A small metallic sphere of mass m and density is dropped into a...

When the sphere reaches terminal velocity, the net force acting on it is zero. This means the downward weight ($W$) is balanced by the sum of the upward buoyant force ($F_b$) and the viscous force ($F_v$).

$W = F_b + F_v$

Given that the viscous force is one-fourth of the weight ($F_v = \dfrac{mg}{4}$):

$mg = F_b + \dfrac{mg}{4}$

$F_b = mg - \dfrac{mg}{4} = \dfrac{3mg}{4}$

The buoyant force is also given by the weight of the displaced liquid: $F_b = V \sigma g$, where $V is the volume of the sphere and$\sigma$is the density of the liquid.<br><br>The volume of the sphere can be written as$V = \dfrac{m}{\rho}$. Substituting this into the buoyant force equation:<br><br>\left(\dfrac{m}{\rho}\right) \sigma g = \dfrac{3mg}{4}$

$\dfrac{\sigma}{\rho} = \dfrac{3}{4}$

$\sigma = \dfrac{3\rho}{4}$

Answer: $\dfrac{3\rho}{4}$