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A sphere is dropped under gravity through a fluid of viscosity h. If the average acceleration is half of the initial acceleration, the time to attain the terminal velocity is (ρs=densityofthesphere,r=radius).


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Solution

Step 1: Given data:

The viscosity of the fluid is h.

The density of the sphere is ρs.

The radius of the sphere is r.

The average acceleration is half of the initial acceleration, i.e,aavg=a2, where, a is the initial acceleration of the sphere.

Step 2: The downward force on a body in a liquid:

  1. We know, the downward force on a body when it is placed in a liquid is Fd=Bodyweight-buoyancyforceonthebody.
  2. So, the downward force exerted on the body is Fd=43πr3×ρs×g-43πr3×ρw×g,
  3. where, 43πr3 is the volume of the solid and ρw is the density of the liquid, g is the acceleration due to gravity.

Step 3: Terminal velocity:

  1. The terminal velocity is the constant velocity when a body freely falls through a liquid.
  2. The terminal velocity is defined by the form, vt=2r2ρs-ρwg9η,
  3. where ρw is the density of the liquid, ρs is the density of the body, η is the coefficient of viscosity and r is the radius of the solid body.

Step 4: Finding the downward force:

As we know, the downward force on the body is

Fd=43πr3×ρs×g-43πr3×ρw×g

So, Fd=43πr3gρs-ρw.................(1)

Step 5: Finding the acceleration:

We know that acceleration, a=Fm, where, F is the force on a body and m is the mass of the body.

So,

a=Fm=43πr3gρs-ρw43πr3×ρs=ρs-ρwgρsa=ρs-ρwgρs......................(2)

Step 6: Finding the time:

Let the time to attain the terminal velocity is t.

AS we know, from the kinematics formulae, v=u+at, v and u are the final and initial velocity and a is the acceleration of the body.

So,

vt=0+a2t;since,aavg=a2t=2vta=2×2r2ρs-ρwg9ηρs-ρwgρs=4ρsr29ηsince,terminalvelocity,vt=2r2ρs-ρwg9ηort=4ρsr29η.

Therefore, the time to attain the terminal velocity is 4ρsr29η.


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