Question

What is the velocity v of a metallic ball of radius r falling in a tank of liquid at the instant when its acceleration is one half of a freely falling body?( the density of metal and of liquid sigma and rho respectively, the viscosity of the liquid is n)

Answer 1

Given:

A metallic ball of radius r , density Dm

A tank of liquid of density Dl and viscosity n .

To Find:

The velocity v of metallic ball, when its acceleration is one half of a freely falling body.

Solution:

The Forces acting on the ball are:

• In the Upward direction: Viscous force (Fv) and Buoyant force(Fb).
• In the downward direction: Weight due to gravity(mg).

Let Net force = m a  = Dm x 4/3 x πr³ a = 4Dmπr³a/3

Let V be the velocity of the metallic ball.

Force eqution:

• mg - Fv - Fb = ma
• 4πDmr³g/3 - 6πnrV -  4πDlr³g/3= 4Dmπr³/3 x g/2
• 6πnrV = 4πr³g/3 ( Dm -  Dl - Dm/2 ) = 4πr³g/6( Dm - 2Dl)
• V = (Dm - 2Dl)r²g/9n

Given Dm =  $\sigma$ and Dl = $\rho$

The velocity v of metallic ball, when its acceleration is one half of a freely falling body is V = $\frac{1}{9n} (\sigma - 2\rho)r^{2} g$.