1. At constant velocity, net force on the ball is zero

When the ball falls through glycerine and attains terminal velocity, the forces balance:

• Weight downward: $W = mg$
• Buoyant force upward: $B = \rho_f V g$
• Viscous force upward: $F_v$

So,

$mg = B + F_v$

Hence,

$F_v = mg - B$

2. Given data

• Mass of ball: $m = 0.3\,\text{g} = 0.3 \times 10^{-3}\,\text{kg} = 3 \times 10^{-4}\,\text{kg}$
• Density of ball: $\rho_b = 8\,\text{g/cc}$
• Density of glycerine: $\rho_f = 1.3\,\text{g/cc}$
• $g = 10\,\text{m/s}^2$

3. Find volume of the ball

Using

$\rho_b = \frac{m}{V} \Rightarrow V = \frac{m}{\rho_b}$

In cgs units:

$V = \frac{0.3}{8} = 0.0375\,\text{cc}$

Now convert to SI:

$1\,\text{cc} = 10^{-6}\,\text{m}^3$

So,

$V = 0.0375 \times 10^{-6} = 3.75 \times 10^{-8}\,\text{m}^3$

4. Calculate weight of the ball

$W = mg = 3 \times 10^{-4} \times 10 = 3 \times 10^{-3}\,\text{N}$

5. Calculate buoyant force

Convert fluid density to SI:

$1.3\,\text{g/cc} = 1.3 \times 10^3\,\text{kg/m}^3$

Then,

$B = \rho_f V g$

$B = 1.3 \times 10^3 \times 3.75 \times 10^{-8} \times 10$

$B = 4.875 \times 10^{-4}\,\text{N}$

6. Find viscous force

$F_v = W - B$

$F_v = 3 \times 10^{-3} - 4.875 \times 10^{-4}$

$F_v = 2.5125 \times 10^{-3}\,\text{N}$

Write in the form $x \times 10^{-4}\,\text{N}$:

$2.5125 \times 10^{-3} = 25.125 \times 10^{-4}\,\text{N}$

So,

$x \approx 25$

7. Comparison with stored answer

Derived answer: $25$

Stored correct answer: $25$

They agree.