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Question

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

Vedclass · Accepted Answer

$\frac{{4\rho {r^2}}}{{9\eta }}$

Vedclass · Answer

Let $r$ be the radius of the sphere and $ho$ be its density.

It is dropped in a flui of density $ho_{0} .$ On just entering into the fluid, the effective downward froce acting on the sphere is

$F=m a s s 	imes$ acceleration $=\frac{4}{3} \pi r^{3}\left(ho-ho_{0}ight) g$

Let a be the initial ac celeration produced. then,

$a=\frac{	ext { force }}{	ext { mass }}=\frac{\frac{4}{3} \pi r^{3}\left(ho-ho_{0}ight) g}{\frac{4}{3} \pi r^{3} ho}=\left(\frac{ho-ho_{0}}{ho}ight) g$

When the sphere attains the terminal velocity $v$ its ac celeration becomes zero.

average ac celeration,

$a_{1}=\frac{a+0}{2}=\frac{a}{2}=\left(\frac{ho-ho_{0}}{2 ho}ight) g$

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 $\frac{2 r^{2}\left(ho-ho_{0}ight) g}{9 \eta}=0+\left(\frac{ho-ho_{0}}{2 ho}ight) g t$

or $t=\frac{2 r^{2}\left(ho-ho_{0}ight) g}{9 \eta} \cdot \frac{2 ho}{\left(ho-r h_{0}ight) g}=\frac{4 ho r^{2}}{9 \eta}$

From this relation, it is clear that time $t$ is independent of the density $ho_{0}$ of the given fluid.

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)

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