscosity \( \eta \) is given by \( F = 6\pi \eta r v \). What are the SI units of viscosity \( \eta \)?"* **Explanation:** We are given the formula for drag force and need to find the SI units of viscosity (\( \eta \)). --- ## Step 2: Solve for the SI Units Given: - \( F = 6\pi \eta r v \) - \( F \) (Force) has SI units of \( \text{kg} \cdot \text{m/s}^2 \) - \( r \) (radius) has SI units of \( \text{m} \) - \( v \) (speed) has SI units of \( \text{m/s} \) Rearrange the equation for \( \eta \): \[ \eta = \frac{F}{6\pi r v} \] Substitute the units: \[ [\eta] = \frac{[\text{kg} \cdot \text{m/s}^2]}{[\text{m}] \cdot [\text{m/s}]} = \frac{\text{kg} \cdot \text{m/s}^2}{\text{m}^2/\text{s}} = \frac{\text{kg} \cdot \text{m/s}^2 \cdot \text{s}}{\text{m}^2} = \frac{\text{kg}}{\text{m} \cdot \text{s}} \] **Explanation:** By dividing the SI units of each quantity and simplifying, we get the SI unit for viscosity. --- ## Final Answer The SI units of viscosity (\( \eta \)) are \[ \boxed{\text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-1}} \]