Questions involving Viscosity and Buoyancy

The following simple question involving viscous force and force of buoyancy is meant for checking whether you have a good understanding of basic points:

A wooden ball of relative density 0.5 is released from the bottom of a still water reservoir. Its acceleration while moving up will

(a) go on decreasing initially

(b) go on increasing initially

(c) remain constant at g/2 (in magnitude)

(d) remain constant at g (in magnitude)

(e) be zero throughout

The net force acting on the ball at the moment of releasing is its apparent weight which is upwards. [The apparent weight = Real weight – up thrust = Vρg – Vσg. Since the density of wood (ρ) is less than that of water (σ), the apparent weight is negative which means it is directed upwards]

The sphere therefore moves up with an acceleration. But, when the velocity (v) increases from zero, the opposing viscous force (6πrηv) also increases thereby reducing the net upward force. The upward acceleration therefore goes on decreasing. So, the correct option is (a).

The following question which appeared in Kerala Engineering Entrance 2008 question paper also involves viscous force and force of buoyancy; but the latter is negligible:

Eight drops of a liquid of density ρ and each of radius ‘a’ are falling through air with a constant velocity of 3.75 cm s^–1. When the eight drops coalesce to form a single drop, the terminal velocity of the new drop will be

(a) 1.5×10^–2 ms^–1

(b) 2.4×10^–2 ms^–1

(c) 0.75×10^–2 ms^–1

(d) 25×10^–2 ms^–1

(e) 15×10^–2 ms^–1

The viscous force acting on a sphere of radius ‘a’ is 6πahv where h is the coefficient of viscosity of the fluid and ‘v’ is the velocity of the sphere. Terminal velocity is attained when the opposing viscous force is equal in magnitude to the apparent weight of the sphere. Therefore,

6πahv = (4/3)πa³(ρ – σ)g where ρ is the density of the sphere, σ is the density of the fluid and ‘g’ is the acceleration due to gravity.

The above equation shows that the terminal velocity ‘v’ is directly proportional to the square of the radius of the sphere.

In the above problem, eight identical drops coalesce to form a single drop. The new drop thus formed has eight times the volume of each small drop. The radius ‘R’ of the new drop is given by

(4/3)πR³ = 8×(4/3)πa³

Therefore, R =2a

Since the radius of the new drop is twice that of each small drop, the terminal velocity of the new drop must become four times. The correct option therefore is 15×10^–2 ms^–1.