At terminal velocity, net force on the ball is zero

So the viscous force equals: $F_v = \text{weight of ball} - \text{buoyant force}$

Hence, $F_v = (\rho_s - \rho_l) V g$ where:

• $\rho_s = 10.5\,\text{g/cc}$ = density of sphere
• $\rho_l = 1.5\,\text{g/cc}$ = density of glycerine
• $V$ = volume of sphere

Convert units to SI

Since: $1\,\text{g/cc} = 1000\,\text{kg/m}^3$

Therefore, $\rho_s = 10.5 \times 1000 = 10500\,\text{kg/m}^3$ $\rho_l = 1.5 \times 1000 = 1500\,\text{kg/m}^3$

So, $\rho_s - \rho_l = 9000\,\text{kg/m}^3$

Radius of the sphere

Given: $r = 1\,\text{mm} = 10^{-3}\,\text{m}$

Volume of the sphere

$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (10^{-3})^3 = \frac{4}{3}\pi \times 10^{-9}\,\text{m}^3$

Compute viscous force

$F_v = (9000)\left(\frac{4}{3}\pi \times 10^{-9}\right)(9.8)$

First, $9000 \cdot \frac{4}{3} = 12000$

So, $F_v = 12000\pi \times 10^{-9} \times 9.8$ $F_v = 117600\pi \times 10^{-9}$

Using $\pi = \frac{22}{7}$, $F_v = 117600 \times \frac{22}{7} \times 10^{-9}$

$117600 \div 7 = 16800$ $16800 \times 22 = 369600$

Therefore, $F_v = 369600 \times 10^{-9}$ $F_v = 3696 \times 10^{-7}\,\text{N}$

Compare with given form

Given: $F_v = 3696 \times 10^{-x}\,\text{N}$

Therefore, $x = 7$

Comparison with stored answer

Stored correct answer = 7, which matches our result.