$$V_t =\sqrt{\frac{2mg}{\rho AC_D}}$$

or

$$V_t \propto \sqrt{d}$$

Where,

- \(V_t\) is terminal velocity

- \(m\) is mass

- \(g\) is acceleration due to gravity

- \(\rho\) is density of fluid

- \(A\) is cross-sectional area of the particle

- \(C_D\) is drag coefficient

As mass is directly proportional to the volume and volume of a sphere is directly proportional to the cube of its diameter;

$$m\propto d^3$$

$$A\propto d^2$$

We can see that the diameter appears in the denominator with a larger exponent as compared to the numerator. Therefore larger spheres will have lower terminal velocity.