A small metal sphere is dropped from rest into a long column of a...

The terminal velocity of a sphere falling through a viscous liquid is given by $v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta}$.

When the temperature of the liquid increases from $20^\circ\text{C}$ to $60^\circ\text{C}$, its viscosity $\eta$ decreases. Since $v_t$ is inversely proportional to $\eta$, the terminal velocity increases, meaning $v_2 > v_1$.

The initial acceleration of the sphere at the moment it is dropped ($v=0$) is determined only by gravity and the buoyant force. It is given by $a = g\left(1 - \dfrac{\sigma}{\rho}\right)$, where $\rho$ is the density of the sphere and $\sigma$ is the density of the liquid. Since the densities do not change significantly, the initial acceleration remains unchanged, meaning $a_1 = a_2$.

Answer: $v_2 > v_1$ and $a_1 = a_2$