In the measurement of viscosity of liquids using terminal velocity...

The terminal velocity $v$ of a spherical ball of radius $r$ and density $\sigma$ falling through a liquid of density $\rho$ and viscosity $\eta$ is given by Stokes' law:

$v = \dfrac{2r^2g}{9\eta}(\sigma - \rho)$

Rearranging the terms to express $v$ as a function of the ratio $\dfrac{\sigma}{\rho}$:

$v = \dfrac{2r^2g\rho}{9\eta} \left( \dfrac{\sigma}{\rho} - 1 \right)$

Let $y = v$ and $x = \dfrac{\sigma}{\rho}$. The equation becomes:

$y = k(x - 1)$

where $k = \dfrac{2r^2g\rho}{9\eta}$ is a positive constant.

This represents the equation of a straight line with a positive slope $k$ and an x-intercept at $x = 1$ (i.e., $v = 0$ when $\dfrac{\sigma}{\rho} = 1$).

Looking at the given graphs, only the second graph shows a straight line with a positive slope that intersects the horizontal axis at $\dfrac{\sigma}{\rho} = 1$.

Answer: