A solid steel ball of diameter 3.6 mm acquired terminal velocity $$2.45 \times 10^{-2}$$ m/s while falling under gravity through an oil of density $$925 \text{ kg m}^{-3}$$. Take density of steel as $$7825 \text{ kg m}^{-3}$$ and g as $$9.8 \text{ m/s}^2$$. The viscosity of the oil in SI unit is

For a small sphere moving slowly through a viscous liquid, Stokes’ law gives the viscous drag force as $$F_{v} = 6 \pi \eta r v$$, where $$\eta$$ is the viscosity, $$r$$ the radius of the sphere and $$v$$ its speed.

At terminal velocity, the sphere moves with constant speed, so the net force on it is zero. The downward forces (weight) are balanced by the upward forces (buoyant force + viscous drag):

$$\text{Weight}\;-\;\text{Buoyant force}\;=\;\text{Viscous drag}$$
$$\bigl(\rho_{s} V g - \rho_{o} V g\bigr)\;=\;6 \pi \eta r v_{t}$$

Here
$$\rho_{s} = 7825\;\text{kg m}^{-3}$$ (density of steel)
$$\rho_{o} = 925\;\text{kg m}^{-3}$$ (density of oil)
$$V = \tfrac{4}{3}\pi r^{3}$$ is the volume of the sphere
$$v_{t} = 2.45 \times 10^{-2}\;\text{m s}^{-1}$$ is the terminal velocity.

Cancelling the common factor $$V g$$ and simplifying:

$$(\rho_{s}-\rho_{o})\;\tfrac{4}{3}\pi r^{3} g = 6 \pi \eta r v_{t}$$

Divide both sides by $$\pi r$$ to isolate $$\eta$$:

$$\eta = \frac{2 r^{2} g\,(\rho_{s}-\rho_{o})}{9 v_{t}}\quad -(1)$$

Now substitute the numerical data.
Diameter of ball $$= 3.6\;\text{mm} \Rightarrow r = 1.8\;\text{mm} = 1.8 \times 10^{-3}\;\text{m}$$

Density difference:
$$(\rho_{s}-\rho_{o}) = 7825 - 925 = 6900\;\text{kg m}^{-3}$$

Using $$g = 9.8\;\text{m s}^{-2}$$ and $$v_{t} = 2.45 \times 10^{-2}\;\text{m s}^{-1}$$ in equation $$(1)$$:

$$\eta = \frac{2\,(1.8 \times 10^{-3})^{2}\,(9.8)\,(6900)}{9\,(2.45 \times 10^{-2})}$$

Step-wise calculation:
$$r^{2} = (1.8 \times 10^{-3})^{2} = 3.24 \times 10^{-6}\;\text{m}^{2}$$
$$2 r^{2} = 2 \times 3.24 \times 10^{-6} = 6.48 \times 10^{-6}$$
$$6.48 \times 10^{-6} \times 6900 = 4.471 \times 10^{-2}$$
$$4.471 \times 10^{-2} \times 9.8 = 0.438\;(\text{approximately})$$
Denominator $$= 9 \times 2.45 \times 10^{-2} = 0.2205$$

Finally,
$$\eta = \frac{0.438}{0.2205} \approx 1.99\;\text{Pa·s}$$

The viscosity of the oil is therefore $$\mathbf{1.99\;Pa\cdot s}$$.

Hence, the correct option is Option D.