Question

Question #f9140

Answer 1

`10.9~~11" Poise"`
(A)

Explanation:

A bubble when placed in a viscous liquid experiences three forces:

1. Weight of bubble. Acting downwards due to force of gravity.
2. Buoyancy. Upwards force equal to the weight of the viscous liquid displaced. As the bubble goes up it expands due to decrease in liquid pressure.
3. Drag. Downwards force. As predicted by Stokes' law. This force is dependent velocity of bubble.

When all three forces are in equilibrium, net force acting on the bubble is zero. Velocity of the bubble does not change as It has reached its terminal velocity.

For the purpose of air bubble in this problem first force has been ignored as this is very small in comparison to Drag.

Assuming that movement of bubble is laminar, bubble has smooth surface, it moves in homogeneous solution and bubble and solution do not interfere with each other, the drag force `F_d` is given by the expression

`F_d = 6 π μ R v`
where `μ` is coefficient viscosity, `R` is the radius of bubble and `v` is its velocity relative to solution.

Also force of buoyancy is given by the expression

`F_b=4/3πR^3ρg`
Where ρ is the density of the liquid and `g` is acceleration due to gravity`=981cmcdot s^-2`.

To find out terminal velocity `v_t` both forces must be equal to each other.
`:.6πμRv_t=4/3πR^3ρg`
`=>v_t=2/9(ρgR^2)/μ`
Inserting given values in CGS units above we get
`0.35=2/9xx((1.750)xx981xx(0.1)^2)/μ`
`=>mu=2/9xx((1.750)xx981xx(0.1)^2)/(0.35)`
`=>mu=10.9" Poise"`