1. For a small solid sphere falling through a viscous liquid, the terminal velocity is given by Stokes’ law:

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

where:

• $r$ = radius of sphere
• $\rho_s$ = density of sphere
• $\rho_l$ = density of liquid
• $\eta$ = viscosity of liquid

So,

$v_t \propto \frac{r^2(\rho_s-\rho_l)}{\eta}$

2. For sphere $P$:

• Density of sphere = $8\,\text{g cm}^{-3}$
• Diameter = $1\,\text{cm}$, so radius $r_P = 0.5\,\text{cm}$
• Liquid density = $0.8\,\text{g cm}^{-3}$
• Viscosity $\eta_P = 3$ poise

Thus,

$\rho_s - \rho_l = 8 - 0.8 = 7.2$

Hence,

$v_P \propto \frac{(0.5)^2(7.2)}{3}$

3. For sphere $Q$:

• Density of sphere = $8\,\text{g cm}^{-3}$
• Diameter = $0.5\,\text{cm}$, so radius $r_Q = 0.25\,\text{cm}$
• Liquid density = $1.6\,\text{g cm}^{-3}$
• Viscosity $\eta_Q = 2$ poise

Thus,

$\rho_s - \rho_l = 8 - 1.6 = 6.4$

Hence,

$v_Q \propto \frac{(0.25)^2(6.4)}{2}$

4. Now compute the ratio:

$\frac{v_P}{v_Q} = \frac{\dfrac{(0.5)^2(7.2)}{3}}{\dfrac{(0.25)^2(6.4)}{2}}$

Since

$\frac{(0.5)^2}{(0.25)^2} = \left(\frac{0.5}{0.25}\right)^2 = 2^2 = 4$

we get

$\frac{v_P}{v_Q} = 4 \cdot \frac{7.2}{6.4} \cdot \frac{2}{3}$

Now,

$\frac{7.2}{6.4} = \frac{72}{64} = \frac{9}{8}$

So,

$\frac{v_P}{v_Q} = 4 \cdot \frac{9}{8} \cdot \frac{2}{3}$

$= \frac{36}{8} \cdot \frac{2}{3}$

$= \frac{9}{2} \cdot \frac{2}{3} = 3$

5. Therefore, the ratio of terminal velocities of $P$ and $Q$ is:

$\boxed{3}$

6. Comparison with stored correct answer:

• Derived answer = $3$
• Stored correct answer = $3$

They agree.