Match List-I with List-II for a small solid sphere of volume V ,...

(A) Buoyant force is the weight of the displaced fluid. For a sphere of volume $V$ in a fluid of density $\sigma$, $F_b = V\sigma g$. This matches (I).

(B) When just released, the initial velocity is zero, so the viscous force is zero. The net downward force is $W - F_b = V\rho g - V\sigma g$.
Initial acceleration $a = \dfrac{\text{Net force}}{\text{Mass}} = \dfrac{V(\rho - \sigma)g}{V\rho} = g\left(1 - \dfrac{\sigma}{\rho}\right)$. This matches (III).

(C) At terminal velocity, the sphere moves with constant velocity, so the net force is zero.
Viscous force $F_v = W - F_b = V\rho g - V\sigma g = V(\rho - \sigma)g$. This matches (II).

(D) The terminal velocity of a sphere of radius $r$ falling in a viscous fluid of coefficient of viscosity $\eta$ is given by Stokes' law as $v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta}$. This matches (IV).

Therefore, the correct matching is (A) - (I), (B) - (III), (C) - (II), (D) - (IV).

Answer: (A) - (I), (B) - (III), (C) - (II), (D) - (IV)