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Stokes’ law describes sedimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve terminal velocity quickly. One can measure the time it takes for a particle to fall a certain distance and then use Stokes’ law to calculate the viscosity of the liquid. Suppose a steel ball bearing (density 7.8 \times 10^3~\text{kg/m}^37.8×10 ​3 ​​ kg/m ​3 ​​ and diameter 3.0~\text{mm}3.0 mm) is dropped in a container of motor oil. It takes 12 s to fall a distance of 0.60 m. Calculate the viscosity of the oil.

Answer :

okpalawalter8

Answer:

The viscosity is  [tex]\eta  = 0.76243 \ kg/ m \cdot s [/tex]  

Explanation:

From the question we are told that

  The  density is  [tex]\rho  =  7.80 *10^{3} \  kg/m^3[/tex]

  The diameter is  [tex]d =  3.0 \  mm =0.003 \ m[/tex]

    The  time taken is  [tex]t  =  12 \ s[/tex]

   The  distance covered is  [tex]d =  0.60 \ m[/tex]

Generally the velocity of the ball is  

      [tex]v  =  \frac{d}{t}[/tex]

=>    [tex]v  =  \frac{0.60}{12}[/tex]

=>    [tex]v  =  0.05 \ m/s [/tex]

Generally the mass of the steel ball is  

    [tex]m  =  \rho  *  V[/tex]

Here  V  is the volume and this is mathematically represented as

     [tex]V  =  \frac{4}{3} *  \pi  * [\frac{d}{2}  ]^3[/tex]

=>    [tex]V  =  \frac{4}{3} *  3.142  * [\frac{0.003}{2}  ]^3[/tex]

=>      [tex]V  = 1.414 *10^{-8} \  m^3[/tex]

So

     [tex]m  = 7.80 *10^{3}  *   1.414 *10^{-8}[/tex]

     [tex]m  = 0.00011 \  kg [/tex]

Generally the viscosity is mathematically represented as  

     [tex]\eta  =  \frac{m  *  g}{6\pi  *  r  *  v }[/tex]

Here  r is the radius represented as

      [tex]r =  \frac{d}{2}[/tex]

=>   [tex]r =  \frac{0.003}{2}[/tex]

[tex]r =  0.0015  \  m [/tex]

So

  [tex]\eta  =  \frac{0.00011  *  9.8}{6 *  3.142   * 0.0015  * 0.05 }[/tex]    

=> [tex]\eta  = 0.76243 \ kg/ m \cdot s [/tex]

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