Question

A solid sphere of radius R is
falling in a viscous medium.
The terminal velocity attained
by the falling body will be
proportional to
(A) R². (C)1/R
(B)R. (D)1/R*R​

Answer 1

Terminal velocity is directly proportional to the square of its radius.

Option (A) is correct.

Explanation:

Since sphere is moving with constant velocity, there is no acceleration in it.

When the sphere of radius R is falling in a liquid of density σ and coefficient of viscosity η it attains a terminal velocity v, under two forces

Effective force acting downward

=V(ρ−σ)g = 4 / 3πR^3(ρ−σ)g

where ρ is density of sphere.

(ii) Viscous forces acting upwards =6πηRv

Since the sphere is moving with a constant velocity v, there is no acceleration in it, the net force acting on it must be zero. That is

6πηRv = 4 / 3πR^3(ρ−σ)g

v = 2 / 9 R^2(ρ−σ)gη

v ∝ R^2  

Thus, terminal velocity is directly proportional to the square of its radius.

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