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A sphere is dropped under gravity through a fluid of viscosity $\eta$ . If the average acceleration is half of the initial acceleration, the time to attain the terminal velocity is ($\rho$ = density of sphere ; $r$ = radius)
$\frac{{4\rho {r^2}}}{{9\eta }}$
$\frac{{9\rho {r^2}}}{{4\eta }}$
$\frac{{4\rho \,r}}{{9\eta }}$
$\frac{{9\rho \,r}}{{4\eta }}$
Solution
Let $r$ be the radius of the sphere and $\rho$ be its density.
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.
