It is dropped in a flui of density $\rho_{0} .$ On just entering into the fluid, the effective downward froce acting on the sphere is

$F=m a s s \times$ acceleration $=\frac{4}{3} \pi r^{3}\left(\rho-\rho_{0}\right) g$

Let a be the initial ac celeration produced. then,

$a=\frac{\text { force }}{\text { mass }}=\frac{\frac{4}{3} \pi r^{3}\left(\rho-\rho_{0}\right) g}{\frac{4}{3} \pi r^{3} \rho}=\left(\frac{\rho-\rho_{0}}{\rho}\right) g$

When the sphere attains the terminal velocity $v$ its ac celeration becomes zero.

average ac celeration,

$a_{1}=\frac{a+0}{2}=\frac{a}{2}=\left(\frac{\rho-\rho_{0}}{2 \rho}\right) g$

Let sphere takes time t to attain the terminal velocity $v$ when dropped in the fluid. then $u=0,$ using the relation, $v=u+a_{1} t,$ we get $\frac{2 r^{2}\left(\rho-\rho_{0}\right) g}{9 \eta}=0+\left(\frac{\rho-\rho_{0}}{2 \rho}\right) g t$

or $t=\frac{2 r^{2}\left(\rho-\rho_{0}\right) g}{9 \eta} \cdot \frac{2 \rho}{\left(\rho-r h_{0}\right) g}=\frac{4 \rho r^{2}}{9 \eta}$

From this relation, it is clear that time $t$ is independent of the density $\rho_{0}$ of the given fluid.