Question

The velocity of a ball of mass m and density ${{d}_{1}}$ is dipped in a container containing some liquid of density ${{d}_{2}}$ becomes constant after some time. The viscous force acting on the ball will be?
(A) $mg(1-\dfrac{{{d}_{2}}}{{{d}_{1}}})$
(B) $mg(\dfrac{{{d}_{1}}+{{d}_{2}}}{{{d}_{1}}})$
(C) $mg(\dfrac{{{d}_{1}}+{{d}_{2}}}{{{d}_{2}}})$
(D) None of the above

Answer 1

Hint: In this problem a ball of some mass and given density is dropped in a container containing some liquid and after some time its velocity becomes constant. This means its acceleration is zero. So, the net force acting on the body is zero. We need to find out all the forces acting on the body and we can equate them to arrive at the solution.

Complete step by step answer:
Mass of the ball$=m$
Density of the ball$={{d}_{1}}$
Density of the liquid$={{d}_{2}}$
Weight of the block$=mg$
Volume of the ball$=\dfrac{m}{{{d}_{1}}}$
Upthrust on the ball$=\dfrac{mg}{{{d}_{1}}}{{d}_{2}}$
Downward force acting on the ball thus, $=mg-\dfrac{m}{{{d}_{1}}}{{d}_{2}}g$
Also, the upward force acting on the ball is the viscous force due to the liquid and is given by $=6\pi \eta rv$,
where $\eta $ is the coefficient of viscosity of the liquid inside the container, $r$is the radius of the ball and v is its velocity.
So, these two forces must be equal, thus
$6\pi \eta rv=$$mg-\dfrac{m}{{{d}_{1}}}{{d}_{2}}g$
So, viscous force $=mg(1-\dfrac{{{d}_{2}}}{{{d}_{1}}})$

So, the correct option is A.

Note: In this problem since the ball is moving with constant velocity, its acceleration comes out to be zero. The viscous force is the force between a body and a fluid that is liquid or gas. Motion is must for viscous force to come into play.Sometimes we may get confused between the centre of buoyancy and the centre of gravity. Centre of Gravity is the point in a body where the gravitational force acts on the body.