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(A) Let $r$ be the radius of the sphere and $\rho$ be its density. Let $\rho_0$ be the density of the fluid.Initial acceleration $a$ is given by:$a =

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $r$ be the radius of the sphere and $ho$ be its density. Let $ho_0$ be the density of the fluid.Initial acceleration $a$ is given by:$a = \frac{	ext{Net Force}}{	ext{Mass}} = \frac{\frac{4}{3}\pi r^3(ho - ho_0)g}{\frac{4}{3}\pi r^3ho} = \left(\frac{ho - ho_0}{ho}ight)g$Terminal velocity $v_t$ is given by Stokes' Law:$v_t = \frac{2r^2(ho - ho_0)g}{9\eta}$Given that the average acceleration $a_{avg} = \frac{a}{2}$,and using the kinematic relation $v_t = u + a_{avg}t$ with $u=0$:$v_t = \left(\frac{a}{2}ight)t$Substituting the values:$\frac{2r^2(ho - ho_0)g}{9\eta} = \frac{1}{2} \left(\frac{ho - ho_0}{ho}ight)g \cdot t$Solving for $t$:$t = \frac{2r^2(ho - ho_0)g}{9\eta} \cdot \frac{2ho}{(ho - ho_0)g} = \frac{4ho r^2}{9\eta}$

$A$ 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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