JEE Physics 2025 Question with Solution

Given: diameter of the steel ball $d = 3.6 \times 10^{-3} \, \text{m}$, terminal velocity $v_T = 2.45 \times 10^{-2} \, \text{m/s}$, density of steel $\rho_s = 7825 \, \text{kg/m}^3$, density of oil $\rho_l = 925 \, \text{kg/m}^3$, and $g = 9.8 \, \text{m/s}^2$.

Find: the viscosity $\eta$ of the oil.

Use Stokes' law for terminal velocity of a sphere:

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

The radius is

$$r = \frac{d}{2} = \frac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3} \, \text{m}$$

Rearranging for viscosity,

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

Substitute the values:

$$\eta = \frac{2 (1.8 \times 10^{-3})^2 (7825 - 925)(9.8)}{9(2.45 \times 10^{-2})}$$

Now,

$$7825 - 925 = 6900 \, \text{kg/m}^3$$

and

$$(1.8 \times 10^{-3})^2 = 3.24 \times 10^{-6} \, \text{m}^2$$

So,

$$\eta = \frac{2 \times 3.24 \times 10^{-6} \times 6900 \times 9.8}{9 \times 2.45 \times 10^{-2}}$$

This gives

$$\eta \approx 1.99 \, \text{Pa s}$$

Therefore, the viscosity of the oil is $1.99 \, \text{Pa s}$ and the correct option is D.

Given: all quantities are already in SI units except the diameter, which must first be converted to radius.

Find: the viscosity $\eta$.

Write Stokes' law directly in the form

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

Take

$$r = 1.8 \times 10^{-3} \, \text{m}, \qquad \rho_s - \rho_l = 6900 \, \text{kg/m}^3$$

Then substitute:

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

Evaluating gives approximately

$$\eta \approx 1.98 \text{ to } 1.99 \, \text{Pa s}$$

Hence the most appropriate choice is D.