1. Use Stokes' law for terminal velocity of a bubble

Since the bubble rises steadily, it has attained terminal velocity. For a small spherical bubble in a viscous liquid (neglecting air density), the upward terminal speed is taken as

$v = \frac{2 r^2 \rho g}{9\eta}$

where:

• $r$ = radius of bubble,
• $\rho = 1750\,\text{kg m}^{-3}$ is density of solution,
• $g \approx 9.8\,\text{m s}^{-2}$,
• $\eta$ = coefficient of viscosity.

2. Convert given quantities to SI units

Initial diameter:

$d = 2\,\text{mm} = 2 \times 10^{-3}\,\text{m}$

Hence radius:

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

Velocity:

$0.35\,\text{cm s}^{-1} = 0.35 \times 10^{-2}\,\text{m s}^{-1} = 3.5 \times 10^{-3}\,\text{m s}^{-1}$

3. Solve for viscosity

From

$v = \frac{2 r^2 \rho g}{9\eta}$

we get

$\eta = \frac{2 r^2 \rho g}{9v}$

Substitute values:

$\eta = \frac{2(10^{-3})^2(1750)(9.8)}{9(3.5\times 10^{-3})}$

$\eta = \frac{2 \times 10^{-6} \times 1750 \times 9.8}{3.15 \times 10^{-2}}$

$2 \times 1750 \times 9.8 \times 10^{-6} = 34300 \times 10^{-6} = 3.43 \times 10^{-2}$

So,

$\eta = \frac{3.43 \times 10^{-2}}{3.15 \times 10^{-2}} \approx 1.089\,\text{Pa·s}$

4. Convert SI unit to poise

We know:

$1\,\text{Pa·s} = 10\,\text{poise}$

Therefore,

$\eta \approx 1.089 \times 10 = 10.89\,\text{poise}$

Nearest integer:

$\boxed{11}$

5. Compare with stored answer

Derived answer = $11$. Stored correct answer = $11$.

So they agree.