# A Sphere Is Dropped Under Gravity Through A Fluid Of Viscosity . If The Average Acceleration Is Half Of The Initial Acceleration. What Is The Time To Attain The Terminal Velocity?

Here,

The coefficient of viscosity of the fluid = $$\eta$$

Density of the fluid = $$\sigma$$

Density of the material of the sphere = $$\rho$$

The radius of the sphere = r

Now mass of the sphere, m = $$\frac{4}{3}\Pi r^{3}\rho$$

The mass of the fluid displaced,$${m}’ = \frac{4}{3}\Pi r^{3}\sigma$$

The net downward force on the sphere,

F = weight of the sphere – the weight of the fluid displaced

$$\Rightarrow mg – {m}’g = \frac{4}{3}\Pi r^{3}\rho g – \frac{4}{3}\Pi r^{3}\sigma g$$

OR

F = $$\frac{4}{3}\Pi r^{3} (\rho – \sigma )g$$—[1]

As the sphere falls with an initial acceleration a. Then, according to Newton’s second law of motion,

F = $$ma \left ( \frac{4}{3}\Pi r^{3} \rho \right )a$$—[2]

From equations [1] and [2] $$\frac{4}{3}\Pi r^{3} (\rho -\sigma) g = (\frac{4}{3}\Pi r^{3}\rho)a$$

OR

$$a = (\frac{\rho – \sigma}{\rho})g$$

The terminal velocity of the ball is

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

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