A spherical ball of radius $$1 \times 10^{-4} \text{ m}$$ and density $$10^5 \text{ kg/m}^3$$ falls freely under gravity through a distance $$h$$ before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of $$h$$ is approximately: (The coefficient of viscosity of water is $$9.8 \times 10^{-6} \text{ N s/m}^2$$)

The ball moves in two stages:
(i) free fall through air for a height $$h$$ (ignore air drag),
(ii) motion in water with constant speed (terminal velocity).

Stage (ii): terminal velocity in water
For a sphere of radius $$r$$ moving slowly through a viscous liquid, Stokes’ law gives the viscous force
$$F_{\text{viscous}} = 6\pi \eta r v$$.
At terminal velocity all forces balance:
weight $$= \text{buoyancy} + \text{viscous force}$$

Weight: $$W = \rho_s \, V \, g$$, Buoyant force: $$B = \rho_w \, V \, g$$,
where $$V = \frac{4}{3}\pi r^{3}$$ is the volume.
Hence
$$\rho_s V g = \rho_w V g + 6\pi \eta r v$$

Simplify (take $$V$$ common):
$$(\rho_s - \rho_w)V g = 6\pi \eta r v$$

Substitute $$V = \frac{4}{3}\pi r^{3}$$:
$$(\rho_s - \rho_w)\left(\frac{4}{3}\pi r^{3}\right)g = 6\pi \eta r v$$

Cancel $$\pi$$ and rearrange for $$v$$:
$$v = \frac{2}{9}\,\frac{(\rho_s - \rho_w) g r^{2}}{\eta}$$ $$-(1)$$

Insert the given data:
$$\rho_s = 1.0\times10^{5}\,\text{kg m}^{-3}$$,
$$\rho_w = 1.0\times10^{3}\,\text{kg m}^{-3}$$,
$$r = 1.0\times10^{-4}\,\text{m}$$,
$$\eta = 9.8\times10^{-6}\,\text{N s m}^{-2}$$,
$$g = 9.8\,\text{m s}^{-2}$$.

Compute the difference in densities:
$$(\rho_s - \rho_w) = 1.0\times10^{5} - 1.0\times10^{3} = 9.9\times10^{4}\,\text{kg m}^{-3}$$

Now evaluate $$v$$ from $$(1)$$:
$$v = \frac{2}{9}\,\frac{(9.9\times10^{4})(9.8)(1.0\times10^{-4})^{2}}{9.8\times10^{-6}}$$

$$r^{2} = (1.0\times10^{-4})^{2} = 1.0\times10^{-8}$$,
so
$$v = \frac{2}{9}\,\frac{(9.9\times10^{4})(9.8)(1.0\times10^{-8})}{9.8\times10^{-6}}$$

The factor $$9.8$$ in numerator and denominator cancels, giving
$$v = \frac{2}{9}\,\frac{9.9\times10^{4}\times1.0\times10^{-8}}{1.0\times10^{-6}}$$

$$9.9\times10^{4}\times1.0\times10^{-8}=9.9\times10^{-4}$$;
dividing by $$1.0\times10^{-6}$$ multiplies by $$10^{6}$$:
$$\frac{9.9\times10^{-4}}{1.0\times10^{-6}} = 9.9\times10^{2}=990$$

Therefore
$$v = \frac{2}{9}\times 990 \approx 220\,\text{m s}^{-1}$$

Stage (i): free fall in air
For free fall without air resistance
$$v^{2} = 2 g h \;\Rightarrow\; h = \frac{v^{2}}{2g}$$ $$-(2)$$

Insert $$v \approx 220\,\text{m s}^{-1}$$ and $$g = 9.8\,\text{m s}^{-2}$$ into $$(2)$$:
$$h = \frac{(220)^{2}}{2\times9.8} = \frac{48400}{19.6} \approx 2.47\times10^{3}\,\text{m}$$

The nearest value among the options is $$\boxed{2518\ \text{m}}$$ (Option B).